Plus One Chemistry Previous Year Question Paper 2018

Kerala Plus One Chemistry Previous Year Question Paper 2018

Time Allowed: 2 hours
Cool off time: 15 Minutes
Maximum Marks: 60

General Instructions to Candidates

• There is a ‘cool off time’ of 15 minutes in addition to the writing time.
• Use the ‘cool off time’ to get familiar with questions and to plan your answers.
• Read the instructions carefully.
• Read questions carefully before answering.
• Calculations, figures and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provied.
• Give equations wherever necessary.
• Electronic devices except non programmable calculators are not allowed in the Examination Hall.

Answer all questions from question numbers 1 to 7. Each carry one score.

Question 1:
Name the quantum number which gives the spacial orientation of an orbital with respect to standard as et of coordinate axes.

Question 2:
Among N^3-, O², F^–, Na⁺ and Al³⁺, which one has the smallest size?

Question 3:
Which pollutant in water causes brown mottling of teeth ?

Question 4:
Examine the following graph and name the gas law corresponding to it.

Question 5:
If Z – axis is the internuclear axis, name the type of covalent bond formed by the overlapping of two Py – orbitals.

Question 6:
Which among the following measurements, contains the highest number of significant figures?

Question 7:
Write the formula of the basic structural unit of silicates.

Answer any ten from question numbers 8 to 20. Each carries two scores.

Question 8:
Write any two limitations of octet rule.

Question 9:
Draw the ‘sawhorse’ projections of the eclipsed and staggered conformations of ethane.

Question 10:
Give the relation between Kp and Kc for the reaction given below.

Question 11:
State and illustrate the law of multiple proportions.

Question 12:
a. Define ‘ normal boiling point’ of a liquid.
b. Give a reason for the use of pressure cooker to cook food, at high altitudes.

Question 13:
Write two important results observed during the photoelectric effect.

Question 14:
Differentiate state functions from path functions and give one example for each.

Question 15:
Define the terms, Biochemical Oxygen Demand (BOD) and Eutrophication.

Question 16:
Give the chemical equation for the conversion of hexane to benzene. Write the name of the process.

Question 17:
H₂0 and HSO₄, can act both as Bronsted acids and bases. For each case give he corresponding conjugate acid and conjugate base.

Question 18:
What is metamerism ? Write the metamers of C₄H₁₀O.

Question 19:
What are Zeolites ? Give any two uses of zeolites.

Question 20:
First law of thermodynamics can be stated as Δu = q + w. How can this equation be expressed for:
a. An isothermal reversible change ?
b. A process carried out at constant volume ?

Answer any seven from question numbers 21 to 29. Each carries three scores.

Question 21:
Predict the products :

Question 22:
Redox reactions are classified into four types. Describe any three of them with suitable examples.

Question 23:
Enthalpies of formation of some compounds are given below:

Using these data, calculate the enthalpy of reaction for N₂O_4(g) + 3CO_(g) → N₂ O_(g) + 3CO_2(g)

Question 24:
Sketch the structures of graphite and diamond. What is the impact of structure on physical properties of these allotropes?

Question 25:
Briefly explain the different types of hydrides.

Question 26:
Calculate the amount of CO₂(g) produced by the reaction of 32g of CH₄ and 32g of O₂.

Question 27:
Give reasons for the following :
a. ‘O’ has lower ionization enthalpy than N and F.
b. Cl has higher negative electron gain enthalpy than F.

Question 28:
Write the postulates of kinetic molecular theory of gases.

Question 29:
The ionization constant of nitrous acid is 4.5 i x 10^-4. Calculate the pH of 0.04M solution of j nitrous acid in water.

Answer any three from question numbers 30 to 33. Each carries four scores.

Question 30:
Name the commercial process used to prepare sodium carbonate and write the chemical equations of the steps involved in it.

Question 31:
What is ‘sodium fusion extract’? How the presence of N, S and halogens inorganic compounds are detected?

Question 32:
The diatomic species Ne₂ does not exist, but Ne₂ can exist. Explain on the basis of molecular orbital theory.

Question 33:
Explain how the different series of lines are formed in the hydrogen spectrum. Derive an equation to find the waW number of a line in the hydrogen spectrum.

