Plus One Chemistry Notes Chapter 1 Some Basic Concepts of Chemistry

Students can Download Chapter 1 Some Basic Concepts of Chemistry Notes, Plus One Chemistry Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Notes Chapter 1 Some Basic Concepts of Chemistry

INTRODUCTION
Chemistry is the branch of science which deals with the composition, properties and transformation of matter. These aspects can be best understood in terms of basic constituents of matter: atoms and molecules. That is why chemistry is called the sci-ence of atoms and molecules.

IMPORTANCE OF CHEMISTRY
Chemistry plays an important role in almost all walks of life. In recent years chemistry has tackled with a fair degree of success some of the pressing aspects of environmental degradation. Safer alternatives to environmentally hazardous refrigerants like CFCs, responsible for ozone depletion in the stratosphere, have been successfully synthesised.

NATURE OF MATTER
Matter is anything which has mass and occupies space. Matter can exist in three physical states: solid, liquid and gas.

At the macroscopic or bulk level, matter can be clas-sified as mixtures or pure substances.
A material containing only one substance is called a pure substance. Materials containing more than one substance are called mixtures. Pure substances are further classified into two types: elements and com¬pounds. Mixtures are also of two types: homoge¬neous mixtures and heterogeneous mixtures. A mix¬ture is said to be homogeneous if it has same com¬position throughout. Some examples of homoge¬neous mixtures are air, gasoline, kerosene, milk, alloys, etc. Heterogeneous mixtures are the mixtures which have different composition in different parts. Some examples of heterogeneous mixtures are iron and sulphur, muddy water, etc.

PROPERTIES OF MATTER AND THEIR MEASUREMENT
Every substances has characteristic properties. These are classified into two categories – physical properties and chemical properties.
Physical properties are those properties which can be measured or observed without changing the iden¬tity or the composition of the substance. Some ex¬amples of physical properties are colour, odour, melt¬ing point, boiling point, density, etc. chemical properties are characteristic reactions of different substances; these include acidity or basic¬ity, combustibility, etc.

THE INTERNATIONAL SYSTEM OF UNITS (SI UNITS)
A unit may be defined as the standard of reference chosen to measure any physical quantity. There are many different systems of units.
The improved metric system of units accepted inter-nationally is called International System of Units or SI units. The SI system has seven base units from which all other units are derived.

UNCERTAINTY IN MEASUREMENT
Scientific measurements involving some measuring devices have some degree of uncertainty. The magnitude of uncertainty depends on the accuracy of the measuring device and also on the skill of its operator.
The closeness of a set of values obtained from identical measurements of a quantity is known as precision of the measurement.
The term accuracy is defined as the closeness of a measurement or a set of measurements to its true value.

SIGNIFICANT FIGURES
The total number of digits in a measurement is called the number of significant figures. It includes the num¬ber of figures that are known with certainty plus the last uncertain digit, beginning with the first non-zero digit.
1) All non-zero digits are significant. For example, in 285 cm, there are three significant figures and in 0.25 mL, there are two significant figures.

Plus One Chemistry Notes Chapter 1 Some Basic Concepts of Chemistry

2) Zeros preceding to first non-zero digit are not significant. Such zero indicates the position of decimal point. Thus, 0.03 has one significant figure and 0.0052 has two significant figures.
3) Zeros between two non-zero digits are significant. Thus, 2.005 has four significant figures.

4) Zeros at the end or right of a number are significant provided they are on the right side of the decimal point. For example,0.200 g has three significant figures. But, if otherwise, the zeros are not significant. For example, 100 has only one significant figure.

5) Exact numbers have an infinite number of significant figures. For example, in 2 balls or 20 eggs, there are infinite significant figures as these are exact numbers and can be represented by writing infinite number of zeros after placing a decimal i.e.,2 = 2.000000 or 20 = 20.000000

LAWS OF CHEMICAL COMBINATIONS
Law of Conservation of Mass
It states that matter can neither be created nor destroyed.
This law was put forth by Antoine Lavoisier in 1789.

Law of Definite Proportions
This law was given by, a French chemist, Joseph Proust. He stated that a given compound always contains exactly the same proportion of elements by weight. It is sometimes also referred to as Law of definite composition.

Plus One Chemistry Notes Chapter 1 Some Basic Concepts of Chemistry

Law of Multiple Proportions
This law was proposed by Dalton in 1803. According to this law, if two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in the ratio of small whole numbers.

Gay Lussac’s Law of Gaseous Volumes
This law was given by Gay Lussac in 1808. He observed that when gases combine or are produced in a chemical reaction they do so in a simple ratio by volume provided all gases are at same temperature and pressure.

Avogadro Law
In 1811, Avogadro proposed that equal volumes of gases at the same temperature and pressure should contain equal number of molecules Avogadro made a distinction between atoms and molecules which is quite understandable in the present times.

DALTON’S ATOMIC THEORY
The main postulates of the theory are:-

  1. Matter is made up of extremely small, indivisible particles called atoms.
  2. Atoms of the same element are identical in all respects i.e. size and mass.
  3. Atoms of different elements are different, i.e., they possess different sizes, shapes, masses, and chemical properties.
  4. Atoms of different elements may combine with each other in a simple whole number ratio to form compound atoms or molecules.
  5. Atoms can neither be created nor destroyed, i.e., atoms are indestructible.

ATOMIC AND MOLECULAR MASSES
ATOMIC MASS
The mass of an atom is extremely small. These are very inconvenient for calculations. This difficulty was overcome by expressing atomic masses as relative masses, i.e., with respect to the mass of an atom of a standard substance. The scale in which the relative masses are expressed is called atomic mass unit scale or amu scale. One atomic mass unit (amu) is equal to one-twelfth (1/ 12) the mass of an atom of carbon-12. Recently the symbol ‘u’ (unified mass) is used in place of amu.
Plus One Chemistry Notes Chapter 1 Some Basic Concepts of Chemistry image 1

AVERAGE ATOMIC MASS
The atomic masses of many elements have fractional values because they exist as mixture of isotopes.
In the case of such elements, the atomic mass is taken as the average of the atomic masses of the various isotopes. For example, ordinary chlorine is a mixture of two isotopes with atomic masses 35 u and 37 u and they are present in the ration 3:1. Therefore,
Atomic mass of Chlorine = \(\frac{35 \times 3+37 \times 1}{3+1}\) = 35.5u

Plus One Chemistry Notes Chapter 1 Some Basic Concepts of Chemistry

MOLECULAR MASS
Molecular mass is the sum of atomic masses of the elements present in a molecule.
For example, molecular mass of water
= 2 x atomic mass of hydrogen + 1 x atomic mass of oxygen
= 2 × (1.008u) + 1 × 16.00 u
= 18.02 u

FORMULA MASS
In ionic compounds, we use formula mass instead of molecular mass. Formula mass of an ionic compound is the sum of the atomic masses of all atoms in a formula unit of the compound.

MOLE CONCEPT AND MOLAR MASSES
‘Mole’ was introduced as the seventh base quantity for the amount of substance in SI system. One mole is the amount of a substance that contains as many particles or entities as there are atoms in exactly 12 g (or 0.012 kg) of the 12C isotope. This number is known as ‘Avogadro constant’ (NA = 6.022 × 1023). The mass of one mole of a substance in grams is called its molar mass. The molar mass is numerically equal to atomic/ molecular/ formula mass in u.

PERCENTAGE COMPOSITION
One can check the purity of a sample by analysing its mass percentage
Mass % of an element in a compound =
Plus One Chemistry Notes Chapter 1 Some Basic Concepts of Chemistry image 2

EMPIRICAL FORMULA FOR MOLECULAR FORMULA
An empirical formula represents the simplest whole number ratio of various atoms present in a compound whereas the molecular formula shows the exact number of different types of atoms present in a molecule of a compound.

Problem 1.2
A compound contains 4.07 % hydrogen, 24.27 % carbon and 71.65 % chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulae?
Solution:
Step 1. Conversion of mass per cent to grams.
Since we are having mass per cent, it is convenient to use 100 g of the compound as the starting material. Thus, in the 100 g sample of the above compound, 4.07g hydrogen is present, 24.27g carbon is present and 71.65 g chlorine is present.

Step 2. Convert into number moles of each element
Divide the masses obtained above by respective atomic masses of various elements.
Plus One Chemistry Notes Chapter 1 Some Basic Concepts of Chemistry image 3

Step 3. Divide the mole value obtained above by the smallest number
Since 2.021 is smallest value, division by it gives a ratio of 2:1:1 for H:C:CI.
In case the ratios are not whole numbers, then they may be converted into whole number by multiplying by the suitable coefficient.

Step 4. Write empirical formula by mentioning the numbers after writing the symbols of respective elements.
CH2Cl is, thus, the empirical formula of the above compound.

Step 5. Writing molecular formula
a) Determine empirical formula mass. Add the atomic masses of various atoms present in the empirical formula.
ForCH2Cl, empirical formula mass is 12.01 + 2 1.008 + 35.453 = 49.48 g

b) Divide Molar mass by empirical formula mass
Plus One Chemistry Notes Chapter 1 Some Basic Concepts of Chemistry image 4
c) Multiply empirical formula by ‘n’ obtained above to get the molecular formula
Empirical formula = CH2Cl, n = 2.
Molecular formula = 2 × [CH2Cl]=C2H4Cl2

STOICHIOMETRY AND STOICHIOMETRIC CALCULATIONS
‘Stoichiometry’ deals with the calculation of masses (sometimes volumes also) of the reactants and prod¬ucts involved in a chemical reaction. The coefficients of reactants and products in a balanced chemical equation is called the stoichiometric coefficients.

LIMITING REAGENT
The amount of the product obtained in a chemical reaction is determined by the amount of the reactant that is completely consumed in the reaction. This reactant is called the limiting reagent. Thus, limiting reagent may be defined as the reactant which is com¬pletely consumed in a reaction containing two or more reactants.

REACTIONS IN SOLUTIONS
1. Mass per cent :
It is obtained by the following relation:
Plus One Chemistry Notes Chapter 1 Some Basic Concepts of Chemistry image 5
2.Mole Fraction :
It is the ratio of number of moles of a particular component to the total number of moles of the solution. If a substance ‘A’ dissolves in substance ‘B’ and their number of moles are nA and nB respectively; then the mole fractions of A and B are given as
Plus One Chemistry Notes Chapter 1 Some Basic Concepts of Chemistry image 6
3. Molarity :
It is the most widely used unit and is denoted by M. It is defined as the number of moles of the solute in 1 litre of the solution.
Plus One Chemistry Notes Chapter 1 Some Basic Concepts of Chemistry image 7
4. Molality :
It is defined as the number of moles of solute present in 1 kg of solvent. It is denoted by ‘m’.
Plus One Chemistry Notes Chapter 1 Some Basic Concepts of Chemistry image 8

Note :
Molarity of a solution changes with temeprature. But molality of a solution does not change with temperature since mass remains unaffected with temperature.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 14 Environmental Chemistry

Students can Download Chapter 13 Hydrocarbons Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 14 Environmental Chemistry

Plus One Chemistry Environmental Chemistry One Mark Questions and Answers

Question 1.
The greatest affinity for haemoglobin is shown by
a) NO
b) CO
c) O2
d) CO2
Answer:
b) CO

Question 2.
London smog is ___________ in nature.
Answer:
reducing

Question 3.
Addition of phosphate fertilizers into water leads to
a) Increased growth of decomposers
b) Reduced algal growth
c) Increased algal growth
d) Nutrient enrichment
Answer:
d) Nutrient enrichment

Plus One Chemistry Chapter Wise Questions and Answers Chapter 14 Environmental Chemistry

Question 4.
Which of the following is a viable particulate?
a) Algae
b) Smoke
c) Mist
d) Fumes
Answer:
a) Algae

Question 5.
Which of the following is not a greenhouse gas?
a) CO2
b) CH4
c) O2
d) Water vapour
Answer:
c) O2

Question 6.
Methyl isocyanate is prepared by the action of ___________ on methyl amine
Answer:
COCl2

Question 7.
In a photo chemical smog the gas that causes eye irritation is ___________ .
a) CO2
b) CH4
c) PAN
d) Acrolein
e) both (c) and (d)
Answer:
PAN

Question 8.
Super sonic jet planes can contribute to zone depletion by introduction of the gas straight to the stratosphere
Answer:
NO

Plus One Chemistry Chapter Wise Questions and Answers Chapter 14 Environmental Chemistry

Question 9.
The difference between the amounts of dissolved oxygen in a sample of water saturated with oxygen and that after incubation for a period of 5 days is known as.
Answer:
BOD

Question 10.
Excess of nitrate ions in drinking water causes.
Answer:
Methenoglobinemia

Plus One Chemistry Environmental Chemistry Two Mark Questions and Answers

Question 1.
Write the equation for the combustion of ethane.
a) Complete combustion
b) Incomplete combustion
Answer:
a) 2C2H6(g) +7O2 → 4CO2(g) + 6H2O(v)
b) 2C2H6(g)+5O2 → 4CO(g) + 6H2O(v)

Question 2.
Write the three common pollutants.
Answer:
1. Gases such as CO and SO2
2. Compounds of metals like lead, mercury, zinc.
3. Radioactive substance

Plus One Chemistry Chapter Wise Questions and Answers Chapter 14 Environmental Chemistry

Question 3.
Three equations are given below. After studying it, write about Acid rain.
(1) CO2(g) + H2O(l) → 2H+ (aq) + CO32-(aq)
(2) 2SO2(g) + O2(g) +2H2O(l) → 2H2SO4(aq)
(3) 4NO2(g) + O2(g) + 2H2O(l) → 2HNO3 (aq)
Answer:
Oxides of Sulphur and Nitrogen, mist of HCI and Phosphoric acid, etc. present in polluted airdissolve in rain water making it more acidic. This is known as acidic rain. SO2 and NO2 present in polluted air are the major contributors of acid rain.
2SO2(g) +O2(g) + 2H2O(l) → 2H2SO4(aq)
4NO2(g) +O2(g) + 2H2O(l) → 2HNO3(aq)

Question 4.
Marble of Taj Mahal reacts with acid rain. What is its result?
Answer:
The marble with which Taj Mahal is made is continuously eaten away by acid rain. It is due to the presence of chemical factories in Agra. The acids present in acid rain react with marble and making the marble dull and rough.
CaCO3 + H2SO4 → CaSO4 +H2O + CO2

Question 5.
Why usage of chlorofluorocarbons being discouraged? or Explain ozone layer depletion?
Answer:
The decomposition of CFC’s destroying ozone. CF2Cl2(g) + hv → Cl(h) + CF2Cl(g)
The reactive Cl atom reacts with O3 to form ClO radical. Cl(g) + O3 → ClO(g) + O2(g)
Thus each Cl atom produced, can destroy many O3 molecules. This leads to ozone depletion.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 14 Environmental Chemistry

Question 6.
What is meant by BOD?
Answer:
It is biochemical oxygen demand. It is the amount of dissolved oxygen required by micro-organisms to oxidise organic and inorganic matter present in polluted water.

Question 7.
1. How we can control air pollution?
2. How does the soil pollution occur?
Answer:
1. Air pollution can be controlled by the following ways.
• Exhaust to oxidize CO to CO2.
• CO2 level can be maintained by planting trees.
• Hydrogen gas is looked upon as pollution less future fuel.
• The large amount of nitrogen oxides emitted from power plant can be removed by scrubbing it with H2SO4.
2. It occurs due to
• Indiscriminate use of fertilizers, pesticides etc.
• Dumping of waste materials.
• Deforestation

Question 8.
What do you mean by greenhouse effect?
Answer:
Global warming is caused by the excess amount of CO2 in the atmosphere. When CO2 accumulate due to the decrease of trees it will increase the temperature of earth. This leads to melting of ice. So the sea level rises.

Question 9.
What do you mean by global warming?
Answer:
As more and more infrared radiations are trapped, the atmosphere becomes hotter and the global temperature rises up. This is known as global warming. There has been a marked increase in the levels of CO2 in the atmosphere due to severe deforestation and burning of fossil fuels.

Question 10.
Give two examples in which green chemistry has been applied.
Answer:
1) Oxidative cracking process in the formation of ethylene is a significant step.
2) Fuel cells for cellular phones have been developed. This cell last for the full life time of the phone.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 14 Environmental Chemistry

Question 11.
Oxygen play a key role in the troposphere while ozone in the stratosphere.
i) How is ozone formed in the atmosphere?
ii) What are the causes of depletion of ozone layer?
Answer:
i) By electric discharge of O2 during lightning.
ii) UV radiations enter to the earth surface and it causes skin diseases.

Question 12.
No new industries are allowed in thickly populated cities by Government order. Why such an order is issued?
Answer:
The city will be destroyed due to acid rain caused by industrial pollution. The Govt, order was to protect the environment (orthe city).

Question 13.
Discuss the importance of dissolved oxygen in water.
Answer:
It helps living organisms in water. They inhale dissolved oxygen in water.

Question 14.
Define environmental chemistry.
Answer:
Environmental chemistry is defined as the branch of science which deals with the chemical processes occurring in the environment. It involves the study of origin, transport, reactions, effects and the fates of chemical species in the environment.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 14 Environmental Chemistry

Question 15.
Explain tropospheric pollution.
Answer:
Tropospheric pollution occurs due to the presence of gaseous and the particulate pollutants.
• Gaseous air pollutants. These include mainly oxides of sulphur (SO2, SO3), oxides of nitrogen (NO, NO2) and oxides of carbon (CO, CO2) in addition , to hydrogen sulphide (H2S), Hydrocarbons, ozone and other oxidants.
• Particulate pollutants. These include dust, mist, fumes, smoke, smog, etc.

Question 16.
Which gases are responsible for greenhouse effect? List some of them.
Answer:
The main gas responsible for greenhouse effect is CO2. Other greenhouse gases are methane, nitrous oxide, water vapours, chlorofluorocarbons (CFC’s) and ozone.

Question 17.
A large number of fish are suddenly found floating dead on a lake. There is no evidence of toxic dumping but you find an abundance of phytoplankton. Suggest a reason for the fish kill.
Answer:
Excessive phytoplankton (organic pollutants such as leaves, grass, trash, etc) present in water is biodegradable. A large population of bacteria decomposes this organic matter in water. During this process they consume the oxygen dissolved in water. Water has already limited dissolved oxygen (≈10 ppm) which gets is further depleted. When the level of dissolved oxygen falls below 6 ppm, the fish cannot survive. Hence, they die and float dead in water.

Question 18.
How can domestic waste be used as mannure?
Answer:
Domestic waste comprises two types of materials, biodegradable such as leaves, rotten food, etc, and non-biodegradable such as plastics, glass, metal scrap, etc. The non-biodegradable waste is sent to industry for recycling. The biodegradable waste should be deposited in the land fills. With the passage of time, it is converted into compost manure.

Question 19.
For your agricultural field or garden, you have developed a compost producing pit. Discuss the process in the light of bad odour, flies and recycling of wastes for a good produce.
Answer:
The compost producing pit should be set up at a suitable place or in a tin to protect ourselves from bad odour and flies. It should be kept covered so that flies cannot make entry into it and the bad odour is minimized. The recyclable material like plastics, glass, newspapers, etc., should be sold to the vendor who further sells it to the dealer. The dealer further supplies it to the industry involved in recycling process.

Plus One Chemistry Environmental Chemistry Three Mark Questions and Answers

Question 1.
Fish grow in cold water as well as in warm water.
1. Do you agree? What is the reason?
2. pH of rain water is 5.6. Is it true? What is the reason?
Answer:
1. No. Fish grow in cold water. Warm water contains less dissolved O2 than cold water,

2. Normally rain water contains dissolved CO2 and hence it shows acidic pH of 5.6.
H2O + CO2 \(\rightleftarrows \) H2CO3 or 2H+ + CO2
When pH of rain water became below 5.6, it becomes acid rain.

Question 2.
Write the mechanism of formation of photochemical smog.
Answer:
At high temp, the petrol and diesel engines, N2 & O2 combine to form NO which is emitted into atmosphere. NO is then oxidised in airto form NO2 which absorbs sunlight and form NO and free O atom.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 14 Environmental Chemistry 1
The O atoms being reactive and combine with O2 to form O3.
O2 (g) + O(g) → O3(g)
The O3 react with NO formed by the photochemical decomposition of NO2.
NO(g) + O3(g) → NO2(g) + O2(g)
NO2 and O3 are good oxidising agents and they react with unburnt hydrocarbons in the polluted airto form substances such as acrolein and formaldehyde. These are the main substances of photochemical smog.

Question 3.
1. What are the major pollutants of water?
2. What is meant by eutrophication?
Answer:
1. Micro-organism present in domestic sewages, organic wastes, plant nutrients, toxic metals, sediments, pesticides and radioactive substances,

2. Addition of P to water as PO43- ion encourages the formation of algae which reduces the dissolved oxygen content of water. This is called eutrophication.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 14 Environmental Chemistry

Question 4.
1. Acid rain causes extensive damage to vegetation and aquatic life. )
i) What do you mean by acid rain?
ii) Name the chemicals responsible for acid rain.
2. List gases which are responsible for greenhouse effect.
Answer:
1. i) When the pH of the rain water drops below 5.6, it is called acid rain.
ii) Oxides of nitrogen and sulphur mist of hydrochloric acid and phosphoric acid etc.

2. Such as carbon dioxide, methane, ozone, chlorofluorocarbon compounds (CFC).

Question 5.
Statues and monuments in India such as Tajmahal are affected by acid rain. How?
Answer:
The statues of monuments in India are affected by acid rain. For example, the air in the vicinity of Taj Mahal contains very high levels of oxides of sulphur and nitrogen. This results in acid rain which reacts with marble of Taj Mahal causing pitting.
CaCO3 + H2SO4 → CaSO4 + H2O + CO2
CaCO3 + 2HNO3 → Ca(NO3)2 +H2O + CO3.
As a result, the monument is being slowly eaten away and the marble is getting decolourised and lustreless. Thus, acid rain is considered as a threat to Taj Mahal.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 14 Environmental Chemistry

Question 6.
1. Writethree major consequences of air pollution.
2. Write any two suitable methods to control air pollution.
Answer:
1. We cannot get pure oxygen for inspire.
Air pollution causes global warming.
It leads to acid rain.
It leads to diseases.

2. Plant trees.
Reduce the use of motor vehicles.
Do not burn plastics.

Question 7.
Carbon monoxide gas is more dangerous than carbon dioxide gas. Why?
Answer:
CO combines with haemoglobin to forms a complex entity, carboxyhaemoglobin which is about 300 times more stable than oxy-haemoglobin. In blood, when the concentration of carboxyhaemoglobin reaches 3 -4%, the oxygen-carrying capacity of the blood is significantly reduced. In other words, the body becomes oxygen-starved. This results into headache, nervousness, cardiovascular disorder, weak eyesight etc.

Plus One Chemistry Environmental Chemistry Four Mark Questions and Answers

Question 1.
Classify the following pollutants:
(a) Carbon monoxide (CO)
(b) Detergents
(c) Plastic
(d) DDT
(e) Sewage
(f) Cigarette smoke
Answer:

Type of Pollution
Air Water Soil
CO
Cigarette
smoke
DDT
Detergent
Sewage
DDT
Plastic

Question 2.
As a result of stratospheric pollution, the ozone layer is destructed.
a) What is ozone layer?
b) How is it useful to us?
c) Write a harmful effect of ozone layer depletion.
Answer:
a) The layer of ozone seen in stratosphere is called ozone layer.
b) Ozone layer plays a significant role in protecting earth from UV rays.
c) Due to ozone layer depletion agricultural crops were found to give reduced yields. Small aquatic organisms which are very sensitive are destroyed due to the increase in the level of UV radiation.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 14 Environmental Chemistry

Question 3.
a) What is smog?
b) What are the adverse effects of photochemical smog?
c) Write any two methods to control photochemical smog.
Answer:
a) Smog is a mixture of smoke and fog. This is the most common example of air pollution that occurs in many cities throughout the world. There are two types of smog:
1) Classical smog
2) Photochemical smog

b) Adverse effects of photochemical smog:
• Eye irritants
• Nose irritation
• Head ache
• Chest pain
• Dryness of throat
• Cough
• Difficulty in breathing

c) 1) Use catalytic converters in automobiles.
2) Plant certain plants (e.g. Pinus) which can metabolise nitrogen oxide.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons

Students can Download Chapter 13 Hydrocarbons Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons

Plus One Chemistry Hydrocarbons One Mark Questions and Answers

Question 1.
Which of the following cannot be prepared by Wurtz reaction?
a) CH4
b) C2H6
c) C3H8
d) C4H8
Answer:
a) CH4

Question 2.
The cyclic polymerization of propane produces __________ .
Answer:
1, 3, 5-trimethylbenzene

Question 3.
The reaction
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 1
a) Hydration
b) Dehydration
c) Dehydrogenation
d) Dehalogenation
Answer:
b) Dehydration

Question 4.
Say TRUE or FALSE.
Calcium carbide on hydrolysis gives ethylene.
Answer:
False

Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons

Question 5.
3-Hexyne reacts with Na/liquid NH3 to produce
a) cis-3-Hexene
b) trans-3-Hexene
c) 3-Hexylamine
d) mm2-Hexylamine
Answer:
b) trans-3-Hexene

Question 6.
Fill in the blanks after finding the correct relationship
CH3 – O – CH3 : Ether, CH3-CH2-OH: …………….
Answer:
Alcohol

Question 7.
Choose the correct answer from the brackets given below:
1) General formula of alkene (CnH2n, CnH4n-2)
2) The 2 different forms of elemental carbon (Bitumen, Diamond, Charcoal, Coke, Led, Graphite)
3) Find the odd man out. Give reason (C2H4, C3H6, C4H8, C2H6)
Answer:
1) CnH2n
2) Diamond, graphite
3) C2H6. It is an alkane. All others are alkenes.

