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Students can Download Chapter 3 Classification of Elements and Periodicity in Properties 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 3 Classification of Elements and Periodicity in Properties

Plus One Chemistry Classification of Elements and Periodicity in Properties One Mark Questions and Answers

Plus One Chemistry Chapter 3 Questions And Answers Question 1.
Which of the following is not a Dobereiner triad?
a) Cl, Br, I
b) Ca, Sr, Ba
c) Li, Na, K
d) Fe, Co, Ni
Answer:
d) Fe, Co, Ni

Plus One Chemistry Chapter 3 Question 2.
The elements of s-block and p-block are collectively called ___________
Answer:
Representative elements

Classification Of Elements And Periodicity In Properties Questions And Answers Question 3.
The cause of periodicity of properties is
a) Increasing atomic radius
b) Increasing atomic weights
c) Number of electrons in the valence shell
d) The recurrence of similar outer electronic configuration
Answer:
d) The recurrence of similar outer electronic configuration

An online valence electrons calculator finds the abbreviated or condensed electron configuration of an element with these instructions.

Classification Of Elements And Periodicity In Properties Important Questions Question 4.
Halogen with highest ionization enthalpy is ___________ .
Answer:
Fluorine

Classification Of Elements And Periodicity In Properties Question 5.
Which of the following represents the most electropositive element?
a) [He]2s¹
b) [He]2s²
c) [Xe]6s¹
d) [Xe]6s²
Answer:
c) [Xe]6s¹

Periodic Classification Of Elements Class 10 Extra Questions With Answers Question 6.
Second electron gain enthalpy is
Answer:
always positive

Classification Of Elements And Periodicity In Properties Class 11 Pdf Question 7.
Correct order of polarising power is
a) Cs⁺ < K⁺ < Mg²⁺ < Al³⁺
b) Al³⁺ < Mg² + K⁺ < Cs⁺
c) Mg²⁺ < Al³⁺ < K⁺ < Cs⁺
d) K⁺ < Cs⁺ < Mg²⁺ < Al³⁺
Answer:
a) Cs⁺ < K⁺ < Mg²⁺ < Al³⁺

Chemistry Chapter 3 Class 11 Important Questions Question 8.
The IUPAC name of the element with atomic number is 109 is ___________
Answer:
Une

Important Questions Of Periodic Classification Of Elements Class 11 Question 9.
The size of iso electronic species F~, Ne and Na+ is affected by
a) Nuclear charge
b) Principal quantum number.
c) Electron – electron interaction in outer orbitals.
d) None of the factors because their size is the same.
Answer:
Nuclear charge as nuclear charge is high the size is small.

Chapter 3 Chemistry Class 11 Question 10.
In transition elements the differentiating electron occupies (n-1)d sublevel in preference to ______________
Answer:
np level

Plus One Chemistry Classification of Elements and Periodicity in Properties Two Mark Questions and Answers

Important Questions For Class 11 Chemistry Chapter 3 Question 1.
The arguments made by two students are as given:
Student 1: ‘Hydrogen belongs to Group 1.’
Student 2: ‘Hydrogen belongs to Group 17.’
1. Who is right?
2. What is your opinion?
Answer:
1. Nobody is right.

2. Hydrogen has a one s-electron and hence can be placed in group 1 (alkali metals). It can also • gain an electron to achieve a noble gas arrangement and hence it can behave similar to a group 17 (halogen family) element. Because it a special case, hydrogen is placed separately at the top of the Periodic Table.

Class 11 Chemistry Chapter 3 Important Questions With Answers Question 2.
Match the following:

Sodium	f-block
Oxygen	s-block
Uranium	d-block
Silver	p-block

Answer:
Sodium – s-block
Oxygen – p-block
Uranium – f-block
Silver – d-block

Periodic Classification Of Elements Important Questions Question 3.

1. Which one has greater size, Na or K?
2. Justify your answer.

Answer:
1. K

2. K comes below Na in the Periodic Table. The atomic size increases down the group due to the fact that the inner energy levels are filled with electrons, which serve to shield the outer electrons from the pull of the nucleus.

Questions On Periodic Classification Of Elements Class 11 Question 4.
The general characteristics of a particular block of elements is as given:
They are highly electropositive, soft metals. They are good reducing agents. They lose the outermost electron(s) readily to form 1+ ion of 2+ ion.

1. Which block has this general characteristics?
2. Write down two general characteristics of p-block.

Answer:
1. s-block

2. The p-block contains metals, non-metals and metalloids. They form ionic as well as covalent compounds.

Class 11 Chemistry Chapter 3 Important Questions Question 5.
The variation in electron gain enthalpies of elements is less systematic than for ionization enthalpies. Out of oxygen and sulphur which has greater negative value for electron gain enthalpy? Justify.
Answer:
Sulphur has greater negative value (-200 kJ mol1) for electron gain enthalpy compared to that of oxygen (-141 kJ mol¹). This is because, due to smaller size of oxygen the added electron experiences much repulsion from the electrons present in the shell. Due to large size, the electron-electron repulsion is much less in sulphur.

Question 6.
Some elements are given. Li, Cs, Be, C, N, O F, I, Ne, Xe.

1. Arrange the above elements in the increasing order of ionization enthalpy.
2. Arrange the given elements in the decreasing order of negative electron gain enthalpy.

Answer:

1. Cs < I < Li < Xe < Be < C < N < O < F < Ne
2. Cl > F > O > N > C > Be > Li > I >C s > Xe > Ne

Plus One Chemistry Classification of Elements and Periodicity in Properties Three Mark Questions and Answers

Question 1.
Analyse the given figure and answer the questions that follow.

1. What is meant by atomic radius?
2. Explain covalent radius.
3. Write down another two types of terms expressed as atomic size.

Answer:

1. Atomic radius is defined as the distance from the centre of the nucleus of the atom to the outermost shell of electrons.
2. Covalent radius is defined as one half of the distance between the centre of nuclei of two similar atoms bonded by a single covalent bond.
3. Vander Waals’ radius, Metallic radius

Question 2.
Consider the statement: The element with 1s2
configuration belongs to the p-block.’

1. Identify the element.
2. Do you agree with this statement?
3. Justify.

Answer:

1. Helium
2. Yes
3. Strictly, helium belongs to the s-block but its positioning in the p-block along with other group 18 elements is justified because it has a completely filled valence shell (1s²) and as a result, exhibits characteristic of other noble gases.

Question 3.
The properties of elements are a periodic function of
their atomic weights.

1. Who proposed this law?
2. Can you see anything wrong in this law? If yes, justify your answer.
3. State modem periodic law.

Answer:

1. Mendeleev
2. Yes, atomic number is the more fundamental property of an element than atomic mass.
3. The physical and chemical properties of the elements are periodic functions of their atomic numbers.

Question 4.
1. Define ionisation enthalpy.
2. IE₁ <IE₂ <IE₃
What is meant by this? Justify.
Answer:
1. Ionisation enthalpy is the amount of energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state.

2. It shows the increasing order the magnitude of successive ionization enthalpies. As the positive charge of the ion increases, it becomes more difficult to remove the valence electron due to increase in effective nuclear charge.

Question 5.
Say whether the following are true or false:

1. On moving across a period ionization enthalpy decreases.
2. Mg is biggerthan Cl.
3. Ionization enthalpy of Li is less than that of K.

Answer:

1. False
2. True
3. False

Question 6.
Analyze the graph given below:

1. Identify the graph.
2. Account forthe following observations:
i) ‘Ne’ has the maximum value of ∆_iH.
ii) In the graph from Be to B, ∆_iH decrease.
Answer:
1. Graph showing the variation of ionisation enthalpy (∆_iH) with atomic number (Z) of the elements of second period.

2. i) ‘Ne’ is an inert gas and it has closed electron shell with stable octet electronic configuration. Hence it has the maximum ionisation enthalpy in the second period.
ii) Be has completely filled electronic configuration and a more stable. So ionization enthalpy is high

Question 7.
Study the graph and answer the questions that follow:

1. On moving down a group what happens to electron gain enthalpy?
2. Why chlorine shows more negative electron gain enthalpy than fluorine? ,

Answer:
1. On moving down a group,electron gain enthalpy becomes less negative because the size of the atom increases and the added electron would be farther from the nucleus.

2. Due to small size of F, the added electron goes to the smaller 2p quantum level and suffers significant replusion from the other electrons present in this level. But due to big size of Cl, the added electron goes to the n = 3p quantum level and occupies a larger region of space and the electron-electron repulsion is much less.

Question 8.
Electron gain enthalpy is an important periodic property.
1. What is meant by electron gain enthalpy?
2. What are the factors affecting electron gain enthalpy?
3. How electron gain enthalpy varies on moving across a period? Justify.

Answer:
1. Electron gain enthalpy(∆_egH) is the enthalpy change accompanying the process of addition of an electron to a neutral gaseous atom to convert it into a negative ion.

2. Effective nuclear charge, atomic size, electronic configuration

3. Electron gain enthalpy becomes more negative with increase in the atomic number across a period. The effective nuclear charge increases from left to right across a period and consequently, it will be easier to add an electron to a smaller atom since the added electron on an average would be closer to the nucleus. Thus more energy is released.

Question 9.
1. The electron gain enthalpies of Be and Mg are positive. What is your opinion? Justify.
2. Electron gain enthalpies of nobles gases have large positive values. Why?
Answer:
1. The statement is correct. Electron gain enthalpies of Be (+240 kJ/mol) and Mg (+230 kJ/mol) are positive because they have stable electronic configurations with fully filled s-orbitals (Be – 2s², Mg – 3s²). Hence, the gain of electron is highly endothermic.

2. Noble gases have stable octet electronic configuration of ns2 np6 (except He -1 s²) and they have practically no tendency to accept additional electron. They have large positive electron gain enthalpies because the electron has to enter the next higher principal quantum level leading to a very unstable electronic configuration. Hence, energy has to be supplied to add an extra electron.

Question 10.
During a group discussion, a student argued that ionization enthalpy depends only upon electronic configuration.
1. Do you agree? Comment.
2. Define shielding effect/screening effect.
3. Is there any relation between ionization enthalpy and shielding effect/screening effect? Explain.
Answer:
1. No. In addition to electornic configuration, ionization enthalpy depends upon other factors like atomic size, nuclear charge and shielding effect/screening effect.

2. In multi-electron atoms, the nuclear charge experienced by the valence electron will be less than the actual charge on the nucleus because it is shielded by inner core of electrons. This is called shielding effect or screening effect.

3. Yes. Ionization enthalpy decreases with increase in shielding effect/screening effect of inner electrons. This is because as a result of shielding of the valence electron by the intervening core of electrons it experiences a net positive charge which is less than the actual charge of the nucleus. In general, shielding is effective when the orbitals in the inner shells are completely filled.

Question 11.
a) Which of the following has higher first ionization enthalpy, N or O? Justify.
b) Which one is bigger, For F ? Why?
Answer:
1. N. This is because in N, three 2p-electrons reside in different atomic orbitals in accordance with Hund’s rule (\(2p_{ x }^{ 1 }2{ p }_{ y }^{ 1 }2{ p }_{ z }^{ 1 }\)) whereas in O, two the four 2p-electrons must occupy the same 2p-orbital resulting in an increased electron-electron repulsion (\(2p_{ x }^{ 2 }2{ p }_{ y }^{ 1 }2{ p }_{ z }^{ 1 }\)). Consequently, it is easier to remove the fourth 2p-electron from O than it is, to remove one of the three 2p-electrons from N.

2. F (136 pm) is bigger than F (72 pm). An anion is bigger than its parent atom. Addition of one electron in F results in increased repulsion among the electrons and a decrease in effective nuclear charge. Thus, attraction between nucleus and the electrons decreases and hence size increases.

Question 12.
Electronegativity differs from electron gain enthalpy.

1. Do you agree?
2. What do you mean by electronegativity?

Answer:

1. Yes.
2. Electronegativity is defined as the tendency of an atom in a chemical compound to attract the shared pair of electrons to itself.

Question 13.
Ionization enthalpy is an important periodic property.
1. What is the unit in which it is expressed?
2. What are the factors influencing ionization enthalpy?
3. How ionisation enthalpy varies in the periodic table?
Answer:
1. kJ mol_-1.

2. Atomic/ionic radius, nuclear charge, shielding effect/screening effect, penetration effect and electronic configuration.

3. Ionization enthalpy generally increases with increase in atomic number across a period due to regular increase in nuclear charge and decrease in atomic size. Thus, the attractive force between the nucleus and the electron cloud increases. Consequently, the electrons are more and more tightly bound to the nucleus.

Ionisation enthalpy generally decreases from top to bottom a group: This is because the effect of increased nuclear charge is cancelled by the increase in atomic size and the shielding effect. Consequently, the nucelar hold on the valence electron decreases gradually and ionization enthalpy decreases.

Question 14.
The second period elements show anomalous ‘ behaviour.
1. Give reason.
2. What are the anomalous properties of second period elements?
Answer:
1. The first element is each group belong to the second period. The difference in behaviour of these elements from the other elements of the same group can be attributed to the following factors:

• Small atomic size
• Large charge/radius ratio
• High electronegativity
• Absence of d-orbtials in the valence shell

2. The important anomalous properties of second period elements are: diagonal relationship, maximum covalence of four and ability to form pπ —pπ multiple bonds.

Question 15.
Atomic size, valency, ionization enthalpy, electron gain enthalpy and electronegativity are the important periodic properties of elements.
1. What do you mean by periodicity?
2. Periodic properties are directly related to ___________
Answer:
1. The periodical repetition of elements with similar properties after certain regular intervals when the elements are arranged in the order of increasing atomic numbers is called periodicity,

2. Electronic configuration.

Question 16.
1. Which is the element among alkali metals having lowest ionization enthalpy?
2. What is meant by valence of an element? How it varies in the periodic table?
3.

Identify the elements A, B, C and D, if the graph represents halogens.
Answer:
1. Fr

2. Valence of an element is the combining capacity of that element. In the case of representative elements the number of valence electrons increases from 1 to 8 on moving across a period, the valence to the element with respect to H and Cl increases from 1 to 4 and then decreases from 4 to zero. On moving down a group, the number of valence electrons remains same and, therefore, all the elements in a group exhibit same valence.

3. A = F, B = Cl, C = Br, D = I

Question 17.
Li and Mg belonging to first and second group in periodic table respectively resemble each other in many respects.
1. Name the relationship.
2. B can only form [BF₄]^– ion while Al can form [AIF₆]^3-, though both B and Al belong to group 13. Justify.
Answer:
1. Diagonal relationship.

2. This is because B, being a second period element has a maximum covalence of 4. It cannot expand its covalence beyond 4 due to absence of d-orbitals. But Al, being a third period element has vacant d-orbitals in its valence shell and hence can expand its covalence beyond 4.

Question 18.
During a group discussion a student argues that both oxidation state and valence are the same.
1. Do you agree?
2. Justify taking the case of [AICI(H₂O)₅]²⁺.
Answer:
1. No. Valence refers to the combining capacity of an element whereas oxidation state is the charge assigned to an element in a compound based on the assumption that the shared electron in a covalent bond belongs entirly to the more electronegative element.

2. In [AICI(H₂O)₅]²⁺the valence of Al is 6 while its oxidation state is +3.

Question 19.
Among the elements of the third period, identify the element

1. With highest first ionization enthalpy.
2. That is the most reactive metal.
3. With the largest atomic radius.

Answer:

1. Ar
2. Na
3. Na

Question 20.
A cation is smaller than the corresponding neutral atom while an anion is larger. Justify.
Answer:
A cation is smaller than its parent atom because it has fewer number of electrons while its nuclear charge remains the same. The size of an anion will be larger than that of the parent atom because the addition of one or more electrons would result in increased . repulsion among the electrons and a decrease in effective nuclear charge.

Question 21.
1. How the metallic character varies in the periodic table?
2. Categorize the following oxides into acidic, basic, neutral and amphoteric:
Al₂O₃, Na₂O, CO₂, Cl₂O₇, MgO, CO, As₂O₃, N₂O
Answer:
1. The metallic character decreases from left to right across the period due to increase in ionization enthalpy along a period which makes loss of electrons difficult. From top to bottom a group metallic character increases due to decrease in ionization enthalpy. Thus, metallic character decreases diagonally from left bottom to right top of the periodic table.

2. Acidic oxides: CO₂, Cl₂O₇
Basic oxides: Na₂O, MgO
Neutral oxides: CO, N₂O
Amphoteric oxides: Al₂O₃, As₂O₃

Question 22.
A group of ions are given below:
Na⁺, Al³⁺, O^2-, Ca²⁺, Mg²⁺, F^–, N^3-, Br^–
1. Find the pair which is not isoelectronic.
2. Arrange the above ions in the increasing order of size.
Answer:
1. Ca²⁺ and Br
2. Al³⁺ < Mg²⁺ < Na⁺ < F^– < O^2- < N^3- < Ca²⁺ < Br^–

Plus One Chemistry Classification of Elements and Periodicity in Properties Four Mark Questions and Answers

Question 1.
Statement 1: ‘Atomic mass is the fundamental property of an element.’
Statement 2: ‘Atomic number is a more fundamental property of an element than its atomic mass.’
1. Which statement is correct? Justify your answer.
2. Name the scientist who proposed this statement? What observation led him to this conclusion?
Answer:
1. Statement 2. Atomic number indicates the number of electrons present in an element. Most of the chemical properties of an element depend on its electronic configuration,

2. Henry Moseley. He observed regularities in the characteristic X-ray spectra of the elements. A plot of √υ (where u is the frequency of the X-rays emitted) against atomic number (Z) gave a straight line and not the plot of √υ vs atomic mass.

