Question Set A

CBSE 11th Chemistry Chemical Bonding and Molecular Structure Model Questions Papers Study Material - QB365

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Questions:

Generic 3

5 x 3 = 15

Zinc can displace Cu from aqueous CuSO₄ solution but Ag cannot.

What is the wavelength for the electron accelerated by 1.0 x 10⁴ volts?

Name different blocks of elements in the periodic table. Give general electronic configuration of each block.

What is resonance? what are the guidelines for writing resonating structures. Out of the following four resonance structure for the CO₂ molecule, which are important for describing in the molecules and why?

What volume of 0.250 M HCl (aq) is required to react completely with 22.6 g of sodium carbonate according to the reaction ?
\({ Na }_{ 2 }{ CO }_{ 3 }(s)+2HCi(aq)\longrightarrow 2NaCl(aq)+{ H }_{ 2 }o(l)+{ CO }_{ 2 }(g)\)

Generic 2

5 x 2 = 10

Give reasons for the following:
\(\mathbf{H}_{2}^{+} \text {and } \mathbf{H}_{2}^{-}\) ions have same bond order but \(\mathbf{H}_{2}^{+}\) are more stable than \(\mathbf{H}_{2}^{-}\)

Calculate the number of moles of NaOH in 27 cm³ of 0.15 M NaOH solution.

The electronic energy of the ground state of hydrogen atom works out to be -1.312 x 10⁶J mol^-1. What change will occur in the position of electron in this atom if energy of 9.84 x 10⁵J mol^-1 is added to the hydrogen atom?

In the reactions given below, identify the species undergoing oxidation and reduction:
\( 3 \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{~s})+8 \mathrm{Al}(\mathrm{s}) \rightarrow 9 \mathrm{Fe} (\mathrm{s}) +4 \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{~s}) \)

10.

Name the species that will be isoelectronic with the following atoms or ions
(i) Na
(ii) Cl^-
(iii) Ca²⁺
(iv) Rb⁺

Generic 5

2 x 5 = 10

11.

(a) Explain the formation of ionic bond with two examples.
(b) Discuss the conditions which favour the formation of ionic bond

12.

Balance the following equations by the oxidation number method.
\({ MnO }_{ 2 }+{ C }_{ 2 }{ O }_{ 4 }^{ 2- }\longrightarrow { Mn }^{ 2+ }+{ CO }_{ 2 }\)

Multiple Choice Question

10 x 1 = 10

13.

Which of the following order of energies of molecular orbitals of N₂ is correct?

(a)

\( \left(\pi 2 p_{y}\right)<\left(\sigma 2 p_{z}\right)<\left(\pi^{*} 2 p_{x}\right) \approx\left(\pi^{*} 2 p_{y}\right) \)

(b)

\(\left(\pi 2 p_{y}\right)>\left(\sigma 2 p_{z}\right)>\left(\pi^{*} 2 p_{x}\right) \approx\left(\pi^{*} 2 p_{y}\right) \)

(c)

\(\left(\pi 2 p_{y}\right)<\left(\sigma 2 p_{z}\right)>\left(\pi^{*} 2 p_{x}\right) \approx\left(\pi^{*} 2 p_{y}\right) \)

(d)

\(\left(\pi 2 p_{y}\right)>\left(\sigma 2 p_{z}\right)<\left(\pi^{*} 2 p_{x}\right) \approx\left(\pi^{*} 2 p_{y}\right)\)

14.

Ionic radii vary in ______.

(a)

inverse proportion to the effective nuclear charge.

(b)

inverse proportion to the square of effective nuclear charge.

(c)

direct proportion to the screening effect.

(d)

direct proportion to the square of screening effect.

15.

1 g of M₂CO₃ on treatment with excess HCl produces 0.01186 moles of CO₂. The molar mass of M₂CO₃ in g mol^-1is _______.

(a)

1186

(b)

84.3

(c)

118.6

(d)

11.86

16.

If a reaction is carried out in acidic medium then which is used-to balance the equation?

(a)

H⁺ ions

(b)

OH^- ions

(c)

H^- ions

(d)

O^2- ions

17.

Lewis dot structure of CO, \(\mathrm{NO}_{2}^{-}\) and \(\mathrm{CO}_{3}^{2-}\)are I, II and III respectively

Which of the above structure(s) is/are wrong?

(a)

Only I

(b)

Only II

(c)

Only III

(d)

All of these

18.

The given reaction, X(g) + e^- ⟶ X^-(g) is an example of ______.

(a)

electron gain enthalpy

(b)

electron loss enthalpy

(c)

ionisation enthalpy

(d)

Both (a) and (c)

19.

Which of the following options does not represent ground state electronic configuration of an atom?

(a)

\(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{8} 4 s^{2}\)

(b)

\(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{9} 4 s^{2}\)

(c)

\(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{1}\)

(d)

\(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{5} 4 s^{1}\)

20.