Question 8.
The structural formula of a compound and the name given by a student to it is given below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 2
Answer:
The name given by the student is wrong. The correct name of the compound is 3-Ethyl -2-methyl nonane.

Question 9.
Which is not the isomer of CH3-CH2-O-CH2-CH3? (1)
a) CH3-O-CH2-CH2-CH3
b) CH3-CH2-CH2-CH2-OH
c) CH3-CH2-CO-CH3
d) CH3-CH2(OH)-CH-CH3
Answer:
c) CH3CH2COCH3

Question 10.
Gammexane has the formula
Answer:
C6H6Cl6

Question 11.
Bayer’s reagent is __________ .
Answer:
dil. alkaline KMnO4

Question 12.
The electrophile attacking benzene during nitration is __________ .
Answer:
NO2+

Question 13.
The compound that is least readily nitrated is __________ .
a) phenol
b) Toluene
c) Ethylbenzene
d) Benzoic acid
e) Xylene
Answer:
d) Benzoic acid

Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons

Question 14.
The hydrocarbon formed when Beryllium carbide is treated with water is __________ .
Answer:
Methane

Plus One Chemistry Hydrocarbons Two Mark Questions and Answers

Question 1.
The lUPAC names of 2 compounds with their structural formulae is given below. Make out the errors and correct them.
A. 2, 2-Dimethyl-3-hexyne
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 3
B. 1-Butyne
CH3-CH2-CH=CH3
Answer:
A is wrong. The strucutral formula of 2,2-Dimethyl -3- hexyne is
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 4

Question 2.
Is there an organic compound named 2-Ethylpentane? Why? If no, write the correct answer.
Answer:
No. When an ethyl group comes with the second carbon atom, the longest chain will have 6 carbon atoms. Hence its correct name will be 3-Methyl hexane.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 5

Question 3.
Write down 2 similarities and 2 difference of the hydrocarbons given below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 6
Answer:

Similarities Dissimilarities
Both are unsaturated compounds Both have different functional group
Both hydrocarbons have same word root One hydrocarbon is alkene and the other is alkyne

Question 4.
An equation for combined Chemical reaction of acetylene is given below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 7
Find the products X’ and ‘Y’?
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 8

Question 5.
In a Chemistry class, teacher asked students to write the geometrical form of dicarboxylic acid with the formula C4H4O4. For this question, one student wrote
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 9
Both of them argued fortheir answers. Hearing this argument teacher told that both of them are right and she also explained the reason for it. Can you write the answer given by the teacher?
Answer:
Two geometrical isomers are possible for dicarboxylic acid. They are cis and transform. In the cis form, similar groups are present on the same side of the double bond whereas in transform, identical groups are present on different side of the double bond.

Question 6.
Match the following:

Inductive effect CH3-CH = CH2
Electrometric effect C6H5-NO2
Hyper conjugation CH3-CH2 – Br
Resonance effect CH3-CH2(+)

Answer:

Inductive effect CH3 – CH2 (+)
Electrometric effect CH3 – CH2 – Br
Hyper conjugation CH3-CH = CH2
Resonance effect C6H5-NO2

Question 7.
Addition of HBr to propene yields 2-bromopropane, what happens if benzoyl peroxide is added to the above reaction.
Answer:
When propene is allowed to react with HBr in the presence of peroxide, 2 bromo propane is obtained as the minor product and this is called peroxide effect. Anti markonikov’s rule.

Plus One Chemistry Hydrocarbons Three Mark Questions and Answers

Question 1.
Reactions
CH3-C ≡ C-CH3 + H2
Reaction: 2
X + HCl → Y
a) What type of reaction is reaction 1 ?
b) Find X and Y.
c) Which are the different products obtained from the reaction of X with oxygen?
Answer:
a) Addition reaction
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 10

Question 2.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 11
Here are some functional groups. You are supposed to form 3 structures of hydrocarbons using 3 different functional groups and try to find their names.
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 12

Question 3
CH3 – CH2 – CH2 – CH2 – OH, CH3 – CH2 – CH – CH3
i)For which isomerism can the example above be considered?
ii) Define it.
Answer:
i)Position isomerism
ii) Position isomerism arises as a result of the difference in the position of double bond, triple bond or functional group.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons

Question 4.
Explain the following with necessary chemicals equations.
i) Wurtz reaction
ii) Kolbe’s reaction
iii) Ozonolysisof alkenes
Answer:
i) Wurtz reaction – when alkyl halide is allowed to react with metallic sodium in presence of dry ether an alkane is obtained.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 13
ii) Kolbes reaction – When a solution of sodium acetate is electrolized, ethane is obtained.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 14
iii) Ozonolysis ofalkene: When an alkene is allowed to react with ozone, an ozonide is obtained. This on hydrolysis gives Aldehyde or ketone. The whole process is called ozonolysis.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 15

Question 5.
b) Complete the reaction:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 15
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 17

Question 6
1. Explain the reaction between sodium metal and bromoethane in dry ether. (3)
2. Draw Sawhorse and Newman’s projections of the different conformers of ethane.
Answer:
1. 2CH3CH2Br + 2Na → CH3-CH2-CH2-CH3 + 2NaBr
2.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 18

Question 7.
Analyse the given reactions, give the major products. Justify your answer.
a) HBris added to 1-Butene two products are obtained.
b) Action of excess chlorine with benzene in dark.
c) Addition of chlorine to benzene in uv light.
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 19
c) When Benzene is allowed to react with chlorine in presence of sunlight. Benzene hexa chloride (BHC) is obtained.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 20

Question 8.
Predict the product in the following reactions and identify the rules:-
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 21
Answer:
a) When propene is allowed to react with HBr in the presence of peroxide, 2-Bromo propane is obtained as the minor product. (Peroxide effect, Kharasch effect or Anti Markownikoffs rule of addition).
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 22

Question 9.
Write any two necessary condition for a compound to be aromatic. Convert Acetylene to benzene.
Answer:
Cyclic, Planar and should contain (4n+2)π electrons.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 23

Plus One Chemistry Hydrocarbons Four Mark Questions and Answers

Question 1.
Match the following:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 24
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 25

Question 2.
Fill in the blanks:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 26
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 27

Question 3.
Given below are the structures of some hydrocarbons. Pick out the correct IUPAC names for them from the box.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 28
Answer:
1) 4 – Ethyl -2, 3-dimethyl heptane
2) 3, 4-Dimethyl hex – 3-ene
3) 4, 5, 5- triethyl 3 methyl 2 heptyne
4) 3ethyl-2, 5, 5, trimethylheptane

Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons

Question 4.
a) I am an unsaturated hydrocarbon.
b) My wordroot is pent.
c) My suffix is ene. The double bond lies between 2nd and 3rd carbon atoms.
d) l have a branch of methyl group on my second carbon atom. Who am I?
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 29

Question 5.
From the given table find out the isomer pairs and which type of isomerism they have?
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 30
Answer:
Isomer pairs
a – ii – functional group isomerism
b – i – chain isomerism
c-iv-position isomerism
d – iii – metamerism Question 6

Qn 6.
Given below is the structural formula of a compound written by a student.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 31
i) Draw all the possible conformers of the compound?
ii) Arrange them in the order of stability?
Answer:
i)
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 32
ii) Eclipsed

Question 7.
A cross word puzzle.
Down
1. Two or more compounds having the same
molecular formula but different physical or chemical properties are known as isomers and the phenomenon is known as
2. Hydrocarbons having the general formula CnH2n.
3. IUPAC name of C6H5CH3.
4. ………….. is added to the word root to show whether the hydrocarbon is saturated or unsaturated.
5. Hydrocarbons contain carbon-carbon triple bond.

Cross
6. From where does the inorganic compounds mainly originate from?
7. Name the alkene with the moelcular formula C10H20
8. Prefix of the functional group carboxylic acid.
9. Compounds with two rings.
10. Compounds having same molecular formula but different arrangements of carbon atoms on either side of the same functional group are called ……..
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 33
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 34

Question 8.
1. Name the product obtained when HBr is added to propene. Why?
2. Acetylene is more acidic than ethylene or ethane. Why?
Answer:
1. When propene is treated with HBr, both 2-bromopnopane and 1-bromopropane are formed as products. The major product is2-Bromo-propane. This is due to the rule known as Markonikoffs rule of addition. It states that when a hydrogen halide is added to an unsymmetrical alkene, the halogen atom will goes to the doubly bonded carbon containing lesser number of hydrogen atoms.

2. In Acetylene, the hybridisation of carbon is sp and hence 50% s-character is present.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons

Question 9.
a) How will you prepare butane in the laboratory using ethyl bromide (CH3CH2Br) as one of the raw materials. Write relevant equation.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 35
Identify the product ‘X’. Statethe law that explains the formation of X.
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 36
The law of Anti-Markownikoff’s rule of addition explains the formation of 1 -Bromopropane.

Question 10.
1. Addition of HBrto propene yields 2-Bromopropane, while in the presence of Benzoyl peroxide. The same reaction yields 1-Bromopropane. Give reason. Justify your answer.
2. Three compounds are given. Benzene, m- dinitrobenzene and toluene. Identify the compound which will undergo nitration most easily and why?
Answer:
1. When propene is allowed to react with HBr in the presence of “peroxide” 2-Bromopropane is obtained as the minor product. (Peroxide effect or Kharach effect or Anti-Markownikoff’s role of addition) Peroxide effect is applicable only in the case of HBr.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 37

2. Toluene. This is because the -CH3 group, being an activating group activates the benzene ring towards electrophilic substitution in toluene.

Question 11.
a) Write IUPAC names of the products obtained by addition reactions of HBr to hex-1-ene:
i) In the absence of peroxide.
ii) In the presence of peroxide.
b) Howwill you convert:
i) Benzene to toluene
ii) Benzene to nitrobenzene
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 38

Question 12.
The equations for two chemical reactions are given below:
i) CH ≡ CH + HCl → A → B
ii) OH4 + O2 → C + D
Which are the products A, B, C, and D?
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 39

Question 13.
The 2 conformations shown below belongs to the compound cyclohexane. Infinite number of conformations are possible for cyclo hexane. But these 2 conformations given below has a peculiarity. Try to find it. Also, define the terms conformers and the phenomenon conformation?
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 40
Answer:
Infinite number of conformations are possible for cyclohexane. Out of it chair conformation is the most stable one and the boat conformation is the least stable form.

The different arrangement of a compound which arises as a result of rotation about carbon single bond are called conformers and the phenomenon is called conformation.

Plus One Chemistry Hydrocarbons NCERT Questions and Answers

Question 1.
How do you account for formation of ethane during chlorination of methane? (3)
Answer:
Chlorination of methane takes place through a free radical chain mechanism as given below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 41
From the above mechanism, it is evident that during chain propagation step, CH3free radicals areproduced. In the chain termination step, the two free CH3 radicals may combine together to form ethane (CH3-CH3) molecule.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons

Question 2.
For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of double or triple bond as indicated: (4)
a) C4H8 (one double bond)
b) C3H8 (one triple bond)
Answer:
a) Isomers of C4H8 having one double bond are:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 47

Question 3.
What are the necessary conditions for any compound to show aromaticaly? (3)
Answer:
The conditions for a compound to show aromaticity are:
i) The molecule must be cyclic
ii) It must have a conjugated system of (4n+2) π- electrones
iii) The molecule must be planar so that delocalization of π-electrones can take place.

Question 4
Explain why the following system are not aromatic?
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 42
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 43
It dose not contain all the π- electrons in the ring. Therefore, it is not an aromatic compound.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 44
It contains only four electrons, therefore, the system is not aromatic because it dose not contain (4n + 2) π- electrones.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 45
Cyclootatetraene is a conjugated system having 8 π-electrons.
Therefore, the molecule is not contain (4n+2) π-electrons.

Question 5.
In the alkane, H2CCH2C(CH3)2 CH2CH(CH3)2, identify 1°, 2°, 3° carbon atoms and give the total number of atoms bonded to each one of these.
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons 46
1° Carban atoms = 5
Hydrogen atoms attached to 1° carbon atoms = 15 2° Carbon atoms = 2
Hydrogen atoms attached to 2° carbon atoms = 4 3° Carbon atom = 1
Hydrogen atoms attached to 3° carbon atom = 1

Plus One Chemistry Chapter Wise Questions and Answers Chapter 13 Hydrocarbons

Question 6.
What effect does branching of an alkane chain has on its melting point?
Answer:
As the branching increases melting point increases.

Question 7.
Why does benzene undergo electrophilic subsitution easily and nucleophilic substitutions with difficulty? (2)
Answer:
Benzene molecule has two n cloud rings, one above and the other below the plane of atoms. Therefore, it is likely to be attached by electrophiles which subsequently brings about substitution.

The nucleophiles would be repelled by the π-electron rings and hence benzene reacts with nucleophiles with difficulty.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry: Some Basic Principles and Techniques

Students can Download Chapter 12 Organic Chemistry: Some Basic Principles and Techniques Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry: Some Basic Principles and Techniques

Plus One Chemistry Organic Chemistry: Some Basic Principles and Techniques One Mark Questions and Answers

Question 1.
Which of the following does not contain fused benzene ring?
a) Naphthalene
b) Anthracene
c) Diphenyl
d) Phenanthrene
Answer:
c) Diphenyl

Question 2.
Hemolytic fission of a covalent bond results in the formation of ___________ .
CH2=CH – C ≡ CH2OH
Answer:
Free radicals

Question 3.
The IUPAC name of is
a) 2-pentyl-4-en-1-ol
b) 4-penten-2-yn-1-ol
c) 1-pentene-3-yn-5-ol
d) 5-Hydroxy-1-pentene-3-yne
Answer:
b) 4-penten-2-yn-1-ol

Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry: Some Basic Principles and Techniques

Question 4.
Lassigne’s solution on treating with sodium nitro prusside solution gives a violet colour indication the presence of ___________ in the organic compound.
Answer:
sulphur

Question 5.
In Kjeldahl’s method, nitrogen present is estimated as ___________ .
Answer:
NH3

Question 6.
In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue colour is obtained due to the formation of:
a) Na4[Fe(CN)6]
b) Fe4[Fe(CN)6]3
c) Fe2[Fe(CN)6]
d) Fe3[Fe(CN)6]4
Answer:
The Prussian blue colour i§ due to the formation Fe4[Fe(CN)6]2. Thus, option (b) is correct,

Question 7.
Which of the following carbocation is most stable?
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 1
Answer:
The order of stability of carbocation is : 3º>2º>1º
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 2
Thus, option (b) is correct.

Question 8.
The best and latest technique for isolation, purification and separation of organic compounds is:
a) Crystallisation
b) Distillation
c) Sublimation
d) Chromatography
Answer:
Chromatography. Thus, option (d) is correct.

Question 9.
The following reaction is classified as:
CH2CH2l + KOH (aq) → CH3CH2OH + Kl
a) Electrophilic Substitution
b) Nucleophilic substitution
c) Elimination
d) Addition.
Answer:
This is an example of nucleophilic substitution reaction since the nucleophile l – is replaced by the nucleophile OH ion. Thus, option (b) is correct.

Question 10.
Which of the carbocation is more stable?
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 3
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 4

Question 11.
Absolute alcohol cannot be obtained by fractional distillation because
Answer:
constant boiling azeotrope mixture is formed with water

Question 12.
Lassaigne’s test fails in
a) NH2NH2
b) H2NCONH2
c) C6H2NHNH2
d) NH2OH
Answer:
a) Hydrazine NH2NH2

Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry: Some Basic Principles and Techniques

Question 13.
Beilstein test is for the detection of __________ .
Answer:
Halogens

Question 14.
Glycerine can be purified by __________ .
Answer:
Distillation under reduced pressure.

Plus One Chemistry Organic Chemistry: Some Basic Principles and Techniques Two Mark Questions and Answers

Question 1.
2-Butene exhibits geometrical isomerism.
1. Represent the cis-trans isomers of 2-Butene.
2. Explain ozonolysis with a suitable example.
Answer:
1.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 5
2. When an alkene is allow to react with ozone, an ozonide is obtained. This on hydrolysis gives aldehyde/ ketone. The whole process is called ozonolysis. Eg:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 6

Question 2.
Draw the structures of the molecules represented by the IUPAC names, Pent-3-en-1-ol and 2- Nitrocyclohexene.
Answer:
a) CH3 – CH = CH – CH2 – CH2 – OH
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 7

Question 3.
Addition of HBr to propene yields 2-bromopropane, what happens if benzoyl peroxide is added to the above reaction.
Answer:
When propene is allowed to react with HBr in the presence of peroxide, 2 bromo propane is obtained as the minor product and this is called peroxide effect.

Question 4.
Write the IUPAC name of
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 8
Answer:
i) 2-Pentanone
ii) 3-Methyl-1-pentanal

Question 5.
Write any two necessary condition for a compound to be aromatic. Convert Acetylene to benzene.
Answer:
Cyclic, Planar and should contain (4n+2) n electrons.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 9

Question 6.
What are hybridisation states of each carbon atom in the following compounds?
CH2 = C = O, CH3CH = CH2,(CH3)2CO,
CH2 = CHCN, C6H6
Answer:
The hybridisation of each carbon is written as superscript on the carbon atom in the molecule.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 10

Question 7.
Which is expected to be more stable, O2 NCH2CH2O or
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 11
has -l- effect and hence it tends to disperse the -ve charge on the O-atom. In contrast, CH3CH2 exerts + l – effect. It, therefore, tends to intensity the -ve change and hence destabilizes it.

Plus One Chemistry Organic Chemistry: Some Basic Principles and Techniques Three Mark Questions and Answers

Question 1.
a) Write the structure of the following compounds.
i) Hexane-2, 4-dione
ii) 3-Bromo-4 methyl hexane-2-ol
iii) 2-Bromo-4-methyl hept-5-enal
Answer:
a) i) Hexane 2, 4-dione
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 12

Question 2.
Isomers are compounds with same formula and different properties.
a) Write all the possible structural isomers of C5H12.
b) Give the IUPAC name of above isomers.
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 13
b) 1) Pentane
2) 2-Methyl butane
3) 2,2-Dimethyl propane

Question 3.
a) Give the IUPAC names of:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 14
b) How is nitrogen detected by Lassaignes test?
c) Name a suitable technique for separation of the components from a mixture of benzene (b.p.353 K) and aniline (b.p.-457 K)
Answer:
a) (i) 2-Chloro 3-methyl hexane
(ii) 2-Methylbutanal

b) About 2 mL of the extract is boiled with about 1 mL of freshly prepared ferrous sulphate solution. One or two drops of con.H2SO4 are added to the solution. Presence of nitrogen is indicated by the appearance of a blue colouration.

c) Distillation

Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry: Some Basic Principles and Techniques

Question 4.
i) Draw the structure of propanone. Write the hybridisation of each carbon in propanone.
ii) Arrange the following carbocations in the increasing order of their stability. Justify.
CH3+, CH3CH2+, (CH3)2CH+
iii) What is the method used to separate a mixture of o-Nitrophenol from p-Nitrophenol? Which property is utilized for separation?
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 15
Ist carbon of Propanone is sp³ hybridisation.
IInd carbon of Propanone is sp² hybridisation.
IIIrd carbon of Propanone is sp³ hybridisation.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 16
iii) Chromatography; Difference in absorption rate of different substances..

Question 5.
1. Give the IUPAC name of the following compound.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 37
2. How can you detect the presence of nitrogen in an organic compound?
3. Arrange the following in the increasing order of stability.
(CH3)2CH+, CH3-CH2+, (CH3)3C+
Answer:
1. 2-chloro, 2-methyl propane

2. Presence of nitrogen can be detected by Lassigne’s test. The sodium fusion extract is boiled with FeSO4 and then acidified with con.H2SO4. The formation of Prussian blue colour conforms the presence of nitrogen.
3.Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 17

Question 6.
Predict the product in the following reactions and identify the rules:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 18
Answer:
a) When propene is allowed to react with HBr in the presence of peroxide, 2-Bromo propane is obtained as the minor product. (Peroxide effect, Kharasch effect or Anti MarkownikofFs rule of addition).
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 19

Question 7.
a) Draw the structures of the following compounds.
i) 3-hexenoic acid
ii) 2-chloro-2-methyl butanol
iii) 4-nitro-1-pent-l-yne
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 20

Question 8.
Write the IUPAC names of the following:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 21
Answer:
i) 2,2,4-Trimethyl pentane
ii) 2-Methyl 1-butene
iii) Propyl benzene

Question 9.
Draw the structure of the following molecules.
i) 3, 4-Dimethylhept-3-ene
ii) Neo-pentane
iii) 3-Nitrocyclohexene
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 22

Question 10.
What is the relationship between the members of following pairs of structures? Are they structural or geometrical isomers or resonance contributors?
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 23
Answer:
a) Structural isomers (actually position isomers as well as metamers)
b) geometrical isomers,
c) resonance contributors because they differ in the position of electrons but not atoms:

Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry: Some Basic Principles and Techniques

Question 11.
Expain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and phosphorus.
Answer:
The organic compound is fused with sodium metal to convert these elements (which are present in the covalent form) to ionic form. For example, Sulphur is changed to Na2S, nitrogen to NaCN and phosphorus to Na3PO4 The presence of sulphide ions, cyanide ions and phosphate ions can thus be confirmed by using suitable reagents.

Plus One Chemistry Organic Chemistry: Some Basic Principles and Techniques Four Mark Questions and Answers

Question 1.
1) Write the structural formula of
a) 4-Ethyl-1-fluoro-2-nitrobenzene
b) 2,3,6-Trimethyl octane.
c) 1,2-Dibromo benzene.
2) Categorize the following as nucleophile and electrophile
a) HS
b) BF3
c) NO2+
d) C2H5O
e) (CH3)3N
f)NH2
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 24

Question 2.
1. What is chromatography? Name the different types of chromatography.
2. What is Lassaigne’s test?
Answer:
1. Chromatography is a valuable method for the separations, purification and identification of the constituents of a mixture. Chromatography is classified into 2 types.
a) Adsorption chromatography
b) Partition chromatography

2. Lassigne’s test is used to determine the presence of nitrogen, halogens and sulphur present in the organic compound.

Question 3.
1. Write IUPAC names of the products obtained by addition of HBr to Hex-1-ene.
i) in the absence of peroxide
ii) in the presence of peroxide
2. Complete the following reaction:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 25
Answer:
1.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 38
2. 2HCHO + H2O2

Question 4.
1. What is metamerism? Give example for metamers.
2. What are free radicals? How are they formed?
Answer:
1. It is the isomerism which arises due to different alkyl chains on either side of the functional group in the molecule.
e.g. Methoxypropane(CH3OC3H7) and ethoxyethane (C2H5OC2H5) are metamers.

2. Free radicals are highly reactive species containing unpaired electrons. They are formed by homolytic cleavage of covalent bond. e.g. \(\dot { C } \)H3

Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry: Some Basic Principles and Techniques

Question 5.
1. How will you prepare butane?
2. Explain Markownikoff’s rule for the addition reaction using a suitable example.
Answer:
1. When an alkyl halide (ethyl chloride) is allowed to react with metalic sodium in presence of dry ether,
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 26
2. Markownikoff’s rule of addition:
When a hydrogen hallide is added to an unsymmetrical alkane the halogen atom will goes to the double bond carbon containing lesser number of hydrogen atom, eg:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 27

Question 6.
Identify the isomerism exhibited by the following compounds.

  1. CH3CH2CH2OH and (CH3)2CHOH
  2. CH3CH2CHO and CH3COCH3
  3. CH3CH22CH2CH3andCH3OCH2CH2CH3
  4. CH3(CH2)3 CH3 and (CH3)4C

Answer:

  1. Position isomerism
  2. Functional group isomerism
  3. Metamerism
  4. Chain isomerism

Question 7.
i) Predict the products of
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 28

ii) Classify the following compounds into aromatic and non-aromatic.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 29

(iii) Which of the following compounds will show geometrical isomerism?
a) CH3 CH=CHCH3
b) (CH3)23 C=CH2
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 30

Question 8.
Write the IUPAC name
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 31
b) Write the structure of the following.
i) 2-Chloro-2-methyl butanol
ii) 4-Nitro-1-pentene
Answer:
a) i) 4-Methyl pentanal
ii) Cyclohexanol
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 32

Question 9.
a) Write IUPAC names of the products obtained by addition reactions of HBrto hex-1-ene:
i) In the absence of peroxide.
ii) In the presence of peroxide.
b) How will you convert:
i) Benzene to toluene.
ii) Benzene to nitrobenzene
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 33

Question 10
a) Explain the term
(i) Inductive effect
(ii) Nucleophile
b) Write the structural formula of the following.
i) 2,5,6-Trimethyloctane
ii) 2,4-Dimethylpentane
c) Suggest the suitable technique for separation of organic compounds given in the data.
i) Aniline-water mixture
ii) Glycerol from spent-lye
Answer:
a) Inductive effect:
If more electronegative atom X linked to a carbon atom then the bonded electron pair will be shifted more towards X. So X acquires a small negative charge and carbon get a small charge. So this carbon atom attracts the a bonded electron pair towards its from nearest carbon atom. Therefore the change is transferred from one carbon to other.