Question 2.
Atoms possessing stable configuration have less tendency to loss electrons and consequently will have high value of ionization enthalpy.
1. Justify this statement by taking the case of half-filled and completely filled electronic configurations.
2. The noble gases have highest ionization enthalpies in each respective periods. Why?
Answer:
1. Atoms with half-filled and completely filled electronic configurations have extra stability due to symmetric distribution of electrons and maximum exchange energy. Hence, more energy is required for the removal of their electrons. Elements like N (1s² 2s² \(2p_{ x }^{ 1 }2{ p }_{ y }^{ 1 }2{ p }_{ z }^{ 1 }\)), P (1s² 2s² 2p⁶ 3s² \(3p_{ x }^{ 1 }3{ p }_{ y }^{ 1 }3{ p }_{ z }^{ 1 }\)) etc. possessing half-filled shells have high ionization enthalpies.
Elements like Be ((1s² 2s²), Mg(1s² 2s² 2p⁶ 3s²) etc. having completely filled shells show high values of ionization enthalpy.

2. Noble gases have closed electron shells and very stable octet electronic configurations (except He). Hence, maximum amount of energy is required to remove their valence electron.

Question 3.
Mendeleev arranged the elements in the order of increasing atomic weights.
a) Write down the merits of Mendeleev’s periodic table.
b) What are the demerits of Mendeleev’s periodic table?
Answer:
a) Merits of Mendeleev’s periodic table:

• Study of elements – Elements are classified into groups with similar properties, thus facilitating the study of properties of elements.
• Prediction of new elements – Mendeleev left certain vacant places in his table which provided a clue for the discovery of new elements, e.g. Eka-AI (for Ga), Eka-Si (for Ge).
• Determination of correct atomic weights – With the help of this table, doubtful atomic weights of some elements were corrected.

b) Demerits of Mendeleev’s periodic table:

• Position of hydrogen was not certain.
• Anomalous pairs of elements – Certain elements of higher atomic weight preceed those with lower atomic weight, e.g. I (at.wt. 127) was placed after Te (at.wt. 128).
• Lanthanides and Actinides are not given proper places in this periodic table.
• No proper position for isotopes.

Question 4.
1. Name any three numerical scales of electronegativity.
2. How electronegativity varies in the periodic table? Justify.
Answer:
1. Pauling scale, Mullimen-Jaffe scale, Allred- Rochow scale.
2. Electronegativity generally increases from left to right a period and decreases from top to bottom in a group. This is because, from left to right across a period atomic size decreases and attraction between the valence electrons and the nucleus increases. From top to bottom in a group atomic size increases and attraction between the valence electrons and the nucleus decreases.

Question 5.
1. First ionization enthalpy of Na is lower than that of Mg. But its second ionization enthalpy is higher than that of Mg. Explain.
2. Which one is smaller, Na or Na⁺? Give reason.
Answer:
1. By the removal of one electron from Na it gets
the stable octet configuration of Ne. But when the first electron is removed from Mg it gets the unstable configuration of Na. It requires more energy due to small size and greater nuclear charge of Mg. In the case of Na the second electron is to be removed from a stable octet configuration which requires more energy than the removal of second electron from Mg.

2. Na⁺ (95 pm) is smaller than Na (186 pm). Acation is smaller than its parent atom. Na⁺ has fewer number of electrons (10 electrons) compared to Na (11 electrons). But the nuclear charge remains the same in both. Thus, effective nuclear charge per electron is greater in Na⁺. Thus, the attraction between nucleus and the remaining electrons increases and size decreases.

Question 6.
Removal of electron becomes easier on moving down the group.
1. Comment the above statement based on ionization enthalpy.
2. How electronic configuration influences the ionization enthalpy value?
Answer:
1. On moving from top to bottom in a given group, size of the atom increases and ionisation enthalpy decreases. Hence, it becomes easier to remove the valence electron.

2. Atoms with octet configuration, half-filled and completely filled configurations have extra stability and hence have higher values of ionization enthalpy.

Question 7.
The energy released during the addition of an electron to an isolated neutral atom is called electron gain enthalpy.
1. Explain how electron gain enthalpy differ from electronegativity.
2. The second ionisation enthalpy of an element is always greater than the first ionisation enthalpy. Give reason.
Answer:
1. Electron gain enthalpy(AegH) is the enthalpy change accompanying the process of addition of an electron to a neutral gaseous atom to convert it into a negative ion. It is a quantitative property of an isolated gaseous atom, which can be measured. Whereas, electronegativity is a qualitative measure of the ability of an atom in a chemical compound to attract shared electrons to itself. It is not a measureable quantity.

2. This is because to remove second electron from a positively charged ion more amount of energy is required due to increase in effective nuclear charge.

Question 8.
The physical and chemical properties of elements are periodic functions of their atomic numbers.
1. The atomic number of an element ‘X’ is 19. Write the group number, period and block to which X’ belong in the periodic table.
2. Name the element with
i) highest electronegativity and
ii) highest electron gain enthalpy
Answer:
1. The element is K.
₁₉K= 1s² 2s² 2p⁶ 3s² 3p6 4s¹
Group number = 1
Period number = 4
Block = s-block

2. i) Fluorine
ii) Chlorine

Plus One Chemistry Classification of Elements and Periodicity in Properties NCERT Questions and Answers

Question 1.
What is the basic theme of organisation in the periodic table? (2)
Answer:
The basic theme of organisation of elements in the periodic table is to facilitate the study of the properties of all the elements and their compounds. On the basis of similarities in chemical properties, the various elements have been divided into different groups. This has made the study of elements simple because their properties are now studied in the form of groups rather than individually.

Question 2.
What is the basic difference in approach between Mendeleev’s Periodic Law and the Modern Periodic Law? (2)
Answer:
Mendeleev Periodic Law states that the properties of the elements are a periodic function of their atomic weights whereas Modern Periodic Law states that the properties of elements are a periodic function of their atomic numbers. Thus, the basic difference in approach between Mendeleev’s Periodic Law and Modern Periodic Law is the change in basis of arrangements of elements from atomic weight to atomic number.

Question 3.
Consider the following species. N^3-, O^2-, F^–, N²⁺, Mg²⁺ and Al³⁺ (2)
1. What is common in them?
2. Arrange them in order of increasing ionic radii.
Answer:
1. Each one of these ions contains 10 electrons and
hence these are isoelectronic ions,

2. The ionic radii of isoelectronic ions decrease with the increase in the magnitude of the nuclear charge. Among the isoelectronic ions: N^3-, O^2-, F^–, Na⁺, Mg²⁺ and Al³⁺, nuclear charge increase in the order:
N^3-< O^2- < F^– < Na⁺ < Mg²⁺ < Al³⁺
Therefore, the ionic radii decrease in the order:
N^3- > O^2- > F^– > Na⁺ > Mg²⁺ > Al³⁺

Question 4.
Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell?
(a) Valence principal quantum number (n)
(b) Nuclear charge (Z)
(c) Nuclear mass
(d) Number of core electrons (1)
Answer:
Nuclear mass does not affect the valence shell. Thus, option (c) is the correct answer.

Question 5.
Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidising property is: (1)
a) F > Cl > O > N
b) F > O > Cl > N
c) Cl > F > O > N
d)0>F>N>CI
Answer:
a) F > Cl > O > N
Across a period, the oxidising character increases from left to right. Therefore, among F, O and N, oxidising power decreases in the order: F > O > N. However, within a group, oxidising power decreases from top to bottom. Thus, F is a stronger oxidising agent than Cl. Thus, overall decreasing order of oxidising power is F > Cl > O > N and the choice (a) is correct.

Students can Download Chapter 7 Equilibrium 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 7 Equilibrium

Plus One Chemistry Equilibrium One Mark Questions and Answers

Question 1.
Equilibrium in a system having more than one phase is called _________
Answer:
heterogeneous

Question 2.
Addition of a catalyst to a chemical system at equilibrium would result in
a) Increase in the rate of forward reaction
b) Increase in the rate of reverse reaction
c) A new reaction path
d) Increase in the amount of heat evolved in the reaction
Answer:
c) A new reaction path

Question 3.
With increase in temperature, equilibrium constant of a reaction
a) Always increases
b) Always decreases
c) May increase or decrease depending upon the number of moles of reactants and products
d) May increase or decrease depending upon whether reaction is exothermic or endothermic
Answer:
d) May increase or decrease depending upon whether reaction is exothermic or endothermic

Question 4.
Water is a conjugate base of ____________ .
Answer:
H₃O⁺

Question 5.
Which of the following substances on dissolving in water will give a basic solution?
a) Na₂CO₃
b) Al₂(SO₄)₃
c) NH₄Cl
d) KNO₃
Answer:
a) Na₂CO₃

Ged exam books to download, equilibrium concentration calculator, Finding the Least Common Denominator, combining like expressions.

Question 6.
Choose the correct answer for the reaction,
N₂(g) + 3H₂(g) \(\rightleftharpoons \) 2NH₃(g); ∆_rH = -91.8 kJ mol^-1 The concentration of H₂(g) at equilibrium can be increased by
1) Lowering the temperature
2) Increase the volume of the system.
3) Adding N₂ at constant volume.
4) Adding H₂ at constant volume.
Answer:
2) and 4) are correct.

Question 7.
Conjugate base of a strong acid is a
Answer:
Weak base

Question 8.
The expression forostwald dilution law is
Answer:
Ka = Cα²

Question 9.
The hydroxyl ion concentration in a solution having p^H = 4 will be
Answer:
10^-14

Question 10.
A mono protic acid in 1M solution is 0.01 % ionized the dissociation constant of this acid is
Answer:
10^-8

Question 11.
A certain buffer solution contains equal concentration of x^– and Hx. The Kafor Hx is 10^-6 p^H of buffer is
Answer:
6

Question 12.
In the equilibrium reaction
CaCO_3(s) → CaO_(s) + CO_2(g) the equilibrium constant is given by —–
Answer:
PCO₂

Question 13.
Congugate base of a strong acid is a _________ .
a) Strong base
b) Strong acid
c) Weak acid
d) Weak base
e) Salt
Answer:
d) Weak base

Question 14.
The species acting both as bronsted acid and base is _________ .

Answer:
b) HSO₄^–

Question 15.
P^H of .01 M KOH solution will be _________ .
Answer:
12

Plus One Chemistry Equilibrium Two Mark Questions and Answers

Question 1.
“High pressure and low temperature favours the formation of ammonia in Haber’s process.” Analyse the statement and illustrate the conditions using Le-Chatliers principle?
Answer:
The given statement is correct.
N₂(g) + 3H₂(g) \(\rightleftharpoons \) 2NH₃(g); ∆_rH =-91.8 kJ mol^-1 Since the number of moles decreases in the forward reaction a high pressure of ~ 200 atm is applied. Since the forward reaction is exothermic the optimum temperature of ~ 700 K is employed for maximum yield of ammonia.

Question 2.
“Chemical equilibrium is dynamic in nature”. Analyse the statement and justify your answer.
Answer:
At equilibrium the reaction does not stop. Both forward and backward reactions are taking place at equal rates. Thus, at equilibrium two exactly opposite changes occur at the same rate. Hence, chemical equilibrium is dynamic in nature.

Question 3.
Pressure has no influence in the following equilibrium: N₂(g) + O₂(g) \(\rightleftharpoons \) 2NO(g)

1. Do you agree with this?
2. What is the reason for this?

Answer:

1. Yes.
2. Here the total number of moles of the reactants is equal to that of the products. Hence pressure is having no influence in this equilibrium.

Question 4.
During a class room discussion a student is of the view that the value of equilibrium constant can be influenced by catalyst.

1. Do you agree with the statement?
2. Justify the role of catalyst in an equilibrium reaction?

Answer:

1. No.
2. Catalyst does not affect the equilibrium composition of a reaction mixture. It does not appear in the balanced chemical equation or in the equilibrium constant expression. It only helps to attain the equilibrium state in a faster rate.

Question 5.
What is the equilibrium constant (K) in the following cases?

1. Reaction is reversed.
2. Reaction is divide by 2.
3. Reaction is multiplied by 2.
4. Reaction is splitted into two.

Answer:

1. 1/K
2. √K
3. K²
4. K₁K₂

Question 6.
1. What is homogeneous equilibrium?
2. Suggest an example for this.
Answer:
1. The equilibrium in which the reactants and products are in the same phase,

2. N₂(g) + 3H₂(g) \(\rightleftharpoons \) 2NH₃(g)
In this equilibrium, the reactants and products are in the gaseous phase.

Question 7.
The equilibrium constants for two reactions are given. In which case the yield of product will be the maximum?
For first reaction: K₁ = 3.2 × 10^-6
For second reaction: K₂ = 7.4 × 10^-6
Answer:
Higher the value of K, greater will be the yield of product. So maximum yield will be in the second case.

Question 8.
Write an expression for equilibrium constant, Kc forthe ‘ reaction, 4NH₃(g) + 5O₂(g) \(\rightleftharpoons \) 4NO(g) + 6H₂O(g)
Answer:

Question 9.
1. State Henry’s law.
2. Suggest an example fora gas in liquid equilibrium.
Answer:
1. The mass of a gas dissolved in a given mass of a solvent at any temperature is directly proportional to the pressure of the gas above the solvent.

2. Equilibrium between the CO₂ molecules in the. gaseous state and the CO₂ molecules dissolved in water under pressure,
CO₂(g) \(\rightleftharpoons \) CO₂(in solution)

Question 10.
1. What is heterogeneous equilibrium?
2. Suggest an example forthis.
Answer:
1. Equilibrium in a system having more than one phase is called heterogeneous equilibrium

2. Equilibrium between solid Ca(OH)₂ and its saturated solution:
Ca(OH)₂(s) + (aq) \(\rightleftharpoons \) Ca₂+(aq) + 2OH^–(aq)

Question 11.
For the equilibrium 2SO₃(g) → 2SO₂(g) + O₂(g), K_c at 47 °C 3.25 × 10^-9 mol per litre. What will be the value of K_p at this temperature (R = 8.314 J K^-1mol^-1).
Answer:
R = 8.314 J K^-1 mol^-1 ∆n = 3 – 2 = 1
T = 47 °C = 273 + 47 = 320 K
K_c = 3.25 × 10^-9
K_p = K_c (RT)^∆n
= 3.25 × 10^-9 (8.314 × 320)¹
= 3.25 × 10^-9 × 8.314 × 320 = 8.65 × 10^-12

Question 12.
The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ions in it.
Answer:
pH= -log[H⁺] = 3.76
log[H⁺] = – 3.76
[H⁺] = antilog of (- 3.76) = 1.738 × 10^-4 mol L^-1

Question 13.
The equilibrium constant can be expressed in terms of partial pressure as well as concentration.
1. Give the relation between K_p and K_c.
2. What is the relation between K_p and K_c for the reaction, N₂(g) + O₂(g) \(\rightleftharpoons \) 2NO(g)?
Answer:
1. K_p = K_c(RT)^∆n, where ∆n = (number of moles of gaseous products) – (number of moles of gaseous reactants).

2. Here, ∆n = 2 – 2 = 0
K_p = K_c(RT)^∆n , K_p = K_c (RT)°
∴ K_p = K_c

Question 14.
1. Explain Arrhenius concept of acids and bases with suitable examples.
2. How proton exists in aqueous solution? Give reason.
Answer:
1. According to Arrhenius theory, acids are substances that dissociates in water to give hydrogen ions, H⁺(aq) and bases are substances that produce hydroxyl ions, OH^–(aq). Forexample, HCl is an Arrhenius acid and NaOH is an Arrhenius base.

2. In aqueous solution the proton bonds to the oxygen atom of a solvent water molecule to give trigonal pyramidal hydronium ion, H₃O⁺(aq). This is because a bare proton, H+ is very reactive and cannot exist freely in aqueous solutions.

Question 15.
1. What is an acidic buffer?
2. Suggest an example for an acidic buffer.
Answer:
1. An acidic buffer is a buffer solution having pH less than 7. It is prepared by mixing a weak acid and its salt formed with a strong base.

2. Mixture of acetic acid (CH₃COOH) and sodium acetate (CH₃COONa) is an example for an acidic buffer. Its pH is around 4.75.

Question 16.
1. What is a basic buffer?
2. Suggest an example for basic buffer.
Answer:
1. A basic buffer is a buffer solution having pH greater than 7. It is prepared by mixing a weak base and its salt formed with a strong acid.

2. Mixture of ammonium hydroxide (NH₄OH) and ammonium chloride (NH4CI) is an example for a basic buffer. Its pH is around 9.25.

Plus One Chemistry Equilibrium Three Mark Questions and Answers

Question 1.
The concentration of reactant and products for the reaction, H₂(g) + l₂(g) \(\rightleftharpoons \) 2Hl(g) are recorded as follows:

Reactant or Product	Molar Concentration
H2	0.080
l2	0.060
HI	0.490

a) Write down the expressions for equilibrium constant of the above reaction.
b) Calculate the equilibrium constant at the temperature 298 K if [Hl] = 0.49 M, [H₂]=0.08 M and [l₂]=0.06 M.
Answer:

Question 2.
1. When equilibrium is reached in a chemical reaction?
2. What is the influence of molar concentration in a reaction at equilibrium?
3. Write the expression for equilibrium constant for the decomposition of NH4CI by the reaction,
NH₄Cl \(\rightleftharpoons \) NH₃+HCl
Answer:
1. When the rate of forward reaction is equal to rate of backward reaction, the chemical reaction is said to be in equilibrium.