Answers

Answer 1:
Magnetic quantum number

Answer 2:
Al³⁺

Answer 3:
Fluoride (P)

Answer 4:
Gay-Lussac’s law

Answer 5:
π bond

Answer 6:
a

Answer 7:
SiO₄^4-

Answer 8:
The incomplete octet of the central atom
In some compounds, the number of electrons surrounding the central atom is less than eight. This is especially the case with elements having less than four valence electrons.

Odd-electron molecules
In molecules with an odd number of electrons like nitric oxide, NO and nitrogen dioxide, N02, the octet rule is not satisfied for all the atoms

Answer 9:

Answer 10:
K_p = K_c (RT)^∆n
∆n = 3 – 2 = 1
K_p = K_c (RT)¹= K_c(RT)

Answer 11:
Law of Multiple Proportions
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

Here the masses of oxygen which com¬bine with a fixed mass of hydrogen bear a simple ratio, i.e., 16: 32 or 1: 2.

Answer 12:
a. The temperature at which a real gas obeys ideal gas law over an appreciable range of pressure is called Boyle temperature or Boyle point.
b. At high altitudes atmospheric pressure is low. Therefore liquids at high altitudes boil at lower temperatures in comparison to that at sea level. Since water boils at low temperature on hills, the pressure cooker is used for cooking food.

Answer 13:
i. There is no time lag between this striking of light beam and ejection of an electron from the metal surface.
ii. The no.of electron ejected is proportional to the intensity or brightness of light

Answer 14:
A state function is a property of the system whose values depends upon the initial and final states of the system and independent of the path by which the state is reached.
Path function is a property of the system whose values depend on the path followed,
eg. Heat (q), Work (w).

Answer 15:
BOD is Biochemical Oxygen Demand. It is defined as the amount of oxygen in milligrams dissolved in water needed to break down the organic matter present in one litre of water for five days at 20 °C.Pure water contains BOD of 0-3 ppm. 5 ppm means water is contaminated.
The process in which the nutrient enriched water bodies support a dense plant population, which kills animal life by depriving it of oxygen and results in subsequent loss of biodiversity is known as eutrophication

Answer 16:
Alkanes having six or more carbon atoms on heating to 773 K at 10^-20 atmospheric pressure in presence of Cr₂O₃ or V₂O₅ or MO₂O₃ catalyst give aromatic compounds. This process is known as aromatization.

Answer 17:
H₂O : Conjugate acid – H₃O⁺ Conjugate base – OH^–
HS0₄^– : Conjugate acid – HSO₄^–
Conjugate base – SO₄^2-

Answer 18:
Metamerism: It arises due to different alkyl chains on either side of the functional group in the molecule. For example, C₄H₁₀O represents methoxypropane (CH₃OC₃H₇) and ethoxyethane (C₂H₅OC₂H₅).

Answer 19:
Zeolites : If aluminum atoms replace few silicon atoms in three-dimensional network of silicon dioxide, overall structure known as aluminosilicate, acquires a negative charge. Cations such as Na⁺, K⁺ or Ca²⁺ balance the negative charge. Examples are feldspar and zeolites. Zeolites are idely used as a catalyst in petrochemical industries for cracking of hydrocarbons and isomerisation, e.g., ZSM- 5 (A type of zeolite) used to convert alcohols directly into gasoline. Hydrated zeolites are used as ion exchangers in softening of “hard” water.

Answer 20:
a. q = -w = 2.303 RT log \(\frac { { V }_{ f } }{ { V }_{ i } }\)
q = -w
b. ∆u = ∆w

Answer 21:
a. CH₃– CH₂– CH₂– Br 1 – bromopropane

Answer 22:
1. Combination reactions. The reaction involves combination of a compound from two or more components atleast one of which must be in the elemental form or both the components are in the elemental state.
eg., CO + O₂ \(\underrightarrow { \Delta }\) C0₂

2. Decomposition reactions. The reaction involves decomposition of a compound into two or more components atleast one of which must be in the elemental state.
eg., 2NaH \(\underrightarrow { \Delta }\) 2Na + H₂

3. Displacement reactions. In a displace-ment reaction, an ion or an atom in a com-pound is replaced by an ion or an atom of another element,
eg., A + BC → AB + C

a. Metal displacement reaction.
A metal in a compound can be displaced by another metal in the uncombined state.

b. Non-metal displacement reaction.
This include hydrogen displacement and a rarely occuring reaction involving oxygen displacement.