Nucleophile :
Nucleus loving species negatively charge species like \(\overline { OH } \), \(\overline { X } \) etc. are called nucleophiles.
b) 2,5,6-Trimethyloctane
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 39

c) i) Fractional distillation.
ii) Distillation under reduced pressure.

Question 11.
Compounds having the same molecular formula exhibit different properties is called isomerism. Explain different types of isomerism with examples.
Answer:
Chain isomerism: Consider the molecular formula C5H12 the following 3 isomers are possible for this compound.
They are
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 34
Position isomerism:
This isomerism arises as a result of the difference in the position of double bond triple bond and functional group.
The following chain isomers are possible for the molecular formula C4H8.
CH2 = CH – CH2 – CH3 1-butene
CH3 – CH = CH – CH3 2-butene

Metamerism:
This isomerism arises as a result of the difference in the alkyl group present on either side of the functional group.
Consider the molecular formula C4H10O. The following two metamers are possible for this molecular formula.
CH3 – O – CH2 -CH2 -CH3 Methyl propyl ether
CH3 – CH2 – O – CH2 – CH3 Diethyl ether

Functional group isomerism:
Consider the molecular formula C2H6O the following two functional isomers are possible for this compound.
1) CH3-CH2-OH Ethanol
2) CH3 – O – CH3 Di methyl ether

Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry: Some Basic Principles and Techniques

Question 12.
Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?
Answer:
Sodium extract is boiled with nitric acid to decompose NaCN and Na2S, if present,
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 35
If the sodium extract is not boiled with nitric acid then NaCN and Na2S formed will react with AgN03 and hence will interfere with the test as shown below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry Some Basic Principles and Techniques 36

Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements

Students can Download Chapter 11 The p Block Elements Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements

Plus One Chemistry The p Block Elements One Mark Questions and Answers

Question 1.
The aqueous solution of borax is
a) Acidic
b) Alkaline
c) Neutral
d) Amphoteric
Answer:
b) Alkaline

Question 2.
Say TRUE or FALSE.
Boron in aqueous solution forms B3+ ion.
Answer:
False

Question 3.
Which of the halide of group 14 does not exist?
a) CF4
b) Cl4
c) SiF4
d) Pbl4
Answer:
d) Pbl4

Question 4.
Orthoboric acid, H3BO3 is a
a) Protonic acid
b) Arrhenius acid
c) Lewis acid
d) Bronsted-Lowery acid
Answer:
c) Lewis acid

Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements

Question 5.
The zeolite used as a catalyst in petrochemical industries for cracking of hydrocarbons and isomerization is _________ .
Answer:
ZSM-5

Question 6.
Dry ice is __________ .
Answer:
Solid CO2

Question 7.
Thermodynamically most stable allotrope of carbon is __________ .
Answer:
Graphite

Question 8.
The alkalimetal used in solar cells is __________ .
Answer:
Cs (Caesium)

Question 9.
Acidity is in the order
Answer:
BBr3 > BCl3 > BF3

Question 10.
AlCl3 fumes in most air because
Answer:
HCl is formed due to hydrolysis in moist air

Plus One Chemistry The p Block Elements Two Mark Questions and Answers

Question 1.
Silicon belongs to the carbon family. Graphite is an important allotrope of carbon. But Si does not form an analogue of graphite. What are the possible reasons?
Answer:
Silicon atom is much bigger in size than carbon. Si-Si bond energy is less than C-C bond energy and also Si does not form compounds in sp² hybrid state.

Question 2.
Boron is an element with atomic number 5.
a) Write down the electronic configuration of boron.
b) Mention any two uses of boron.
c) Write down some compounds of boron.
Answer:
a) 1s²2s²2p¹
b) To increase the hardness of the steel.
Used as the semiconductor in electronic devices.
c) Diborane, borax, orthoboric acid, boron trifluoride

Question 3.
Answer the questions, with the help of the following:

  • Hard solid
  • Melting point above 450 K.
  • Four allotropic forms are known.
  • Low electrical conductivity.
  • Mass number! 3

1. Which is the element?
2. Write any two uses of this element.
3. Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements 1
4. Write any three compounds of this element.
Answer:
1. Boron
2. To increase hardness of steel
As semiconductor in electric devices
3. 2B
4. Diborane, Borax, Boric acid

Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements

Question 4.
1. Write any two allotropic forms of carbon.
2. C + ½ O2 → …………….
3. Write any two uses of carbon monoxide.
4. How is carbon dioxide produced?
5. What is the hardest element/form of an element in the world?
Answer:
1. Diamond, Graphite
2. CO (Carbon monoxide)
3. Reducing agent in metallurgy
Used in the manufacture of methanol
4. Carbon dioxide is formed by the complete combustion of carbon.
C + O2 → CO2
5. Diamond

Question 5.
From the compounds of group 14 elements write an appropriate example for each of the following:

No. Type of compound Name/ Formulae of example
1. A strong reducing oxide
2. A giant covalent oxide
3. A strongly reducing chloride
4. A covalent chloride not hydrolysed by water

Answer:
1-CO
2-SiO2
3-SnCl2
4-CCl4

Question 6.
Diborane has an unusual structure. Justify the statement with figure.
Answer:
In diborane, each Batom uses sp³ hybrid orbitals for bonding. Out of the four sp³ hybrid orbitals on each B atom, one is without an electron. The terminal B-H bonds are normal 2-centre-2-electron bonds but the two bridge bonds are 3-centre-2-electron bonds. The 3-centre-2-electron bridge bonds are also referred to as banana bonds.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements 2

Question 7.
Match the following:

A B
CO i. A semi conductor in electronic devices
H2 ii. Reducing agent in metallurgy
O3 iii. Thermal decomposition of Ammonia
Boron iv. Used as a chemical reagent in organic chemistry

Answer:
CO – ii,
N2 – iii,
O3 – iv,
Boron – i

Question 8.
BCl3 fumes in moist air.
1. Give reason.
2. Write down a balanced equation that can reveal the answer.
Answer:
1. When water of moist air reacts with boron halide hydrogen chloride is formed which causes fumes,

2. BCl3 + 3H2O → H3BO3 + 3HCl

Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements

Question 9.
‘Generally, non-metal oxides are basic.’
1. Do you agree?
2. What do you meant by oxides?
3. Which are the different types of oxides?
4. Give examples for each type of oxides.
Answer:
1. No. Generally, non metallic oxides are acids.
2. The binary compounds formed by the combination of oxygen with metals or nonmetals are called oxides.
3. Acidic oxides, basic oxide, amphoteric oxide, neutral oxide.
4. Acidic oxide → SO2, NO2
Basic oxide → MgO
Amphoteric oxide → Al2O3, ZnO
Neutral oxide → CO, N2O

Question 10.
Match the following:
1. Borane – H3BO3
2. Boric acid – Na2B4O7.10H2O
3. Borax – Amphoteric oxide
4. Al2O3 – Boron hydride
Answer:
1. Borane – Boron hydride
2. Boric acid – H3BO3
3. Borax – Na2B4O7.10H20
4. Al2O3 – Amphoteric oxide

Question 11.
How diborane reacts with
1. Oxygen?
2. Water?
Answer:
1. Diborane catches fire spontaneously upon exposure to air. It burns in oxygen releasing an enormous amount of energy.
B2H6 + 3O2 → B2O2 + 3H2O; ∆rHΘ -1976 kJ mol-1

2. Diborane is readily hydrolysed by water to give boric acid.
B2H6(g) + 6H2O(l) → 2B(OH)3(aq) + 6H2(g)

Question 12.
1. CCl4 cannot be hydrolysed. Give reason.
2. Draw the structure of the dimer of AlCl3.
Answer:
1. Carbon has no d orbitals to accommodate the lone pair of electrons from oxygen atom of H20.
2.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements 3

Question 13.
1. Draw the structure of boric acid.
2. Starting from borax how will you prepare boric acid? (Write the chemical equation).
Answer:
1.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements 4
2. Na2B4O7 + 2HCl + 5H2O → 2NaCl + 4B(OH)3.

Question 14.
CO2 is a gas but SiO2 is a solid. Give reason.
Answer:
CO2 exists as discrete molecules due to the formation of pπ -pπ double bond between carbon and oxygen. But SiO2 has a three dimensional network structure in which each Si atom is covalently bonded in a tetrahedral manner to four oxygen atoms.

Question 15.
1. What are zeolites?
2. What is ZSM-5?
Answer:
1. Zeolites are aluminosilicates with three-dimensional network structure in which Al atoms replace few Si atoms. Cations such as Na+, K+ or Ca2+ balance the negative charge of aluminosilicate anion.

2. ZSM-5 is a zeolite catalyst used in petrochemical industry to convert alcohols directly into gasoline.

Plus One Chemistry The p Block Elements Three Mark Questions and Answers

Question 1.
Explain the following:
1. Allotropy
2. Coke and Charcoal
Answer:
1. The phenomenon of existence of an element in two or more forms, which have different physical properties but almost similar chemical properties, is called allotropy and the different forms are called allotropes.

2. Coke and Charcoal are amorphous forms of carbon. Coke is formed-by the destructive distillation of coal. It is used as a fuel and also as a reducing agent in metallurgy.

Question 2.
Give justifications.
1. The first ionisation enthalpy of carbon is greater than that of boron, whereas the reverse is correct for the second ionisation enthalpy.
2. Graphite is a better lubricant on moon than that on earth.
Answer:
1. This is because carbon has greater nuclear charge. For second ionisation enthalpy, an electron is to be removed from 2p of carbon while in boron the second electron is in 2s orbital. Removal of a 2s electron is more difficult due to the high penetrating power of 2s orbital.

2. Graphite contains hexagonal sheets of carbon held together by weak van der Waals’ forces and low density. In the moon, the different layers experience less weight due to less gravitational force and the density is still reduced. The lighter layers can easily slide over another to make graphite more lubricating on the moon than on earth.

Question 3.
1. Identify the compound with the following structure:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements 5
2. Mention the angles a & b.
3. Write down the corresponding elements in the figure.
1……………..
2……………..
3……………..
4……………..
5……………..
6……………..
Answer:
a) Diborane(B2H6).
b) a-97°
b-120°
c) 1-H, 2-H, 3-B, 4-B, 5-H, 6-H

Question 4.
The simplest boron hydride is diborane.
1. Write down the molecular formula of diborane.
2. Distinguish between terminal hydrogen and bridging hydrogen atoms of diboran.
Answer:
1. B2H6

2. In diborane, four hydrogen atoms and two boron atoms are present in a single plane. These hydrogen atoms are called terminal atoms. The terminal B-H bonds are regular two centre-two electron bonds. The remaining two hydrogen atoms above and below this plane are called bridging hydrogen atoms. The two bridge bonds (B-H-B) are three centre-two electron bonds.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements

Question 5.
1. How is orthoboric acid prepared?
2. Account for the acidic nature of orthoboric acid.
Answer:
1. Orthoboric acid is prepared by acidifying an
aqueous solution of borax.
Na2B4O7 + 2HCl + 5H2O → 2NaCl + 4B(OH)3

2. Orthoboric acid is a weak nonobasic acid. It acts as a Lewis acid by accepting electrons from a hydroxyl ion.
B(OH)3 + 2HOH → [B(OH)3] + H3O+

Question 6.
1. How is diborane prepared in the laboratory?
2. BCl3 is a good Lewis acid. Why?
Answer:
1. Diborane can be conveniently prepared in the laboratory by the oxidation of sodium borohydride with iodine.
2NaBH4 + l2 → B2H6 + 2Nal + H2

2. In BCl3, the central boron atom contains only six electrons. Hence, it has the tendency to accept electrons and acts as a Lewis acid.

Question 7.
1. Name the allotropes of carbon.
2. Carbon monoxide is highly poisonous. Do you agree? Justify.
Answer:
1. Graphite, diamond and fullerene.

2. I agree with this statement. Because, CO has strong and reversible binding with haemoglobin resulting in the formation of carboxyhaemoglobin. This reduces the amount of haemoglobin available in blood for oxygen transport. This causes laboured respiration, muscle weakness and even death.

Question 8.
1. Diamond is hard and non conducting while graphite is soft and conducting. Why?
2. Explain the action of heat on boric acid.
3. What is inorganic benzene? How is it formed?
Answer:
1. In diamond, all the carbon atoms are in sp³ hybridised state. Due to this closely packed arrangement, it is hard. Since, there are no free electrons it is an insulator. But in graphite, the carbon atoms are in sp² hybridised state. It is a good conductor due to the presence of delocalised electrons between the layers. Graphite cleaves easily between the layers. Therefore, It is very soft and slippery.

2. On heating, orthoboric acid above 370 K forms metaboric acid, HBO2 which on further heating yields boric oxide, B2O3.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements 6

3. B3N3H6 is called Inorganic benzene. It is obtained by heating diborane with ammonia.

Question 9.
1. Explain the difference in properties of diamond and graphite on the basis of their structures.
2. How do you explain the lower atomic radius of gallium as compared to aluminium?
Answer:
1. In diamond, all the carbon atoms are in sp³ hybridised state. As a result of this hybridisation, a closely packed arrangement is present in diamond and it explains the hardness of diamond. Due to the absence of electrons, diamond is an insulator.

In graphite, the carbon atoms are in sp² hybridised state. As a result of this hybridisation, graphite has a layered structure with hexagonal rings. Each layer can slide over the other which explains the lubricating property of graphite.

2. This can be explained on the basis of the variation in the inner core of the electronic configuration. The presence of additional 10 d-electrons offer only poor screening effect for the outer electrons from the increased nuclear charge in gallium. Consequently, the atomic redius of gallium (135 pm) is less than that of aluminium (143 pm).

Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements

Question 10.
1. Boron resembles silicon in many of its properties. What is this resemblance generally known as?
2. What is dry ice? What is it used for?
3. What are silicones? How do they differ from silicates?
Answer:
1. Diagonal relationship

2. Carbon dioxide can be obtained as a solid in the form of dry ice, by allowing the liquified CO2 to expand rapidly.
Dry ice is used as a refrigerant for ice-cream and frozen food.

3. Silicones are a group of organosilicon polymers containing R2SiO repeating units.
Silicates are minerals with SiO44- as the basic structural unit. In silicates either the discrete unit is present or a number of such units are joined together via corners by sharing 1, 2, 3 or 4 oxygen atoms per silicate units.

Plus One Chemistry The p Block Elements Four Mark Questions and Answers

Question 1.
Match the following:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements 7
Answer:
1 – c – f
2 – a – h
3 – b – g
4 – d – e

Question 2.
a) Elements of group 13 are
1) Al, Cr, Cd, Ga, Ti
2) B, C, Si, Ga, Ti
3) B, Al, Ga, Ti, In
4) B, Al, Cr, Ca, Ti
b) Which of the following is not a mineral of boron? (borax, bauxite, colemanite, tincal)
c) In the reaction between NH3 and BF3 ammonia acts as
(Lewis base, Lewis acid, brownsted base)
d) Diamond and Graphite are carbon’s
(isotropic forms, allotropic forms, isotopic forms, amorphous form)
Answer:
a) 3) B, Al, Ga, TI, In
b) Bauxite
c) Lewis base
d) allotropic forms

Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements

Question 3.
The hydrides of boron are called boranes.
1. How diborane reacts with ammonia?
2. Account for the exceptional hardness of diamond.
Answer:
1. Diborane reacts with ammonia to give B2H6.2NH3
initially, which can be formulated as [BH2(NH3)2]+[BH4]. Further heating gives borazine, B3N3H6 known as ‘inorganic benzene’ in view of its ring structure with alternate B-H and N-H groups,

2. In diamond, all the carbon atoms are in sp3 hybridised state. As a result of this hybridisation, a rigid three dimensional network of carbon atom is generated with directional covalent bonds throughout the lattice. It is very difficult to break this network structure.

Plus One Chemistry The p Block Elements NCERT Questions and Answers

Question 1.
Why does boron trifluoride behave as a Lewis acid? (2)
Answer:
The B atom in BF3 has only 6 electrons in the valence shell and thus needs two more electrons to complete its octet. Therefore, it easily accepts a pair of electrons from nucleophiles such as F, NH3, (C2H5)2O, RCH2OH etc. and thus behaves as a Lewis acid.

Question 2.
Explain why is there a phenomenal decrease in ionisation enthalpy from carbon to silicon? (2)
Answer:
Due to increase in atomic size and screening effect the force of attraction of the nucleus for the valence electron decreases considerably in Si as compared to C. As a result, there is a phenomenal decrease in ionisation enthalpy from carbon to silicon.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements

Question 3.
Consider the compounds, BCl3 and CCl4. How will they behave with water? (2)
Answer:
The B atom in BCl3 has only six electrons in the valence shell and hence is an electron-deficient molecule. It easily accepts a pair of electrons donated by water and hence BCl3 undergoes hydrolysis to form boric acid (H3BO3) and HCl.
BCl3 + 3H2O → H3BO3 + 3HCl

In contrast, C atom in CCl4 has 8 electrons in the valence shell. Therefore, it is an electron-precise molecule. As a result, it neither accepts nor donates a pair of electrons. In simple words, it does not accept a pair of electrons from H2O molecule and hence CCl4 does not undergo hydrolysis in water.

Question 4.
Aluminium trifluoride is insoluble in anhydrous HF but dissolves on addition of NaF. Aluminium trifluoride precipitates out of the resulting solution when gaseous BF3 is buddled through. Give reasons. (3)
Answer:
Anhydrous HF is a covalent compound and is strongly H-bonded. Therefore, it does not give free F ions and hence AlF3 does not dissolve in HF. In contrast, NaF is an ionic compound and hence F ions are easily available. As a result, it combines with AlF3 to form the soluble complex.
3NaF + AlF3 → Na3[AlF3]

On bubbling gaseous BF3, AlF3 is precipitated. It is because BF3 is a stronger Lewis acid than AlF3. This is attributed to smaller size and higher electronegativity of Boron. As a result of this B has much higher tending to form complexes than Al. Therefore, when BF3 is added to the above solution, AlF3 gets precipitated.
Na3[AlF6] + 3BF3 → 3Na[BF4] + AlF3(s)

Plus One Chemistry Chapter Wise Questions and Answers Chapter 11 The p Block Elements

Question 5.
In some of the reactions, thallium resembles aluminium, whereas in others it resembles with group 1 metals. Support this statement by giving some evidences. (3)
Answer:
Aluminium shows a uniform oxidation state of+3 in its compounds. Like aluminium, thallium also shows +3 oxidation state in some of its compounds like TlCl3, Tl2O3, etc. Al is known to form octahedral complexes like [AlF6]3-. Similarly, Tl also forms octahedral complexes as [TlF6]3-.

Thallium also resembles group 1 metals. Like group 1 metals which show a stable oxidation state of +1 in their compounds Tl, due to inert pair effect, also shows +1 oxidation state in some of its compounds such as Tl2O, TlCl, TlClO4, etc. Similarly, like group 1 oxides, Tl2O is strongly basic.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 10 The s Block Elements

Students can Download Chapter 10 The s Block Elements Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 10 The s Block Elements

Plus One Chemistry The s Block Elements One Mark Questions and Answers

Question 1.
The element placed at the bottom of the alkali metal family is expected to
a) Have maximum ionisation enthalpy
b) Be the least reducing agent
c) Be the least electropositive element
d) Be the most easily ionisable
Answer:
d) Be the most easily ionisable

Question 2.
Which of the following have lowest thermal stability?
a) Li2CO3
b) Na2CO3
c) K2O3
d) Rb2CO3
Answer:
a) Li2CO3

Plus One Chemistry Chapter Wise Questions and Answers Chapter 10 The s Block Elements

Question 3.
The chemical formula of Plaster of Paris is
a) Ca2SiO4.½H2O
b) Ca2SiO4.2H2O
c) CaSO4.½H2O
d) CaSO4.2H2O
Answer:
c) CaSO4.½H2O

Question 4.
A mixture of NaOH and CaO is known as _________ .
Answer:
Soda lime

Question 5.
Which of the following is least soluble in water?
a) BeSO4
b) BaSO4
c) CaSO4
d) SrSO4
Answer:
BaSO4

Question 6.
Pick out the odd one and write reason for it.
Ca(OH)2, Mg(OH)2, Ba(OH)2, Be(OH)2 Sp
Answer:
Be(OH)2. The others hydroxides are basic but Be(OH)2 is amphoteric.

Question 7.
The stability of alkaline earth metal carbonate
Answer:
Increases from Be to Ba

Question 8.
Mg2C3 on hydrolysis gives _________
Answer:
Propyne

Plus One Chemistry Chapter Wise Questions and Answers Chapter 10 The s Block Elements

Question 9.
The raw materials required for the manufacture of cement clinker are
Answer:
limestone & clay

Question 10.
A sodium potassium alloy is used as a _________ .
Answer:
coolant in nuclear reactor.

Question 11.
Li2CO3 decomposes at a lower temperature whereas Na2CO3 at highertemperature _________ .
Answer:
Li+ is very small

Plus One Chemistry The s Block Elements Two Mark Questions and Answers

Question 1.
Choose the false statements and rewrite them.
a) Alkali metals possess both +1 & +2 oxidation states.
b) Lithium is found to be the strongest reducing agents among the alkali metals.
c) Manufacturing of rayon is known as viscose process.
d) Washing soda is used to remove temporary hardness of water.
Answer:
a) False. Alkali metals possess only +1 oxidation state.
b) True
c) True
d) False. Washing soda is used to remove the permanent hardness.

Question 2.
a) Write the scientific name of slaked lime.
b) What is its role in white wash?
c) What do you mean by slaking of lime?
d) What happens when slaked lime is treated with dry chlorine?
Answer:
a) Calcium hydroxide
b) It is disinfectant hence it is used in white wash.
c) Calcium hydroxide is obtained by adding water to quick lime. This process is called slaking of lime.
d) Calcium hydroxide reacts with dry chlorine to form calcium hypochlorite, a constituent of bleaching powder.
2Ca(OH)2 + 2Cl2 → CaCl2 + CaOCl2 + 2H2O

Question 3.
1. Which metal is present in chlorophyll?
2. What is the role of calcium in our body?
Answer:
1. Mg
2. About 90% of body calcium is present in bones and teeth. It also plays important roles in neuromuscular function, intemeuronal transmission, cell membrane integrity and blood coagulation.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 10 The s Block Elements

Question 4.
A compound is used as drying agent as such or as soda lime with NaOH and it is used on a very large scale as a building material.
a) Which is this compound?
b) How can we prepare this compound?
c) What are the properties of this compound?
Answer:
a) Calcium oxide
b) It can be prepared by heating lime stone in a rotary kiln at 1070-1270 K.
c) Pure calcium oxide is an amorphous white solid of very high melting point. It is sparingly soluble in water. Calcium oxide readily absorbs moisture and carbon dioxide.

Question 5.
Listen the chemical reaction,
Plus One Chemistry Chapter Wise Questions and Answers Chapter 10 The s Block Elements 1

  1. Write the common name of the reactant.
  2. What is the temperature corresponding to A?
  3. What is B (product)? Write its chemical formula.

Answer:

  1. Gypsum
  2. 393 K
  3. Plaster of Paris (CaSO4. ½H2O) or (CaSO4)2.H2O

Question 6.
Explain with the help of chemical equations what happens when

  1. lime stone is heated?
  2. water is dropped on quick lime?
  3. gypsum is heated to 393 K?

Answer:

1. When lime stone is heated, it is converted into CaO and CO2.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 10 The s Block Elements 2
2. When water is dropped on quick lime calcium hydroxide is formed.
CaO + H2O → Ca(OH)2
3. When gypsum is heated to 393 K, Plaster of Paris is obtained.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 10 The s Block Elements 3

Question 7.
Alkali metal halides are all high melting, colourless crystalline solids.

  1. Write any other physical property of alkalimetal halides.
  2. How alkali metal halides are prepared?

Answer:
1. All the alkali metal halides are soluble in water except LiF. They have high negative enthalpies of formation. The melting and boiling points always follow the trend: fluoride > chloride > bromide > iodide.

2. Alkali metal halides are prepared by the reaction of the appropriate oxide, hydroxide or carbonate with aqueous hydrohalic acid.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 10 The s Block Elements

Question 8.
What is Plaster of Paris? How is it prepared? What is its use?
Answer:
The chemical formula of plaster of Paris is (CaSO4)2.H2O.
It is prepared by heating gypsum at 393 K.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 10 The s Block Elements 4
It is a hemihydrate of calcium sulphate.
It is a white powder. On mixing with an adequate quantity of water it forms a plastic mass that gets into a hard solid in 5 to 10 minutes.

Plaster of Paris is used

  • for making moulds for pottery, ceramics, etc.
  • for making models, statues and decorative materials.
  • for immobilising the affected part of organ where there is a bone fracture or sprain.

Question 9.
Give the biological importance of Na and K.
Answer:
Biological importance of sodium: The Na+ ions participate in the transmission of nerve signals, in regulating the flow of water across cell membranes and in the transport of sugars and amino acids into cells. Biological importance of potassium: The K+ ions activate many enzymes, participate in the oxidation of glucose to produce ATP and, with sodium, are responsible for the transmission of nerve signals.