2. Rate of chemical reaction is directly proportional to the product of molar concentration of the reactants.

3. \(\mathrm{K}=\frac{\left[\mathrm{NH}_{3}\right] \mathrm{HCl}}{\left[\mathrm{NH}_{4} \mathrm{Cl}\right]}\)

Question 3.
1. What is meant by K_p?
2. How K_p is related to K_c?
Answer:
1. K_p is the equilibrium constant in terms of the partial pressures of the reactants and products (Pressure should be expressed in bar as standard state is 1 bar). It is used for reactions involving gases.

2. K_p = K_c (RT)^∆n
where R = universal gas constant, T = absolute temperature and ∆n = number of moles of gaseous product(s) – number of moles of gasesous reactant(s).

Question 4.
a) What do you mean by equilibrium constant?
b) Write any two characteristics of equilibrium constant.
c) Write an expression for equilibrium constant of the reaction, 2SO₂(g) + O₂(g) \(\rightleftharpoons \) SO₃(g).
Answer:
a) Equilibrium constant at a given temperature is the ratio of product of molar concentrations of the products to that of the reactants, each concentration term being raised to the respective stoichiometric coefficients in the balanced chemical equation.

b) 1. The value of equilibrium constant is independent of the initial concentrations of the reactants and products.
2. Equilibrium constant is temperature dependent having one unique value for a particular reaction represented by a balanced equation at a given temperature.

Question 5.
2NO₂(g) \(\rightleftharpoons \) N₂O₄(g); ∆H = -52.7 kJ mol^-1
1. What change will happen if we increase the temperature?
2. What is the effect of increase in pressure in the above equilibrium?
3. What happens when N₂O₄ is removed from the reaction medium?
Answer:
1. Since the forward reaction is exothermic, on increasing temperature the rate of backward reaction (endothermic reaction) increases.

2. Since the number of moles decreases in the forward reaction, on increasing pressure, the rate of forward reaction increases.

3. Rate of forward reaction increases.

Question 6.
Consider this reaction:
CO(g) + 2H₂(g) \(\rightleftharpoons \) CH₃OH(g); ∆_rH = -92 kJ mol^-1
Explain the influence of the following on the basis of
Le Chatelier’s principle.
1. Decrease in pressure.
2. Increase in temperature.
3. Increase in the partial pressure of hydrogen.
Answer:
1. On decreasing pressure the reaction shifts in the direction in which there is increase in the number of moles. Thus, the rate of backward reaction increases on decreasing pressure.

2. On increasing temperature, the rate of endothermic reaction increases. Here, backward reaction is endothermic. Hence, on increasing temperature the rate of backward reaction increases.

3. Hydrogen, being a reactant increase in its partial pressure increases the rate of forward reaction.

Question 7.
The equilibrium showing dissociation of phosgene gas is given below:
COCl₂(g) \(\rightleftharpoons \) CO(g) + Cl₂(g)
When a mixture of these three gases at equilibrium is compressed at constant temperature, what happens to
1. The amount of CO in mixture?
2. The partial pressure of COCl₂?
3. The equilibrium constant for the reaction?
Answer:
1. Amount of CO decreases, because the system favours the reaction in which number of moles decreases with increase of pressure i.e., backward reaction.

2. Increases.
3. Equilibrium constant remains the same since temperature is constant.

Question 8.
The equilibrium constant of the reaction H₂(g) + l₂(g) \(\rightleftharpoons \) 2Hl(g) is 57 at 700 K. Now, give the equilibrium constants for the following reactions at the same temperature:

Answer:

Question 9.
1. What are buffer solutions?
2. Which of the following are buffer solutions?
NaCl + HCl
NH₄Cl + NH₄OH
HCOOH + HCOOK
3. What is the effect of pressure on the following equilibria?
i) Ice \(\rightleftharpoons \) Water
ii) N₂(g) + O₂(g) \(\rightleftharpoons \) 2N0(g)
Answer:
1. These are solutions which resist change in pH on dilution or with the addition of small amounts of acid or alkali.

2. NH₄Cl + NH₄OH

3. i) When pressure is increased the melting point of ice decreases and hence the rate of forward reaction will increase.
ii) Pressure has no effect in this equilibrium because there is no change in the number of moles of the gaseous reactants and products.

Question 10.
The aqueous solution of the compounds NaCl, NH₄Cl and CH₃COONa show different pH.

1. Identify the acidic, basic and neutral solution among them.
2. The concentration of hydrogen ion in a soft drink is 4 × 10^-4. What is its pH?

Answer:

1. Acidic-aqueous solution of NH₄Cl Neutral – aqueous solution of NaCl Basic – aqueous solution of CH₃COONa
2. pH = – log[H⁺] = – log[4 × 10^-4] = 3.398

Question 11.
1. What is pH? What is its significance?
2. The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10^-3. What is its pH?
Answer:
1. pH is a logarithmic scale used to express the hydronium ion concenration in molarity more conveniently. The pH of a solution is defined as negative logarithm to the base 10 of the activity of hydrogen ion. pH = — log \({ { a }_{ { H }^{ + } } }\) = —log[H⁺]

2. [H⁺] = 3.8 × 10^-3
pH = -log[H⁺]
= -log [3.8 × 10^-3] = -(-2.42) = 2.42

Question 12.
1. State the Le-Chatelier’s principle.
2. Apply the above principle in the following equilibrium and predict the effect of pressure.
CO(g) + 3H₂(g) \(\rightleftharpoons \) CH₄(g) + H₂O(g)
Answer:
1. The Le Chateliers principle states that a change in any of the factors that determine the equilibrium conditions of a system will cause the system to change in such a manner so as to reduce or to counteract the effect of the change.

2. On increasing pressure the rate of forward reaction increases. This is because number of moles decreases in the forward reaction. In other words, the value of Q_c decreases on increasing pressure. As Q_c < K_c, the reaction proceeds in the forward direction.

Question 13.
1. Explain Lewis concept of acids and bases.
2. Why does BF₃ act as a Lewis acid?
Answer:
1. A Lewis acid can be defined as a species which accepts electron pair and a Lewis base is a species which donates an electron pair.

2. In BF₃, the boron atom is electron deficient and it accepts a lone pair of electron. So it acts as a Lewis acid.

Question 14.
1. How the value of AG influence the direction of an equilibrium process?
2. The equilibrium constant for a reaction is 8. What will be the value of ∆G at 27 °C?
Answer:
1. If ∆G is negative, then the reaction is spontaneous and proceeds in the forward direction.
If ∆G is positive, then reaction is considered non- spontaneous. instead, as reverse reaction would have a negative ∆G, the products of the forward reaction shall be converted to the reactants.
If ∆G is 0, reaction has achieved equilibrium. At this point, there is no longer any free energy to drive the reaction.
2.

Question 15.
1. What is common ion effect?
2. Suggest an example for this effect.
Answer:
1. Common ion effect may be defined as the suppression of the dissociation of a weak electrolyte by the addition of some strong electrolyte containing a common ion.

2. The dissociation equilibrium of NH₄OH is shifted towards left in presence of NH₄Cl having the common ion, NH₄⁺.

Question 16.
1. Predict whether an aqueous solution of (NH₄)₂SO₄ is acidic, basic or neutral?
2. Justify your answer.
Answer:
1. An aqueous solution of (NH₄)₂SC₄ is acidic in nature.

2. (NH₄)₂SO₄ is formed from weak base, NH₄OH, and strong acid, H₂SO₄. In water, it dissociates completely
(NH₄)₂SO₄(aq) → 2NH₄⁺ (aq) + SO₄^2- (aq)
NH₄⁺ ions undergoes hydrolysis to form NH₄OH and H⁺ ions.
NH₄+(aq) + H₂O(l) \(\rightleftharpoons \) NH₄OH(aq) + H⁺(aq)
NH₄OH is a weak base and therefore remains almost unionised in solution. This results in increased H⁺ ion concentration in solution making the solution acidic.

Question 17.
1. What are sparingly soluble salts? Suggest an example.
2. Define solubility product constant, K_sp.
3. Obtain the relation between solubility product constant (K_sp) and solubility (S), of a solid salt of general formula M_x^p+ X_y^q-.
Answer:
1. Sparingly soluble salts are those salts with solubility less than 0.01 M.
e.g. BaSO₄

2. The solubility product of a sparingly soluble salt at a given temperature is defined as the product of the concentrations of its ions in the saturated solution, with each concentration term raised to the power equal to the number of times the ion occurs in the equation representing the dissociation of the electrolyte.

3. The equilibrium in the saturated solution of the salt can be represented as,

Question 18.
The Le Chatelier’s principle is applicable to physical and chemical equilibria.
1. What are the factors which can influence the equilibrium state of a system?
2. Explain the factors affecting the chemical equilibrium on the basis of Le Chatelier’s principle taking Haber’s process for the manufacture of ammonia as an example.
Answer:
1. The following factors can influence the equilibrium state of a system:

• Change in concentration of the reactants or products.
• Change in temperature.
• Change in pressure.
• Addition of inert gas.
• Presence of catalyst.

2. N₂(g) + 3H₂(g) \(\rightleftharpoons \) 2NH₃(g); ∆_rH = -91.8 kJ mol^-1
When concentration of N₂ or H₂ is increased, a good yield of NH₃ can be achieved. The rate of forward reaction can also be increased by removing NH₃ from the reaction mixture.

When pressure is increased, the system will try to decrease pressure and for this system will proceed in that direction where there is minimum number of moles i.e., forward reaction. Thus, a good yield of NH₃ can be achieved by increasing pressure.

Since the formation of NH₃ is an exothermic reaction, a good yield of NH₃ can be achieved by decreasing the temperature. But if the temperature is decreased to very low value the reactant molecules do not have sufficient energy to interact. Hence, an optimum temperature of 500°C is used.

Question 19.
1. Soda water is prepared by dissolving CO₂ in water under high pressure. What is the principle involved in this process?
2. At 1000 K, equilibrium constant Kc for the reaction 2SO₃(g) \(\rightleftharpoons \) 2SO₂(g) + O₂(g) is 0.027. What is the value of K_p at this temperature?
Answer:
1. Henry’s law
2. K_p =K_c(RT)^∆n
∆n = 3.2 = 1
K_p = 0.027 × (0.0831 × 1000)¹ = 2.2437

Question 20.
1. For the reaction PCL \(\rightleftharpoons \) PCl₃ +Cl_c
i) Write the expression of K_c.
ii) What happens if pressure is increased?
2. Write the conjugate acid and base of the following species:
i) H20 ii) HCO;
3. Name the phenomenon involved in the preparation of soap by adding NaCI.
Answer:
1. i) \(\kappa_{c}=\frac{\left[\mathrm{PC}_{3}\right]\left[\mathrm{Cl}_{2}\right]}{\left[\mathrm{PCl}_{5}\right]}\)
ii) If we increase the pressure the system will try to decrease the pressure. For this system will proceed in the direction where there is minimum number of moles, i.e., rate of backward reaction increases by decreasing the pressure.

2. i) Conjugate acid of H₂O is H₃O⁺
Conjugate base of H₂O is OH”
ii) Conjugate acid of HCO,” is H₂CO₃ Conjugate base of HCO₃^– is CO₃^2-

3. Common ion effect.

Question 21.
a) The pH of black coffee is 5.0. Calculate the hydrogen ion concentration.
b) The K_sp of barium sulphate is 1.5 × 10^-9. Calculate the solubility of barium sulphate in pure water.
Answer:

Question 22.
1. What is conjugate acid-base pair?
2. Illustrate with an example.
Answer:
1. The acid-base pairthat differs only by one proton is called conjugate acid-base pair. Such acid-base pairs are formed by loss or gain of a proton.

2. Consider the ionization of hydrochloric acid in water.

HCl(aq) acts as an acid by donating a proton to H₂O molecule which acts as a base because it accepts the proton. The species H₂O⁺ is produced when water accepts a proton from HCl. Therefore, Cl^– is the conjugate base of HCl and HCl is the conjugate acid of Cl^–. Similarly, H₂O is the conjugate base of H₂O⁺ and H₃O⁺⁺ is the conjugate acid of the base H₂O.

Question 23.
1. What are the applications of equilibrium constant?
2. What is meant by reaction quotient, Q_c?
3. Predict the direction of net reaction in the following cases:
i) Q_c < K_c
ii) Q_c > K_c
iii) Q_c = K_c
Answer:
1. The applications of equilibrium constant are:
• To predict the extent of a reaction on the basis of its magnitude.
• To predict the direction of the reaction.
• To calculate equilibrium concentrations.

2. Reaction quotient, Q_c at a given temperature is defined as the ratio of the product of concentrations of the reaction products to that of the reactants, each concentration term being raised to their individual stoichiometric coefficients in the balanced chemical equation, where the concentrations are not necessarily equilibrium values.

3. i) When Q_c > K_c, the reaction will proceed in the
direction of reactants (reverse reaction), i.e., net reaction goes from right to left.
ii) When Q_c < K_c, the reaction will proceed in the direction of products (forward reaction), i.e., net reaction goes from left to right.
iii) When Q_c = K_c, the reaction mixture will be at equilibrium, i.e., no net reaction occurs.

Question 24.
Solubility product helps to predict the precipitation of salts from solution.
1. Find the relation between solubility (S) and solubility product (K_sp) of calcium fluoride and zirconium phosphate.
2. The solubility product of two sparingly soluble salts XY₂ and AB are 4 × 10^-15 and 1.2 × 10^-16 respectively. Which salt is more soluble? Explain.
Answer:
1. The equilibrium in the saturated solution of calcium fluoride can be represented as,
CaF₂(s) \(\rightleftharpoons \) Ca²⁺(aq) + 2F^–(aq)
K_sp = [Ca²⁺][F^–]² = S.(2S)₂ = 4S³
The equilibrium in the saturated solution of zirconium phosphate can be represented as,
Zr₃(PO₄)₄(s) \(\rightleftharpoons \) 3Zr⁴⁺(aq) + 4PO₄^3-(aq)
K_sp = [Zr⁴⁺]³[PO₄^3-]⁴ = (3S)³.(4S)⁴ = 6912S⁷

2. XY₂ is more soluble than AB.

Ionic Strength Calculator … Ionic strength of a solution indicates the concentration of ionic charge in the solution.

Question 25.
a) How common ion effect can influence the solubility of ionic salts?
b) What is the application of common ion effect in gravimetric estimation?
Answer:
1. In a salt solution, if we increase the concentration of any one of the ions, according to Le Chatelier’s principle, it should combine with the ion of its opposite charge and some of the salt will be precipitated till K_sp = Q_sp. Similarly, if the concentration of one of the ions is decreased, more salt will dissolve to increase the concentration of both the ions till K_sp = Q_sp.

2. The common ion effect is used for almost complete precipitation of a particular ion as its sparingly soluble salt, with very low value of solubility product for gravimetric estimation.

Plus One Chemistry Equilibrium Four Mark Questions and Answers

Question 1.
Match the following:

Answer:

Question 2.
In Contact process, SO₃ is prepared by the oxidation of SO₂ as per the following reaction:
2SO₂(g) + O₂(g) \(\rightleftharpoons \) 2SO₃(g); ∆H = -189.4
a) What happens to the rate of forward reaction when i) temperature is increased?
ii) pressure is decreased?
iii) a catalyst V₂O₅ is added?
b) Calculate the pH of 0.01 M H₂SO₄ solution. Also, calculate the hydroxyl ion concentration in the above solution.
Answer:
1. D When temperature is increased, the rate of forward reaction decreases since it is exothermic.
ii) When pressure is decreased the rate of forward reaction decreases since it is associated with decrease in number of moles.
iii) When a catalyst V₂O₅ is added the rate of both forward and backward reactions are increased by the same extent and equilibrium is reached earlier.
2.

The method presented in our buffer pH calculator allows you to compute the pH of both arterial and venous blood.

Question 3.
Calculate the [H⁺] in the following biological fluids whose pH are given in brackets.
i) Human muscle fluid (6.83)
ii) Human stomach fluid (1.22)
iii) Human blood (7.38)
iv) Human saliva (6.4)
Answer:
i) pH =-log[H+] = 6.83
log[H⁺] = – 6.83
[H⁺] = antilog (- 6.83) = 1.48 × 10^-7 mol L^-1
ii) [H⁺] = antilog (- 1.22) = 6.03 × 10^-2 mol L^-1
iii) [H⁺] = antilog (- 7.38) = 4.17 × 10^-8 mol L^-1
iv) [H⁺] = antilog (- 6.4) = 3.98 × 10^-7 mol L^-1

Question 4.
The pH value of a solution determines whether it is acidic, basic or neutral in nature.
1. The concentration of hydrogen ion in the sample of a soft drink is 3.8 × 10^-3 mol/L. Calculate its pH. Also predict whether the above solution is acidic, basic or neutral.
2. The dissociation constants of formic acid (HCOOH) and acetic acid (CH₃COOH) are 1.8 × 10^-4and 1.8 × 10^-4 respectively. Which is relatively more acidic? Justify your answer.
Answer:
1. pH = – log[H⁺] = – log[3.8 × 10^-3] = 2.42
Since pH is less than 7, it is an acidic solution,

2. HCOOH is more acidic.
K_a value is directly proportional to the acid strength, i.e., greater the K_a value, stronger is the acid.