Answer 23:
Enthalpy of reaction
(∆_r H)= [81 +3(-393)] -[9.7 +3(-110)]
= [81 – 1179] – [9.7 – 330] = -778kJmol^-1

Answer 24:
Graphite. Graphite has layered structure. Distance between 2 layers is 340 pm. C-C bond length within the layer is 141.5 pm. Each carbon atom in hexagonal ring undergoes sp² hybridisation with 3 neigh-bouring carbon atoms. The 4^th electron present in the unhybridised p- orbital is used for forming a π bond.

Diamond: Purest form of carbon. In dia-mond, each carbon atom is in sp3 hybridised state and is tetrahedrally bonded to 4 other carbon atom. The C-C bond length is 154 pm and C-C-C bond angle is 109°28. This 3 dimensional net work structure makes diamond harder and a bad conducor of electricity.

Answer 25:
Binary compounds of hydrogen with other elements are called hydrides. It can be ex-pressed as EH_x(MgH₂) or E_mH_n (B₂H₆).

ionic or Saline Hydrides
There are stoichiometric compounds of H₂ formed with most of the s-block elements j which are highly electropositive in character. The ionic hydrides are crystalline, nonvolatile and non-conducting in solid state.
In molten state, ionic hydrides conduct I electricity. Salts like hydrides are strong bases such as LiH, BeH₂, MgH₂.

Covalent or Molecular Hydrides
Covalent hydrides are formed by most of the p-block elements. It have very low con¬ductivity and are acidic in nature. Covalent hydrides are soft and have very low melting and boiling points,
eg., CH₄, NH₃

Molecular hydrides are further classified j according to the relative numbers of electrons and bonds in their lewis structure,

a. Electron-deficient hydrides. These hydrides are deficient in electrons to write their conventional lewis structure.
eg: group 13 elements (B₂H₆).

b. Electron-precise hydrides. These hydrides have the required no: of electro¬ns to write their conventional lewis structure,
eg., group 14 elements (CH₄).

c. Electron-rich hydrides. These hydrides have excess which are present as lone pairs, eg. group 15-17 elements (NH₃ has 1 lone pair, H₂O has 2 lone pairs).

Metallic or Interstitial Hydrides (Non- stoichiometric)
These hydrides are formed by many d- block and f-block elements. Metallic hydrides are hard, possess metaliclusture, conduct electricity and show magnetic properties. eg., LaH_2.87, Y₆H_2.55. These hydrides are non-stoichiometric being deficient in hydrogen.

Answer 26:
To solve such questions you must write a balanced reaction first.
CH₄ (g)+2O₂ (g) → CO₂ (g) +2H₂O (g)
The molecular mass of methane is 16
32 grams of Methane mean 2 mole
32 grams of Oxygen mean 1 mole
2 moles of Oxygen for every mole of methane. So that makes oxygen the limiting reagent. So 1 moles of 02 will form 0.5 moles of C0₂.
0.5 mole of CO₂ = 0.5*44 = 22 grams
So 22 grams of CO₂will be produced

Answer 27:
a. Oxygen has outer electronic configuration 2s² 2p⁴ removal of one electron from the 2p subshell will make its configuration2s² 2p³ thus it attains a half-filled configuration thus lesser energy is required to remove an electron from oxygen where-as in nitrogen the electronic configuration 1s² 2s² 2p³ it has a stable half filled configuration and thus more energy is required to remove an electron salts.

from nitrogen, in case of flourine, the electronic configuration is 2s²2p⁵ it has to gain one more electron to attain noble gas configuration and become stable thus more energy is required to remove an I electron from flourine thus ionisation enthalpy of oxygen is lesser than nitrogen or fluorine.

b. Cl has more negative electron gain enthalpy than F. F has less negative electron gain enthalpy because in it the added electron goes to the smaller { energy level (n=2) and hence suffers ‘ significant repulsion from the electrons already present in this shell also flurine possesses high charge density.

Answer 28:
All gases are made up of a large number of extremely small particles called molecules.

• The molecules are separated from one another by large distances so that the actual volume of the molecules is negligible as compared to the total volume of gas.
•  The molecules are in a state of continuous rapid motion in all directions. During their motion, they keep on colliding with one another and also with the walls of the: container.
• Molecular collisions are perfectly elastic, there is no net loss or gain of energy in their collisions. However, there may be a redistribution of energy during such collisions,
• There are no attractive forces between the molecules. The move completely independent of each other.