Question 10.
1. Name the element showing anomalous behaviour in group 2.
2. Give reason forthis anomalous behaviour,
3. List any two similarities between Be and Al.
Answer:
1. Be

2. Be has exceptionally small atomic and ionic size, it has high ionisation enthalpy, it cannot exhibit coordination number more than four as in its valence shell there are only four orbitals. There are no d-orbitals.

2. i) Both Be and Al are not readily attacked by acids because of the presence of an oxide film on the surface of the metal.
ii) The chlorides of both Be and Al have Cl bridged chloride structure in vapour phase.

Plus One Chemistry The s Block Elements Three Mark Questions and Answers

Question 1.
A compound of calcium is used for immobilising the fractured bones of body.

  1. Write down the common name and molecular formula of the compound.
  2. Which property of the compound helps to make plaster?
  3. What do you mean by dead burnt plaster? How does it form?

Answer:
1. Plaster of Paris (CaSO4. ½H2O) or (CaSO4)2.H2O

2. When we add water to Plaster of Paris, there is slight increase in volume and this helps Plaster of Paris to take the shape of any mould in which it is present.

3. When Plaster of Paris is heated above 393 K, Plaster of Paris loses water of crystallization to form anhydrous calcium sulphate, CaSO4. It is known as dead burnt plaster which does not set in presence of water.

Question 2.
Lithium is group 1 element. It shows some similarities with group 2 element magnesium.
1. Write the name of the relationship.
2. Explain this relationship.
3. What is the reason for this relationship?
4. Give other example for this relationship.
Answer:
1. Diagonal relationship.

2. The first element of alkali group shows some similarities with the second element of the second group, which is diagonally present. This relation is called diagonal relationship.

3. Diagonal relationship is due to the following reasons:

  • Similarity in their polarising powers due to similar charge/radius ratio.
  • Almost similar electronegativity of the 2 elements.

4.

  • Berylium and Aluminium
  • Boron and Silicon

Plus One Chemistry Chapter Wise Questions and Answers Chapter 10 The s Block Elements

Question 3.
Mentionafewimportarrtusesofthefollowing compounds.

  1. Epsom salt
  2. Marbl
  3. Sodium hydroxide

Answer:
1. Epsom salt:
It is used as a mordant in dyeing. Used in medicine in purgative. Used in tanning industry.

2. Marble:
It is used to make quick lime, cement etc. Used as a raw material in the Solvay process. Used as a building material.

3. Sodium hydroxide:
It is used as a laboratory reagent. Used in petroleum refining. Used in viscose process.

Question 4.
Cement is a complex mixture of silicates and aluminates.
1. What is the function of gypsum in cement?
2. Explain setting of cement.
Answer:
1. The purpose of adding gypsum is only to slow down the process of setting of the cement so that it gets sufficiently hardened.

2. When the cement is mixed with water, it forms a gelatinous mass which sets slowly to a hard mass having 3 dimensional network structure with -Si- O-Si and -Si-O-AI- chains. This is an exothermic process and is called setting of cement.

Question 5.
The alkali metals and their salts impart characteristic colour to an oxidising flame.
1. What is the reason for this?
2. Give the flame colour of Na and K.
3. Name the alkali metal which imparts crimson red colour to the oxidizing flame.
Answer:
1. When the heat energy is supplied to alkali metal or its salt, the electrons are excited to higher energy levels. When these excited electrons come back to their original energy levels, they emit radiations which fall in the visible region of the electromagnetic spectrum and they appear coloured.

2. Na-Yellow, K-Violet

3. Li

Question 6.
a) How will you prepare Ca(OH)2 and CaCO3 from quicklime.
b) Give any two uses of quick lime.
Answer:
a) By adding water to quick lime we can prepare Ca(OH)2. This process is called slaking.
CaO + H2O → Ca(OH)2
When CO2 is passed thruogh lime water we can prepare CaCO3.
Ca(OH)2 + CO2 → CaCO3 + H2O

b) 1. As an important primary material for manufacturing cement.
2. In the manufacture of Na2CO3 from NaOH.

Question 7.
Lithium shows similarities in properties with Magnesium.
a) Namethe above phenomenon.
b) Give any two similarities of Lithium with Magnesium.
c) How is bleaching powder prepared?
Answer:
a) Diagonal relationship

b) 1) Both Lithium and Magnesium are harder and lighter than other elements in the respective groups.
2) The oxides of both (Li2O and MgO) do not combine with excess O2 to give any superoxide.

c) It is prepared by passing Cl2 gas through dry slaked lime.
2Ca(OH)2 + 2Cl2 → CaCl2 + CaOCl2 + 2H2O

Plus One Chemistry The s Block Elements Four Mark Questions and Answers

Question 1.
1. What is the name of the method that is used in manufacturing of sodium hydroxide?
2. Explain the method.
3. Write the equations of the reactions involved in this process.
Answer:
1. Castner-Kellner method

2. The Castner-Kellner cell consists of a large tank. The bottom of the tank is filled with mercury. The tank is made into 3 compartments by 2 partitions. The outer compartments are filled with NaCl solution and the middle compartment is filled with a very dilute solution of NaOH. The mercury layer of the bottom of cell in the outer compartments acts as cathode and the mercury in the middle compartments acts as anode. As a result of electrolysis Na and Cl2 are produced in the outer compartments. Na gets coated with mercury, and sodium amalgam is produced in the outer compartments. Due to the action of eccentric wheels at the bottom of the tank, the tank can be tilted. As a result of this Na is carried into the middle compartment and is then allowed to react with water in the middle compartment. As a result of this reaction NaOH is produced. When the concentration of NaOH in the middle compartment is raised to 40%, it is taken out from the tank. On cooling, crystals of NaOH separate out.

3. NaCl → Na+ + Cl
At cathode: Na+ + e → Na
Na + Hg → Na/Hg (Sodium amalgam)
At anode: Cl → Cl + e
Cl + Cl → Cl2 (g)

Question 2.
Match the following:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 10 The s Block Elements 5
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 10 The s Block Elements 6

Question 3.
A piece of metallic sodium is added to liquid ammonia.

  1. What is the observation?
  2. What is the reason for this?
  3. What happens when the solution is kept for some time?
  4. What happens if the solution is concentrated?

Answer:

  1. The solution turns into blue.
  2. The blue colour of the solution is due to the presence of solvated electrons.
  3. On standing, the solution slowly liberates hydrogen resulting in the formation of amide.
  4. When the concentration of the solution increases the colour turns bronze and the solution becomes diamagnetic.

Question 4.
1. Some compounds of sodium and calcium are given below. Match them.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 10 The s Block Elements 7
2. On passing CO2 through lime water, milkiness appears. On further passing CO2 milkiness disappears. What is the Chemistry behind it?
Answer:
1.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 10 The s Block Elements 8
2. When CO2 is passed through lime water it turns milky due to the formation of insoluble calcium carbonate.
Ca(OH)2 + CO2 → CaCO3 +H2O
When CO2 is passed through lime water for a long time, it becomes colourless due to the formation of water soluble calcium bicarbonate.
CaCO3 + H2O + CO2 → CaOHCO3)2

Question 5.
Cement is an important building material employed in different kinds of construction works.

  1. What are the major raw materials for making cement?
  2. How is cement prepared?
  3. What are the important ingredients of Portland cement?
  4. Explain the Chemistry involved in the setting of cement.

Answer:
1. Limestone, Clay

2. When clay and lime are strongly heated together they fuse and react to form cement clinker. This clinker is mixed with 2-3% by weight of gypsum to form cement.

3. Dicalcium silicate (Ca2SiO4) – 26%, tricalcium silicate (Ca3SiO5) – 51% and tricalcium aluminate (Ca3Al2O6) – 11%

4. When mixed with water, the setting of cement takes place to give a hard mass. This is due to the hydration of the molecules of the constituents and their rearrangement.

Question 6.
1. Match the following :
Plus One Chemistry Chapter Wise Questions and Answers Chapter 10 The s Block Elements 9
2. Study the list of elements given below:
Na, Mg, Li, B, C
Select the pairs showing diagonal relationship.
3. Write the chemical equation showing the reaction of sodium metal with water.
Answer:
1.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 10 The s Block Elements 10
2. Li, Mg
3. 2Na + 2H2O → 2NaOH + H2

Question 7.
a) Arrange the following compounds in the increasing order of solubility in water:
i) Mg(OH)2, Be(OH)2, Ba(OH)2, Ca(OH)2, Sr(OH)2
ii) SrSO4, MgSO4, BeSO4, BaSO4, CaSO4,
b) i) Write the difference between lime water and milk of lime.
ii) Complete the following reaction.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 10 The s Block Elements 11
Answer:
a) i) Be(OH)2 < Mg(OH)2< Ca(OH)2< Sr(OH)2< Ba(OH)2
ii) BaSO4 < SrSO4< CaSO4< MgSO4< BeSO4

b) i) A clear solution of calcium hydroxide in water is called lime water. A suspension of calcium hydroxide in water is called milk of lime,
Plus One Chemistry Chapter Wise Questions and Answers Chapter 10 The s Block Elements 12

Plus One Chemistry The s Block Elements NCERT Questions and Answers

Question 1.
Explain why alkali and alkaline earth metals cannot be obtained by chemical reduction methods? (2)
Answer:
Alkali and alkaline earth metals are themselves very strong reducing agents and reducing agents stronger than them are not easily available. Therefore, these metals cannot be obtained by chemical reduction methods.

Question 2.
Why are potassium and caesium, rather than lithium used in photoelectric cells? (2)
Answer:
Potassium and caesium have much lower ionisation enthalpy than that of lithium. As a result, these metals on exposure to light, lose electrons much more easily than lithium. Hence, potassium and caesium rather than lithium are used in photoelectric cells.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 10 The s Block Elements

Question 3.
When an alkali metal dissolves in liquid ammonia the solution can acquire different colours. Explain the reasons for this type of colour change. (2) Answer:
When an alkali metal is dissolved in liqiud ammonia it produces a blue coloured conducting solution due to formation of ammoniated cation and ammoniated electron as given below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 10 The s Block Elements 13
When the concentration is above 3 M, the colour of solution is copper-bronze. This colour change is because the concentrated solution contains clusters of metal ions and hence possess metallic lustre.

Question 4.
Beryllium and magnesium do not give colour to flame whereas other alkaline earth metals do so, why? (2)
Answer:
Due to the small size, the ionisation enthalpies of Be and Mg are much higher than those of other alkaline earth metals. This means that the valence electrons in beryllium and magnesium are more tightly held by the nucleus. Therefore, they need large amount of energy for excitation of electrons to higher energy levels. Since such a large amount of energy is not available in bunsen flame, therefore, these metals do not impart any colour to the flame.

Question 5.
Potassium carbonate cannot be prepared by Solvay process. Why? (2)
Answer:
Potassium carbonate cannot be prepared by Solvay process because potassium biocarbonate being highly soluble in water does not get precipitated in carbonation tower when CO2 is passed through a concentrated solution of KCI saturated with ammonia.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 10 The s Block Elements

Question 6.
Why Li2CO3 decomposes at lower temperature whereas Na2CO3 at higher temperature? (2)
Answer:
Li+ ion is smaller in size. It forms more stable lattice with smaller anion oxide, O2- than with CO32- ion. Therefore, Li2CO3decomposes into Li2O at lower temperature on the other hand, Na+ ion being larger in size forms more stable lattice with larger anion CO32- than with O2- ion. Therefore, Na2CO3 is quite stable and decomposes into Na20 at very high temperature.

Question 7.
How would you explain? (3)
1. BeO is insoluble but BeSO4 is soluble in water.
2. BaO is soluble but BaSO4 is insoluble in water.
3. Lil is more soluble than Kl in ethanol.
Answer:
1. BeO has higher lattice enthalpy than hydration enthalpy and hence is insoluble in water. BeSO4 on the contrary, is soluble because the lattice enthalpy is less due to bigger sulphate ion.

2. In barium oxide, BaO, due to larger size of barium ion the lattice enthalpy is less than hydration enthalpy and hence it is soluble in water. On the other hand, in BaSO4 dueto largersize of barium as well as sulphate ion, the magnitude of hydration enthalpy is much smallerthan the lattice enthalpy and hence it is insoluble in water.

3. In Kl the chemical bond is ionic in character. On the other hand, due to small size of lithium ion and its high polarising power, the bond in Li is predominantly covalent in character. Hence, Li is more soluble than Kl in ethanol.

Question 8.
Which of the alkali metals is having least melting point? Justify. (2)
a) Na
b) K
c) Rb
d) Cs
Answer:
d) Cs.
As the atomic size of the metal increases, the strength of metallic bonding decreases and hence its melting point decreases. Since the size of Cs is the largest, therefore its melting point is the lowest.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Students can Download Chapter 9 Hydrogen Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Plus One Chemistry Hydrogen One Mark Questions and Answers

Question 1.
In which of the following compounds does hydrogen have an oxidation state of -1?
a) CH4 b) NH3 C) HCl d) CaH2
Answer:
d) CaH2

Question 2.
The radio active isotope of hydrogen is __________ .
Answer:
Tritium

Question 3.
Temporary hardness of water is due to the presence of
a) MgSO4
b)Ca(HCO3)2
c) CaSO4
d) NaHCO3
Answer:
b) Ca(HCO3)2

Question 4.
D2O is used as __________ in nuclear reactors.
Answer:
Moderator

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 5.
30 volumes of H2O2 means
a) 30% H2O2 solution
b) 30 cm³ of the solution contains 1 g of H2O2
c) 1 cm³ of the solution liberates 30 cm3 of O2 at STP
d) 30 cm³ of the solution contains 1 mole of H2O2
Answer:
c) 1 cm³ of the solution liberates 30 cm3 of O2 at STP

Question 6.
Name the three isotopes of hydrogen.
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 1

Question 7.
Dihydrogen is prepared on industrial scale from syngas by
Answer:
Water gas shift reaction

Question 8.
Among the following elements which does not make a hydride is _________
a) Ti
b) Mg
c) Co
d) Pd
Answer:
c)Co

Question 9.
Hydrogen is purified by _________ .
Answer:
Oculusion on pd

Question 10.
H2O2 is _________ .
Answer:
Diamagnetic

Question 11.
Ortho & para hydrogen are __________ .
Answer:
Nuclear spin isomers

Plus One Chemistry Hydrogen Two Mark Questions and Answers

Question 1.
Write one method each for the laboratory preparation of dihydrogen from
i) mineral acid
ii) aqueous alkali.

  1. Which is the catalyst used for the reaction?
  2. Which is the product in this reaction?

Answer:
1.By the reaction of granulated zinc with dilute HCl.
Zn + 2HCl → ZnCl2 + H2

2. By the reaction of zinc with aqueous alkali.
Zn + 2NaOH → Na2ZnO2 + H2

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 2.
Classify the following into those causing temporary hardness and permanent hardness:
[Mg(HCO3)2, MgCl2, CaCO3, CaSO4, NaCl, NaHCO3, Ca(HCO3)2]
Answer:
Temporary hardness – Mg (HCO3)2, Ca(HCO3)2
Permanent hardness – MgCl2, CaSO4

Question 3.
The bond angle in water is different from the tetrahedral bond angle.

  1. What is the bond angle and shape of water?
  2. Justify.

Answer:
1. 104.5°

2. There are three types of repulsions in water. They are: Ip – Ip repulsion, Ip – bp repulsion and bp – bp repulsion. In order to minimize the stronger Ip- Ip repulsion, the bond angle is reduced to 104.5° from 109.5°. Thus, the shape of water is distorted tetrahedral or angular.

Question 4.
Among NH3, H2O and HF which would you expect to have highest magnitude of hydrogen bonding? Why?
Answer:
Strength of H-bond depends upon the atomic size and electronegativity of the other atom to which H- atom is covalently bonded. Smaller size and higher electronegativity favour H-bonding. Among N, F and O atoms, F is the smallest and its electronegativity is highest. Hence, HF will have highest magnitude of H-bonding.

Question 5.
1. Soap does not give lather with hard water. Why?
2. What are the disadvantages of hard water?
Answer:
1. Hard water contains Ca+, Mg2+ions in the form of
their bicarbonates, chlorides or sulphates. Hard water forms scum/precipitate with soap. So soap does not give lather with hard water,

2. Hard water is unsuitable for laundry. It is harmful for boilers because of deposition of salts in the form of scale. This reduces the’ efficiency of the boiler.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 6.
Write one example each for the oxidising action of H2O2 in acidic medium and basic medium.
Answer:
In acidic medium, H2O2 oxidises PbS to PbSO4.
PbS(s) + 4H2O2(aq) → PbSO4(s) + 4H2O(l)
In basic medium, H2O2 oxidises Fe2+ to Fe3+.
2Fe2+ + H2O2 → 2Fe3+ + 2OH

Question 7.
Write one example each for the reducing action of H2O2 in acidic medium and basic medium.
Answer:
In acidic medium H2O2 reduces MnO4 to Mn2+.
2MnO4 + 6H+ + 5H2O2 → 2Mn2+ + 8H2O + 5O2
In basic medium H2O2 reduces l2 to l.
l2 + H2O2 + 2OH → 2l + 2H2O + O2

Question 8.
Write two examples for redox reactions involving water.
Answer:
1. Water can be easily reduced to dihydrogen by highly electropositive metals like Na. Here, Na is oxidised to NaOH.
2H2O(I) + 2Na(s) → 2NaOH(aq) + H2(g)

2. With F2, water is oxidised to O2. Here, F2 is reduced to F-.
2F2(g) + 2H2O(l) → 4H+(aq) + 4F(aq) + O2(g)

Question 9.
Distinguish between

  1. Hard and Heavy water
  2. Temporary and permanent hardness of water.

Answer:
1. A sample of water said to be hard water if it does not give lather with soap. Heavy water is the oxide of deuterium (D2O).

2. Temporary hardness is due to the presence of bicarbonate of calcium or magnesium. It can be removed by boiling.

Permanent hardness is due to the presence of chlorides and sulphates of calcium and magnesium. It is not removed by boiling.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 10.
Hydrogen combines with elements to give binary compounds known as hydrides.
1. Name the three categories of hydrides.
2. Classify the given hydrides into different categories: NH3 LiH, TiH, CH4, NaH
Answer:
1. Ionic hydrides, Covalent hydrides and Metallic/ Interstitial hydrides

2. NaH, LiH – Ionic hydrides
NH3, CH4 – Covalent hydrides
TiH – Interstitial hydride

Plus One Chemistry Hydrogen Three Mark Questions and Answers

Question 1.
A chart prepared by a student based on the similarities of hydrogen with alkali metals and halogens is as shown below. Correct the mistakes in it.

Similarities with alkali metals Similarities with halogen
-1 oxidation state Reducing agent + 1 oxidation state High Ionisation energy
Diatomic state During electrolysis, both of them are produced at the cathode
Non metals

Answer:

Similarities with alkali metals Similarities with halogen
+1 oxidation state Reducing agent -1 oxidation state High Ionisation energy
During electrolysis, both of them are produced at the cathode Diatomic state
Non metals

Question 2.
Prepare a short note on different types of hydrides.
Answer:
Hydrogen reacts with metals or non metals to form binary compounds known as hydrides.
Hydrides are mainly classified into the following 3 types:
1. Ionic or salt like hydrides
These are stoichiometric compounds of dihydrogen formed with most of the s-block elements which are highly electropositive in character. These are crystalline, non-volatile and non-conducting in solid state. But their melts conduct electricity and on electrolysis liberate dihydrogen gas at anode. They react violently with water producing dihydrogen gas. e.g. NaH, KH.

2. Covalent hydrides or Molecular hydrides
These are formed by the action between dihydrogen and non metals (p-block elements). Covalent hydrides are classified into electron-deficient (e.g. B2H6), electron-precise (e.g. CH4) and electron-rich hydrides (e.g. NH3).

3. Metallic or Non-stoichiometric or Interstitial hydrides
These are formed by the reaction of dihydrogen with many d-block and f-block elements. These hydrides conduct heat and electricity. They are almost always non- stoichiometric, being deficient in hydrogen, e.g.
LaH2.87,YbH2.55,VH0.56 etc.

Question 3.
Consider the chemical equation and fill in the blanks:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 2

  1. This method is used for the preparation of ……………
  2. The electrode used is …………
  3.  ……….. is produced at cathode.

Answer:

  1. Hydrogen
  2. Platinum
  3. Hydrogen

Question 4.
Analyse the equation: 2Na + H2 → 2NaH

  1. In this chemical equation, H2 reacts with …………
  2. Hydrogen reacts with metals to form …………
  3. Hydrogen is in ………… oxidation state in NaH.

Answer:

  1. Na
  2. Metal hydrides
  3. -1

Question 5.
a) How is hydrogen of high purity prepared?
b) Dihydrogen is relatively inert at room temperature. Give reason.
c) Write any two uses of hydrogen.
Answer:
a) High purity (>99.95%) dihydrogen is obtained by electrolysing warm aqueous barium hydroxide . solution bewteen nickel electrodes.
b) This is due to high H-H bond enthalpy.
c) 1. In oxy-hydrogen torches.
2. Liquid hydrogen is used as a fuel in rockets.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 6.
A sample of cold river water does not easily give lather with soap, but on boiling it does.

  1. Evaluate and write the chemical equation involved.
  2. In some cases, the water does not give ready lather even if it is boiled. Why?

Answer:
1. If the sample of water has temporary hardness, it is removed during boiling. Here, the bicarbonates of magnesium is precipitated as Mg(OH)2 and that of calcium is precipitated as CaCO3.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 3

2. The sample of water might have permanent hardness due to the presence of dissolved chlorides and sulphates of Ca and Mg. This cannot be removed just by boiling.

Question 7.
Hydrogen has three isotopes – Protium, Deuterium and Tritium.
a) Of these which is the radio active one?
b) Name a compound which contains the isotope deuterium.
c) Make a table which shows number of protons, neutrons and electrons in each isotope.
Answer:
a) Tritium
b) Heavy water – D2O
c)

Isotope No. of Protons No. of Neutrons No. of Electrons
Protium 1 0 1
Deuterium 1 1 1
Tritium 1 2 1

Question 8.
What are the three types of hydrates? Give examples for each.
Answer:
1. Hydrates with coordinated water, e.g. [Cr(H2O)6]Cl3.
2. Hydrates with interstitial water, e.g. BaCl2.2H2O
3. Hydrates with hydrogen bonded water, e.g. CuSO4.5H2O.

Question 9.
1. What is meant by amphoteric nature of water?
2. Suggest an example to show this property.
Answer:
1. The amphoteric nature of water means that it has the ability to act as an acid as well as a base.

2. Water acts as an acid with NH3 and a base with H2S.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 4

Question 10.
a) What are the advantages of dihydrogen as a fuel?
b) What are the disadvantages of using dihydrogen as a fuel?
c) What is meant by the term ‘hydrogen economy’?
Answer:
a) 1. It is abundantly available in the combined state as water.
2. Use of dihydrogen as fuel provides pollution free atmosphere because its combustion product is only water.
3. Heat of combustion per gram of dihydrogen is more than twice that of jet fuel.

b) 1. Dihydrogen does not occur in free state in nature.
2. Hydrogen gas has explosive flammability which causes problem to its storage and transportation.
3. A cylinder of compressed dihydrogen weighs about 30 times as much as a tank of petrol containing the same amount of energy.

c) Hydrogen economy refers to the use of dihydrogen as an alternative source of energy. The basic principle of hydrogen economy is storage and transportation of energy in the form of dihydrogen instead of fossil fuels or electric power.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 11.
What do you mean by temporary and permanent hardness? Explain one chemical method each for removing temporary hardness and permanent hardness.
Answer:
Temporary hardness is due to the presence of bicarbonates of calcium or magnesium. Permanent hardness is due to the presence of chlorides and sulphates of calcium or magnesium ions.
Clark’s method :
This method is used for removing temporary hardness. In this method, calculated amount of lime is added to hard water. It precipitates out calcium carbonate and magnesium hydroxide which can be filtered off.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 5
Treatment with washing soda: This method is used to remove permanent hardness. Washing soda reacts with soluble calcium and magnesium chlorides and sulphates in hard water to form insoluble carbonates.

Question 12.
a) Write a method for the preparation of H2O2.
b) What is 100 volume H2O2?
c) Draw the structure of H2O2 in gas phase.
Answer:
a) H2O2 is prepared by acidifying barium peroxide and removing excess water by evaporation under reduced pressure.
BaO2.8H2O(s) + H2SO2(aq) → BaSO4(s) + H2O2(aq) + 8H2O(I)

b) A 30% solution of H2O2 is called 100 volume H2O2. It means that one mL of 30% H2O2 solution will give 100 V of oxygen at STP.

c) H2O2 is has non-planar structure as shown below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 7

Question 13.
lon exchange process is commonly employed for large scale production of soft water.

  1. What is the basic principle involved in ion exchange method of softening water?
  2. What are inorganic cation exchangers? Give example.

Answer:
1. Adsorption

2. These are complex inorganic salts like sodium- aluminium silicate NaAlSiO4 which can exchange cations such as Ca2+ and Mg2+ ions in hard water for Na+ ions. e.g. Zeolite.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 14.
1. What is heavy water? Mention one use of heavy water.
2. Explain why hydrogen peroxide is not stored in glass vessels.
3. What is calgon? What is its use?
Answer:
1. D2O. Used as moderator in nuclear reactors.