Question 5.
a) Write the expression for Henderson – Hasselbalch equation for i) An acidic buffer & ii) A basic buffer.
b) Calculate the pH of a solution which is 0.1 M in
CH₃COOH and 0.5 M in CH₃COONa. Ka for CH₃COOH is 1.8 × 10^-6.
Answer:

Plus One Chemistry Equilibrium NCERT Questions and Answers

Question 1
A liquid is in equilibrium with its vapour in a sealed containerat a fixed temperature. The volume of the container is suddenly increased. (3)
a) What is the initial effect of the change on vapour pressure?
b) How do rates of evaporation and condensation change initially?
c) What happens when equilibrium is restored finally and what will be the final vapour pressure?
Answer:
a) Vapour pressure decreases due to increase in volume.
b) Rate of evaporation remains same and rate of condensation decreases.
c) Finally the same vapour pressure is restored and the rate of evaporation becomes equal to the rate

Question 2.
The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10^-3 M. What is its pH? (2)
Answer:
pH = -log[H⁺]
= -log (3.8 × 10^-3) = 2.42

Question 3.
The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it. (2)
Answer:
H= -log [H⁺]
or log [H⁺] = -3.76 = -4.24
[H⁺] = antilog (-3.76) = 1.74 × 10^-4M

Question 4.
The ionization constants of HF, HCOOH and HCN at 298 K are 6.8 × 10^-4, 1.8 × 10^-4 and 4.8 × 10^-9 respectively. Calculate the ionization constants of the corresponding conjugate base. (3)
Answer:
The relation between ionization constant of an acid and that of its conjungate base is Ka x Kb = Kw

Question 5
The ionization constant of nitrous acid is 4.5 x 10-4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis. (2)
Answer:
Sodium nitrite is a salt of strong base and weak acid. Its degree of hydrolysis, h is given by the relation.

Question 6
A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine. (2)
Answer:
Pyridinium hydrochloride is a salt of a weak base (pyridine) and a strong acid (HCl). The pH of an aqueous solution of this salt is given by the relation:

Students can Download Chapter 1 Some Basic Concepts of Chemistry 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 1 Some Basic Concepts of Chemistry

Plus One Chemistry Some Basic Concepts of Chemistry One Mark Questions and Answers

Plus One Chemistry Chapter Wise Questions And Answers Question 1.
Which of the following is a mixture?
a) Graphite
b) Sodium chloride
c) Distilled water
d) Steel
Answer:
d) Steel

Plus One Chemistry Chapter 1 Questions And Answers Question 2.
1 µ g = __________ g
[10^-3, 10^-6, 10^-9, 10^-12] ‘
Answer:
10^-6

Plus One Chemistry First Chapter Questions And Answers Question 3.
The number of significant figures in 0.00503060 is __________ .
Answer:
6

Plus One Chemistry Chapter Wise Questions And Answers Pdf Question 4.
The balancing of chemical equations is based on which of the following law?
a) Law of multiple proportions
b) Law of conservation of mass
c) Law of definite proportions
d) Gay-Lussac law
Answer:
b) Law of conservation of mass

By using the percent/actual yield calculator, you can get the accurate amount of percent, actual, and theoretical yield produced in a chemical reaction.

Plus One Chemistry Questions And Answers Question 5.
Which among the following is the heaviest?
a) 1 mole of oxygen
b) 1 molecule of sulfur trioxide
c) 100 u of uranium
d) 44 g carbon dioxide
Answer:
d) 44 g carbon dioxide

Plus One Chemistry Previous Year Question Papers And Answers Chapter Wise Question 6.
Calculate the number of atoms in 48 g of He?
Answer:
Gram atomic mass of He = 4 g.
Thus, numberofatomsin4g (1 mol) He = 6.02 × 10²³
So number of atoms in 48 g of He = \(\frac{48}{4}\) × 6.02 × 10²³
=12 × 6.02 × 10²³
= 7.224 × 10²⁴

Hsslive Chemistry Previous Questions And Answers Chapter Wise Plus One Question 7.
One mole of CO₂ contains how many gram atoms?
Answer:
3 gram atoms.

Plus One Chemistry Previous Year Questions And Answers Chapter Wise Question 8.
The ratio of gram atoms of Au and Cu in 22ct gold is __________
Answer:
7 : 2

Plus One Chemistry Chapter Wise Questions And Answers Pdf Download Question 9.
A compound contains 69.5% oxygen and 30.5% nitrogen and its molecular weight is 92. The compound will be
Answer:
N₂O₄

Plus One Chemistry Chapter 1 Question 10.
The total number of electrons present in 1 mole of water is
Answer:
6 × 10²⁴.

Hsslive Plus One Chemistry Chapter Wise Questions And Answers Question 11.
40g NaOH is present in 100 ml of a solution. Its molarity is __________
Answer:
10 M

Plus One Chemistry Some Basic Concepts of Chemistry Two Mark Questions and Answers

Plus One Chemistry Some Basic Concepts Of Chemistry Questions Question 1.
Classify the following substances into homogeneous and heterogeneous mixtures.

• Milk
• Iron
• Air
• Gasoline
• Kerosene
• Muddy water

Answer:

Homogeneous	Heterogeneous
Milk, Iron Gasoline Air Kerosene	Muddy Water

Plus One Chemistry Previous Year Questions Chapter Wise Question 2.
Calculate the volume occupied by 4.4 g of CO₂ at STP?
Answer:
1 mole CO₂ = 44 g
4.4 CO₂ = 0.1 mole CO₂
Volume occupied by 1 mol CO₂ at STP = 22.4 L
∴ Volume occupied 0.1 mol CO₂ at STP = 0.1 × 22.4 L
= 2.24 L

Plus One Chemistry Chapter 1 Previous Questions And Answers Question 3.
During a group discussion a student argued that “the water of sea and river should have different chemical composition”.

1. What is your opinion?
2. Which law would you suggest to support your answer?
3. State the law.

Answer:

1. I can’t join with him.
   The water of sea and water of river must have the same chemical composition.
2. Law of definite proportions.
3. A given compound always contains exactly the same proportion of elements by weight.

Plus One Chemistry Question And Answer Question 4.
“When science developed some theories are also modified”.
Write the modified atomic theory.
Answer:

1. Atom is no longer considered as indivisible, it has been found that atom is made up of sub atomic particles called protons, neutrons and electrons.
2. Atoms of the same element may not be similar in all respects.
3. Atoms of different elements may be similar in one or more respects.
4. The ratio in which atomic unit may be fixed and integral but may not be simple.
5. The mass of atom can be changed into energy.

Plus One Chemistry Chapter Wise Previous Questions And Answers Question 5.
Carbon combines with oxygen to form CO and CO₂.

1. What is the law behind this?
2. State the law.

Answer:

1. Law of multiple proportions.
2. If two elements 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.

Hss Live Plus One Chemistry Chapter Wise Questions And Answers Question 6.
Calculate the volume occupied by 6.02×1025 molecules of oxygen at STP.
Answer:
Volume occupied by 1 mole of oxygen gas at STP = 22.4 l
i.e., Volume occupied by 6.02×1023 molecules of oxygen gas at STP = 22.4 l
Hence the volume occupied by 6.02 × 10²⁵ molecules
of oxygen gas at STP = \(\frac{22.4 \times 6.02 \times 10^{25}}{6.02 \times 10^{23}}\) = 2240 l.

Plus One Chemistry Chapter Wise Questions And Answers Hsslive Question 7.
Calculate the molality of a solution of NaOH containing 20g of NaOH in 400 g solvent.
Answer:

Question 8.
Calculate the mole fraction of NaOH in a solution containing 20 g of NaOH per 360 g of water.
Answer:

Question 9.
12 g of carbon reacts with 32 g of oxygen to form 44g of carbon dioxide.

1. Which law of chemical combination is applicable here?
2. State the law.

Answer:

1. Law of conservation of mass.
2. Matter can neither be created nor destroyed. Or, during any physical or chemical change, the total mass of the products remains equal to the total mass of the reactants.

Question 10.
When hydrogen and oxygen combine to form water, the ratio between volume of reactants and products is 2:1:2.

1. Which law of chemical combination is applicable here?
2. State the law.

Answer:

1. Gay Lussac’s law of gaseous volumes.
2. 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.

Question 11.
Carbon form two oxides, the first contains 42.9% C and the second contains 27.3% carbon. Show that these are in agreement with the law of multiple proportions.
Answer:
In the first compound:
C = 42.9%
O = 100-42.9 = 57.1%
So, the ratio between the masses of C and O = 42.9:57.1 = 1:1.33
In the second compound:
C = 27.3%
O= 100-27.3= 72.7%
So, the ratio between the masses of C and O = 27.3 : 72.7= 1:2.66
Hence, the ratio of masses of oxygen which combines with a fixed mass of carbon is 1.33:2.66 or 1:2, a simple whole number ratio. This illustrates the law of multiple proportions.

Question 12.
Match the following:

A	B
1 amu	1.008 x 1.66 x10–24
Mass of 1 H atom	6.02 x 1023
Molar volume of O2 at STP	11.2L
Volume of 14g of N2 at STP	1.66 x 10–24
Avogadro number	22.4L

Answer:

A	B
1 amu	1.66 x 10–24
Mass of 1 H atom	1.008 x 1.6 x10–24
Molar volume of O2 at STP	22.4L
Volume of 14g of N2 at STP	11.2L
Avogadro number	6.02 x 1023

Question 13.
Calculate the molality of a solution containing 10 g ofNaOH in 200 cm3 of solution. Density of solution is 1.4 g/mL. (Molar mass of NaOH = 40)
Answer:
Mass of the solution = 200 × 1.04 = 208 g
Mass of NaOH (WB) = 10g Molar mass of NaOH (MB) = 40 g mol^-1
Mass of water (WA) = (208 -10) g = 198 g = 0.198 kg

Question 14.
Calculate the mass percentage of oxygen in CaCO₃.
Answer:
Molecular mass of CaCO₃ = 100 g mol^-1
Mass of oxygen in 100 g CaCO₃=3 × 16 g = 48 g
Percentage of oxygen in CaCO₃ =\(\frac{48}{100}\)×100 = 48%

Question 15.
KCIO₃ on heating decomposes to KCI and O₂. Calculate the mass and volume of O₂ produced by heating 50 g of KCIO₃.
Answer:
The reaction is represented as,

According to the equation, 96 g of oxygen is obtained from 245 g of KCIO₃.
Hence mass of oxygen obtained from 50 g KCIO₃ is \(\frac{96 \times 50}{245}=19.6 \mathrm{g}\)
According to the equation 245 g of KCIO₃ gives 3 moles of O₂ at STP which is 3 × 22.4 L = 67.2 L
Volume of oxygen liberated by 50g of KCIO₃
= \(\frac{67.2 \times 50}{245}=13.71 \mathrm{L}\)

Question 16.
Calculate the number of molecules present in

1. 11g of CO₂.
2. 56 mL Of CO₂ at STP.

Answer:
1. \(\frac{11}{44}\) = 0.25 mole
1 mole of CO₂ contains 6.022 × 10²³ molecules.
∴ 0.25 mole CO₂ contains 6.022 × 10²³ × 0.25
= 1.51 × 1023 molecules.

2. 56 mL = 0.056 L
\(\frac{0.056}{22.4}\) =0.0025 mole
= 6.022 × 10²³ × 0.0025=1.5 × 10²¹ molecules

Question 17.
Calculate the number of moles of 02 required to produce 240 g of MgO by burning Mg metal. [Atomic mass: Mg=24, 0=16]
Answer:
2 Mg + O₂ → 2MgO
No. of moles of MgO = \(\frac{240}{40}\)=6
No. of moles of 02 required = 6/2 = 3

Question 18.
Arrange the following in the increasing order of their mass.
(a) 1 g of Ca
(b) 12 amu of C
(c) 6.022 × 10²³ mol-ecules of CO₂
(d) 11.2 L of N₂ at STP
Answer:
a) Mass of 1 g Ca = 1 g
b) Mass of 12 amu C = \(\frac{12}{6.022 \times 10^{23}}\) = 2 × 10^-23
c) Mass of 6.022 × 10²³ molecules of CO₂ = 44 g
d) Mass of 11.2 L of N₂ at NTP = \(\frac{28 \times 11.2}{22.4}\) = 14 g
(b) < (a) < (d) < (c)

Question 19.
Complete the table:

42g N2	1.5 mole N2	33600mLN2 (STP)
16g 0:	– – – – mole O2	11.2 L of O2 (STP)
….g CO2	1 mole CO2	– – – – L of C O2 (STP)
28g CO	1 mole CO	– – – – mLCO (STP)

Answer:

42g N2	1.5 Mole N2	33600mLN2 (STP)
16g O2	0.5 Mole O2	11.2 L of O2 (STP)
44 g CO2	1 mole CO2	22.4 L of CO2 (STP)
28g CO	1 mole CO	22400mLCO (STP)

Question 20.
1. Irrespective of the source, pure sample of H₂0 always contains 88.89% by mass of oxygen and 11.11% by mass of hydrogen.
a) Which law is illustrated here?
b) State the law.
2. Complete the table by filling in the blanks:

48 g O2	1.5 mol O2	……mL O2 (at STP)
…… g Na	2 gram atom Na	2NA Na atoms
…….g CO2	2.5 mol CO2	56 L (at STP)
8.5 g NH3	……mol NH3	11.2 L (at STP)

Answer:
1. a) Law of definite proportions.
b) The same chemical compound always contains the same elements combined in the same fixed proprotion by mass.
2.

48 g O2	1.5 mol O2	33600 mL O2 (at STP)
46g Na	2 gram atom Na	2Na Na atoms
110 g CO2	2.5 mol CO2	56 L (at STP)
8.5 g NH3	0.5 mol NH3	11.2L(atSTP)

Question 21.
Calcium carbonate reacts with aqueous HCl to give CaCl₂ and CO₂ according to the reaction:
CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O
What mass of CaCO₃ is required to react completely with 25 mL of 0.75 M HCl?.
Answer:
Mass of HCI in 0.25 mL 0.75 M HCl \(=\frac{36.5 \times 0.75 \times 0.25}{1000}=0.6844 \mathrm{g}\)
As per reaction 100 g CaCO₃ reacts with 2 × 36.5 = 73 g of HCl
∴ Mass of CaCO₃ reacting with 0.6844 g HCl \(=\frac{100 \times 0.6844}{73}=0.9375 \mathrm{g}\)

Plus One Chemistry Some Basic Concepts of Chemistry Three Mark Questions and Answers

Question 1.
During a Seminar, a student remarked that “Dalton’s atomic theory has some faulty assumptions”.
a) Do you agree with him?
b) What is the present status of Dalton’s atomic theory?
c) Write any two wrong postulates of Dalton’s atomic theory.
Answer:
a) I agree with him. Out of 6 Dalton’s postulates, 5 postulates are faulty and only one is correct.
b) Dalton’s atomic theory has undergone many modifications.
c)

• All substances are made up of small indivisible particles called atoms.
• Atoms of the same elements are identical in mass and other properties.

Question 2.
One gram atom of an element contains 6.023 × 10²³ atoms.

1. Find the number of atoms in 8 g oxygen.
2. Which is heavier, 1 oxygen atom or 10 hydrogen atoms?
3. Define mole and Avogadro number.

Answer:
1. 16 g oxygen contains 6.022 × 10²³ atoms
∴ 8 g oxygen contains \(\frac{6.022 \times 10^{23}}{2}\) = 3.011 × 10²³

2.

3. 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.
Avogadro number – It is the number of discrete particles present in 1 mole of any substsnce. (Avogadro number, N_A = 6.022 × 10²³)

Question 3.

1. Classify the following as homogeneous and heterogeneous mixtures.
   Air, Smoke, Gunpowder, NaCI solution, Petrol, Bronze, Mixture of sugar and sand.
2. State and explain law of multiple proportions with example.

Answer:
1. Homogeneous-Air, NaCI solution, Bronze, Gun powder, Petrol.
Heterogeneous – Mixture of sugar and sand, Smoke.

2. When two elements combine to form more than one compound the different masses of one of the elements combining with fixed mass of the other element bear a simple ratio. ,
eg. Carbon reacts with oxygen to form two compounds viz. CO and CO₂. In CO mass ratio is 12:16. In CO₂ mass ratio is 12:32. Then mass ratio between oxygen in the 2 compounds is 16:32 or 1:2 which is a simple whole number ratio. Hence, the law is verified.

Question 4.
1. One mole of an ideal gas occupies 22.4 L at STP
a) Calculate the mass of 11.2 L of oxygen gas at STP.
b) Calculate the number of atoms present in the above sample.
2. 21 g of nitrogen gas is mixed with 5 g of hydrogen gas to yield ammonia according to the equation.
N₂ + 3H₂ → 2NH₃
Calculate the maximum amount of ammonia that can be formed.
Answer:
1. a) Mass of 22.4 L oxygen at STP = 32 g
∴ Mass of 11.2 L oxygen at STP = 16 g
b) No. of atoms present in 16 g of O₂
\(\frac{6.02 \times 10^{23}}{2}\) ×2 = 6.02 × 10²³ atoms

2. N₂ + 3H₂ → 2NH₃
1 mole N₂ + 3 mole H₂ → 2moles NH₃
1 mole N₂ requires 3 mol H₂
i.e., 28g N2 requires 6 g H₂
Hence, 21 g N2 requires \(\frac{6 \times 21}{28}\) = 4.5 g H₂
21 g N₂ reacts completely and 0.5g H₂ remains unreacted.
Hence, N₂ is the limiting reagent.
28g N₂ gives 2 × 17 g NH₃
∴ 21 g N₂ gives \(\frac{2 \times 17 \times 21}{28}\) = 25.5 g NH₃

Question 5.
When two elements combine to form more than one compound the different masses of one of the elements combining with fixed mass of the other bear a simple ratio.
i) Name the above law.
ii) Explain the above law by taking oxides of carbon.
Answer:
i) Law of multiple proportions.
ii) Carbon reacts with oxygen to form two compounds viz. CO and CO₂.
In CO, mass ratio is 12:16
In CO₂,mass ratio is 12:32
Ratio of the masses of oxygen combining with a fixed mass of carbon in the two compounds is 16:32 or 1:2, which is a simple whole number ratio.