An electron is moving in Bohr's fourth orbit. Its de-Broglie wavelength is λ Calculate the circumference of the fourth orbit.

(a)

2 λ

(b)

4 λ

(c)

4/λ

(d)

2/λ

21.

The empirical formula and molecular mass of a compound are CH₂O and 180 g respectively. What will be the molecular formula of the compound?

(a)

C₉H₁₈O₉

(b)

CH₂O

(c)

C₆H₁₂O₆

(d)

C₂H₄O₂

22.

Which of the following is most powerful oxidizing agent in the following

(a)

O₃

(b)

KMnO₄

(c)

H₂O₂

(d)

K₂Cr₂O₇

Assertion and reason

5 x 1 = 5

23.

Assertion: The ratio by volume of gaseous reactants and products is in agreement with their molar ratio.
Reason: Volume of a gas is inversely proportional to the number of moles of a gas.
Codes:
(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.
(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion
(c) Assertion is correct, reason is incorrect
(d) Assertion is incorrect, reason is correct.

24.

Assertion : Helium is placed in group 18 along with p-block elements.
Reason : It shows properties similar to p-block elements.
Codes:
(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.
(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion
(c) Assertion is correct, reason is incorrect
(d) Assertion is incorrect, reason is correct.

25.

Assertion: The decomposition of hydrogen peroxide to form water and oxygen is an example of disproportionation reaction.
Reason: The oxygen of peroxide is in -1 oxidation state and it is converted to zero oxidation state in O₂ and -2 oxidation state in H₂O.
Codes:
(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) Assertion is true but Reason is false.
(d) Both Assertion and Reason are false.

26.

Assertion: Sodium chloride formed by the action of chlorine gas on sodium metal is a stable compound.
Reason: This is because sodium and chloride ions acquire octet in sodium chloride formation.
Codes:
(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) Assertion is true but Reason is false.
(d) Both Assertion and Reason are false.

27.

Assertion: All isotopes of a given element show the same type of chemical behaviour.
Reason: The chemical properties of an atom are controlled by the number of electrons in the atom.
Codes:
(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) Assertion is true but Reason is false.
(d) Both Assertion and Reason are false.

Case Study Questions

5 x 4 = 20

28.

Lewis dot structure of HCN molecule is given below. Answer questions on the basis of this structure.
1. How many electrons are involved in the sigma bond between the carbon atom and the nitrogen atom?
(A) 1
(B) 0
(C) 2
(D) 6
2. What types of bonds exist between the C atom and the N atom in HCN?
(A) 3 sigma bonds and zero pi bonds
(B) 3 pi bonds and zero sigma bonds
(C) 2 pi bonds and one sigma bond
(D) 1 pi bond and two sigma bonds
3. How many electrons does the N atom contribute to the pi bonds?
(A) 3
(B) 2
(C) 1
(D) 5

29.

A teacher displayed the modern periodic table in the classroom. She explained how elements in the same group have similar chemical properties due to the same number of valence electrons, while elements in a period show trends in atomic radius, ionisation enthalpy, and electronegativity. The class was asked to compare Na, Mg, and Al, and also F, Cl, and Br in terms of these trends.
1. Which of the following shows the correct trend in atomic size across a period (Na → Al)?
(a) Increases
(b) Decreases
(c) Remains the same
(d) First increases then decreases
2. Among F, Cl, and Br, which has the highest electron affinity?
(a) F
(b) Cl
(c) Br
(d) All are equal
3. Which of the following elements has the highest ionisation enthalpy?
(a) Na
(b) Mg
(c) Al
(d) F

30.

In a physics lab, students observe two different experiments. One is J.J. Thomson’s cathode ray experiment, which led to the discovery of the electron, and the other is Goldstein’s canal ray experiment, which led to the identification of protons. Both experiments showed that atoms are not indivisible, as believed by Dalton.
1.What did Thomson conclude about the nature of cathode rays?
(a) They are neutral particles
(b) They are positively charged
(c) They are negatively charged
(d) They do not deflect in electric fields
2.Goldstein’s canal rays led to the discovery of:
(a) Electrons
(b) Neutrons
(c) Protons
(d) Photons
3.Which model of the atom was first to describe the existence of both positive and negative particles within the atom?
(a) Dalton’s Model
(b) Thomson’s Plum Pudding Model
(c) Rutherford’s Model
(d) Bohr’s Model

31.