Answer 29:
NO_-2 + H₂O ⇌ HNO₂ + OH^–
K_h= [ HNO₂ ] [ OH^– ] /NO^– ₂]
Kw / ka = 10^-14 / 4.5 x 10^-4 = 0.22 x 10^-10
Now, If x moles of the salt undergo hydrolysis,  then the concentration of various species present i in the solution will be:
[NO^-2] = 0.04 -x;0.04
[ HN0₂ ] = x [ OH-] = x
K_h = x² / 0.04 = 0.22 x 10^-10 x2
x² = 0.0088 x 10^-10
x = 0.093 x 10^-5
[ OH^– ] = 0.093 x 10^-5M
[H₃0⁺] = 10^-14] / 0.093 x 10^-5] = 10.75 x 10^-9] M
pH = -log( 10.75 x 10^-9] ) = 7.96

Answer 30:
Sodium carbonate is generally prepared by Solvay Process. In this process, advantage is taken of the low solubility of sodium hydrogencarbonate whereby it gets precipitated in the reaction of sodium chloride with ammonium hydrogencarbonate. The latter is prepared by passing C02 to a concentrated solution of sodium chloride saturated with ammonia, where ammonium carbonate followed by ammonium hydro-gencarbonate are formed. The equations for the complete process may be written as :

2NH₃ + H₂0 + C0₂ → (NH₄)₂CO₃
(NH₄)₂CO₃+ H₂O + CO₂ → 2NH₄HCO₃
NH₄HCO₃+ NaCl → NH₄Cl + NaHCO₃

Sodium hydrogencarbonate crystal sepa¬rates. These are heated to give sodium carbonate.

2NaHCO₃ →Na₂CO₃ + C0₂ + H₂0

In this process NH₃ is recovered when the solution containing NH₄CI is treated with Ca(OH)₂. Calcium chloride is obtained as a by product.

Answer 31:
Nitrogen, sulphur, halogens and phosphorus present in an organic compound are detected by “Lassaigne’s test”. The elements present in the compound are converted from covalent form into the ionic form by fusing the compound with sodium metal. Cyanide, sulphide and halide of sodium so formed on sodium fusion are extracted from the fused mass by boiling it with distilled water. This extract is known as sodium fusion extract.

By fusing organic compound with sodium metal, the elements present in the compound forms corresponding water soluble salts.

Detection of Nitrogen:
N exist as NaCN in sodium fusion extract.

CN^– + Fe²⁺ \(\underrightarrow { FeS{ O }_{ 4 } }\) [Fe(CN)₆]^4-

On heating with con.H² SO⁴ , Fe²⁺ oxidised to Fe³⁺ to give prussion blue in colour.

3[Fe(CN)₆]^4- + 4Fe³⁺ \(\underrightarrow { x{ H }_{ 2 }O }\) Fe₄[Fe (CN)₆]₃.xH₂O Prussian blue

Detection of Halogens:
Halogen exist as halides (max.) in fusion extract. When AgNO₃ and HNO₃ is added they form AgX.
X^– + Ag⁺ → AgX ; (X = Cl, Br, I)

AgCI – white ppt
Agl – yellow ppt
AgBr – pale yellow

Detection of Sulphur:
Sulfur as Na2S is fusion extract when sodium nitroprusside added gives violet color.
Na²S + Na²[Fe(CN)⁵NO)] → Na⁴[Fe(CN)⁵ NOS] (violet colour)

Answer 32:
Electronic configuration of Ne2 is

Answer 33:
When electric discharge is passed through hydrogen gas enclosed in a discharge tube under low pressure and the emit¬ted light is analysed by a spectroscope, the spectrum consists of a large number of lines which are grtouped into different series. These series appear in different regions of light and are named after their discoveries. The complete spectrum is known as hydrogen spectrum.
According to Bohr’s model for Hydrogen
atom, E_n = -R_H\((\frac { 1 }{ { n }^{ 2 } } )\)
and ∆E = E_f -E_i
where E_fis the energy associated with final state while E_i is the energy of the initial state.

Here n_f and n_i represent final and initial electron orbits.

The frequency (υ) for the absorption and emission of the photon can be calculated as follows:

Plus One Chemistry Previous Year Question Papers and Answers