Question 6.
A compound contains 80% carbon and 20% hydrogen. If the molecular mass is 30 calculate empirical formula and molecular formula.
Answer:

Question 7.
A compound contains 4.07% of hydrogen, 24.27% of carbon and 71.65% of chlorine. The molar mass is 98.96. What is the empirical and molecular formula?
Answer:

Question 8.
Nitrogen forms various oxides.
1. Identify the law of chemical combination illustrated here. Also state the law.
2. Determine the formula of each oxide from the given data and illustrate the law.

Answer:
1. Law of multiple proportions.
When two elements combine to form more than one compound the different mass of one of the elements which combine with the fixed mass of the other element bear a simple ratio.

2.

In NO and NO₂, the masses of oxygen combining with a fixed mass (14 g) of nitrogen are in the ratio, 16:32 = 1:2. Similarly, in N₂O and N₂O₃, the masses of oxygen combining with a fixed mass (28 g) of nitrogen are in the ratio, 16:48 = 1:3. These are simple whole number ratios. Hence, the law of multiple poportions is verified.

Plus One Chemistry Some Basic Concepts of Chemistry Four Mark Questions and Answers

Question 1.
Which of the following weighs more?
a) 1 mole of glucose
b) 4 moles of oxygen
c) 6 moles of N
d) 5 moles of sodium
Answer:
a) 1 mole glucose = (72 + 12 + 96) g = 180 g
b) 4 moles of oxygen = 4 × 32g = 128g
c) 6 moles of nitrogen = 6 × 14 g = 84 g
d) 5 moles of Na = 5 × 23 g = 115 g.
Thus, 1 mole glucose weighs more.

Question 2.
3 g of H₂ is mixed with 29 g of O₂ to yield water.
1. Which is the limiting reagent?
2. Calculate the maximum amount of water that can be formed.
3. Calculate the amount of the reactants which remains unreacted.
Answer:
1.

According to the equation, 4 g H2 requires 32 g
So 3 g H₂ requires \(\frac{377 \times 33}{44}\) = 24 g O₂.
Here 3 g H₂ is mixed with 29 g of O₂. All H₂ will react. Hence H₂ is the limiting reagent.

2. According to the equation, 4 g H₂ gives 36 g H₂O. Hence 3 g H₂ will give 36 × 3/4 = 27 g H₂O.

3. Amount of O₂ unreacted = (29 – 24)g = 5 g

Question 3.
a) Calculate the mass of oxygen required for the complete burning of 2 g of carbon.
b) Calculate the molar mass of (i) CO₂ (ii) CH₄
Answer:

Question 4.
One gram mole of a substance contains 6.022 x 1023 molecules.
1. 24 g of carbon is treated with 72 g of oxygen to form CO₂. Identify the limiting reagent.
2. Find the number of molecules of CO₂ formed in this situation.
Answer:
1.

2 mol of C requires 2 mol of O₂.
2 mol C completely reacts with 2 mol of O₂ and 0.25 mol O₂ and 0.25 mol O₂ remains unreacted. Hence, C is the limiting reagent.

2. No. of moles of CO₂ formed = 2
∴ No. of molecules of CO₂ formed
= 2 × 6.022 × 10²³ = 1.2044 × 10²⁴

Question 5.
One gram mole of a substance contains 6.022×1023 molecules.
i) Find out the number of molecules in 2.8 g of nitrogen.
ii) Which is the heavier-one SO₂ molecule or one CO₂ molecule?
Answer:
i) No. of molecules in1 mole of N₂ = 6.022 × 10²³
i.e., No. of molecules in 28 g of N₂ = 6.022 × 10²³
∴ No. of molecules in 2.8 g N₂

Question 6.
a) How can you illustrate the law of multiple proportions by using oxides of metals containing 78.7% and 64.5% of the metal?
b) Match the following:
1/12^th the mass of C¹² atom – 1 mole
1 g of hydrogen atom – amu
22.4 L O₂ at NTP – gram mole
180 g of glucose – gram atom
6.022 × 10²³ particles – molar volume
Answer:
a) In 100 g samples of the two oxides, the masses of the metal are 78.7 g and 64.5 g respectively.
First Oxide :
Mass of oxygen = 100 – 78.7 = 21.3 g
No. of parts by mass of oxygen combining with one part by mass of metal =\(\frac{78.7}{21.3}=3.7 \mathrm{g}\)

Second oxide:
Mass of oxygen = 100 – 64.5 = 35.5 g
No. of parts by mass of oxygen combining with one part by mass of metal = \(\frac{64.5}{35.5}=1.9 \mathrm{g}\)
The ratio of masses of oxygen combining with a fixed mass of metal = 3.7 : 1.9 = 2: 1, a simple whole number ratio.

b) 1/12^th the mass C¹² atom – amu
1 g of hydrogen atom – gram atom
22.4 L O₂ at NTP – molar volume
180 g of glucose – gram mole
6.022 × 10²³ particles – 1 mole

Question 7.
Calculate
1. The number of molecules present in 1 g of water.
2. The volume of 0.2 mole of sulphur dioxide at STP.
Answer:
1. Number of moles in 1 g water = \(\frac{1}{8}\)
∴ No. of molecules in 1 g water
\(=\frac{1 \times 6.022 \times 10^{23}}{18}=3.35 \times 10^{22}\)

2. Volume of 0.2 mol S02 at STP = 0.2 × 22.4 litre
= 4.48 litre

Question 8.
“One mole of all substances contain the same number of specified particles.”
a) Justify the statement.
b) Howto connect mole, gram mole, and gram atom?
c) What is the relation between number of moles and volume?
d) Calculate the number of moles of a gas in 11.2 L at • STP.
Answer:
a) This statement is true i.e., one mole of all sub-stances contain the same number of specified particles. According to Avogadro’s law.
1 mole of any substance contains 6.022 × 10²³ specified particles.

b)

1 gram mole is the molecular mass expressed in gram. It is the mass of 1 mole molecules in gram. Thus, 1 gram mole contains 1 mole molecules.
1 gram atom is the atomic mass expressed in gram. It is the mass of 1 mole atoms in gram. Thus, 1 gram atom contains 1 mole atoms.

c) Number of moles is directly proportional to volume (according to Avogadro law).

Plus One Chemistry Some Basic Concepts of Chemistry NCERT Questions and Answers

Question 1.
Calculate the molecular mass of the following : (3)
1. H₂O
2. CO₂
3. CH₄
Answer:
1. Molecular mass of H₂O = 2(1.008 u) +16.00 u
= 18.016u

2. Molecular mass of CO₂ = 12.01 u + 2(16.00 u)
= 44.01 u

3. Molecular mass of CH₄ = 12.01 u + 4(1.008 u)
= 16.042 u

Question 2.
Calculate the mass percent of different elements present in sodium sulphate (Na₂SO₄). (2)
Answer:

Question 3
Calculate the amount of carbon dioxide that could be produced when (3)
1. 1 mole of carbon is burnt in air.
2. 1 mole of carbon is burnt in 16 g of dioxygen.
3. 2 moles of carbon are burnt in 16 g of dioxygen.
Answer:
1. The balanced equation for the combustion of carbon dioxide in dioxygen in air is

In air combustion is complete.
Hence, Amount of CO₂ produced when 1 mole of carbon is burnt in air = 44 g

2. As only 16 g dioxygen is available it is the limiting reagent.
Hence, amount of CO₂ produced = \(\frac{44}{32}\)×16 = 22 g

3. Here again, dioxygen is the limiting reactant. Therefore, amount of CO₂ produced from 16 g dioxygen = \(\frac{44}{32}\)×16 = 22g

Question 4.
Chlorine is prepared in the laboratory by treating manganase dioxide (MnO2) with aqueous hydrochloric acid according to the reaction,
4HCl(aq) + MnO₂(s) → 2H₂O(l) + MnCl₂(aq) + Cl₂(g)
How many grams of HCl react with 5.0 g of manganese dioxide?
(Atomic mass of Mn = 54.94 u) (2)
Answer:
1 mol of MnO₂, i.e., 54.94 + 32 = 86.94 g MnO₂ react with 4 moles of HCl, i.e., 4 × 36.5 g = 146 g of HCl.
∴ Mass of HCl reacting with 5.0 g of MnO₂
\(=\frac{146}{86.94} \times 5 \mathrm{g}=8.4 \mathrm{g}\)

Question 5.
Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040. (2)
Answer:

Students can Download Chapter 6 Thermodynamics 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 6 Thermodynamics

Introduction
The study of energy transformations forms the subject matter or thermodynamics.

Thermodynamic Terms
The system and the surroundings
A system in thermodynamics refers to that part of universe in which observations are made and remain-ing universe constitutes the surroundings. The surroundings include everything other than the system. System and the surroundings together constitute the universe.

Types Of The System
1. Open System:
In an open system, there is exchange of energy and matter between system and surroundings

2. Closed System:
In a closed system, there is no exchange of matter, but exchange of energy is possible between system and the surroundings

3. Isolated System:
In an isolated system, there is no exchange of energy or matter between the system and the surroundings. The presence of reactants in a thermos flask or any other closed insulated vessel is an example of an isolated system.

The State Of The System
The state of a system means the condition of the system when is macroscopic properties have definite values. If any of the macroscopic properties of the system changes, the state of the system will change. A process is said to occur when the state of the system changes.
The measurable properties required to describe the state of a system are called state variables or state functions. Temperature, pressure, volume, composition etc. are state variables.

The Internal Energy as a State Function
1. Work:
The system which can’t exchange heat between the system and surroundings through its boundary is called adiabatic system. The manner in which the state of such a system may be changed will be called adiabatic process. Adiabatic process is a process in which there is no transfer of heat between the system and surroundings.
Internal energy, U, of the system is a state function. The positive sign expresses that wad is positive when work is done on the system. Similarly, if the work is done by the system, wad will be negative.

2. Heat:
A system change its internal energy by ex-change of heat. 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. first law of thermodynamics, which states that The energy of an isolated system is constant.
i. e., ∆U = q + w
It is commonly stated as the law of conservation of energy i.e., energy can neither be created nor be destroyed.

Applications
Work
The work done due to expansion or compression of a gas against an opposing external pressure is called the pressure – volume type of work. It is a kind of mechanical work.

If Y is the initial volume and V_f is the final volume of a certain amount of gas and P_ex is the external pressure, then the work involved in the process is given by
w = – P_ex (V_f – V_i) or w = -P_ex ∆V

The negative sign of this expression is required to obtain conventional sign for w.

It must be noted that the above expression gives the work done by the gas in irreversible expansion or compression

Work done in isothermal reversible expansion (or compression) of a gas is given by the relation
w_rev =-2.303 nRT log \(\frac{V_{f}}{V_{i}}\)
Where n = the number of moles of the gas

Free expansion
Expansion of a gas in vaccum is called free expansion. Since P = 0 in vacuum, work done in free expansion = 0
Isothermal and free expansion of an ideal gas.
1. For isothermal irreversible change q= -w = p_ex (V_f – V_i)
2. For isothermal reversible change
q = -w = nRT In \(\frac{V_{f}}{V_{i}}\)
= 2.303 nRT log \(\frac{V_{f}}{V_{i}}\)
3. For adiabatic change q = 0, ∆U = w_ad

Enthalpy, H
1. A useful new state function:
We know that the heat absorbed at constant volume is equal to change in the internal dnergy i.e., ∆U= q_p

We may write equation as ∆U=q_p – p∆V at constant pressure, where q_p is heat absorbed by the system and -pAV represent expansion work done by the system.

We can rewrite the above equation as U₂ -U₁ = q_p – p(V₂ – V₁)

On rearranging, we get q_p = (U₂ +pV₂) = (U₁ +pV₁) Now we can define another thermodynamic function, the enthalpy H [Greek word enthalpien, to warm or heat content] as :
H=U+PV

so, equation becomes q_p = H₂ – H₁ = ∆H

Although q is a path dependent function, H is a state function because it depends on U, p and V, all of which are state functions.

Therefore, ∆H is independent of path. Hence,q_p is also independent of path.

For finite changes at constant pressure, we can write ∆H = ∆U + ∆pV

It is important to note that when heat is absorbed by the system at constant pressure, we are actually measuring changes in the enthalpy. Remember ∆H = q_p heat absorbed by the system at constant pressure.

∆H is negative for exothermic reactions which evolve heat during the reaction and ∆H is positive for endothermic reactions which absorb heat from the surroundings.

2. Extensive and Intensive Properties:
An extensive property is a property whose value depends on the quantity or size of matter present, in the system. For example, mass, volume, internal energy, enthalpy, heat capacity, etc. Those properties which do not depend on the quantity or size of matter present are known as intensive properties. For example temperature, density, pressure, etc.

3. Heat Capacity:
The heat required to rise the temperature of the system in case of heat absorbed by the system.

The increase of temperature is proportional to the heat transferred q = coeff × ∆T

The magnitude of the coefficient depends on the size, composition, and nature of the system. We can also write it as q = C ∆T

The coefficient, C is called the heat capacity. Water has a large heat capacity i.e., a lot of energy is needed to raise its temperature. C is directly proportional to amount of substance. The molar heat capacity of a substance, C_m = \(\frac{C}{n}\), is the heat capacity for one mole of the substance and is the quantity of heat needed to raise the temperature of one mole by one degree Celsius (or one kelvin).

Specific heat, also called specific heat capacity is the quantity of heat required to raise the temperature of one unit mass of a substance by one degree Celsius (or one kelvin). q = c × m × ∆T

4. The relationship between C_p and C_v for an ideal gas:
At constant volume, the heat capacity, C is denoted by C_v and at constant pressure, this is denoted by C_p. Let us find the relationship between the two. We can write equation for heat, q at constant volume as q_v=C_v ∆T = ∆U at constant pressure as q_p = C_p∆T = ∆H

The difference between C_p and C_v can be derived for an ideal gas as:
For a mole of an ideal gas, ∆H = ∆U + ∆(pV)
= ∆U + ∆(RT)
= ∆U + R∆T
On putting the values ∆H of ∆H and we have
C_p∆T = C_v∆T
C_p = C_v +R
C_p – C_v = R

Measurement Of ∆U And ∆H: Calorimetry
We can measure energy changes associated with chemical or physical processes by an experimental technique called calorimetry.

1. ∆U measurements:
Here, a steel vessel (the bomb) is immersed in a water bath. The whole device is called calorimeter. The steel vessel is immersed in water bath to ensure that no heat is lost to the surroundings. A combustible substance is burnt in pure dioxygen supplied in the steel bomb. Heat evolved during the reaction is transferred to the water around the bomb and its temperature is monitored. Since the bomb calorimeter is sealed, its volume does not change i.e., the energy changes associated with reactions are measured at constant volume. Under these conditions, no work is done as the reaction is carried out at constant volume in the bomb calorimeter. Even for reactions involving gases, there is no work done as ∆V = 0.

2. ∆H measurements:
Measurement of heat change at constant pressure (generally under atmospheric pressure) can be done in a calorimeter at constant p

In an exothermic reaction, heat is evolved, and system loses heat to the surroundings.

Therefore, q_p will be negative and ∆_rH will also be negative. Similarly in an endothermic reaction, heat is absorbed, q_p is positive and ∆_rH will be positive.

Enthalpy Change, ∆_rH Of A Reaction – Reaction Enthalpy
The enthalpy change accompanying a reaction is called the reaction enthalpy.
The reaction enthalpy change is denoted by ∆_rH
∆_rH = (sum of enthalpies of products) – (sum of enthalpies of reactants)

1. Standard enthalpy of reactions:
The standard enthalpy of reaction is the enthalpy change for a reaction when all the participating substances are in their standard states.
The standard state of a substance at a specified temperature is its pure form at 1 bar.

2. Enthalpy changes during phase transformations:
The enthalpy change that accompanies melting of one mole of a solid substance in standard state is called standard enthalpy of fusion or molar enthalpy of fusion ∆_fusH°.

Amount of heat required to vaporize one mole of a liquid at constant temperature and under standard pressure (1bar) is called its standard enthalpy of vaporization or molar enthalpy of vaporization ∆_vapH°. And that of sublimation is called Standard enthalpy of sublimation, ∆_subH°.

3. Standard enthalpy of formation
The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states of aggregation (also known as reference states) is called Standard Molar Enthalpy of Formation( ∆_fH°).

Hess’s Law of Constant Heat Summation Hess’s Law:
If a reaction takes place in several steps then its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature.

Enthalpy Calculator is a free online tool that displays the Enthalpy for the given equation.

Enthalpies For Different Types Of Reactions
1. Standard enthalpy of combustion
Enthalpy of combustion of a substance is defined as the enthalpy change accompanying the complete combustion of one mole of the substance in excess of air or oxygen.

Standard enthalpy of combustion is defined as the enthalpy change accompanying the complete combustion of one mole of the substance in excess of air or oxygen when all the reactants and products are tin their standard states at the specified temperature. It is denoted as ∆_cH°.
For example, the complete combustion of one mole of methane evolves 890.3 kJ of heat. Thus, the enthalpy of combustion of methane is- 890.3 kJ mol^-1.

Combustion reactions are always accompanied by the evolution of heat. Hence enthalpy of combustion is always negative.