2. To prevent its decomposition.

3. Calgon is chemically sodium hexametaphosphate (Na6P6O18). It is used to remove Ca2+ and Mg2+ ions of hard water by converting them into soluble complexes.
M2+ + Na4P6O182- → [Na2MP6O18]2- + 2Na+ (M = Mg,Ca)

Question 15.
1. Name the oxide of isotope of hydrogen used in nuclear reactor.
2. What are cation exchange resins? What is their role in removing permanent hardness of water?
Answer:
1. Heavy water (D2O).

2. Cation exchange resins contain large organic molecule with -SO3H group and are water insoluble. Ion exchange resin (RSO3H) is changed to RNa by treating with NaCl. The resin exchanges Na+ ions with Ca2+ and Mg2+ ions present in hard water to make the water soft.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 16.
1. What is permuitit? How is it useful in removing permanent hardness of water?
2. Compare the structures of water and hydrogen peroxide.
Answer:
1. Hydrated sodium aluminium silicate is called permutit (NaZ). When it is added to hard water it exchanges Ca2+ and Mg2+ ions with Na+ ions.
2NaZ(s) + M2+(aq) —> MZ2(s) + 2Na+(aq) M = Ma, Ca)

2. Water molecule has an angular or bent shape. H2O2 has a non-planar structure.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 8

Question 17.
1. Give the industrial method of preparation of H2O2.
2. How is heavy water prepared?
3. How is pure de-mineralised water obtained?
Answer:
1. Industrially H2O2 is prepared by the auto-oxidation of 2-ethylanthraquinol.

2. Heavy water is prepared by exhaustive electrolysis of water or as a byproduct in some fertilizer industries.

3. Pure de-mineralised water is obtained by passing water successively through a cation exchange and an anion exchange resins.

In the cation exchange process, the H+ exchanges for Na+, Ca2+, Mg2+ and other cations present in water. This process results in proton release and makes the water acidic.
2RH(s) + M2+(aq) \(\rightleftharpoons \) MR2(s) + 2H+(aq)

In the anion exchange process, the OH exchanges for anions like Cl, HCO3, SO42- etc. present in water.
RNH3+OH(S) + X(aq) \(\rightleftharpoons \) RNH3+X(s) + OH(aq)

The OH ions, thus liberated neutralise the H+ ions set free in the cation exchange process to get pure de-mineralised water.
H+(aq) + OH(aq) → H2O(l)

Question 18.
1. Explain why hydrogen peroxide is stored in coloured plastic bottles.
2. Write any two uses of H2O2.
Answer:
1. In the presence of metal surfaces or traces of alkali (present in glass conatiners), the decomposition of H2O2 (2H2O2 → 2H2O + O2) is catalysed. It is therefore stored in wax-lined glasses or plastic vessels in dark.

2. 1. As a hair bleach and as a mild disinfectant.
2. In the syntheseis of hydroquinone, tartaric acid and certain food products and pharmaceuticals.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 19.
1. A sample of hard water was found to loose its hardness on boiling. Name the type of hardness associated with this sample. 0
2. Write the name and formula of four minerals which cause permanent hardness of water.
3. What are electron deficient hydrides? Whether they behave as Lewis acids or Lewis bases? Why?
Answer:
1. Temporary hardness

2. CaCl2, MgCl2, CaSO4, MgSO4

3. These are covalent hydrides having too few electrons for writing conventional Lewis structure. They act as Lewis acids because they can accept electrons, e.g. B2H6

Plus One Chemistry Hydrogen Four Mark Questions and Answers

Question 1.
Match the following:

A B
Temporary hardness p-block elements
Hydrides Reducing agent
Permanent hardness Chloride
Alkali metals Bicarbonate

Answer:

A B
Temporary hardness Bicarbonate
Hydrides p-block elements
Permanent hardness Chloride
Alkali metals Reducing agent

Question 2.
Certain samples of water do no produce easy lather with soap.

  1. What is this condition of water called?
  2. Which are two types of this condition?
  3. Suggest two methods to change this condition of water.
  4. What are the problems caused by this condition of water?

Answer:

  1. Hardwater.
  2. Temporary hardness and permanent hardness.
  3. By boiling water and by adding Na2CO3
  4. Wastage of soap and boiler explosion.

Question 3.
Match the following:

1. D2O a) Hendry Cavendish
2. Hydrogen b) Water gas
3. Ca + H2 c) 31H
4. C2H4 + H2 d) Heavy water
5. Tritium e) 21H
6. Deuterium f) CaH2
7. CO + H2 g) C2H6
8. CO + Z2 h) Producer gas

Answer:
1) – d)
2) – a)
3) – f)
4) – g)
5) – c)
6) – e)
7) – b)
8) – h)

Question 4.
1. A list of compounds are given below:
H2O, HCl, CH4
Construct chemical reactions to show the preparation of H2 from each of the above compounds.
2. What is syn gas? How is it prepared?
3. What is coal gasification?
4. Explain water gas shift reaction.
Answer:
1.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 9
2. Syn gas is a mixture of CO and H2. It is prepared by passing steam over red hot coke.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 10

3. The process of production of syngas from coal is called coal gasification.

4. The production of dihydrogen can be increased by reacting CO of syngas mixtures with steam in the presence of iron chromate as catalyst. This is called water gas shift reaction.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 11

The CO2 is removed by scrubbing with sodium arsenite solution.

Plus One Chemistry Hydrogen NCERT Questions and Answers

Question 1.
Write the names of isotopes of hydrogen. What is the mass ratio of these isotopes? (2)
Answer:
The various isotopes of hydrogen are:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen 12

Question 2.
Why does hydrogen occur in a diatomic form rather than in a monoatomic form under normal conditions? (2)
Answer:
Hydrogen atom has only one electron and thus, to achieve stable inert gas configuration of helium, it shares its single electron with electron of another hydrogen atom to form a stable diatomic molecule. The stability of H2 is further confirmed by the fact, that formation of one mole of gaseous H2 molecules results in the release of 435.8 kJ of energy.
H(g) + H(g) H2(g); ∆H = – 435.8 kJ mol-1

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 3.
How do you expect the metallic hydrides to be useful for hydrogen storage? Explain. (2)
Answer:
In some of the transition metal hydrides, hydrogen is absorbed as H-atoms. Due to the inclusion of H- atoms, the metal lattice expands and thus becomes less stable. Therefore, when such metallic hydride is heated, it decomposes to release hydrogen gas and very finely divided metal. The hydrogen evolved in this manner can be used as a fuel. Thus, transition metals or their alloys can act as sponge and can be used to store and transport hydrogen to be used as a fuel.

Question 4.
What causes the temporary and permanent hardness of water? (2)
Answer:
Temporary hardness is caused by presence of bicarbonates of calcium and magnesium, i.e., Ca(HCO3)2 and Mg(HCO3)2 in water whereas permanent hardness is caused by presence of soluble chlorides and sulphates of calcium and magnesium, i.e., CaCl2, CaSO4, MgCl2 and MgSO4 in water.

Question 5.
Write chemical reactions to show amphoteric nature of water. (2)
Answer:
Water is amphoteric in character. It means that it can act as proton donor as well as proton acceptor. When it reacts with acids, (stronger than itself), it behaves as a base. When it reacts with bases (stronger than itself) it acts as an acid.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 9 Hydrogen

Question 6.
Is demineralised or distilled water useful for drinking purposes? If not, how can it be made useful? (2)
Answer:
Demineralised or distilled water is not useful for drinking purposes because it does not contain even useful minerals. Therefore, to make it useful for drinking purposes, useful minerals in proper amounts should be added to demineralised or distilled water.

Question 7.
How does H2O2 behave as a bleaching agent? (2)
Answer:
The bleaching action of H2O2 is due to the nascent oxygen which is liberates on decomposition.
H2O2 → H2O + [O]
The nascent oxygen oxidise the colouring matter to colourless products. Hence, H2O2 is used for the bleaching of delicate maerials like ivory, feather, silk, wool, etc.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions

Students can Download Chapter 8 Redox Reactions Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions

Plus One Chemistry Redox Reactions One Mark Questions and Answers

Question 1.
In which of the following, oxidation number of chlorine is +5?
a) Cl
b) ClO
c) ClO2
d) ClO3
Answer:
d) ClO3

Question 2.
An oxidising agent is a substance which can
a) Gain electrons
b) Lose an electronegative radical
c) Undergo decrease in the oxidation number of one of its atoms
d) Undergo any one of the above changes
Answer:
d) Undergo any one of the above changes

Question 3.
The arrangement of metals in the order of decreasing tendency to lose electrons is called _________ .
Answer:
Activity series

Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions

Question 4.
When KMnO4, reacts with acidified FeSO4
a) Only FeSO4 is oxidised
b) Only KMnO4 is oxidised
c) FeSO4 is oxidised and KMnO4 is reduced
d) KMnO4 is oxidised and FeSO4 is reduced
Answer:
c) FeSO4 is oxidised and KMnO4 is reduced

Question 5.
In the disproportionation reaction, which of the following statements is not true?
a) The same species is simultaneously oxidised as well as reduced
b) The reacting species must contain an element having at least three oxidation states
c) The element in the reacting species is present in the lowest oxidation state
d) The element in the reacting species is present in the intermediate oxidation state
Answer:
c) The element in the reacting species is present in the lowest oxidation state

Question 6.
Find the oxidation state of oxygen in OF2.
Answer:
The oxidation number of fluorine in its compounds is always taken as -1.
In OF2
X+ (-1 × 2) = 0
X = +2

Question 7.
The oxidation numbers of chlorine atoms in bleaching powder is _________ .
Answer:
-1

Question 8
SO2 can act as
a) Oxidising agent only
b) Reducing agent only
c) Both oxidising and reducing agents
d) Acid and a reducing agent only
Answer:
c) Both oxidising and reducing agents

Question 9.
In the reaction
2KMnO4 +16HCl → 5Cl2 + MnCl2 + 2KCl + 8H2O the reduction product is _________ .
Answer:
MnCl2

Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions

Question 10.
The strongest reducing agent is .
a) K
b) Ba
c) Li
d) Na
Answer:
c) Li

Question 11.
Oxidation state of oxygen in H2O2 is _________ .
Answer:
-1

Plus One Chemistry Redox Reactions Two Mark Questions and Answers

Question 1.
Balance the following equation using oxidation number method:
MnO2 + Cl → Mn2+ + Cl2
Answer:
Assigning oxidation numbers:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 1
Equating the increase and decrease in oxidation number:
MnO2 + 2Cl → Mn2+ + Cl2
Balancing hydrogen and oxygen atoms:
MnO2 + 2Cl + 4H+ → Mn2+ + Cl2 + 2H2O

Question 2.
Balance the following equation using half reaction method:
Cu + NO3 → Cu2+ + NO2
Answer:
Separating into half reactions:
Oxidation half: Cu → Cu2+
Reduction half: NO3 → NO2
Balancing oxygen and hydrogen atoms:
NO3 + 2H+ → NO2 + H2O
Balancing charge by adding electrons and making the number of electrons equal in the two half reactions:
Cu → Cu2+ + 2e
2NO3 + 4H+ + 2e → 2NO2 + 2H2O
Adding the two half reactions to achieve the overall reaction:
Cu + 2NO3 + 4H+ → Cu2+ + 2NO2 + 2H2O

Question 3.
Complete the following ionic equations:

  1. Al3+ + 3e → …………….
  2. MnO42- → + e
  3. K → K+ + ……………
  4. Fe2+ → Fe3+ +

Answer:

  1. Al3+ + 3e → Al
  2. MnO42- → MnO4+ e
  3. K → K+ + e
  4. Fe2+ → Fe3+ + e

Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions

Question 4.
Find the oxidation number of P in the following compounds:

  1. Na2PO4
  2. H3P2O7
  3. PH3
  4. H3PO4

Answer:

  1. Na2PO4, Oxidation state of P = +6
  2. H4P2O7, Oxidation state of P = +5
  3. PH3, Oxidation state of P = -3
  4. H3PO4, Oxidation state of P = +5

Question 5.
Choose the correct oxidation number of sulphur in the compounds in column Afrom column B.

Column A Column B
Na2SO4 -2
H2sO3 +7
H2S +6
H2S2O7 +4

Answer:

Column A Column B
Na2SO4 +6
H2SO3 +4
H2S -2
H2S2O8 +7

Question 6.
Explain oxidation number and valency.
Answer:
Valency of an atom is its combining capacity and is denoted by a number without sign. The valency of an element is always a whole number.

Oxidation number is a net charge which an atom has or appears to have when the other atoms from the molecule are removed as ions assuming that the shared pair of electrons is with more electronegative atom.

Question 7.
Some rules related to oxidation number are given below. Correct the mistakes.

  • Oxidation number of alkali metals and alkaline earth metals is +2.
  • Oxidation number of hydrogen is always +1.
  • Algebraicsum of oxidation number of all the atoms in an ion is not equal to the charge on the ion.

Answer:

  • Oxidation number of alkali metals is +1.
  • Oxidation number of alkaline earth metals is +2.
  • Oxidation number of H is +1 except in metallic hydrides.

Question 8.
Match the following:

Oxidation number of Cl in Cl2O7 Cu
Oxidant Zn
Stannous Chloride, SnCl2 +7
Oxidation number of C in diamond Get reduced easily
The metal which can’t displace H from dil.HCl Zero
Reducing agent for mercuric chloride

Answer:

Oxidation number of Cl in Cl2O7 +7
Oxidant Get reduced I easily
Stannous Chloride, SnCl2 Reducing agent for mercuric chloride
Oxidation number of C in diamond Zero
The metal which can’t displace H from dil.HCl Cu

Question 9.
1. Calculate the oxidation number of oxygen in OF2 and KO2.
2. When Zn rod is dipped in blue CuSO4 solution ‘ the blue colour of CuSO4 fades due to displacement reaction. Write the reaction and identify the following:
i) The substance oxidised and the substance reduced.
ii) The oxidant and the reductant.
Answer:
1. OF2: x + (-1 × 2) = 0
x – 2 = 0
x = +2
KO2: (+1 × 1) + 2x = 0
1 + 2x = 0
2x = -1
x = –\(\frac{1}{2}\)

2. Zn(s) + CuSO4 (aq) → ZnSO4(aq) + Cu(s)
i) Substance oxidised – Zn
Substance reduced – Cu
ii) Oxidant-Cu
Reductant – Zn

Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions

Question 10.
a) Calculate the oxidation number of C in CH4 and in CH3Cl.
b) The sum of oxidation numbers of all atoms in a molecule is …………
Answer:
a) CH4:
x + (1 × 4) = 0
x + 4 = 0
x = -4
Oxidation number of C in CH4 is -4.

CH3Cl:
x + (3 × 1) + -1 = 0
x + 3 – 1= 0
x + 2 = 0
x = -2
Oxidation number of C in CH3Cl is -2.

b) Zero

Question 11.
1. Write the oxidation state of each element and identify the oxidising agent and reducing agent in the following reaction:
H2S(g) + Cl2(g) → 2HCl(g) + S(s)
2. Fill in the blanks and classify the following reactions into oxidation and reduction:
i) Mn7+ + 5e → ……………
ii) Sn4+ + …………… → Sn2+
iii) Na → Na+ + ……………
iv) Fe3+ +…………… → Fe2+
Answer:
1.Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 2
Reducing agent – H2S
Oxidising agent -Cl2

2. i) Mn7+ + 5e → Mn2+
ii) Sn4+ + 2e → Sn2+
iii) Na → Na+ + e
iv) Fe3+ + e → Fe2+
Oxidation: Reaction (iii)
Reduction: Reactions (i), (ii) and (iv)

Question 12.
Dihydrogen undergoes redox reactions with many metals at high temperature.
a) Write the reaction between hydrogen with sodium.
b) Comment, whether the product formed, is covalent compound or ionic compound.
c) Which is the reducing agent in this reaction?
Answer:
1. 2Na + H2 → 2NaH
2. Ionic compound is formed. When alkali metals react with hydrogen ionic hydrides are formed.
3. Na is the reducing agent.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions

Question 13.
1. Is it possible to keep copper sulphate solution in zinc pot? Why?
2. Assign oxidation numbers of the underlined elements.
i) NaH2\(\underline { P } \)O4
ii) NaH\(\underline { S } \)O4
Answer:
1. No. Zn being more reactive will displace Cu from CuSO4. Thus Cu will be deposited on the vessel.

2. i) NaH2\(\underline { P } \)O4
+1 +(+1 × 2) + x +(-2 × 4) = 0
1 + 2 + x – 8 = 0
x – 5 = 0
x = +5
ii) NaH\(\underline { S } \)O4
+1 + 1 + x +(-2 × 4) = 0
+1 + 1 + x – 8 = 0
x – 6 = 0
x = +6

Question 14.
Identify the substance oxidised, reduced, oxidising agent and reducing agent in the reaction:
2Cu2O + Cu2S → 6Cu + SO2
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 3
In this reaction, Cu is reduced from +1 state to zero. oxidation state and S is oxidised from -2 state to +4 state. Cu2O helps S in Cu2S to increase its oxidation number. Therefore, Cu(l) is the oxidising agent. S of Cu2S helps Cu both in Cu2S itself and Cu2O to decrease its oxidation number. Therefore, S of Cu2S is the reducing agent.

Question 15.
Explain the following in terms of electron transfer concept:

  1. Oxidation
  2. Reduction
  3. Oxidising agent
  4. Reducing agent

Answer:

  1. Oxidation: Loss of electron(s) by any species.
  2. Reduction: Gain of electron(s) by any species.
  3. Oxidising agent: Any species which accepts electrons).
  4. Reducing agent: Any species which donates electron^).

Question 16.
Represent the following compounds using Stock notation:
Cu2O, SnCl4, MnO, Fe2O3, V2O5
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 4

Question 17.
In a redox reaction, oxidation and reduction occur simultaneously.
a) Write the classical concept of oxidation and reduction.
b) Identify the species undergoing oxidation and reduction in the following reaction:
H2S(S) + Cl2(g) → 2HCl(g) + S(s)
Answer:
1. Oxidation:
addition of oxygen/electronegative element to a substance or removal of hydrogen/ electropositive element from a substance.

Reduction:
removal of oxygen/electronegative element from a substance or addition of hydrogen/ electropositive element to a substance.

2.Oxidised species:
H2S. This is because a more electronegative element, Cl is added to H or a more electro positive element, H has been removed from S.

Reduced species:
Cl. This is due to addition of more electropositive element H to it.

Plus One Chemistry Redox Reactions Three Mark Questions and Answers

Question 1.
An equation is given below:
HNO3+ l2 → HlO3 + NO2 + H2O

  • Find the oxidising agent and reducing agent.
  • Balance the equation using half reaction method.

Answer:
Oxidising agent = HNO3
Reducing agent = l2
Skeletal equation:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 5
Balancing the charge on the half reactions by adding electrons and equalising the number of electrons:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 6

Question 2.
1. Define redox reactions.
2. Predict whether the following reaction is a redox reaction or not? Justify.
Cr2O72- + H2O → 2CrO42- + 2H+
Answer:
1. Redox reactions are those reactions are those reactions in which reduction and oxidation occur simultaneously. These reactions involve change in oxidation state of the interacting species.

2. No.
Because no element undergoes change in oxidation number.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions

Question 3.
a) Find out the oxidising agent and reducing agent in the following reaction:
Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)
b) Balance the following redox reaction in acid medium using oxidation number method.
Cr2O72- + Fe2+ → Cr3+ + Fe3+
Answer:
1. Oxidising agent – Ag
ReducingAgent – Cu

2. Assigning oxidation number:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 7

Question 4.
Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO4, Cr2O72- and NO3.
Answer:
H2SO4
(2 × +1) + x+ (4 × -2) = 0
+2 + x – 8 = 0
x – 6 = 0
x = +6

Cr2O72-
2x + 7 ×-2 = -2
2x = -2 + 14
2x = 12
∴ x = +6

NO3
x + 3 × -2 = -1
x = -1 + 5
x = +4

Question 5.
1. Assign oxidation numbers
(i) P in NaH2PO4
(ii) Mn in KMnO4
(iii) B in NaBH4
(iv) S in H2SO4
2. Identify the oxidising and reducing agents in the following reaction:
CuO + H2 → Cu + H2O
Answer:
1. i) NaH2PO4
Na+1H2+1PO4-2
1+2 + x- 8 = 0
3 + x – 8 = 0
x – 5 = 0
x =+ 5

ii) K+1MnO4-2
1+ x – 8 = 0
x – 7 = 0
x = +7

iii) Na+1BH4+1
1 + x + 1 × 4 = 0
x + 5 = 0
x = -5

iv) H2+1SO4-2
2 + x – 8 = 0
x – 6 = 0
x = +6

2.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 8

Question 6.
A copper rod is dipped in silver nitrate solution.

  1. What are the observations?
  2. Write the displacement reaction.
  3. Identify the species getting oxidised and reduced.

Answer:

  1. The colour of the solution changes to blue. Silver is deposited on the copper rod.
  2. Cu(s) +2AgNO3(aq) → Cu(NO3)2(aq) + 2Ag(s)
  3. Oxidised species – Cu Reduced species – Ag+

Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions

Question 7.
1. Identify the oxidising and reducing agent in the reaction:
CuS + O2 → Cu + SO2
2. Determine the oxidation number of the underlined element in the following:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 9
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 10

Question 8.
1. Identify the substance oxidised, substance reduced, oxidising agent and reducing agent in the reaction:
Cl2 + 2l → 2Cl +l2

2. Calculate the oxidation number of underlined elements in the following compounds:
i) K2\(\underline { Cr } \)2O7
ii) H\(\underline { H } \)O3
Answer:
1. Cl2 is reduced, therefore Cl2 is the oxidising agent. I’ is oxidised, therefore I” is the reducing agent.

2. i) K2\(\underline { Cr } \)2O7
(+1 × 2) + 2x +(-2 × 7) = 0
+2 + 2x – 14 =0
2x – 12 =0
2x = 12
x = +6
ii) H\(\underline { H } \)O3
(+1 × 1) + x + (-2 × 3) = 0
1+ x – 6 = 0
x – 5 = 0
x = +5

Question 9.
Determine oxidation number of the elements underlined in each of the following.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 12
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 13

Plus One Chemistry Redox Reactions Four Mark Questions and Answers

Question 1.
Permanganate ion (MnO4) reacts with bromide ion (Br) in basic medium to give manganese dioxide
(MnO2) and bromate ion (BrO3).
a) Write the balanced ionic equation for this reaction.
b) Identify the oxidising agent and reducing agent in this reaction.
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 14

Question 2.
A redox reaction involves oxidation and reduction.
a) What do you understand by electrode potential?
b) Define a redox couple.
c) Explain the set-up for Daniell cell with a diagram.
d) Write the electrode reactions and overall cell reaction which occur in the Daniel cell.
Answer:
a) The potential difference between metal and its own ion is called electrode potential.

b) A redox couple is defined as the combination of oxidised and reduced forms of a substance taking part in an oxidation or reduction half reaction.

c) Take copper sulphate solution in a beaker and put a copper strip. Take zinc sulphate solution in another beaker and put a zinc rod. The two redox couples are represented as Zn2+/Zn and Cu2+/Cu. Put the beaker containing copper sulphate solution and beaker containing zinc sulphate side by side. Connect two solution by a salt bridge. The Zn and Cu rods are connected by a metalic wire with a provision for ammeter and switch. Transfer of electrons now does not take place directly from Zn to Cu2+, but through metallic wire. The electricity from solution in one beaker to solution in the other beaker flows by the migration of ions through the salt bridge.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 15

Question 3.
Redox reactions are those in which oxidation and reduction takes place. Explain the different types of redox reactions with suitable examples.
Answer:
Combination Reactions: The reactions in which two substances combine together to form a new compound are called combination reactions. These can be denoted as A+ B → C where either A or B or both A and B should be in the elemental form.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 16
Decomposition reactions:
The reactions in which a compound breaks up into two or more substances at least one of which is in elemental form are called decompositions reactions.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 17

Displacement reactions:
The reactions of the type X + YZ → XZ + Y in which an atom or ion in a compound is displaced by an ion (atom) of another element, such that X and Y are in elemental form are called displacement reactions. They are of two categories:
1. Metal displacement reactions: Reactions in which a more electropositive metal displaces a less electropositive metal from its compound.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 18

2. Non-metal displacement reactions: These are reactions in which a non-metal is displaced by another metal or non-metal.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 19
Disproportionation reactions: These are special type of redox reactions in which an element in one oxidation state is simultaneously oxidised and reduced. Here one of the reactants should contain an element that should exist in at least three oxidation states. The element in the form of reacting substance is in the intermediate oxidation state; and both higher and lower oxidation states of that element are formed in the reaction.
e.g. The decomposition of hydrogen peroxide.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 20
Here the oxygen of peroxide, which is present in -1 state, is converted to zero oxidation state in O2 and to -2 state in H2O.

Plus One Chemistry Redox Reactions NCERT Questions and Answers

Question 1.
Fluorine reacts with ice and results in the change :
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 21
Justify that this reaction is a redox reaction.
Answer:
In the given reaction O.N. of F2 changes from zero to -1 in HF and HOF whereas O.N. of oxygen change from -2 in H2O to zero in HOF. Thus, F2 is reduced, whereas oxygen is oxidised and, therefore, it is a redox reaction.