2. Enthalpy of atomization (symbol: ∆_aH°)
It is the enthalpy change on breaking one mole of bonds completely to obtain atoms in the gas phase.

3. Bond Enthalpy (symbol: ∆_bondH°)
The bond dissociation enthalpy is the change in enthalpy when one mole of covalent bonds of a gaseous covalent compound is broken to form products in the gas phase.

Lattice Enthalpy
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.

Spontaneity
A process which has an urge or a natural tendency to occur under a given set of conditions is known as a spontaneous process.
Some of the spontaneous process need no initiation, i.e., they take place by themselves. Dissolution of common salt in water, evaporation of water in an open vessel, combination of NO and oxygen to form NO₂, neutralisation reaction between NaOH and HCl, etc. are examples of such processes. But some other spontaneous processes need initiation. For example, hydrogen reacts with oxygen to form water only when initiated by passing an electric spark. Once initiated, it occurs by itself.

1. Is decrease in enthalpy a criterion for spontaneity?
By analogy, we may be tempted to state that a chemical reaction is spontaneous in a given direction, because decrease in energy has taken place, as in the case of exothermic reactions.lt becomes obvious that while decrease in enthalpy may be a contributory factor for spontaneity, but it is not true for all cases.

2. Entropy and spontaneity
Entropy(S) is the measure of the degree of randomness or disorder in the system. The greater the disorder in an isolated system, the higher is the entropy. The crystalline solid state is the state of lowest entropy (most ordered), The gaseous state is state of highest entropy. ∆S is independent of path.

Dilution factor formula. After the first tube, each tube is the dilution of the previous dilution tube.

∆S is related with q and T for a reversible reaction as: ∆S = \(\frac{q_{\text {rev }}}{T}\)

When a system is in equilibrium, the entropy is maximum, and the change in entropy, ∆S = 0.

3. Gibbs energy and spontaneity
we define a new thermodynamic function the Gibbs energy or Gibbs function, G, as G = H – TS

Gibbs energy change is a better parameter to determine the spontaneity or feasibility of a process. It can be summarised as follows.
i) If ∆G is negative (i.e., <0) the precess will be spontaneous. ii) If ∆G is zero, the precess is in equilibrium state. iii) If ∆G is positive (i.e., >0), the process is non- spontaneous in the forward direction. The reverse process may be spontaneous.

Ncert Supplementary Syllabus

Enthalpy of Dilution
It is known that 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. This argument can be applied to any solvent with slight modification. Enthalpy change for dissolving one mole of gaseous hydrogen chloride in 10 mol of water can be represented by the following equation.
HCl(g) + 10 aq. → HCl. 10 aq. ∆H = -69.01 kJ/mol

Let us consider the following set of enthalpy changes:
(S- 5) HCl(g) + 25 aq. → HCl.25 aq. ∆H = -72.03 kJ/mol
(S-2) HCtlgi + 40 aq. → HCl.40 aq. ∆H = -72.79 kJ/mol
(S-3) HCl(g) + ∞ aq. → HCl. ∞aq. ∆H = -74.85 kJ/mol

The values of ∆H show general dependence of the enthalpy of solution on amount of solvent. As more and more solvent is used, the enthalpy of solution approaches a limiting value, i.e, the value in infinitely dilute solution. For hydrochloric acid this value of AH is given above in equation (S-3).
If we subtract the first equation (equation S-1) from the second equation (equation S-2) in the above set of equations, we obtain

This value (-0.76kJ/mol) of ∆H is enthalpy of dilution. It is the heat withdrawn from the
HCl.25 aq. + 15 aq. → HCl.40 aq.
∆H = [ -72.79 – (-72.03)] kJ/mol
= -0.76 kJ/mol

This value (-0.76kJ/mol) of ∆H is enthalpy of dilution. It is the heat withdrawn from the surroundings when additional solvent is added to the solution. The enthalpy of dilution of a solution is dependent on the original concentration of the solution and the amount of solvent added.

Theoretical Yield Calculator is a free online tool that displays the amount of product predicted with the complete utilisation of the limiting reactant.

Entropy and Second Law of Thermodynamics
We know that for an isolated system the change in energy remains constant. Therefore, increase in entropy in such systems is the natural direction of a spontaneous change. This, in fact, is the second law of thermodynamics. Like first law of thermodynamics, second law can also be stated in several ways. The second law of thermodynamics explains why spontaneous exothermic reactions are so common. In exothermic reactions, heat released by the reaction increases the disorder of the surroundings and overall entropy change is positive which makes the reaction spontaneous.

Absolute Entropy and Third Law of Thermodynamics
Molecules of a substance may move in a straight line in any direction, they may spin like a top and the bonds in the molecules may stretch and compress. These motions of the molecule are called translational, rotational and vibrational motion respectively. When temperature of the system rises, these motions become more vigorous and entropy increases. On the other hand, when temperature is lowered, the entropy decreases. The entropy of any pure crystalline substance approaches zero as the temperature approaches absolute zero. This is called third law of thermodynamics. This is so because there is perfect order in a crystal at absolute zero.

The statement is confined to pure crystalline solids because theoretical arguments and practical evidences have shown that entropy of solutions and super cooled liquids is not zero at 0 K. The importance of the third law lies in the fact that it permits the calculation of absolute values of entropy of pure substance from thermal data alone. For a pure substance, this can be done by summing \(\frac{q_{\text {rev }}}{T}\) increments from 0 K to 298 K. Standard entropies can be used to calculate standard entropy changes by a Hess’s law type of calculation.

Students can Download Chapter 4 Chemical Bonding and Molecular Structure 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 4 Chemical Bonding and Molecular Structure

Introduction
Matter is made up of different type of elements. The attractive force which holds the constituents together is called a chemical bond.

Kossel-Lewis Approach To Chemical Bonding
The bond between constituents are formed by the sharing of a pair of electrons or their transfer. G.N. Lewis introduced simple notations to represent these outer shell electrons in an atom. These notations are called Lewis symbols.

Significance of Lewis Symbols :
The number of dots around the symbol represents the number of valence electrons. This number of valence electrons helps to calculate the common or group valence of the element. The group valence of the elements is generally either equal to the number of dots in Lewis symbols or8 minus the number of dots or valence electrons.

How to Calculate Bond Order? Introduction to Bond Order.

Kossel, in relation to chemical bonding, drew attention to the following facts:
The bond formed, as a result of the electrostatic attraction between the positiveand negative ions was termed as the electrovalent bond. The electrovalence is thus equal to the number of unit charge (s) on the ion.
In terms of Lewis structures

1. According to electronic theory of chemical bond¬ing, atoms can combine either by transfer of valence electrons from one atom to another (gaining or losing) or by sharing of valence electrons in order to have an octet in their valence shells. This is known as octet rule.

2. Covalent Bond, Langmuir in 1919 refined the Lewis postulations by abandoning the idea of the stationary cubical arrangement of the octet, and by introducing the term covalent bond.

By Lewis – Langmuir theory the formation of chlorine molecule is as follows :

In water molecule covalent bond is as follows:

when two atoms share one electron pair they are said to be joined by a single covalent bond. If two atoms share two pairs of electrons, the covalent bond between them is called a double bond. And when combining atoms share three electron pairs as in the case of N₂ molecule a triple bond a triple bond is formed.

• The total number of electrons required for writing the structures are obtained by adding the valence electrons of the combining atoms.
• For anions, each negative charge would mean addition of one electron. For cations, each positive charge would result in subtraction of one electron from the total number of valence electrons.
• The least electronegative atom occupies the central position in the molecule/ion.

Formal charge
Formal charge (F.C.) on an atom in a Lewis structure = total number of valence electrons in the free atom— total number of non bonding (lone pairjelectrons—(1/2) total number of bonding(shared)electrons.

Let us consider the ozone molecule (O₃).
The Lewis structure of O₃ may be drawn as:

The atoms have been marked as 1,2 and 3. The formal charge on:

• The central O atom marked 1 = 6 – 2- \(\frac{1}{2}\)(6) = +1
• The end O atom marked 2 = 6 – 4 – \(\frac{1}{2}\)(4) = o
• The end O atom marked 3 = 6 – 6 – \(\frac{1}{2}\)(2) = -1

Hence, we represent O₃ along with the formal changes as follows:

Limitations of the Octet Rule
There are three types of exceptions to the octet rule. 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.
Some compounds are BCl₃, AlCl₃ and BF₃.

Odd-electron molecules
In molecules with an odd number of electrons like nitric oxide, NO and nitrogen dioxide, NO₂, the octet rule is not satisfied for all the atoms.

The expanded octet
In a number of compounds of these elements there are more than eight valence electrons around the central atom.Some examples are given below.

Ionic Or Electrovalent Bond
The formation of a positive ion involves ionization, i.e., removal of electrons from the neutral atom and that of the negative ion involves the addition of electron(s) to the neutral atom.

A qualitative measure of the stability of an ionic compound is provided by its enthalpy of lattice formation and not simply by achieving octet of electrons around the ionic species in gaseous state.

Lattice Enthalpy
The Lattice Enthalpy of an ionic solid is defined as the energy required to completely separate one mole of a solid ionic compound into gaseous constituentions.

Bond Parameters

Bond Length
It may be defined as the equilibrium distance between the centres of the nuclei of the two bonded i atoms in a molecule. Bond length are measured by spectroscopic, X-ray diffraction and electron diffraction techniques. It is usually expressed in Angstrom units (A°) or picometres (pm)
1 A° = 10^-10m and 1 pm = 10^-12m

Bond Angle
It is defined as the angle between the orbitals containing bonding electron pairs around the central atom in a molecule.

Bond Enthalpy
It is defined as the amount of energy required to break one mole of bonds of a particular type between two atoms in a gaseous state.

Bond Order:
In the Lewis description of covalent bond, the bond order is given by the number of bonds between the two atoms in a molecule. For example, the bond order in H₂ is one, in O₂ is two and in N₂ is three. Isoelectronic molecules and ions have identical bond orders. For example N₂, CO and NO⁺ have bond order 3. It is found that as the bond order increases, bond enthalpy increases and bond length decreases.

Resonance Structures
It is often observed that a single Lewis structure is inadequate for the representation of a molecule in conformity with its experimentally determined parameters. As in the case of O₃.

O₃ is represented by the above 3 structures. These are called canonical structures.

Experimentally determined oxygen-oxygen bond lengths in the O₃ molecule are same (128 pm). Thus the oxygen-oxygen bonds in the O₃ molecule are intermediate between a double and a single bond. According to the concept of resonance, the canonical structures of the hybrid describes the molecule accurately.

Some of the other examples of resonance structures are provided by the carbonate ion and the carbon dioxide molecule.

Polarity of Bonds
In reality no bond or a compound is either completely covalent or ionic. Even in case of covalent bond between two hydrogen atoms, there is some ionic character. As a result of polarisation, the molecule possesses the dipole moment. Which can be defined as the product of the magnitude of the charge and the distance between the centres of positive and negative charge. It is usually designated by a Greek letter Mathematically, it is expressed as follows: Dipole moment (µ) = change (Q) X distance of separation (r)

In case of polyatomic molecules, the dipole moment not only depend upon the individual dipole moments of bonds known as bond dipoles but also on the spatial arrangement of various bonds in the molecule.lt is due to the shifting of electrons to the side of more eletro negative element. For example,

The shifting of electrons is represented by an arrow. In case of H₂O the resultant dipole moment is given by the following figure:

Fajans Rules:
Just as all the covalent bonds have some partial ionic character, the ionic bonds also have partial covalent character. The partial covalent character of ionic bonds was discussed by Fajans in terms of the following rules:

• The smaller the size of the cation and the larger the size of the anion, the greater the covalent character of an ionic bond.
• The greater the charge on the cation, the greater the covalent character of the ionic bond.
• For cations of the same size and charge, the one, with electronic configuration (n-1)d”ns°, typical of transition metals, is more polarising than the one with a noble gas configuration, ns2 np6, typical of alkali and alkaline earth metal cations.

The cation polarises the anion, pulling the electronic charge toward itself and thereby increasing the electronic charge between the two. This is precisely what happens in a covalent bond, i.e., buildup of electron charge density between the nuclei. The polarising power of the cation, the polarisability of the anion and the extent of distortion (polarisation) of anion are the factors, which determine the per cent covalent character of the ionic bond.

The Valence Shell Electron Pair Repul-Sion (Vspert) Theory
The main postulates of VSEPR theory are as follows:

• The shape of a molecule depends upon the number of valence shell electron pairs (bonded or nonbonded) around the central atom.
• Pairs of electrons in the valence shell repel one another since their electron clouds are negatively charged.
• These pairs of electrons tend to occupy such positions in space that minimise repulsion and thus maximise distance between them.
• The valence shell is taken as a sphere with the electron pairs localising on the spherical surface at maximum distance from one another.
• A multiple bond is treated as if it is a single electron pair and the two or three electron pairs of a multiple bond are treated as a single super pair.
• Where two or more resonance structures can represent a molecule, the VSEPR model is applicable to any such structure.

The repulsive interaction of electron pairs de-crease in the order:
Lone pair (lp) – Lone pair (lp) > Lone pair (lp) – Bond pair (bp) > Bond pair (bp) – Bond pair (bp)

Valence Bond Theory
Valence bond theory was introduced by Heitlerand London (1927) and developed further by Pauling and others. A discussion of the valence bond theory is based on the knowledge of atomic orbitals, electronic configurations of elements, the overlap criteria of atomic orbitals, the hybridization of atomic orbitals and the principles of variation and superposition. First, we consider the formation of H2. When the attractive forces become greater than the repulsive forces, the molecule is formed and the system gets minimum energy. Because energy is released when a bond is formed. The energy so released is called bond enthalpy.

Orbital Overlap Concept
When two atoms approach each other, their atomic orbitals undergo partial interpenetration. This partial interpenetration of atomic orbitals is called overlapping of atomic orbitals. The electrons belonging to these orbitals are said to be shared and this results in the formation of a covalent bond. The main ideas of orbital of overlap concept of formation of covalent bonds are

• Covalent bonds are formed by the overlapping of half filled atomic orbitals present in the valence shell of the atoms taking part in bonding.
• The orbitals undergoing overlapping must have electrons with opposite spins.
  Overlapping of atomic orbitals results in decrease of energy and formation of covalent bond.
• The strength of a covalent bond depends upon the extent of overlapping. The greater the overlapping, the stronger is the bond formed.

The above treatment of formation of covalent bond involving the overlap of half-filled atomic orbitals is called valence bond theory.

Types of Overlapping and Nature of Covalent Bonds
The covalent bond may be classified into two types depending upon the types of overlapping:
(i) Sigma(σ) bond, and
(ii) pi(π) bond

(i) Sigma( σ) bond :
This type of covalent bond is formed by the end to end (hand-on) overlap of bonding orbitals along the internuclear axis. This is called as head on overlap or axial overlap. This can be formed by any one of the following types of combinations of atomic orbitals.
s-s overlapping:
In this case, there is overlap of two half filled s-orbitals along the internuclear axis as shown below:

s-p overlapping:
This type of overlap occurs between half filled s-orbitals of one atom and half filled p-orbitals of another atom.

p-p overlapping :
This type of overlap takes place between half filled p-orbitals of the two approaching atoms.

(ii) pi(π) bond :
In the formation of n bond the atomic orbitals overlap in such a way that their axes remain parallel to each other and perpendicular to the internuclear axis. The orbitals formed due to side wise overlapping consists of two saucer type charged clouds above and below the plane of the participating atoms.

Strength of Sigma and pi Bonds
Basically the strength of a bond depends upon the extent of overlapping. In case of sigma bond, the overlapping of orbitals takes place to a larger extent. Hence, it is stronger as compared to the pi bond where the extent of overlapping occurs to a smaller extent. Further, it is important to note that pi bond between two atoms is formed in addition to a sigma bond. It is always present in the molecules containing multiple bond (double ortriple bonds).

Hybridisation
Hybridisationis defined as the process of intermixing of the orbitals of slightly different energies so as to redistribute their energies, resulting in the formation of new set of orbitals of equivalent energies and shape.

The number of hybrid orbitals is equal to the number of the atomic orbitals that get hybridised.

These hybrid orbitals are stable due to their arrangement which provides minimum repulsion between electron pairs. Therefore, the type of hybridisation indicates the geometry of the molecules.

It is not necessary that only half filled orbitals participate in hybridisation. In some cases, even filled orbitals of valence shell take part in hybridisation.

Types of Hybridisation
There are various types of hybridisation involving s, p and d orbitals. The different types of hybridisation are as under:

(I) sp hybridisation:
This type of hybridisation involves the mixing of one s and one p orbital resulting in the formation of two equivalent sp hybrid orbitals. Each sp hybrid orbitals has 50% s-character and 50% p-character. Such a molecule in which the central atom is sp- hybridised and linked directly to two other central atoms possesses linear geometry.The two sp hybrids point in the opposite direction which provides more effective overlapping resulting in the formation of stronger bonds.

Example of molecule having sp hybridisation BeCl₂:
The ground state electronic.configuration of Be is 1s²2s². In the exited state one of the 2s-electrons is promoted to vacant 2p orbital to account for its divalency. One 2s and one 2p-orbitalsget hybridised to form two sp hybridised orbitals. These two sp hybrid orbitals are oriented in opposite direction forming an angle of 180°. Each of the sp hybridised orbital overlaps with the 2p-orbital of chlorine axially and form two Be-Cl sigma bonds.

II) sp² hybridisation :
In this hybridisation there is involvement of one s and two p-orbitals in orderto form three equivalent sp² hybridised orbitals. For example, in BCl₂ 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. The three hybrid orbitals so formed are oriented in a trigonal planar arrangement and overlap with 2p orbitals of chlorine to form three B-Cl bonds. Therefore, in BCl₃ the geometry is trigonal planar with ClBCl bond angle of 120°

III) sp³ hybridisation:
This type of hybridisation can be explained by taking the example of CH₄ molecule in which there is mixing of one s-orbital and three p-orbitals of the valence shell to form four sp³ hybrid orbital of equivalent energies and shape.

There is 25% s-character and 75% p-character in each sp³ hybrid orbital. The four sp3 hybrid orbitals so formed are directed towards the four corners of the tetrahedron.

The angle between sp³ hybrid orbital is 109.5°

The structure of NH₃ and H₂O molecules can also be explained with the help of sp3hybridisation. In NH₃, the valence shell (outer) electronic configuration of nitrogen in the ground state is 2s²\(p_{x}^{1} 2 p_{y}^{1} 2 p_{z}^{1}\) having three unpaired electrons in the sp³ hybrid orbitals and a lone pair of electrons is present in the fourth one. These three hybrid orbitals overlap with 1s orbitals of hydrogen atoms to form three N-H sigma bonds. Due to the force of repulsion, the molecule gets distorted and the bond angle is reduced to 107° from 109.5°. The geometry of such a molecule will be pyramidal.

Other Examples of sp³, sp² and sp Hybridisation:
sp³ Hybridisation in C₂H₆ molecule:
In ethane molecule both the carbon atoms assume sp3 hybrid state. One of the four sp³ hybrid orbitals of carbon atom overlaps axially with similar orbitals of other atom to form sp³-sp³ sigma bond while the other three hybrid orbitals of each carbon atom are used in forming sp³-s sigma bonds with hydrogen atoms Therefore in ethane C-C bond length is 154 pm and each C-H bond length is 109 pm.

sp² Hybridisation in C₂H₄:
In the formation of ethene molecule, one of the sp² hybrid orbitals of carbon atom overlaps axially with sp² hybridised orbital of another carbon atom to form C-C sigma bond. While the other two sp² hybrid orbitals of each carbon atom are used for making sp²-s sigma bond with two hydrogen atoms. The unhybridised orbital (2p_x or 2p_y) of one carbon atom overlaps sidewise with the similar orbital of the other carbon atom to form weak π bond, which consists of two equal electron clouds distributed above and below the plane of carbon and hydrogen atoms.

Thus, in ethene molecule, the carbon-carbon bond consists of one sp²-sp² sigma bond and one pi (π) bond between p orbitals which are not used in the hybridisation and are perpendicular to the plane of molecule; the bond length is 134 pm. The C-H bond is sp^s sigma with bond length 108 pm. The H-C-H bond angle is 117.6° while the H-C-C angle is 121°.

sp Hybridisation in C₂H₂:
In the formation of ethyne molecule, both the carbon atoms undergo sp- hybridisation having two unhybridised orbital i.e., 2p_y and 2p_x. 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 σ bonds. Each of the two unhybridised p orbitals of both the carbon atoms overlaps sidewise to form two π bonds between the carbon atoms. So the triple bond between the two carbon atoms is made up of one sigma and two pi bonds

Hybridisation of Elements involving d-Orbitals
The elements present in the third period contain d orbitals in addition to s and p orbitals. The energy of the 3d orbitals are comparable to the energy of the 3s and 3p orbitals. The energy of 3d orbitals are also comparable to those of 4s and 4p orbitals. As a consequence the hybridisation involving either 3s, 3p, and 3d or 3d, 4s and 4p is possible. However, since the difference in energies of 3p and 4s orbitals is significant, no hybridisation involving 3p, 3d and 4s orbitals is possible.

1. Formation of PCl₅ (sp³d hybridisation):
The ground state and the excited state outer electronic configurations of phosphorus (Z=15) are represented below.

Now the five orbitals (i.eone s, three p, and one d orbitals) are available for hybridisation to yield a set of five sp3d hybrid orbitals which are directed towards the five comers of a trigonal bipyramidal.

Three sigma bond known as equatorial bonds lie in one plane and make an angle of 120° with each other.
The remaining two P-Cl bonds(called axial bonds)-one lying above and the other lying 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, therefore axial bonds have been found to be slightly longer and hence slightly weaker than the equatorial bonds; which makes PCl₅ molecule more reactive.

2. Formation of SF₆ (sp³d² hybridisation):
In SF₆ the central sulphur atom has the ground state outer electronic configuration 3s²3p⁴. In the exited state the available six orbitals i.e., one s, three p and two d are singly occupied by electrons. These orbitals hybridise to form six new sp³d² hybrid orbitals, which are projected towards the six corners of a regular octahedron in SF₆. These six sp³d² hybrid orbitals overlap with singly occupied orbitals of fluorine atoms to form six S-F sigma bonds. Thus SF₆ molecule has a regular octahedral geometry.

The structure of SF₆ is given below.

Molecular Orbital Theory
Molecular orbital (MO) theory was developed by F. Hund and R.S. Mulliken in 1932. The salient features of this theory are :

• The electrons in a molecule are present in the various molecular orbitals as the electrons of atoms are present in the various atomic orbitals.
• The atomic orbitals of comparable energies and proper symmetry combine to form molecular orbitals.
• While an electron in an atomic orbital is influenced by one nucleus, in a molecular orbital it is influenced by two or more nuclei depending upon the number of atoms in the molecule. Thus, an atomic orbital is monocentric while a molecular orbital is polycentric.
• The number of molecular orbital 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.
• Just as the electron probability distribution around a nucleus in an atom is given by an atomic orbital, the electron probability distribution around a group of nuclei in a molecule is given by a molecular orbital.
• The molecular orbitals like atomic orbitals are filled in accordance with the aufbau principle obeying the Pauli’s exclusion principle and the Hund’srule.

Formation of Molecular Orbitals
Linear Combination of Atomic Orbitals (LCAO)
The atomic orbitals of these atoms may be represented by the wave functions ψ_A and ψ_B. The formation of molecular orbitals is the linear combination of atomic orbitals that can take place by addition and by subtraction of wave functions of individual atomic orbitals as shown below.

Energy Level Diagram for Molecular orbitals

The increasing order of energies of various molecular orbitals for O₂ and F₂ is given below:
σ1s < σ*1s < σ2s < σ*2s < σ2p_x <(π2p_x = π2p_y) <(π*2p_x = π*2p_z) < σ*2p_x

This sequence of energy levels of molecular orbitals is not correct for the remaining molecules Li₂, Be₂, B₂, C₂, N₂. For molecules such as B₂, C₂, N₂ etc. the increasing order of energies of various molecular orbitals is
σ1s < σ*1s < σ2s < σ*2s <(π2p_x = π2p_y) < σ2p_x <(π*2p_x = π*2p_z) < σ*2p_z

The important characteristic feature of this order is that the energy of σ 2p_z molecular orbital is higher than that of π2p_x and π2p_y molecular orbitals. If the bonding influence is stronger a stable molecule results and if the antibonding influence is stronger,the molecule is unstable.

Bonding In Some Homonuclear Diatomic Molecules
Bond Order:
Bond order is defined as half of the difference between the number of electrons in the bonding molecular orbitals and that in the antibonding molecular orbitals.
Nh – N
i.e. Bond Order= \(\frac{N_{b}-N_{a}}{2}\). Where N_b is the number of electrons in the bonding molecular orbitals and Na is the number of electrons in the antibonding mo-lecular orbitals.

Significance of bond order:
Bond order conveys the following important informations about a molecule.
i) If the value of bond order is positive, it indicates a stable molecule and if the value of bond order is negative or zero, the molecule is unstable and is not formed.
ii) Bond dissociation energy of a diatomic molecule is directly proportional to the bond order of the molecule. The greater the bond order, the higher is the bond dissociation energy.
iii) Bond order is inversely proportional to the bond length. The higherthe bond ondervalue, smaller is the bond length. For example, the bond length in N₂ molecule (having bond order 3) is less than that in O₂ molecule (having bond order 2).

Magnetic character:
If all the electrons in the mol-ecules of a substance are paired, the substance will be diamagnetic. On the other hand, if there are un-paired electrons in the molecule, the substance will be paramagnetic.

Hydrogen Bonding
Hydrogen bond can be defined as the attractive force which binds hydrogen atom of one molecule with the electronegative atom (F, O orN) of another molecule. When hydrogen is bonded to strongly electronegative element ‘X’, the electron pair shared between the two atoms moves far away from hydrogen atom. As a result the hydrogen atom becomes, highly electropositive with respect to the other atom ‘X’. Since there is displacement of electrons towards X, the hydrogen acquires fractional positive charge (δ⁺) while ‘X’ attain fractional negative charge (δ^–). This results in the formation of a polar molecule having electrostatic force of attraction which can be represented as: H^δ+ – X^δ-

The magnitude of H-bonding depends on the physical state of the compound. It is maximum in the solid state and minimum in the gaseous state. Thus, the hydrogen bonds have strong influence on the structure and properties of the compounds.

Types of Hydrogen Bonds
There are two types of hydrogen bonds

1. Intermolecular hydrogen bond
2. Intramolecular hydrogen bond

1. Intermolecular hydrogen bond:
It is formed between two different molecules of the same or different compounds. For example, H-bond in case of HF molecule, alcohol or water molecules, etc.

2. Intramolecular hydrogen bond:
It is formed when hydrogen atom is in between the two highly electronegative (F, O, N) atoms present within the same molecule. For example, in o-Nitrophenol the hydrogen is in between the two oxygen atoms as shown below:

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Kerala Plus One Chemistry Notes Chapter 2 Structure of Atom

Plus One Chemistry Chapter 2 Notes Pdf Introduction
The atomic theory of matter was first proposed by John Dalton. His theory, called Dalton’s atomic theory, regarded the atom as the ultimate particle of matter.

Sub-Atomic Particles
Discovery Of Electron
The experiments of Michael Faraday in discharge tubes showed that when a high potential is applied to a gas taken in the discharge tube at very low pres-sures, certain rays are emitted from the cathode. These rays were called cathode rays.

The results of these experiments are summarised below:
1. The cathode rays start from cathode and move towards the anode.

2. In the absence of electrical or magnetic field, these rays travel in straight lines.ln the presence of electrical or magnetic field, they behave as negatively charged particles, i.e.,they consist of negatively charged particles, called electrons.

3. The characteristics of cathode rays (electrons) do not depend upon the material of electrodes and the nature of the gas present in the cathode ray tube.Thus, we can conclude that electrons are the basic constituent of all the atoms.

Charge To Mass Ratio Of Electron
In 1897, the British physicist J.J. Thomson measured the ratio of electrical charge (e) to the mass of electron (m_e) by using cathode ray tube and applying electrical and magnetic field perpendicular to each other as well as to the path of electrons.

From the amount of deviation of the particles from their path in the presence of electrical or magnetic field, the value of e/m was found to be 1.75882 × 10¹¹ coulomb per kg or approximately 1.75288 × 10⁸ cou-lomb per gram. The ratio e/m was found to be same irrespective of the nature of the gas taken in the dis-charge tube and the material used as the cathode.

Structure Of Atom Class 11 Notes Charge Of The Electron
Millikan (1868-1953) devised a method known as Oil drop experiment (1906-14), to determine the charge on the electrons. He found the charge on the electron to be – 1.6 × 10-¹⁹C.

Mass of the electron (m)

Discovery Of Protons And Neutrons
Electrical discharge earned out in the modified cathode ray tube led to the discovery of canal rays. The characteristics of these positively charged particles are listed below:

• unlike cathode rays, the e/m ratio of the particles depend upon the nature of gas present in the cathode ray tube.
• Some of the positively charged particles carry a multiple of the fundamental unit of electrical charge.
• The behaviour of these particles in the magnetic or electrical field is opposite to that observed for cathode rays.

The smallest and lightest positive ion was obtained from hydrogen and was called proton. Later, electrically neutral particles were discovered by Chadwick (1932) by bombarding a thin sheet of beryllium by α – particles when electrically neutral particles having a mass slightly greater than that of the protons was emitted. He named these particles as neutrons.

Atomic Models

Structure Of Atom Class 11 Notes Hsslive Thomson Model Of Atom
J.J. Thomson was the first to propose a model of the atom. According to him, the atom is a sphere in which positive charge is spread uniformly and the electrons are embedded in it so as to make the atom electrically neutral. This model is also known as “plumpudding model’. But this model was soon discarded as it could not explain many of the experimental observations.

Hsslive Structure Of Atom Notes Rutherford’s Nuclear Model of Atom
Rutherford and his students (Hans Geiger and Ernest Marsden) bombarded very thin gold foil with α – particles. The experiment is known as α -particle scattering experiment. On the basis of the observations, Rutherford drew the following conclusions regarding the structure of atom :

1. Most of the space in the atom is empty as most of the α -particles passed through the foil undeflected.

2. A few α – particles were deflected. Since the α – particles are positively charged, the deflection must be due to enormous repulsive force showing that the positive charge of the atom is not spread throughout the atom as Thomson had presumed. The positive charge has to be concentrated in a very small volume that repelled and deflected the positively charged α – particles.

3. Calculations by Rutherford showed that the volume occupied by the nucleus is negligibly small as compared to the total volume of the atom.

On the basis of above observations and conclusions, Rutherford proposed the nuclear model of atom (after the discovery of protons). According to this model:
1.The positive charge and most of the mass of the atom was densely concentrated in extremely small region. This very small portion of the atom was called nucleus by Rutherford.

2. The electrons move around the nucleus with a very high speed in circular paths called orbits. Thus, Rutherford’s model of atom resembles the solar system in which the nucleus plays the role of sun and the electrons that of revolving planets.

3. Electrons and the nucleus are held together by electrostatic forces of attraction.

Chemistry Notes For Class 11 Chapter 2 Atomic Numberand Mass Number
’ Knowing the atomic number Z and mass number A of an element, we can calculate the number of protons, electrons and neutrons present in the atom of the element.
Atomic Number (Z) = Number of protons = Number of electrons
Mass Number (A) – Atomic number (Z) = Number of neutrons

Isotopes, Isobars And Isotones
Isotopes are atoms of the same element having the same atomic number but different mass numbers. They contain different number of neutrons. For ex-ample, there are three isotopes of hydrogen having mass numbers 1,2 and 3 respectively. All the three isotopes have atomic number 1. They are represented as \(_{ 1 }^{ 1 }{ H }\), \(_{ 1 }^{ 2 }{ H }\) and \(_{ 1 }^{ 3 }{ H }\) and named as hydrogen or protium, deuterium (D) and tritium (T) respectively. Isobars are atoms of different elements which have the same mass number. For example, \(_{ 6 }^{ 14 }{ C }\) and \(_{ 7 }^{ 14 }{ N }\) are isobars.
Isotones may be defined as atoms of different elements containing same number of neutrons. For example \(_{ 6 }^{ 13 }{ C }\) and \(_{ 7 }^{ 14 }{ N }\) are isotones.

Developments Leading To The Bohr’S Model Of Atom
Neils Bohr improved the model proposed by Rutherford. Two developments played a major role in the formulation of Bohr’s model of atom. These were:

1. electromagnetic radiation possess both wave like and particle like properties(Dual character)
2. Experimental results regarding atomic spectra which can be explained only by assuming quantized electronic energy levels in atoms.

Wave Nature Of Electromagnetic Ra-Diation
Light is the form of radiation and it was supposed to be made of particles known as corpuscules.
As we know, waves are characterised by wavelength (λ), frequency (υ) and velocity of propagation (c) and these are related by the equation
c = vλ or v = \(\frac { c }{ \lambda } \)

The wavelengths of various electro magnetic radia-tions increase in the order.
γ rays < X-rays< uv rays < visible < IR < Microwaves < Radio waves

Particle Nature Of Electro Magnetic Radiation: Planck’S Quantum Theory
Planck suggested that atoms and molecules could emit (or absorb) energy only in discrete quantities and not in a continuous manner, a belief popular at that time. Planck gave the name quantum to the smallest quantity of energy that can be emitted or absorbed in the form of electromagnetic radiation. The energy (E) of a quantum of radiation is proportional to its frequency (υ) and is expressed by the equation E = hυ

Class 11 Chemistry Chapter 2 Notes Photoelectric Effect
When a metal was exposed to a beam of light, electrons were emitted. This phenomenon is called photoelectric effect. Obseravations of the photoelectric effect experiment are the following:

• There is no time lag belween the striking of light beam and the ejection of electrons from the metal surface.
• The number of electrons ejected is proportional to the intensity or brightness of light.
• For each metal, there is a characteristic minimum frequency, u₀ (also known as threshold frequency) below which photoelectric effect is not observed. At a frequency u>u₀, the ejected electrons come out with certain kinetic energy.

The kinetic energies of these electrons increase with the increase of frequency of the light used.

Using Plank’s quantum theory Einstein explained photoelectric effect. When a light particle, photon with sufficient energy strikes an electron instantaneously to the electron during the collision and the electron is ejected without any time lag. Greater the energy of photon greater will be the kinetic energy of ejected electron and greater will be the frequency of radiation.

If minimum energy to eject an electron is hv0 and the photon has an energy equal to hv. Then kinetic en-ergy of photoelectron is given by, hv=hv₀ + 1/2 m_ev² where m_e is the mass of electron and hv₀ is called the work function.

Duel Behaviour Of Electromagnetic Ra-Diation
Light has dual behaviour that is it behaves either as a wave or as a particle. Due to this wave nature, it shows the phenomena of interference and diffraction.