Question 2.
Write formulas for the following compounds:

  1. Mercury (II) chloride
  2. Nickel (II) sulphate
  3. Tin (IV) oxide
  4.  Thallium (I) sulphate
  5. Iron (III) sulphate
  6. Chromium (III) oxide

Answer:

  1. Hg(II)Cl2
  2. Ni(II)SO4
  3. Sn(IV)O2
  4. Tl2(I)SO4
  5. Fe2(III)(SO4)3
  6. Cr2(III)O3

Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions

Question 3.
The compound AgF2 is unstable. However, if formed, the compound acts as a very strong oxiding agent. Why?
Answer:
In AgF2, oxidation state of Ag is + 2 which is very unstable. Since Ag can exist in a stable state of + 1 it quickly accepts an electron to form the more stable + 1 oxidation state.
Ag2+ + e → Ag+

Question 4.
Consider the reactions:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 22
Why does the same reductant, thiosulphate react differently with iodine and bromine?
Answer:
The average O.N. of S in S2O32- is + 2 while in S4O62- it is + 2.5. The O.N. of S in SO42- is+6. Since Br2 is a stronger oxidising agent that l2, it oxidises S of S2O32- to a higher oxidation state of + 6 and hence forms SO42- ion. l2, however, being a weaker oxidising agent oxidises S of S2O32- ion to a lower oxidation of + 2.5 in S4O62- ion.

Question 5.
Why does the following reaction occur?
XeO64-(aq) + 2F(aq) + 6H+(aq) → XeO3(s) + F2(g) + 3H2O(I)
What conclusion about the compound Na4XeO6 (of which XeO64- is a part) can be drawn from the reaction?
Answer:
The balanced equation along with O.N. of the elements above their symbols will be as:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions 23
In the equation the, O.N. of Xe decreases from + 8 in XeO64- to + 6 in XeO3 while that of F increases from – 1 in F to 0 in F2. Therefore, XeO64- is reduced while F is oxidised. This reaction occurs because Na4XeO6 (0r XeO64-) is stronger oxidising agent than F2.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Students can Download Chapter 6 Thermodynamics Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Plus One Chemistry Thermodynamics One Mark Questions and Answers

Question 1.
Hot coffee in a thermos flask is an example of system.
Answer:
Isolated

Question 2.
Which of the following statements is incorrect about internal energy?
a) The absolute value of internal energy cannot be determined
b) The internal energy of one mole of a substance is same at any temperature or pressure
c) The measurement of heat change during a reaction by bomb calorimeter is equal to the internal energy change
d) Internal energy is an extensive property
Answer:
b) The internal energy of one mole of a substance is same at any temperature or pressure

Question 3.
For which of the following the standard enthalpy is not zero?
a) C (Diamond)
b) C (Graphite)
c) Liquid mercury
d) Rhombicsulphur
Answer:
a) C (Diamond)

Question 4.
Say TRUE or FALSE?
Any spontaneous process must lead to a net increase in entropy of the universe.
Answer:
TRUE

Question 5.
The ∆H fora reaction is-30 kJ. On the basis of this fact, we can conclude that the reaction
a) Gives off thermal energy
b) Is fast
c) Is slow
d) Is spontaneous
Answer:
a) Gives off thermal energy

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 6.
Write the type of system in each of the following:

  1. Hot water taken in an open vessel
  2. Hot water taken in a closed metallic vessel
  3. Hot water taken in a thermos flask

Answer:

  1. Open system
  2. Closed system
  3. Isolated system

Question 7.
In a reversible process the total change in entropy is ∆s(universe) is
Answer:
Zero

Question 8.
For the reaction Ag2O \(\rightleftharpoons \) 2Ag + \(\frac{1}{2}\)O2(g) ∆S and ∆H are 66J K-1mol-1 and 30.56 Kg mol respectively. The reaction will not be spontaneous at.
Answer:
463K

Question 9.
One mole of methane undergoes combustion to form CO2 and water at 25°C. The difference between ∆U & ∆H will be
Answer:
-2RT

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 10.
A gas expands from 1 l to 6 l against a constant pressure of 1 atm and it absorbs 500J of heat ∆μ is
Answer:
-6.5J

Question 11.
Born Haber cycle is to find out __________
Answer:
lattice energy

Plus One Chemistry Thermodynamics Two Mark Questions and Answers

Question 1.
1. Explain enthalpy of fusion.
2. Give illustration of fusion of ice.
Answer:
1. It is the enthalpy change when one mole of a solid is converted into its liquid at its melting point.

2. Enthalpy of fusion of ice is 6 kJ/mol. From this it is clear that 6 kJ of energy is required to convert one mole of ice (18 g) into water at 0°C.

Question 2.
a) What do you meant by enthalpy of vapourisation?
b) Explain enthalpy of sublimation.
Answer:
a) It is the enthalpy change when one mole of a liquid is converted into its vapour at its boiling point.
b) It is the enthalpy change when one mole of a solid is converted into its vapour at its transition temperature.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 3.
One equivalent of an acid reacts completely with one equivalent of a base in dilute solution.
1. Which type reaction is this?
2. HCl + NaOH → NaCl + H2O
On the basis of above equation, explain enthalpy of neutralisation.
Answer:
1. Nneutralisation.

2. When one equivalent of HCl (36.5 g) reacts completely with one equivalent of NaOH (40 g), 57.1 kJ energy is liberated.

Question 4.
1. What is the difference between system and surroundings?
2. There are different types of systems. What are they? Explain.
3. Give example for different types of systems.
Answer:
1. A system in thermodynamics refers to that part of universe in which observations are made. The remaining part of the universe other than the system constitutes the surroundings.

2. System is classified into the following three types. Open system: This is a system in which there is exchange of energy and matter between system and surroundings.
Closed system:
This is a system in which there is no exchange of matter, but exchange of energy is possible between system and the surroundings.

Isolated system:
This is a system in which there is no exchange of energy or matter between the • system and the surroundings.

3. Open system
Presence of reactants in an open beaker
Closed system
Presence of reactants in a closed vessel made of conducting material
Isolated system
Presence of reactants in a • thermos flask or any other closed insulated vessel

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 5.
Match the following:

A B
1. Isothermal Temperature varies
2. Adiabatic Temperature constant
3. Isobaric Volume constant
4. Isochoric Pressure constant

Answer:

A B
1. Isothermal Temperature constant
 2.Adiabatic Temperature varies
3. Isobaric Pressure constant
4. Isochoric Volume constant

Question 6.
1. What is meant by enthalpy?
2. Derive an equation for enthalpy change.
3. What is enthalpy change?
Answer:
1. Enthalpy is the sum of internal energy and pressure volume energy.
i.e.,H = U + pV

2. ∆H = ∆U + ∆pV)
∆H = ∆U + p∆V + V∆p
At constant pressure, ∆p=0
∆H = ∆U+p∆V
But ∆U=q+w
∆H=q+w+p∆V
w = -p∆V
i.e; ∆H= q – p∆V + p∆V
∆H = qp

3. Enthalpy change is heat absorbed or released at constant pressure.

Question 7.
1. Find the enthalpy of the reaction,
C(graphite) + O2(g) → CO2(g)
Given,
i) C(graphite)+ ½ O2(g) → CO(g); ∆H =-110.5 kJ mol-1
ii) CO(g) + ½ O2(g) → CO2(g); ∆H =-283.0 kJ mol-1
2. Melting of ice is a spontaneous process. What are the criteria for spontaneity of a process?
Answer:
1. Considerthe reaction,
C(grahite) + O2(g) → CO2 (g); ∆H = x
CO2 can also be prepared through the following two steps:
i) C(graphite)+ ½ O2(g) → CO(g); ∆H =110.5 kJ mol-1
ii) CO(g) + ½ O2(g) → CO2(g); ∆H =-283.0 kJ mol-1
Then by Hess’s law, x = (-110.5+-283.0) kJ=-393.5 kJ

2. Certain endothermic process are found to be spontaneous in nature. Hence, spontaneous behaviour of a process cannot be explained only on the basis of energy consideration.
For a spontaneous process ∆STotal is +ve.
For a nonspontaneous process ∆STotal is -ve.

Question 8.
Explain the following:

  1. Enthalpy of atomization
  2. Enthalpy of solution at infinite dilution

Answer:

  1. It is the enthalpy change on breaking one mole of bonds completely to obtain atoms in the gas phase.
  2. It is the enthalpy change observed on dissolving the substance in an infinite amount of solvent when the interactions between the ions or solute molecules are negligible.

Question 9.
The enthalpy change for the reaction,
N2(g) + 3H2(g) → 2NH3(g) is -92.38 kJ at 298 K.
What is ∆U at 298 K?
Answer:
∆U = ∆H — ∆ngRT
= -92.38 × 10³ J – [-2 × 8.314J K-1 mol-1 × 293 K)]
= -92.38 × 10³J +4.872 × 10³J
= -87.51 × 10³J
= – 87.51 kJ

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 10.
What are the two types of heat capacities? How they are related?
Answer:
The two types of heat capacities are heat capacity at constant pressure (Cp) and heat capacity at constant volume (Cv). These two are related as Cp – Cv = R, where R is the universal gas constant.

Question 11.
Enthalpy and Entropy changes of two reactions are given below: Find out whether they are spontaneous or not at 27°C. Justify.
1. ∆H = 26 kJ/mole, ∆S = 8.3 J/K/mole
2. ∆H = -393.4 kJ/mole, ∆S = 6 J/K/mole
Answer:
1. ∆G = ∆H -T∆S
= 26000 – 300 × 8.3 = 23.510
Since ∆G is positive, the process is non-spontaneous.

2. ∆G = ∆H -T∆S
= -393400 – 300 × 6 = -391600
Since ∆G is negative, the process is spontaneous.

Question 12.
1. What is enthalpy of solution?
2. What is enthalpy of dilution?
Answer:
1. Enthalpy of solution is the enthalpy change associated with the addition of a specified amount of solute to the specified amount of solvent at a constant temperature and pressure.

2. Enthalpy of dilution is the heat withdrawn from the surroundings when additional solvent is added to the solution. It is dependent on the original concentration of the solution and the amount of solvent added.

Question 13.
What is the significance of the second law of thermodynamics in the spontaneity of exothermic and endothermic reactions?
Answer:
The second law of thermodynamics provides explanation for the spontaneity of chemical reactions. In exothermic reactions heat released by the reaction increases the disorder of the surroundings and overall 1 entropy change is positive which makes the reaction spontaneous.

In the case of endothermic reactions, since heat is ‘ absorbed by the system from the surroundings, the entropy change of the surroundings becomes negative (∆Ssurr < 0). In this case the process will be spontaneous only if the entropy change of the reacting system is postive (∆Ssys > 0) and is also greater than ASsurr in magnitude so that the overall entropy change (∆Stotal) is positive.

Question 14.
Explain the importance of third law of thermodynamics.
Answer:
The importance of third law of thermodynamics lies in the fact that it permits the calculation of absolute values of entropy of pure substances from thermal data alone. For a pure substance, this can be done by summing \(\frac { { q }_{ rev } }{ T } \) increments from 0 K to 298 K.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics 1

Question 15.
C12H22O11 + 12O2 → 12CO2 + 11H2O
Consider this equation and answer the following questions.
a) Thermodynamically, which type reaction is this?
b) What is enthalpy of combustion?
c) Give another example.
Answer:
a) Combustion.
b) It is the enthalpy change when one mole of a substance undergoes complete combustion in excess of air or oxygen.
c) C6H12O6 + 6O2 → 6CO2 + 6H2O

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 16.
Bond dissociation energies of hydrogen and nitrogen are 430 kJ and 41.8 kJ respectively and the enthalpy of formation of NH3 is – 46 kJ. What is the bond energy of N-Hbond?
Answer:
3H2 + N2 → 2NH3; AH = -46 kJ
3 × ½H2 + ½N2 → 2 × ½NH3
[(3× ½ × 430) + (½ × 941 .8)] – (3N-H) = – 46
[(3 × 215) + (470.9) + (46)] → [3N-H]
[645 + 470.9 + 46] = 3N-H
N-H = 387.3 kJ

Plus One Chemistry Thermodynamics Three Mark Questions and Answers

Question 1.
In 1840, G.H.Hess (a Russian chemist) proposed an important generalisation of thermochemistry which is known after his name as Hess’s law.
1. State Hess’s law.
2. Give illustration of Hess’s law.
Answer:
1. Enthalpy change in a chemical reaction is same whether it takes place in one step or in more than one step.

2. Considerthe formation of CO2.
C + O2 → CO2; ∆H = x
CO2 can be prepared through the following two steps:
C + ½O2 → CO ; ∆H = y
CO + ½O2 → CO2; ∆H = Z
Then by Hess’s law,
x = y + z

Question 2.
∆U = q-p∆V. If the process is carried out at constant
volume, then ∆V=0. Answer the following questions.
1. Give the equation for ∆U.
2. 1000J was supplied to a system at constant volume. It resulted in the increase of temperature of the system from 45 °C to 50 °C. Calculate the change in internal energy.
Answer:
1. ∆U = qv

2. Since the volume kept constant, ∆V=0
∴ ∆U = qv = 1000J

Question 3.
Thermodynamics deals with macroscopic properties.
1. What is the difference between extensive and intensive properties?
2. Classify the following properties into extensive and intensive.
Pressure, Mass, Volume, Temperature, Density, Heat capacity, Viscosity, Surface tension, Internal • energy, Molar heat capacity, Refractive index, Enthalpy, Specific heat capacity
Answer:
1. Extensive properties are those properties whose value depend on the quantity or size of matter present in the system.
Intensive properties are those properties which do not dependent on the quantity or size of matter present in the system.

2. Extensive properties: Mass, Volume, Heat capacity, Internal energy, Enthalpy Intensive: Pressure, Temperature, Density, Viscosity, Surface tension, Molar heat capacity, Refractive index, Specific heat capacity.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 4.
1. What is meant by state of the system and state variables?
2. Give any four examples for state variables/state functions.
Answer:
1. The state of a system refers to the conditions of existence of a system when its macroscopic properties have definite values. If any of the macroscopic properties of the system changes, the state of the system will change. The measurable properties required to describe the state of a system are called state variables or state functions. A state function is a property of a system whose value depends only upon the initial and final states of the system and is independent of the path by which this state has been reached. Properties whose values depend on the path followed are called path functions.

2. State variables/State functions – Temperature, Pressure, Enthalpy, Entropy

Question 5.
1. Explain the Zeroth law of thermodynamics.
2. What are the important modes of transference of energy. Explain.
Answer:
1. The Zeroth law of thermodynamics states that if two bodies say, ‘A’ and ‘B’ are in thermal equilibrium with another body say, ‘C’, then the bodies A’ and ‘B’will also be in thermal equilibrium with each other. It provides the basis for the measurement of temperature.

2. The two important modes of transference of energy are heat and work.
Heat:
The exchange of energy, which is a result of temperature difference between system and surroundings is called heat (q).

Work:
The exchange of energy between system and surroundings can occur in the form of work which can be mechanical work, electrical work or pressure-volume work. The exchange of energy as pressure-volume work can occur if system consists of gaseous substance and there is a difference of pressure between system and surroundings.

Question 6.
1. Explain the symbols and sign conventions of heat and work.
2. Explain internal energy.
Answer:
1. Heat is represented by the symbol ‘q’. The ‘q’ is positive, when heat is transferred from the surroundings to the system and ‘q’ is negative when heat is transferred from system to the surroundings.
Work is represented by the symbol ‘w’. The ‘w’ is positive when work is done on the system and ‘w’ is negative when work is done by the system.

2. Every substance is associated with a definite amount of energy due to its physical and chemical constitution. This is called internal energy. It is the sum of the different types of energies such as chemical, electrical, mechanical etc.

Question 7.
Fill in the blanks.

  1. If heat is released, ‘q’ is ……………
  2. For exothermic process ‘∆H’is ………………
  3. If work is done on the system, ‘w’ is ………………
  4. For endothermic process ‘∆H’ is ………………
  5. If work is done by the system, ‘w’ is ………………

Answer:

  1. Negative
  2. Negative
  3. Positive
  4. Positive
  5. Negative

Question 8.
1. What is meant by enthalpy of formation?
2. What is the value of standard enthalpy of formation (∆fH) of an element?
Answer:
1. Enthalpy of formation is the enthalpy change when one mole of a compound is formed from its elements in their most stable states of aggregation (i.e., reference state).

2. Zeno

Question 9.
First Law of thermodynamics is the law of conservation of energy.

  1. Give the mathematical form of the first law.
  2. Write the Gibb’s equation.
  3. What is the sign for ∆G for a spontaneous process?

Answer:

  1. ∆U=q+w w = work done
    q = heat absorbed
  2. G = H- TS or ∆G= ∆H – T∆S
  3.  In the case of spontaneous process ∆G = -ve

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 10.
1. Predict the sign of ∆S for the reaction,
NH3(g) + HCl(g) → NH4Cl(s)
2. The reaction between gaseous hydrogen and chlorine is
H2(g) + Cl2(g) → 2HCl(g); ∆rH = -1840 kJ
i) What is the enthalpy of formation of HCl?
ii) How much heat will be liberated at 298 K and 1 atm for the formation of 365 g of HCl?
Answer:
1. ∆S is negative

2. i) ∆fH = \(\frac{-1840}{2}\) = -920 kJ mol-1
ii) Heat liberated during the formation of 1 mole (36.5 g) of HCl = 920 kJ
∴ Heat liberated during the formation of 365 g of HCl = 9200 kJ

Question 11.
Derive the Meyer’s relationship.
Answer:
We have, q = C × ∆T
At constant volume, qv = Cv × ∆T = ∆U
At constant pressure, qp = Cp × ∆T = ∆H
For 1 mole of an ideal gas,
∆H = ∆U + ∆(pV) = ∆U + ∆(RT) = ∆U + R∆T
∴ ∆H = ∆U + R∆T
On putting the values of ∆H and ∆U,
Cp∆T = Cv∆T + R∆T
Cp = Cv + R
Cp – Cv = R, which is the Meyer’s relationship.

Question 12.
1. In a process 701J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?
2. What is free expansion? What is the work done during free expansion of an ideal gas?
Answer:
1. ∆U=q+w = 9 + w = 701J – 394 J = 307J

2. Expansion of a gas in vacuum is called free expansion. No work is done during free expansion of an ideal gas whether the process is reversible or irreversible.

Question 13.
1. Name the instrument used for measuring the ∆U of a process.
2. What is the value of ∆G for a reaction at equilibrium?
3. ∆H and ∆S of a reaction are 30.56 and 0.666 kJ/ mol respectively at 1 atm pressure. Calculate the temperature at which the reaction is in equilibrium.
Answer:
1. Bomb calorimeter
2. Zeno
3. ∆H-T∆S = 0 or ∆H = T∆S
Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics 2

Question 14.
Thermodynamic process differ based on the manner
in which it is carried out.
1. Distinguish between reversible and irreversible processes.
2. Calculate the amount of work done when 2 moles of a gas expands from a volume of 2 L to 6 L isothermally and irreversibly against a constant external pressure of 1 atm.
Answer:
1.

Reversible process Irreversible process
1) Which can be reversed 1) Which cannot be reversed spontaneous process
2) Takes place infinitesimally slowly 2) Takes place spontaneous
3) Work done maximum 3) Work done minimum

2. w = -p∆V = -1 × (6 – 2)
= – 4 L atm

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 15.
1. What are thermochemical equations?
2. Give an example for a thermochemical equation.
Answer:
1. A balanced chem ical equation together with the value of its ∆rH is called a thermochemical equation.
2.

Question 16.
1. Define lattice enthalpy of an ionic compound.
2. What is Born-Haber cycle?
Answer:
1. The lattice enthalpy of an ionic compound is the enthalpy change which occurs when one mole of an ionic compound dissociates into its ions in gaseous state.

2. It is a simplified method developed by Max Born and Fritz Haberto correlate lattice enthanpies of ionic compounds to otherthermodynamic data.

Question 17.
Predict what happens to entropy in the following changes:

  1. Metal is converted into alloy.
  2. Solute crystallizes from solution.
  3. Hydrogen molecule dissociates.

Answer:

  1. The entropy will increase.
  2. The entropy will decrease.
  3. The entropy will increase.

Question 18.
1. Give the relation between change in enthalpy and change in free energy.
2. Name the above relation.
3. What is the significance of the above relation?
Answer:
1. ∆G = ∆H – T∆S

2. Gibbs equation orGibbs-Helmholtz equation.

3. This relation is used to predict the spontaneity of a process based on the value of ∆G . If ∆G is negatve, the process is spontaneous. If ∆G is positive, the process is non-spontaneous.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 19.
1. Predict in each of the following whether entropy increases or decreases.
i) Sublimation of camphor
ii) 4Fe(s) + 3O2(g) → 2Fe2O3(g)
2. The equilibrium constant for a reaction at 30 °C
2.5 x 10-29. What will be the value of ∆G?
Answer:
1. i) entropy increases
ii) entropy increases
2.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics 4

Question 20.
1. Explain the effect of temperature on the spontaneity of a process based on Gibbs equation.
2. For a reaction 2A(g) + B(g) → 2D(g), enthalpy and entropy changes are – 20.5 kJ mol-1 and – 50.4 J K-1mol-1 respectively. Predict whether the reaction occurs at 25 °C.
Answer:
1. If ∆H is -ve and ∆S is +ve, ∆G would certainly be -ve and the process will be spontaneous at all temperatures.
If both ∆H and ∆S are – ve ∆G would be -ve if ∆H > T∆S
If both ∆H and ∆S are + ve ∆G would be -ve if T∆S > ∆H
If ∆H is +ve and AS is -ve, ∆G would certainly be +ve and the process will be non-spontaneous at all temperatures.

2. According to Gibbs equation, ∆G = ∆H – T∆S
∆G = (-20.5 × 10³)-(298 ×-50.4)
= – 20500 + 15019.2 = – 5480.8 J mol-1
Since ∆G is -ve, the process is spontaneous.

Plus One Chemistry Thermodynamics Four Mark Questions and Answers

Question 1.
1. Explain the first, second and third laws of thermodynamics.
2. What do you meant by entropy?
3. Explain the spontaneous process.
Answer:
1. First law:
Energy can neither be created nor be destroyed. Energy in one form can be converted into another form without any loss or gain.

Second law:
Entropy of the universe increases during a spontaneous process.

Third law:
Entropy of a perfect crystalline substance is zero at absolute zero of temperature.

2. Entropy:
Entropy is a measure of randomness or disorder of a system.

3. Spontaneous process:
A spontaneous process is defined as an irreversible process which has a natural tendency to occur either of its own or after proper initiation under the given set of conditions.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 2.
U1, q, w, U2 are given. U1 is internal energy, q is absorbed heat, w is work done and U2 is final energy.
a) Derive an equation for ∆U.
b) Give the equation for w.
c) Calculate the change in internal energy of a system which absorbs 200 J of heat and 315 J of work is done by the system.
Answer:
a) U2 = U1 + q +w
U2 – U1 = q + w
or ∆U = q + w
b) w= -p∆V
c) q = 200J
w = -315J
∆U = ?
∆U = q + w
= 200 + {315}
= 200-315 = -115J

Question 3.
a) Predict whether entropy increases or decreases in the following changes:
i) l2(s) → l2(g)
ii) Temperature of a crystalline solid is raised from 0 Kand 115 K.
iii) Freezing of water
b) Calculate the enthalpy of combustion of methane. Given that standard enthalpies of formation of CH4, CO2 and H2O are -75.2, -394 and -285.6 kJ/mol respectively.
Answer:
a) i) Entropy increases
ii) Entropy increases
iii) Entropy decreases
b) The required equation is,
Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics 5

Plus One Chemistry Thermodynamics NCERT Questions and Answers

Question 1.
In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process? (2)
Answer:
Heat absorbed by the system, (q) = + 701 J
Work done by the system (w) = – 304 J
Change in internal energy (∆U) = q + w
= 701 – 394
= 307 J

Question 2.
The reaction of cyanamide, NH2CN (s) with oxygen was carried out in a bomb calorimeter and AU was found to be – 742.7 kJ mol1 at 298 K. Calculate the enthalpy change for the reaction at 298 K. (2)
Answer:
NH2CN(S) + 3/2O2(g) → N2(g) + CO2(g) + H2O(l)
∆U = 742.7 kJ mol-1;
∆n(g) =2 – 3/2= + 0.5
R = 8.314 × 10-3kJ K-1 mol-1;
T = 298K
According to the relation, ∆H = ∆U + ∆ngRT
∆H = -742.7 kJ + 0.5 mol × 8.314 × 10-3kJ K-1 mol-1 × 298 K
=-742.7 kJ + 1.239kJ
=-741.5 kJ

Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics

Question 3.
Calculate the number of kJ of heat necessary to raise the temperature of 60 g of aluminium from 35 °C to 55 °C. Molar heat capacity of Al is 24 J mol-1 K-1. (2)
Answer:
Moles of Al (n) = \(\frac { 60{ g } }{ 27{ g }{ mol }^{ -1 } } \) = 2.22 mol
Molar heat capacity (Cm) = 24 J mol-1 K-1
∆T = 55 °C – 35 C° = 20C° or 20 K
Now, q = Cm × n × ∆T
= 24.0 J mol-1 K-1 × 2.22 mol × 20 K
= 1065.6 J
= 1.067 kJ

Question 4.
The enthalpy of formation of CO(g), CO2 (g), N2O (g), N2O4 (g) are -110, – 393, 81 and 9.7 kJ mol-1 respectively. Find the value of ∆rH for the reaction:
N2O4 (g) + 3CO(g) → N2O(g) + 3CO2(g)
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics 6

Question 5.
The equilibrium constant for the reaction is 10. Calculate the value of ∆G ; Given R = 8 J K-1 mol-1; T = 300 K.
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics 7

Question 6
Calculate the entropy change in surroundings when 1.0 mol of HzO (I) is formed under standard conditions. Given ∆fH = – 286 kJ mol-1.
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics 8

Question 7.
Comment on the thermodynamic stability of NO(g) and NO2 (g) given :
Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics 9
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 6 Thermodynamics 10

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Students can Download Chapter 5 States of Matter Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Plus One Chemistry States of Matter One Mark Questions and Answers

Question 1.
When a gas is compressed at constant temperature
a) The speeds of the molecules increase
b) The collisions between the molecules increase
c) The speed of the molecules decrease
d) The collisions between the molecules decrease
Answer:
b) The collisions between the molecules increase

Question 2.
The temperature at which a real gas obeys ideal gas law over an appreciable range of pressure is called ___________
Answer:
Boyle temperature or Boyle point

Question 3.
The compressibility factor is given by the expression
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 1
Answer:
a) \(\frac{p V}{n R T}\)

Question 4.
Two flasks of equal volume contain CO2 and SO2 respectively at 298 K and 1.5 atm pressure. Which of the following is equal in them?
a) Masses of the two gases
b) Rates of effusion
c) Number of molecules
d) Molecular structures
Answer:
c) Number of molecules

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Question 5.
With rise in temperature, viscosity of a liquid
a) Increases
b) Decreases
c) Remains constant
d) May increase or decrease
Answer:
b) Decreases

Question 6.
The unit of‘b’ in VanderWaals equation of state.
Answer:
l mol-1

Question 7.
Most probable velocity, average velocity, and root mean square velocity are related by
Answer:
1 : 1.128 : 1.224

Question 8.
The volume of 2.8g of CO at 27°C and 0.821 atm pressure is (R = 0.0821 l atm Km-1 ol-1)
Answer:
3L

Question 9.
The density of gas at 27°C and 1 atm is d. Pressure remaining constant at which of the following temp will its density become 0.75d?
Answer:
400K

Question 10.
The rms velocity of an ideal gas at 27°C is 0.3ms-1. ‘ Its rms velocity at 927°C in (ms-1) is
Answer:
0.6m/s

Plus One Chemistry States of Matter Two Mark Questions and Answers

Question 1.
Find out the relation between the first pair and complete the second pair.
a) Boyle’s law: Temperature
Charles’ law : ……………….
b) Avagadro’slaw: V α n
Ideal gas equation: ……………..
Answer:
a) Pressure
b) pV=nRT

Question 2.
The graphs of Boyle’s law as plotted by Student 1 (Graph 1) and Student 2 (Graph 2) are given below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 2

  1. Which is the correct graph?
  2. Justify your answer.

Answer:

  1. Both the graphs are correct.
  2. According to Boyle’s law, v α 1/p or p α 1/v. This is clear from graph 1. Also, according to Boyle’s law, pv is a constant at constant n and T. This is clear from graph 2.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Question 3.
The rate of diffusion of hydrogen is less than that of oxygen.

  1. Do you agree?
  2. Which law is applied here?
  3. State the law.

Answer:

  1. Yes.
  2. Graham’s law of diffusion.
  3. Rate of diffusion of a gas is inversely proportional to the square root of its molecular mass.

Question 4.
The ideal gas equation has been modified for real gases by applying pressure and volume corrections.

  1. What is the corrected equation known as?
  2. Write the equation and explain the terms.

Answer:

  1. The van derWaals’equation.
  2. \(\left(p+\frac{a n^{2}}{V^{2}}\right)\) (V — nb) = nRT
    where p – pressure, V – volume, n – no. of moles of the gas, a & b – van der Waals’ constants, R – universal gas constant and T – absolute temperature.

Question 5.
‘Moist soil grains are pulled together.’

  1. Name the related phenomenon.
  2. Justify.

Answer:

  1. Surface tension.
  2. This is because the surface area of thin film of water in moist soil is reduced due to surface tension.

Question 6.
1. What is aqueous tension?
2. What is its significance in the determination of pressure of a dry gas?
Answer:
1. The pressure exerted by saturated water vapour at a given temperature is called aqueous tension at that temperature.

2. Pressure of dry gas can be calculated by subtracting aqueous tension from the total presssure of the moist gas.
Pdry gas=PTotal-Aqueous tensi0n

Question 7.
A balloon filled with air, when kept in sunlight bursts after some time.

  1. Name the related law.
  2. Justify.

Answer:

  1. Charles’ law
  2. According to Charles’ law, volume a Temperature. Therefore, the volume increases when temperature increases, When the volume of the gas inside the baloon expanded more than that the balloon could afford, it bursted.

Question 8.
Define surface energy. What is its SI unit?
Answer:
Surface energy is defined as the energy required to rise the surface area of the liquid by one unit. The SI unit of surface energy is J m-2.

Question 9.
a) Based on Boyle’s law how will you show that at a constant temperature, pressure is directly proportional to the density of a fixed mass of the gas?
b) Give the relation between density and molar mass of a gaseous substance.
Answer:
a) According to Boyle’s law,
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 3

Question 10.
The isotherm of carbon dioxide at various temperatures is given below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 4
1. What is the significance of the shaded area?
2. Identify the pressure at which liquid CO2 appears for the first time when cooled form 30.98 °C. What is this pressure called?
Answer:

  1. At any point in the dome shaped shaded area liquid and gaseous CO2 exists in equilibrium.
  2. 73 atm. Critical pressure (pc).

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Question 11.
Certain properties of liquids are given below: Classify them on the basis of effect of temperature on them,

  1. Evaporation
  2. Vapour pressure
  3. Surface tension
  4. Viscosity

Answer:
Properties which increase with increase in temperature: Evaporation & Vapour pressure Properties which decrease with increase in temperature: Surface tension & Viscosity

Question 12.
The size of the water bubbles increases on moving to the surface.
1. Name the law responsible for this.
2. What is your justification?
Answer:
1. Boyle’s law.

2. According to Boyle’s law volume is inversely proportional to pressure. At the bottom of the pond, pressure is greater. So the volume (size) of the bubble was the least. But on coming up, pressure decreases and hence size of the bubble increases.

Question 13.
What are the properties of liquid state?
Answer:
Vapour pressure, Boiling point, Viscosity and Surface tension.

Plus One Chemistry States of Matter Three Mark Questions and Answers

Question 1
Analyse the following graph :
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 5
1. Name the gas law associated with this graph.
2. State the law.
3. Give the mathematical expression of this law.
Answer:
1. Boyle’s law.

2. It states that at constant temperature, pressure of a fixed amount of gas varies inversly with its volume.

3. Mathematically, Boyle’s law can be written as p ∝ \(\frac{1}{V}\) (at constant T and n)
Or p = k × \(\frac{1}{V}\) where k is the proportionality
constant which depends upon the amount of the gas, temperature of the gas and the units in which p and V are expressed.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Question 2.
1. What are van der Waals’ forces?
2. Which are the different types of van der Waals’ forces?
3. Arrange the van der Waal’s forces in the increasing order of their strength.
Answer:
1. The attractive intermolecular forces are known as van der Waals’ forces.

2. Dispersion forces/London forces, Dipole-Dipole forces, Dipole-Induced dipole forces and Hydrogen bonding.

3. Dispersion forces/London forces < Dipole-Induced dipole forces < Dipole-Dipole forces < Hydrogen bonding

Question 3.
A graphical representation of Charles’ law is given below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 6

  1. What is the temperature corresponding to the point‘A’called? .
  2. What will be the temperature at that point A’ in degree Celsius?
  3. What is the significance of this temperature?

Answer:

  1. Absolute zero temperature
  2. -273.15 °C
  3. Absolute zero is the lowest hypothetical or imaginary temperature at which gases are supposed to occupy zero volume.

Question 4.
Assume that two gases X and Y at the same temperature and pressure have the same volume.
1. Which of the following is correct?
No. of moles of X= No.of moles of Y
No. of moles of X ≠ No.of moles of Y
2. Which law helped you to find the answer?
3. State the law.
Answer:
1. No. of moles of X= No.of moles of Y

2. Avogadro’s law

3. Equal volumes of all gases under the same conditions of temperature and pressure contain equal number of molecules.

Question 5.
During a seminar session in the class, the presenter argued that equal amounts of both H2and N2 on heating at constant pressure will expand in the same rate. Another student objected this argument by saying that they will expand differently since their molecular masses are different.

  1. Who is correct in your opinion?
  2. Which law helped you to reach the answer?
  3. State the law and give its mathematical expression.

Answer:
1. The argument of the presenter is correct.

2. Charles’ law

3. Charles’ law states that pressure remaining constant, the volume of a fixed mass of a gas is directly proportional to its absolute temperature.

Mathematically, V ∝ T (at constant n and P) Or \(\frac{V}{T}\) = K, where K is the proportionality constant which depends on the pressure of the gas, its amount and the unit in which V is expressed.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Question 6.
1. What is an ideal gas?
2. Give the ideal gas equation and explain the terms.
3. Derive the ideal gas equation.
Answer:
1. A gas that follows Boyle’s law, Charles’ law and Avogadro law strictly at all conditions is called an ideal gas.

2. The ideal gas equation is pV = nRT where p = pressure, V = volume, n = number of moles, R = universal gas constant and T = absolute temperature.

3. According to Boyle’s law:
V ∝ \(\frac{1}{p}\) at constant T and n.
According to Charles’ law:
V ∝ T. at constant n and p.
According to Avogadro law,
V ∝ n , at constant p and T Combining the above three equations,
V ∝ \(\frac{nT}{p}\)
⇒ V = R\(\frac{nT}{p}\), where R is the proportionality constant known as universal gas constant.
Or pV = nRT, the ideal gas equation.

Question 7.
Partial pressure of a vessel containing Cl2, CO2 and CO is the sum of the partial pressures of Cl2, O2 and CO.
1. If so, is it correct to say partial pressure of a vessel containing NH3 and HCl gases is the sum of their partial
pressures? Justify.
2. Which law helped you to answer this?
3. State the law.
Answer:
1. No. NH3 reacts with HCl to form NH4CI. Since they are not non-interacting gases, their sum of partial pressures may not be equal to the total pressure.

2. Dalton’s law of partial pressures.

3. It states that the total pressure exerted by the mixture of non-reactive gases is equal to the sum of the partial pressures of individual gases.

Question 8.
The average kinetic kenergy of the gas molecules is directly proportional to the absolute temperature.

  1. Which theory is related to this assumption?
  2. Write the other postulates of this theory.

Answer:
1. Kinetic moleculartheory of gases.

2.

  • The volume of a gas molecule is negligible when compared to the whole volume of the gas.
  • There is no force of attraction between the particles of a gas at ordinary temperature and pressure.
  • The gas molecules are in random motion.
  • During motion, they collide with each other and also with the walls of the container.
  • Gravity has no influence in the movement of gas molecules.
  • Pressure of a gas is due to the collision of gas molecules with the walls of the container.
  • At any particular time, different particles in the gas have different speeds and hence different kinetic energies.

Question 9.
Three mateorological baloons filled with equal amount of helium, rising in the atmosphere are shown below: (Assume that temperature remains constant).
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 7

  1. Which of the baloons will be at the lowest altitude?
  2. Which law helped you to find the answer?
  3. State the law.

Answer:

  1. C –
  2. Boyle’s law.
  3. At low altitudes pressure is high. According to Boyle’s law for a given mass of a gas, greater the pressure lower is the volume at constant temperature.

Question 10.
‘All the postulates of the kinetic molecular theory of gases are correct.’

  1. Do you agree with the statement?
  2. If no, write the wrong postulates of this theory.
  3. Give justification.

Answer:
1. No.

2.

  • Volume of the molecules of a gas is negligibly small in comparison to the space occupied by the gas.
  •  There is no force of attraction between the molecules of a gas.

3. If assumption i) is correct, the p vs V graph of experimental data (real gas) and that theoritically calculated from Boyle’s law (ideal gas) should coincide. But this never happens.
If assumption ii) is correct, the gas will never liquify. But gases do liquify when cooled and compressed.

Question 11.
Two gases with equal molecular mass will have the same rate of diffusion.’

  1. Do you agree?
  2. Explain.
  3. Substantiate your answer with an example.

Answer:
1. Yes.

2. According to Graham’s law of diffusion the rate of diffusion depends only on the molecular mass. So if the molecular masses are the same, their rate of diffusion is same.

3. Both CO and N2 have the same molecular mass (28 g mol-1)
Rate of diffusion of CO = Rate of diffusion of N2

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Question 12.
Water can be boiled more quickly on the top of a mountain.

  1. Do you agree?
  2. What is the reason?
  3. What is called boiling point of a liquid?
  4. How normal boiling point and standard boiling point differ?

Answer:
1. Yes.

2. As we move to the top of a mountain atmospheric pressure decreases and hence boiling point decreases. So water boils quickly.

3. Boiling point of a liquid is the temperature at which the vapour pressure of a liquid is equal to the external pressure or atmospheric pressure.

4. The boiling point at 1 atm pressure is called normal boiling point. The boiling point at 1 bar pressure is called standard boiling point.

Question 13.
Ethanol flows faster than honey.

  1. Name the related phenomenon.
  2. Explain this phenomenon.
  3. What is the effect of temperature on this?

Answer:
1. Viscosity.

2. Viscosity is a measure of resistance to flow which arises due to the internal friction between layers of fluid as they slip past one another while liquid flows.

3. Viscosity of liquids decreases as the temperature rises because at high temperature molecules have high kinetic energy and can overcome the intermolecular forces to slip past one another between the layers.

Question 14.
Liquid drops attain spherical shape.

  1. Which property of liquids is responsible for this?
  2. Explain the phenomenon and justify.
  3. Suggest another consequence of this phenomenon.

Answer:
1. Surface tension.

2. Surface tension is defined as the force acting per unit length perpendicular to the line drawn on the surface of liquid. The lowest energy state of the liquid will be when surface area is minimum. Spherical shape satisfies this condition.

3. Fire polishing of glass – On heating, the glass melts and the surface of the liquid tends to take the rounded shape at the edges due to surface tension, which makes the edges smooth.

Question 15.
Vapour pressure is an important property of liquids.

  1. What is vapour pressure?
  2. How boiling point and vapour pressure are related?
  3. Pressure cooker is used for cooking food at higher altitudes. Give reason.

Answer:
1. Vapour pressure of a liquid is the pressure exerted by the vapour which is in equilibrium with liquid at a given temperature.

2. Boiling point of a liquid is the temperature at which the vapour pressure of liquid becomes equal to the atmospheric pressure. Thus, lower the vapour pressure of a liquid higher will be its boiling point and vice-versa.

3. At high altitudes atmospheric pressure is low. Therefore, liquids at high altitudes boil at lower temperatures in comparison to that at sea level. In a pressure cooker, the internal pressure is greater than atmospheric pressure. Hence, in a pressure cooker water boils at a temperature higher than its normal boiling point of 100 °C. Thus, cooking becomes more effective.

Question 16.
Assume that ‘A’, ‘B’ and ‘C’ are three non-reacting gases kept in a vessel at a constant temperature.
Then, PTotal= PA + PB + PC
1. Name the related law.
2. How can you explain the above law on the basis of kinetic molecular theory of gases?
Answer:
1. Dalton’s law of partial pressures.

2. In the absence of attractive forces, the particles of the gas behave independent of one another. The same is true even if there are more than one type of molecules. Thus, the number of molecules colloding the unit area of the wall per second at a given temperature, fora fixed amount of the gas issame.lt implies that the partial pressure of the gas will be unaffected by the presence of the molecules of other gases. But, the total pressure exerted is duet the impact of molecules of all the gases. Hence, the total pressure would be the sum of the partial pressures of the gases.

Question 17.
1. Write the general equation which relates the different variables of a gas used to describe the state of any ideal gas.
2. A flask at 295 K contains a gaseous mixture of N2 and O2 at a total pressure of 1.8 atm. If 0.2 moels of N2 and 0.6 moles of O2 are present, find the partial pressures of N2 and O2.
3. What is meant by Boyle temperature or Boyle point?
Answer:
1. PV = nRT
2.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 8

3. It is the temperature at which a real gas obeys ideal gas law over an appreciable range of pressure.

Question 18.
1. Liquid tries to rise or fall in the capillary. Name the related phenomenon.
2. What is the effect of temperature on the above phenomenon?
3. WhatistheSI unit of the above phenomenon.
Answer:
1. Surface tension.

2. The magnitude of surface tension of a liquid depends on the attractive forces between the molecules. When the attractive forces are large, the surface tension is large. Increase in temperature increases the kinetic energy of the molecules and effectiveness of intermolecular attraction decreases, so surface tension decreases as the temperature is raised.

3. N m-1.

Question 19.
1. Define critical temperature (Tc ).
2. CO2 cannot be liquified above 31.1°C. Why?
3. The critical temperatures of ammonia and carbon dioxide are 405.5 Kand 304.10 K respectively. On cooling, which of these gases will liquify first? Justify.
Answer:
1. It is the highest temperature at which a gas can
be liquified by applying external pressure.

2. The critical temperature (Tc ) of CO2 is 30.98°C. This is the highest temperature at which liquid CO2 is observed. Above this temperature it is gas.

3. Ammonia. This is because, on cooling, critical temperature of ammonia will be reached first. Liquefaction of carbon dioxide will require more cooling.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Question 20.
a) Will water boils at higher temperature at sea level or at top of a mountain. Explain.
b) A vessel of 120 mL capacity contains a certain amount of gas at 35 °C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35 °C. What would be its pressure?
Answer:
a) When atmospheric pressure decreases boiling point of the liquid also decreases. So the boiling point of water at sea level is not same as that at the top of a mountain. Atmospheric pressure decreases from sea level as we go high. Hence, the boiling point at the top of the mountain is less than that at sea level.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 9

Question 21.
Real gases deviate from ideal behaviour.
1. What are the two wrong postulates of kinetic theory of gases, responsible for deviation of real gases from ideal behaviour?.
2. When do real gases deviate from ideal behaviour?
Answer:
1.

  • Volume of the molecules of a gas is negligibly small in comparison to the space occupied by the gas.
  • There is no force of attraction between the molecules of a gas.

2. Real gases deviate from ideal behaviour at high pressure and low temperature, when the gas molecules are very close to each other.

Question 22.
1. What is meant by compressibility factor, Z?
2. What is the significance of compressibility factor?
3. A plot of pV/nRT of oxygen gas against p is as follows:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 10
a) Is the gas ideal or real?
b) Write the equation of state of the above gas.
Answer:
1. Compressibility factor (Z) is the ratio of product pV and nRT. Mathematically, Z = \(z=\frac{p V}{n R T}\).

2. The deviation of real gases from ideal behaviour can be measured in terms of compressibility factor.

3. a) Real gas.
\(\left[p+\frac{a n^{2}}{V^{2}}\right]\)[V-nb] = nRT (van der Waals’ equation)

Question 23.
a) What is the difference between gas and vapour?
b) Analyse the vapour pressure vs temperature curve shown below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 11
Arrange the compounds shown in the graph in the decreasing order of their normal boiling points,

c) The density of a gas was found to be 2.92 g L1 at 27 °C and 2.0 atm. Calculate the molar mass of the gas.
Answer:
a) A gas below its critical temperature can be liquified by applying pressure. Under these conditions, it is called vapour of the substance.

b) water > ethyl alcohol > carbon tetrachloride > diethyl ether
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 12

Question 24.
1. How will you account for the observation that automobile tyre is inflated with lesser air in summer than in winter?
2. A sample of gas occupies 250 mLat27 °C. What volume will it occupy at 35 °C if there is no change in pressure?
Answer:
1. This can be explained on the basis of Gay Lussac’s law, according to which at constant volume, pressure of a fixed amount of a gas varies directly with the temperature. In summer season the temperature will be higher. Hence, pressure will increase and the tyre may burst if filled with more air. But during winter temperature is low and hence pressure will below.

2. According to Charles’ law,
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 13

Question 25.
Real gases behave ideally at low temperature and high pressure.

  1. Is the above statement correct or not?
  2. Justify.
  3. Write the van der Waals’ equation for 1 mole of a real gas.

Answer:
1. The statement is wrong.

2. This is because real gases behave ideally at high temperature and low pressure, when the gas molecules are far apart.

3. \(\left(p+\frac{a}{V^{2}}\right)\)(V – b) = RT

Question 26.
1. Distinguish between real gas and ideal gas.
2. Explain the deviation of the following gases from ideal behaviouron the basis of the pV vs. p plot. CO, CH4, H2 and He.
Answer:
1. Real gas do not follow, Boyle’s law, Charles law, and Avagadro law perfectly under all conditions. Ideal gas follow, Boyle’s law, Charles’ Law and Avagadro law strictly under all conditions.

2.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 14
It can be seen that at constant temperature pV vs p plots for these gases are not straight lines. Two types of curves are seen. In the curves for H2 and He, as the pressure increases the value of pV also increases. These gases show positive deviation from ideal behaviour at all pressures. The second type of plot is seen in the case of CO and CH4. For these gases the pV value decreases with increase in pressure and reaches to a minimum value characteristics of the gas. After that PV value starts increasing. The curve then crosses the line for ideal gas and after that shows positive deviation continuously.

Question 27.
1. What is meant by laminar flow?
2. Derive the expression for the force responsible for flow of layers of a liquid.
Answer:
1. When a liquid flows overa fixed surface, the layer of molecules in the immediate contact of surface is stationary. The velocity of upper layers increases as the distance of layers from the fixed layer increases. This type of flow in which there is a regular gradation of velocity in passing from one layer to the next is called laminar flow.

2. Consider three layers of a flowing liquid as shown below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 15
For any layer, the layer above it accelerates its flow and the layer below this retards its flow. If the velocity of the layer at a distance dz is changed by a value du then velocity gradient is given by \(\frac{du}{dz}\). A force is required to maintain the
flow of layers. This force is proportional to the area of contact (A) of layers and velocity gradient.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 16
where η is a proportionality constant called coefficient of viscosity.

Question 28.
1. What are London forces?
2. What is the relation between London forces and the distance between the particles?
Answer:
1. The attractive force between two temporary dipoles is known as London forces or Dispersion forces.

2. London forces are always attractive and the interaction energy is inversely proportional to the sixth power of the distance between two interacting particles.

Question 29.
1. What is meant by thermal energy and thermal motion?
2. Can oxygen exist as a gas at -273.15°C? Write the significance of this temperature.
Answer:
1. Thermal energy is the energy of a body arising from motion of its atoms or molecules. The movement of particles due to thermal energy is called thermal motion.

2. At – 273.15 °C oxygen will not exist as a gas. In fact all the gases get liquified before this temperature is reached. It is the absolute zero of temperature, which is the lowest hypothetical or imaginary temperature at which gases are supposed to occupy zero volume.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter

Question 30.
Molecues of a gas are in a state of continuous motion
1. What is most probable speed?
2. Give the equation for average speed of molecules.
Answer:
1. Most probable speed is the speed possessed by the maximum fraction of molecules of the gas at a given temperature.

2. If there are ‘n’ molecules in a sample and their individual speeds are u1, u2 ……… un, then, average speed of molecules, uav is given by the equation:
\(u_{a v}=\frac{u_{1}+u_{2}+\ldots+u_{n}}{n}\)

Plus One Chemistry States of Matter Four Mark Questions and Answers

Question 1.
1. State the Avogadro law.
2. Give the mathematical expression of this law.
3. What is the value of molar volume of an ideal gas at 273.15 K and 1 bar?
4. Show that, at constant temperature and pressure, the density of an ideal gas is proportional to its molar volume.
Answer:
1. It states that equal volumes of all gases under the same conditions of temperature and pressure contain equal number of molecules.

2. v ∝ n (at constant P and T)
⇒ V = k × n where k is a proportionality constant.

3. 22.71098 L

4. According to Avogadro law, for n moles of an ideal gas,
V = k × n
But n = \(\frac{m}{M}\) where m is the mass of the gas and M is its molar mass.
Thus, V = k × \(\frac{m}{M}\)
On rearranging the above equation,
M = k × \(\frac{m}{V}\) = k × d
⇒ M ∝ d

Question 2.
The speed of molecules is a measure of their average kinetic energy.
a) What is root mean square speed?
b) Give the equation for root mean square speed.
c) Calculate the following:
i) Root mean square speed of methane molecule at 27°C.
ii) Most probable speed of nitrogen molecule at 25°C
Answer:
a) It is the square root of the mean of the squares of speeds of various molecules of the gas at a given temperature.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 17

Question 3.
The graph A is drawn at high temperature and low pressure and graph B is drawn at low temperature and high pressure.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 18
a) Which graph represents ideal behaviour?
b) Give the equation for combined gas law.
c) A baloon occupies volume of 700 mL at 25°C and 760 mm of pressure. What will be its volume at higher attitude when temperature is 15°C and pressure is 600 mm Hg.
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 19

Question 4.
1. Give the relationship among the three types of molecular speeds.
2. Drawthe Maxwell-Boltzmann distribution showing all the molecularspeeds.
3. Which of the following molecules will have the higher value of most probable speed at the same temperature, N2 or Cl2? Justify.
Answer:
1. Root mean square speed, average speed and the most probable speed have the following relationship:
Urms > Uav > Ump
2.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 20
3. N2. This is because at the same temperature, gas molecules with heavier mass have slower speed than lighter gas molecules. At the same temperature, lighter nitrogen molecules move faster than heavier chlorine molecules. Hence, at any given temperature, nitrogen molecules have higher value of most probable speed than the chlorine molecules.