Evidence For The Quantized Electronic Energy Levels : Atomic Spectra
It is observed that when a ray of white light is passed through a prism, the wave with shorter wavelength bends more than the one with a longer wavelength. Since ordinary white light consists of waves with ail the wave-lengths in the visible range, a ray of white light is spread out into a series of coloured bands called spectrum. In a continuous spectrum light of different colours merges together. For example violet merges into blue, blue into green and soon.

Chapter 2 Class 11 Chemistry Notes Emission and absorption spectra
The spectrum of radiation emitted by a substance that has absorbed energy is called an emission spectrum. Atoms, molecules or ions that have absorbed radiation are said to be “excited”.

A continuum of radiation is passed through a sample which absorbs radiation of certain wavelengths. The missing wavelength which corresponds to the radiation absorbed by the matter, leave dark spaces in the bright continuous spectrum. The study of emission or absorption spectra is referred to as spectroscopy Line spectra or atomic spectra is the spectra where emitted radiation is identified by the appearance of bright lines in the spectra.

Line spectrum of Hydrogen
The hydrogen spectrum consists of several series of lines named after their discoverers. Balmershowed in 1885 on the basis of experimental observations that if spectral lines are expressed in terms of wavenumber (\(\overline { v } \)), then the visible lines of the hydrogen spectrum obey the following formula :
\(\overline { v } \) = 109,677 \(\left[\frac{1}{2^{2}}-\frac{1}{n^{2}}\right] \mathrm{cm}^{-1}\)
where n = 3, 4, 5, ………….
The series of lines described by this formula are called the Balmer series.

The value 109,677cm^-1 is called the Rydberg constant for hydrogen. The first 5 series of lines correspond to n₁ = 1, 2, 3, 4, 5 are known as Lyman, Balmer, Paschen, Bracket and Pfund series respectively. Line specrum becomes more complex for heavier atoms.

Chapter 2 Chemistry Class 11 Notes Bhor’S Model For Hydrogen Atom
Bhors model for hydrogen atom says that
1. the energy of an electron does not change with time.
The diagram shows the Lyman, Balmer and Paschen series of transitions for hydrogen atom.

2. The frequency of radiation absorbed or emitted when transition occurs between two stationary states that differ in energy by ∆E, is given by :
\(v=\frac{\Delta E}{h}=\frac{E_{2}-E_{1}}{h}\)
E₁ and E₂ are the energies of the lower and higher allowed energy states respectively.
The angular momentum of an electron in a given stationary state can be expressed as in equation,

Chemistry Chapter 2 Class 11 Notes Bohr’s theory for hydrogen atom:
1. The stationary states for electron are numbered n = 1,2,3. These integral numbers are known as Principal quantum numbers.
2. The radii of the stationary states are expressed as:
r_n = n² a₀
where a₀ = 52.9 pm

3. The most important property associated with the electron, is the energy of its stationary state. It is
given by the expression, \(E_{n}=-R_{H}\left(\frac{1}{n^{2}}\right)\)
where R_H is called Rydberg constant and its value is 2.18 × 10^-18 J. The energy of the lowest state, also called as the ground state, is
E₁ = -2.18 × 10^-18 \(\left(\frac{1}{1^{2}}\right)\) = -2.18 × 10^-18 J. The energy of the stationary state for n = ∝, will be :
E₂ = -2.18 × 10^-18 J\(\left(\frac{1}{2^{2}}\right)\) = -0.545 × 10^-18 J.

When the electron is free from the influence of nucleus(n = ∞), the energy is taken as zero. When the electron is attracted by the nucleus and is present in orbit n, the energy is emitted and its energy is lowered. That is the reason for the presence of negative sign and depicts its stability relative to the reference state of zero energy and n = ∞

4. Bohr’s theory can also be applied to the ions containing only one electron, similar to that present in hydrogen atom. For example, He⁺ Li²⁺, Be³⁺ and so on. The energies of the stationary states associated with these hydrogen-like species are given by the expression,

Structure Of Atom Class 11 Notes Pdf Explanation of Line Spectrum of Hydrogen
The frequency (v) associated with the absorption and emission of the photon can be evaluated by using equation,

Class 11 Chapter 2 Chemistry Notes Limitations of Bohr’s Model
Bohr’s model was too simple to account for the following points:
1. It fails to account for the finer details (doublet, that is two closely spaced lines) of the hydrogen atom spectrum. This model is also unable to explain the spectrum of atoms other than hydrogen Further, Bohr’s theory was also unable to explain the splitting of spectral lines in the presence of magnetic field (Zeeman effect) or an electric field (Stark effect).
2. It could not explain the ability of atoms to form molecules by chemical bonds.

Towards Quantum Mechanical Model Of The Atom
Two important developments which contributed significantly in the formulation of a more suitable and general model for atoms were:

1. Dual behaviour of matter
2. Heisenberg uncertainty principle

Structure Of Atom Class 10 Notes Pdf Dual Behaviour of Matter
The French physicist, de Broglie proposed that matter, like radiation, should also exhibit dual behaviour i. e., both particle and wavelike properties. This means that just as the photon, electrons should. also have momentum as well as wavelength. de Broglie, from this analogy, gave the following relation between wavelength (λ) and momentum (p) of a material particle.
\(\lambda=\frac{h}{m v}=\frac{h}{p}\)

Heisenberg’s Uncertainty Principle
Werner Heisenberg a German physicist in 1927, stated uncertainty principle which is the consequence of dual behaviour of matter and radiation. It states that it is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron. Mathematically, it can be given as in equation,

∆x is the uncertainty in position and ∆p_x (or ∆v_x) is the uncertainty in momentum (or velocity) of the particle. If the position of the electron is known with high degree of accuracy (∆x is small), then the velocity of the electron will be uncertain ∆v_x is large]. On the other hand, if the velocity of the electron is known precisely ( ∆v_x is small), then the position of the electron will be uncertain (∆x will be large). Thus, if we carry out some physical measurements on the electron’s position or velocity, the outcome will always depict a fuzzy or blur picture.

Significance of Uncertainty Principle
Heisenberg Uncertainty Principle rules out existence of definite paths or trajectories of electrons and other similar particles. The trajectory of an object is determined by its location and velocity at various moments. If we know where a body is at a particular instant and if we also know its velocity and the forces acting on it at that instant, we can tell where the body would be sometime later. We, therefore, conclude that the position of an object and its velocity fix its trajectory. The effect of Heisenberg Uncertainty Principle is significant only for motion of microscopic objects and is negligible for that of macroscopic objects.

Reasons for the Failure of the Bohr Model
In Bohr model, an electron is regarded as a charged particle moving in well defined circular orbits about the nucleus. The wave character of the electron is not considered in Bohr model. Further, an orbit is a clearly defined path and this path can completely be defined only if both the position and the velocity of the electron are known exactly at the same time. This is not possible according to the Heisenberg uncertainty principle. Bohr.model of the hydrogen atom, therefore, not only ignores dual behaviour of matter but also contradicts Heisenberg uncertainty principle. There was no point in extending Bohr model to other atoms. In fact, an insight into the structure of the atom was needed which could account for wave-particle duality of matter and be consistent with Heisenberg uncertainty principle. This came with the advent of quantum mechanics.

Quantum Mechanical Model Of Atom
Quantum mechanics is a theoretical science that deals with the study of motions of microscopic objects such as electrons.

In quantum mechanical model of atom, the behaviour of an electron in an atom is described by an equation known as Schrodinger wave equation. Fora system, such as an atom or molecule whose energy does not change with time, the Schrodinger equation written as Hψ = Eψ where H is a mathematical operator, called Hamiltonian operator, E is the total energy and ψ is the amplitude of the electron wave called wave function.

Hydrogen Atom And The Schrodinger Equation
The wave function ψ as such has no physical significance. It only represents the amplitude of the electron wave. However ψ² may be considered as the probability density of the electron cloud. Thus, by determining ψ² at different distances from the nucleus, it is possible to trace out or identify a region of space around the nucleus where there is high probability of locating an electron with a specific energy.

According to the uncertainty principle, it is not possible to determine simultaneously the position and momentum of an electron in an atom precisely. So Bohr’s concept of well defined orbits for electron in an atom cannot hold good. Thus, in quantum mechanical mode, we speak of probability of finding an electron with a particular energy around the nucleus. There are certain regions around the nucleus where probability of finding the electron is high. Such regions are called orbitals. Thus an orbital may be defined as the region in space around the nucleus where there is maximum probability of finding an electron having a specific energy.

Orbitals and Quantum Numbers
Orbitals in an atom can be distinguished by their size, shape and orientation. An orbital of smaller size means there is more chance of finding the electron near the nucleus. Similarly, shape and orientation mean that there is more probability of finding the electron along certain directions than along others. Atomic orbitals are precisely distinguished by what are known as quantum numbers. Each orbital is designated by three quantum numbers labelled as n, l and m_l

The principal quantum number n’ is a positive integer with value of n= 1, 2, 3 ……………

The principal quantum number determines the size and to large extent the energy of the orbital.

The principal quantum number also identifies the shell. With the increase in the value of ‘n’, the number of allowed orbital increases and are given by ‘n²’ Ait the orbitals of a given value of ‘n’ constitute a single shell of atom and are represented by the following letters
n= 1 2 3 4 ………………
Shell = K LM N ………………

Size of an orbital increases with increase of principal quantum number ‘n’. Since energy of the orbital will increase with increase of n.

Azimuthal quantum number, ‘F is also known as orbital angular momentum or subsidiary quantum number. It defines the three-dimensional shape of the orbital. For a given value of n, l can have n values ranging from 0 to (n – 1), that is, for a given value of n, the possible value of l are: l = 0, 1, 2, ……….. (n – 1)

Each shell consists of one or more subshells or sub-levels. The number of subshells in a principal shell is equal to the value of n. For example h the first shell (n = 1), there is only one sub-shell which corresponds to l = 0. There are two sub-shells (l= 0, 1) in the second shell (n = 2), three l= 0, 1, 2) and so on. Each sub-shell is assigned an azimuths! quantum number (l). Sub-shells corresponding to different values of l are represented by the following symbols.
l : 0 1 2 3 4 5 …………….
Notation for sub-shell : s p d f g h …………….

Magnetic orbital quantum number. ‘m_l’ gives information about the spatial orientation of the or bital with respect to standard set of co-ordinate axis. For any sub-shell (defined by T value) 21+ 1 values of m,are possible and these values are given by:
m, = -l, -(l-1), (l-2)… 0, 1… (l-2), (l-1), l Thus for l = 0, the only permitted value of m,= 0, [2(0) + 1 = 1, one s orbital].

Electron spin ‘s’:
George Uhlenbeck and Samuel Goudsmit proposed the presence of the fourth quantum number known as the electron spin quantum number (m_s). Spin angular momentum of the electron — a vector quantity, can have two orientations relative to the chosen axis. These two orientations are distinguished by the spin quantum numbers ms which can take the values of +½ or -½. These are called the two spin states of the electron and are. normally represented by two arrows, ↑ (spin up) and ↓ (spin down). Two electrons that have different m_s values (one +½ and the other -½) are said to have opposite spins. An orbital cannot hold more than two electrons and these two electrons should have opposite spins.

Shapes of Atomic Orbitals
The orbital wave function or V for an electron in an atom has no physical meaning. It is simply a mathematical function of the coordinates of the electron.

According to the German physicist, Max Bom, the square of the wave function (i.e., ψ²) at a point gives the probability density of the electron at that point.

For 1 s orbital the probability density is maximum at the nucleus and it decreases sharply as we move away from it. The region where this probability I density function reduces to zero is called nodal surfaces or simply nodes. In general, it has been found that ns-orbital has (n – 1) nodes, that is, number of nodes increases with increase of principal quantum number n.

These probability density variation can be visualised . in terms of charge cloud diagrams.

Boundary surface diagrams of constant probability density for different orbitals give a fairly good representation of the shapes of the orbitals. In this representation, a boundary surface or contour surface is drawn in space for an orbital on which the value of probability density |ψ|² is constant. Boundary ‘ surface diagram for a s orbital is actually a sphere centred on the nucleus. In two dimensions, this sphere looks like a circle. It encloses a region in which probability of finding the electron is about 90%. The s-orbitals are spherically symmetric, that is, the probability of finding the electron at a given distance is equal in all the directions.

unlike s-orbitals, the boundary surface diagrams of p orbitals are not spherical. Instead, each p orbital consists of two sections called lobes that are on either side of the plane that passes through the nucleus. The probability density function is zero on the plane where the two lobes touch each other. The size, shape and energy of the three orbitals are identical. They differ, however, in the way the lobes are oriented. Since the lobes may be considered to lie along the x, y or z-axis, they are given the designations 2p_x, 2p_y, and 2p_z. It should be understood, however, that there is no simple relation between the values of m, (-1, 0 and+1) and the x, y and z directions. For our purpose, it is sufficient to remember that, because there are three possible values of m, there are, therefore, three p orbitals whose axes are mutually perpendicular. Like s orbitals, p orbitals increase in size and energy with increase in the principal quantum number

The number of nodes are given by (n -2), that is number of radial node is 1 for 3p orbital, two for 4p orbital and so on.

For l = 2, the orbital is known as d-orbital and the minimum value of principal quantum number (n) has to be 3 as the value of l cannot be greater than n-1. There are five m; values (-2, -1, 0, +1 and +2) for l = 2 and thus there are five d orbitals. The five d-orbitals are designated as d_xy, d_yz, d_xz, d_x²-y² and d_z². The shapes of the first fourd-orbitals are similarto each other, where as that of the fifth one, d_z², is different from others, but all five 3d orbitals are equivalent in energy. The d orbitals for which n is greater than 3 (4d, 5d…) also have shapes similar to 3d orbital, but differ in energy and size.

Besides the radial nodes (i.e., probability density function is zero), the probability density functions for the np and nd orbitals are zero at the plane (s), passing through the nucleus (origin). For example, in case of p_z orbital, xy-plane is a nodal plane, in case of d_xy orbital, there are two nodal planes passing through the origin and bisecting the xy plane containing z-axis. These are called angular nodes and number of angular nodes are given by T, i.e., one angular node for p orbitals, two angular nodes for cf orbitals and so on. The total number of nodes are given by (n-1), i.e., sum of I angular nodes and (n-l-1) radial nodes.

Energies Of Orbitals
The order of energy of orbitals in single electron sys-tem are given below:
1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f The orbitals having same energy are called degenerate.

Filling Of Orbitals In Atom
Aufbau principle: According to this principle in the ground state of an atom, an electron will occupy the orbital of lowest energy and orbitals are occupied by electrons in the order of increasing energy.

Pauli’s exclusiohn principle : Pauli’s exclusion principle states that ‘no two electrons in an atom can have the same values for all the four quantum numbers’

Since the electrons in an orbital must have the same n, I and m quantum numbers, if follows that an orbital can contain a maximum of two electrons provided their spin quantum numbers are different. This is an important consequence of Pauli’s exclusion principle which says that an orbital can have maximum two electrons and these must have opposite spins.

Hund’s rule of maximum multiplicity :
This rule states that electron pairing in orbitals of same energy will not take place until each available orbital of a given subshell is singly occupied (with parallel spin).
The rule can be illustrated by taking the example of carbon atom. The atomic number of carbon is 6 and its electronic configuration is 1s²2s²2p². The two electrons of the 2p subshell can be distributed in the following three ways.

According to Hund’s rule, the configuration in which the two unpaired electron occupying 2px, and 2py orbitals with parallel spin is the correct configuration of carbon.

Exceptional configurations of chromium and copper
The electronic configuration of Cr (atomic number 24) is expected to be [Ar] 4s² 3d⁴, but the actual configuration is [Ar] 4s¹ 3d⁵. Similarly, the actual configuration of Cu (At. No. 29) is [Ar] 4s¹ 3d¹⁰ instead of the expected configuration [Ar] 4s² 3d⁹.

This is because of the fact that exactly half filled or completely filled orbitals (i.e., d⁵, d¹⁰, f⁷, f¹⁴) have lower energy and hence have extra stability.

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Here HSSLive.Guru have given HSSLive Higher Secondary Kerala Plus One Chapter Wise Questions and Answers.

• Plus One Maths Chapter Wise Questions and Answers
• Plus One Physics Chapter Wise Questions and Answers
• Plus One Chemistry Chapter Wise Questions and Answers
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• Plus One Sociology Chapter Wise Questions and Answers
• Plus One Economics Chapter Wise Questions and Answers
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• Plus One Accountancy Chapter Wise Questions and Answers
• Plus One Computer Science Chapter Wise Questions and Answers
• Plus One Computer Application Chapter Wise Questions and Answers
• Plus One English Textbook Answers
• Plus One Hindi Textbook Answers

HSSLive Plus One Chapter Wise Previous Year Questions and Answers Kerala

Here HSSLive.Guru have given HSSLive Higher Secondary Kerala Plus One Chapter Wise Previous Year Important Questions and Answers.

• Plus One Maths Chapter Wise Previous Questions and Answers
• Plus One Physics Chapter Wise Previous Questions and Answers
• Plus One Chemistry Chapter Wise Previous Questions and Answers
• Plus One Botany Chapter Wise Previous Questions and Answers
• Plus One Zoology Chapter Wise Previous Questions and Answers
• Plus One Economics Chapter Wise Previous Questions and Answers
• Plus One Business Studies Chapter Wise Previous Questions and Answers
• Plus One Accountancy Chapter Wise Previous Questions and Answers
• Plus One Computer Science Chapter Wise Previous Questions and Answers
• Plus One Computer Application Chapter Wise Previous Questions and Answers