Plus One Chemistry States of Matter NCERT Questions and Answers

Question 1.
What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 at 30 °C? (2)
Answer:
p1 = 1 bar p2 = ?
V1 = 500 dm³ V2 = 200 dm³
Temperature remains constant.
According to Boyle’s law,
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 21

Question 2.
Using the equation of state pV=nRT show that at a given temperature, the density of the gas is proportional to the gas pressure p. (2)
Answer:
According to ideal gas equation :
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 22

Question 3.
The density of a gas is found to be 5.46 g/dm³ at 27°C and under 2 bar pressure. What will be its density at STP. (3)
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 23

Question 4.
Calculate the volume occupied by 8.8 g of CO2 at 31.1°C and 1 bar pressure (R = 0.083 bar L K-1 mol-1). (2)
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 5 States of Matter 24

Question 5.
Explain the significance of van der Waal parameters. (2)
Answer:
The van der Waal parameter ‘a’ is a measure of the magnitude of intermolecular forces. The van der Waal parameter ‘b’ which is also called co-volume is a measure of effectve size of the gas molecules.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure

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Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure

Plus One Chemistry Chemical Bonding and Molecular Structure One Mark Questions and Answers

Question 1.
The octet rule is not valid for
a) CO2
b)H2O
c) O2
d) CO
Answer:
d) CO

Question 2.
The stability of an ionic crystal depends principally on
a) High electron gain enthalpy of the anion forming species
b) The lattice enthalpy of the crystal
c) Low ionization enthalpy of the cation forming species
d) Low heat of sublimation of cation forming solid
Answer:
b) The lattice enthalpy of the crystal

Question 3.
Which of the following molecules has highest dipole moment?
a) H2S
b)CO2
c) CCl4
d) BF3
Answer:
a) H2S

Question 4.
The d-orbital involved in sp3d hybridization is .
Answer:
dz2

Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure

Question 5.
Which of the following is paramagnetic and has a bond order of ½?
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 1
Answer:
H2+

Question 6.
Dipole moment µ electric charge ‘e’ and bond length ‘d’are related by the equation.
Answer:
M = e x d

Question 7.
In which of the following carbon atom is sp2 hybridised?
a) CO2
b) C2H6
c) C6H6
d) HCN
e) \(C{ H }_{ 3 }-C\equiv CH\)
Answer:
C6H6

Question 8.
AgF is ionic whereas Agcl is covalent. This can be explained by
Answer:
Faja’ens Rule

Question 9.
The shape of covalent molecule CIF3 is _________
Answer:
T – shaped

Question 10.
The C – O bond order in CO32- is
Answer:
1.33

Plus One Chemistry Chemical Bonding and Molecular Structure Two Mark Questions and Answers

Question 1.
The order of repulsion of electron pairs as written by student is given below:
lone pair-lone pair repulsion < lone pair-bond pair repulsion>bond pair-bond pair repulsion.
1. Can you see anything wrong in this?
If yes, correct it.
2. Name the theory behind this.
Answer:
1. Yes.
Repulsion decreases in the order: lone pair-lone pair repulsion>lone pair-bond pair repulsion> bond pair-bond pair repulsion,

2. VSEPR theory

Question 2.
During a small group discussion in the class room a student argued that in acetylene both the carbon atoms are in sp3 hybridised state.

  1. What is your opinion?
  2. What is the bond angle between carbon atoms in acetylene?

Answer:

  1. The student’s argument is wrong. In acetylene both the carbon atoms are triple bonded and are in sp hybridised state,
  2. 180°

Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure

Question 3.
Classify the following compounds according to their hybridisation.
CH4, BF3, C2H4, BeF2, C2H2
Answer:
sp3 hybridisation – CH4
sp2 hybridisation – BF3, C2H4
sp hybridisation – BeF2, C2H2

Question 4.
A student arranged the halide ions in the increasing order of polarisability as: F < l < CI < Br
1. Is this the correct order? If not write it in correct order.
2. Justify.
Answer:
1. No.
Polarisability increases in the order: F < Cl <Br < l

2. Polarisability increases when the size of anion increases.

Question 5.
Give any two differences between sigma and pi bonds.
Answer:
Sigma bond (σ) is formed when two atomic orbitals under head-on overlapping. It is a strong bond. Pi(π) bond is formed when atomic orbitals undergo lateral (sidewise) overlapping. It is a weak bond.

Question 6.
Write the type of hybridisation of each carbon in the compound.
CH3-CH=CH-CN
Answer:
Carbon 1 → sp³
Carbon 2 → sp²
Carbon 3 → sp²
Carbon 4 → sp

Plus One Chemistry Chemical Bonding and Molecular Structure Three Mark Questions and Answers

Question 1.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 2
1. What is meant by the above picture?
2. Which type of bond is present here?
3. Which type of overlapping leads to the formation of π bond?

Answer:

  1. s-s overlapping
  2. A strong sigma bond
  3. This type of covalent bond is formed by the lateral or sidewise overlap of half-filled atomic orbitals.

Question 2.
‘In ethane there are 6 covalent bonds. Five are strong σ bonds and the remaining one is a weak π bond.’

  1. Do you agree with this?
  2. How is a bond different from π bond in the mode of formation?

Answer:

  1. Yes.
  2. Sigma bond is formed by the end to end overlapping of bonding orbitals along the internuclear axis, π bond is formed by the lateral or sidewise overlap of half filled atomic orbitals.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure

Question 3.
Choose the correct molecules from the given clues: H2O, SF6, BF3
1. Clue -1 The central atom is in sp² hybridised state and the molecule has trigonal planar in shape. Clue-2 The bond angle is 120°
2. Clue-1 The number of electron pairs in this molecule is 6.
Clue -2 It has octahedral geometry.
3. Clue-1 The bond angle is reduced from 109° 28′ to 104.5°
Clue-2 It has a bent shape.
Answer:

  1. BF3
  2. SF6
  3. H2O

Question 4.
Give theoretical explanation for the following statements.
1. H2S is acidic while H2O is neutral.
2. Hydrogen chloride gas dissolves in water.
Answer:
1. S-H bond energy is less than that of O-H bond energy. So H+ can be easily generated from H2S.

2. When HCl is treated with H2O it undergoes hydrolysis as per the following reaction and dissolves.
HCl + H2O → H3O+ +Cl

Question 5.
The potential energy level diagram forthe formation of hydrogen molecule as drawn by a student is given below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 3
1. Resketch the graph with correct labelling.
2. How can you determine the radius of one hydrogen atom?
Answer:
1.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 4
2. Bond length in hydrogen molecule is the intermolecular distance between two hydrogen atoms. So, half of the bond length is taken as the radius of hydrogen atom.

Question 6.
Ionisation enthalpy is one of the factors favoring the
formation of ionic bonds.
1. Will you agree with this statement?
2. Explain how?
3. Write anotherfactorfavouring the formation of ionic bonds.
Answer:
1. Yes.

2. In the formation of the ionic bond, a metal atom losses electrons to form cation. This process requires energy equal to the ionisation enthalpy. Lesser the ionisation enthalpy of the metal atom, easier will be the removal of electron from the atom to form cation and hence greater will be the tendency to form ionic bond.

3. Electron gain enthalpy of the element forming anion.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure

Question 7.
Complete the following table:

Sigma bond π -bond
Formation …………. ………….
Strong/Weak …………. ………….
About rotation …………. ………….

Answer:

Sigma bond π -bond
Formation Sigma bond is formed by end to end (or axial) overlap of atomic orbitals π -bond is formed by the sidewise overlap of atomic orbitals
Strong/Weak Strong Weak
About rotation Free rotation Free rotation is not possible

Question 8.
a) The dipole moment of BF3 is zero even though the B – F bonds are polar. Justify.
b) Give the hybridisation involved in the following compounds

  1. NH3
  2. C2H4
  3. SF6
  4. PCl5

c) o-nitro phenol has a lower boiling point than its para isomer. Why?
Answer:
a) In BF3, dut to the symmetric trigonal planar geometry of the molecule, the B – F bond are oriented at an angle of 120°to one another. The three bond moments give a net sum of zero as the resultant of any two is equal and opposite to the third.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 5
b)

  1. sp³
  2. sp²
  3. sp³d²
  4. sp³d

c) In o-nitrophenol, intramolecular hydrogen bonding is present and there is no association of molecules whereas in p-nitrophenol there is inter-molecular hydrogen bonding which causes association of molecules.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 6

Question 9.
1. How many a and n bonds are there in the following molecules i) ethane ii) acetylene?
2. BF3 and NH3 are tetra atomic molecules. But the shape of BF3 is different from that of NH3. Explain this using hybridisation.
Answer:
1. i) Ethane – 7σ bonds
ii) Acetylene-3σ bonds and 2 π bonds

2. In BF3 molecule, the ground state electronic configuration of central boron atom is 1s²2s²2p¹. In the excited state, one of the 2s electrons is promoted to vacant 2p orbital. As a result, boron has three unpaired electrons. These three orbitals (one 2s and two 2p) hybridise to form three sp2 hybrid orbitals oriented in a trigonal planar arrangement and overlap with 2 p orbitals of F to form three B – F bonds. Therefore, BF3 molecule has a planar geometry with FBF bond angle of 120°.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 7
In NH3, the valence shell electronic configuration of N in ground state is 2s² \(2p_{ x }^{ 1 }2{ p }_{ y }^{ 1 }2{ p }_{ z }^{ 1 }\). These four orbitals undergo sp³ hybridisation to form four sp³ hybrid orbitals, three of them containing unpaired electrons and the fourth one containing lone pair. The three hybrid orbitals overlap with 1 s orbitals of hydrogen atoms to form three N – H sigma bonds. Since, the bp-lp repulsion is greater than the bp-bp repulsion, the molecule gets distorted and the bond angle is reduced to 107° from 109.5°. Thus, the geometry of NH3 molecule is trigonal pyramidal.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 8

Question 10.
Covalent bond is formed by the overlaping of atomic orbitals.
1. What is meant by orbital overlapping?
2. What are the 3 types of overlapping?
Answer:
1. Orbital overlapping is the partial interpenetration or merging of atomic orbitals. It results in the pairing of electrons. Greater the overlap the stronger is the bond formed between two atoms.

2. s-s overlapping
s-p overlapping
p-p overlapping

Question 11.
1. Which among the following will exist He2 or He2+? Explain.
2. H2S is a gas at ordinary condition, while H2O is liquid. Account for the above statement.
3. State the hybridisation in the following molecules,
i) PF6
ii) C2H6
Answer:
1. He2+
Helium molecule contains 4 electrons. Out of this 4 electrons, 2 are present in the bonding molecular orbital and the remaining 2 are present in the anti-bonding molecular orbital.
Bond order = ½ (Nb-Na)
= ½ (2-2) = 0
Hence, He2 cannot exist. The molecular orbital diagram is given below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 9
He2+ contains 3 electrons. Out of these 3 electrons, 2 are present in the σ1s level and the remaining one is present in the σ* 1s level.
Bond order = ½ (Nb-Na)
= ½ (2-1) = ½
Since the bond order is half the molecular ion exists but possesses low stability.

2. In H2S, there is no hydrogen bonding whereas in water, molecular association is possible due to intermolecular hydrogen bonding.

3. i) PF5 = sp³d hybridisation
ii) C2H6 = sp³ hybridisation

Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure

Question 12.
Bond order is a term commonly used in MO theory.
1. How is it calculated?
2. How is it related to bond length and bond energy?
Answer:
1. Bond order = ½ (Nb-Na)
where Nb = No. of electrons occupying bonding orbitals and Na = No. of electrons occupying antibonding orbitals.

2. As the bond order increases, bond length decreases and bond energy increases, i.e., bond order is directly proportional to bond energy and inversely proportional to bond length.

Question 13.
1. Explain the hybridisation and geometry of ethyne.
2. What is the difference between bonding molecular orbital and antibonding molecular orbital?
3. How the magnetic nature of a molecule is related to its electronic structure?
Answer:
1. In the formation of ethyne (C2H2), both the carbon atoms undergo sp hybridisation having two unhybridised orbitals (2px and 2py). One sp hybrid orbital of one carbon atom overlaps axially with sp hybrid orbital of the other carbon atom to form C-C sigma bond, while the other hybridised orbital of each carbon atom overlaps axially with the half filled s orbital of hydrogen atoms forming o bonds. Each of the two unbybridised p orbitals of both the carbon atoms overlaps sidewise to form two K bonds between the carbon atoms. Thus, ethyne has a linear geometry with π bond angle of 180°.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 10

2. The molecular orbital which has lower energy than the atomic orbital is called bonding molecular orbital and the molecular orbital which has greater energy than the atomic orbital is called anti bonding molecular orbital.

3. If all the molecular orbitals in a molecule are doubly occupied (i.e., paired), the substance is diamagnetic (repelled by magnetic field). It one or more molecular orbitals are singly occupied (i.e., unpaired) it is paramagnetic (attracted by magnetic field).

Question 14.
Molecular Orbital Theory (MOT) is an advanced theory
of chemical bonding.
1. Write the salient features of MOT.
2. What is meant by LCAO? Illustrate using hydrogen molecule.
3. What are the conditions for the combination of atomic orbitals?
Answer:
1.

  • The electrons in a molecule are present in the
    various molecular orbitals.
  • The atomic orbitals of comparable energies and proper symmetry combine to form molecular orbitals.
  • iii) The electron in a molecular orbital is influenced by two or more nuclei depending upon the number of atoms in the molecule.
  • The number of molecular orbitals formed is equal to the number of combining atomic orbitals. When two atomic orbitals combine, two
    molecular orbitals are formed. One is known as bonding molecular orbital while the other is called antibonding molecular orbital.
  • The bonding molecular orbital has lower energy and hence greater stability than the corresponding antibonding molecular orbital.
  • The electron probability distribution around a group of nuclei in a molecule is given by a molecular orbital.
  • The molecular orbitals are filled in accordance with the Aufbau principle obeying the Pauli’s exclusion principle and the Hund’s rule.

2. LCAO refers to the linear combination of atomic orbitals. It is an approximate method used to explain the formation of molecular obritals. Consider hydrogen molecule consisting of two atoms A and B. Each hydrogen atom has one electron in the 1s orbital. The atomic orbitals of these atoms can be represented by the wave functions ψA and ψB. Mathematically, the formation of molecular orbitals can take place by addition and by subtraction of wave functions of individual atomic orbitals.
ψMO = ψA ± ψB
Therefore, the two molecular orbitals σ and σ* are formed as:
σ* = ψA – ψB
The molecular orbital σ formed by the addition of atomic orbitals is called the bonding molecular orbital while the molecular orbital σ* formed by the subtraction of atomic orbital is called antibonding molecular orbital. The energy level diagram is shown below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 11

3.

  • The combining atomic orbitals must have the same or nearly same energy.
  • The combining atomic orbitals must have the same symmetry about the molecular axis.
  • The combining atomic orbitals must overlap to the maximum extent.

Question 15.
Consider a reaction PCl5(g) → PC3(g) + Cl2(g)
1. What is the change in hybridisation state of phosphorus?
2. Explain why does PCl5 decomposes easily?
Answer:
1. When PCl5 decomposes to PCl3, the hybridisation of P changes from sp³d to sp³.

2. In PCl5, the five sp³d orbitals of P overlap with the singly occupied p orbitals of Cl atoms to form five P-CI sigma bonds. Three P-Cl bonds which lie in one plane and make an angle of 120° with each other are called equatorial bonds. The remaining two P – Cl bonds, called axial bonds, one lie above and the other lie below the equatorial plane, make an angle of 90° with the plane. As the axial bond pairs suffer more repulsive interaction from the equatorial bond pairs, axial bonds are slightly longer and hence slightly weaker than the equatorial bonds. This makes PCl5 molecule more reactive and hence it decomposes easily.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 12

Question 16.
The electron dot structure (Lewis structure) of ammonia molecule is shown below:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 13
1. Write the number of bond pairs of electrons and lone pairs of electrons in ammonia molecule.
2. The structures of o-nitrophenol and p-nitrophenol are shown in the figure. The former is a steam volatile liquid whereas the latter is a solid. Justify your answer giving reason.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 14
Answer:
1. Ammonia molecule contains one lone pair of electrons and 3 bond pair of electrons,

2. In o-nitrophenol, there is intramolecular hydrogen bonding and there is no molecular association. But in p-nitro phenol intermolecular hydrogen bonding is present and hence molecular association is possible.

Plus One Chemistry Chemical Bonding and Molecular Structure Four Mark Questions and Answers

Question 1.
Hydrogen bonding is present in NH3 and H2O.
1. What is hydrogen bond?
2. What are different types of hydrogen bonds?
3. Explain the effect of hydrogen bonding.
Answer:
1. Hydrogen bond is defined as the attractive force which binds hydrogen atom of one molecule with the electronegative atom (F, O, N) of the same or another molecule.

2. Intermolecular hydrogen bond and Intramolecular hydrogen bond.

3. Compounds containing hydrogen bonds show higher melting and boiling points. Compounds whose molecules can form hydrogen bonds with water molecules are soluble in water.

Question 2.
Classify the following compounds according to their
shape.
BeF2, BeCl2, CH4, BF3, PCl5, SF6, SbCl5 NH4+, SiF4, AlCl3.
Answer:
Linear – BeF2, BeCl2
Trigonal planar-AlCl3, BF3
Tetrahedral – CH4, NH4+, SiF4
T rigonal bipyramidal – PCl5, SbCl5

Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure

Question 3.
Benzene is an example of a compound exhibiting resonance.
1. What is meant by resonance?
2. Explain the resonance of ozone.
Answer:
1. When a molecule cannot be represented by a single structure but its characteristic properties can be described by two or more different structures, then the actual molecule is said to be a resonance hybrid of these canonical structures.

2. The resonance in ozone can be represented by the following structures:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 15
According to structures I and II, there is one single bond and one double bond in ozone molecule. But experiments show that both the oxygen- oxygen bonds are equal and the bond length 128 pm) is intermediate between single (148 pm) and double bond (121 pm) lengths. Hence it is assumed that ozone is a resonance hybrid (structure III) of structures I and II.

Question 4.
Match the following:

No. of electrons pairs Shape of molecule Examples
2 Trigonal planar SF6
4 Linear BeF2, BeCl2
3 Tetrahedral BF3, AlCl3
6 Octahedral CH4, SiF4

Answer:

No. of electrons pairs Shape of molecule Examples
2 Linear BeF2, BeCl2
4 Tetrahedral CH4, SiF4
3 Trigonal planar BF3, AlCl3
6 Octahedral SF6

Question 5.
In the formation of methane, carbon undergoes sp³ hybridisation.

  1. What do you mean by sp³ hybridisation?
  2. Give the % s-character and p-character of an sp³ hybrid orbital.
  3. What is the bond angle in methane?
  4. What is the geometry of methane molecule?

Answer:

  1. sp³ hybridisation involves mixing up of one – s and three-p orbitals of the valence shell of an atom to form four sp³ hybrid orbitals of equivalent energies and shape.
  2. Each sp³ hybrid orbital has 25% s-character and 75% p-character.
  3. The angle between the sp3 hybrid orbitals in methane is 109°28′.
  4. Tetrahedron.

Question 6.
Complete the following table:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 16
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 17

Question 7.
Fill in the blanks:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 18
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 19

Question 8.
Dipole moment is used to predict the shape of
molecules.
1. Justify the statement based on the shapes of CO2 and H2O.
2. Which is having high dipole moment? NH3 or NF3? Why?
Answer:
1. Carbon dioxide is a linear molecule in which the two C=0 bonds are oriented in the opposite directions at an angle of 180°. Hence the two C=0 bond dipoles cancel each other and the resultant dipole moment of CO2 is zero. Thus CO2 is non¬polar molecule
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 20
On the other hand, water molecule has a bent structure in which two O-H bonds are oriented at an angle of 104.5°. Therefore, the bond dipoles of two O-H bonds do not cancel each other and the molecule will have a net dipole moment (1.85D).
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 21

2. The dipole moment of NH3 is higher than that of NF3. In both cases, the central N atom has a lone pair whose orbital dipole points away from the N atom. In NH3 the orbital dipole due to the lone pair is in the same direction as the resultant bond dipole of the three N-H bonds. On the other hand, in the case of NF3, the resultant dipole of the three N-F bonds is in the opposite direction to the orbital dipole due to the lone pair. Thus, the orbital dipole due to the lone pair decreases the effect of the resultant N-F bond moments, which results in the low dipole moment of NF3.
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 22

Question 9.
The geometry of a covalent molecule is related to the hybridisation involved in the central atom. Complete the following table:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 23
Answer:
Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure 24

Question 10
Depending upon the type of overlapping, covalent bonds are of two types.
a) Name them and give any two difference between them.
b) Find the total number of these two types of bonds ’ in propane and 2-butene.
Answer:
a) Sigma (σ) bond and pi (π) bond.
Sigma (σ) bond:
This type of covalent bond is formed by the end to end overlapping of half-filled atomic orbitals along the internuclear axis. The overlap is also known as head on overlap or axial overlap. The electrons constituting sigma bond are called sigma electrons.

Pi (π) bond:
This type of covalent bond is formed by the lateral or sidewise overlap of half-filled atomic orbitals. The atomic orbitals overlap in such a way that their axes remain parallel to each other and perpendicularto the internuclear axis.
b) 10 σ bond in propane
11 σ bond and 1π bond in 2-butene

Plus One Chemistry Chemical Bonding and Molecular Structure NCERT Questions and Answers

Question 1.
Explain the formation of a chemical bond. (2)
Answer:
According to Kossel-Lewis approach, the formation of chemical bond between the two atoms takes place either by the transference of electrons or by mutual sharing of electrons. However, according to the modem view the formation of chemical bond between the two approaching atoms occurs only if there is a net decrease of energy because of attractive and repulsive forces.

Question 2.
Write the favourable conditions for the formation of ionic bond. (2)
Answer:
Ionic bond is formed by transference of electrons from one atom to another. The favourable conditions for its formation are:

  • Low ionisation enthalpy of element forming cation.
  • More negative value of electron gain enthalpy of element forming the anion and
  • High value of lattice enthalpy of the compound formed.

Plus One Chemistry Chapter Wise Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure

Question 3.
Although geometries of NH3 and H2O molecules are distorted tetrahedral, bond angle in water is less than that in ammonia. Discuss. (2)
Answer:
The difference in bond angles is due to the different numbers of lone pairs and bond pairs in the two species. In NH3, the N atom has two lone pairs and three bond pairs while in H2O, the O atom has two lone pairs and two bond pairs. The repulsive interactions of lone pairs and bond pairs in water are relatively more than those in NH3. Hence, bond angle around central atom in water is relatively smaller (104.5°) than that in NH3 molecule (107°).

Question 4.
Is there any change in the hybridisation of B and N atoms as a result of the following reaction? (2)
BF3 + NH3 → F3B.NH3
Answer:
During combination of species BH3 and NH3, N atom of NH3 is donor and B atom of BF3 is acceptor. The hybrid state of B in BF3 is sp² and that of N in NH3 is sp³. In the compound F3B+-NH3 both N and B atoms are surrounded by four bond pairs. Thus, the hyrid state of both is sp³. Hence during the reaction the hybrid state of B changes from sp² to sp³ but that of N remains the same.

Question 5.
Define hydrogen bond. Is it weaker or stronger than the van der Waals’ forces? (2)
Answer:
Hydrogen bond can be defined as the attractive force which binds hydrogen atom of one molecule with the electronegative atom (F, O or N) of another molecule. Hydrogen bond is stronger than van der Waals’forces because it is a strong dipole-dipole interaction. The van der Waals’ forces, on the other hand, are weak dispersion forces.

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Adisthana Padavali Malayalam Standard 9 Guide

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Board SCERT, Kerala
Text Book NCERT Based
Class Plus One
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Chapter Plus One Chapter Wise Questions
Category Kerala Plus One

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HSSLive Plus One

HSSLive Plus One Notes Chapter Wise Kerala

HSE Kerala Board Syllabus HSSLive Plus One Notes Chapter Wise Pdf Free Download in both English Medium and Malayalam Medium are part of Plus One Kerala SCERT. Here HSSLive.Guru has given Higher Secondary Kerala Plus One Chapter Wise Quick Revision Notes based on CBSE NCERT syllabus.

HSSLive Plus One Notes Chapter Wise Kerala

Board SCERT, Kerala
Text Book NCERT Based
Class Plus One
Subject All Subjects
Chapter Plus One Notes
Category HSSLive Plus One

We hope the given HSE Kerala Board Syllabus HSSLive Plus One Notes Chapter Wise Pdf Free Download in both English Medium and Malayalam Medium will help you. If you have any query regarding Higher Secondary Kerala Plus One Chapter Wise Quick Revision Notes based on CBSE NCERT syllabus, drop a comment below and we will get back to you at the earliest.

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