The attractive force which holds the two atoms together is called chemical bond. Covalent bond is formed by equal sharing of electrons. Coordinate bond is formed by unequal sharing of electrons. Ionic bond is formed by transfer of electrons from one atom to another. Octet rule, although very useful but it is not universally applicable. According to valence bond theory, covalent bond is formed by overlapping of half filled atomic orbitals resulting in lowering of energy and more stability. Bond order is the number of bonds between atoms in a molecule. Higher the bond order, more will be stability and bond dissociation enthalpy but smaller bond length. Polarity of covalent bond depends upon difference in electronegativity. Covalent character of bond depends upon polarising power, smaller cation and bigger anions have higher polarising power. VSEPR theory helps to predict shapes of molecules.
(a) Write the, electron dot structure of N₂O.
(b) What are ions present in CsI₃?
(c) Out of CN⁺, CN^-, NO, which has highest bond order?
(d) What is correct order of repulsion bp - bp, lp - lp and lp - bp?
(e) Draw the structure of XeOF₄ on the basis of VSEPR theory.
(f) Which out of B₂ ,CO, \(\mathrm{O}_{2}^{2-}\) and NO⁺ are paramagnetic and why?

32.

Chemistry play an important role in human needs for food, health care products and improving life. Cis platin and taxol are used in chemotherapy, AZT (Azidothymidine) is used for AIDS. SI units are international units of measurement. Matter is classified into elements, compounds and mixtures, which can be homogeneous as well as heterogeneous. A mixture can be separated by physical methods, compounds can be separated by chemical methods only. Atomic mass is average of masses of isotopes depending upon their natural abundance. Empirical formula is calculated with the help of percentage composition of elements in a compound and molecular mass helps to calculate molecular formula. A chemical equation must be balanced so as to follow laws of chemical combination.
(a) Express 2.54 mm into S.I units.
(b) Out of milk, diamond, air, petrol which is pure substance?
(c) Balance the equation: NO₂ + H₂ O \(\rightarrow\) HNO₃ + NO
(d) What is percentage of Na in Na₂CO₃ ? (Na = 23u, C = 12, O = 16u)
(e) \({ }_{17}^{35} \mathrm{Cl} \text { and }{ }_{17}^{37} \mathrm{Cl}\) are in ratio of 3 : 1 in nature. What is atomic mass of CI?
(f) What is empirical formula of C₆H₁₂O₆?
(g) Chlorophyll contains 2.68% magnesium atoms. Calculate mass of magnesium atoms in 2 g of chlorophyll.

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Answers Key:

Generic 3

5 x 3 = 15

Zinc is more reactive than Cu but Ag is less reactive than Cu.

Step I. Calculation of the velocity of electron Energy (kinetic energy) of electron =1.0 x 10⁴ volts.
= 1.0 x 10⁴ x 1.6 x 10^-19 J = 1.6 x 10^-15 J
= 1.6 x 10^-15 kg m² S^-2
or 1/2 mv²= 1.6 x 10^-15 kg m² s^-2
or v =\(\left( \frac { 2\times 1.6\times 10^{ -15 }kgm^{ 2 }s^{ -2 } }{ (9.1x10^{ -31 }kg) } \right) ^{ 1/2 }\)= 5.93 x 10⁷ ms^-1
Step II. Calculation of the wavelength of electron According to de Broglie equation,
\(\lambda =\frac { h }{ mv } ;\lambda =\frac { (6.626\times10^{ -34 }kgm^{ 2 }s^{ -1 }) }{ (9.1\times10^{ -31 }kg)\times(5.93\times10^{ 7 }ms^{ -1 }) } \) = 1.22\(\times \)10^-11m.

Elements in the long form of the periodic table have been divided into four blocks i.e., s, p, d and f This division is based upon the name of the orbital which receives the last electron. General electronic configuration of
s-block elements: ns^{1- 2} where n = 2 - 7
p-block elements: ns² np^{1 - 6} where n = 2 - 6
d-block elements: (n -1) d^1-10 ns^{0 - 2} where n = 4 - 7
f-block elements: (n - 2) f^{0 -14} (n -1)d^0-1 ns² where n = 6 - 7

I and II are more important as the I is non-polar and (II) has more number of covalent bonds and negative charge on more electronegative atom and positive charge on more eletropositive element.

1.7 L

Generic 2

5 x 2 = 10

It is because electron present in anti-bonding orbital in H₂ destabilize the molecular ion slightly more than bonding electron stabilizes due to higher energy.

\(\mathbf{M}=n \times \frac{1000}{\text { Volume of solution in } \mathrm{cm}^{3}}\)
\(\Rightarrow \ 0.15=n \times \frac{1000}{27 \mathrm{~cm}^{3}}\)
\(\Rightarrow\) n = 0.00405 moles

\(\Delta \mathrm{E}=-1.312 \times 10^{6} \mathrm{~J}\left(\frac{1}{n_{2}^{2}}-\frac{1}{n_{1}^{2}}\right)\)
9.84 x 10⁵J mol^-1 = +1.312 x 10⁶ J \(\left(\frac{1}{1}-\frac{1}{n_{2}^{2}}\right)\)
\(\Rightarrow \ \frac{9.84}{13.12}=1-\frac{1}{n_{2}^{2}} \Rightarrow \frac{3}{4}=1-\frac{1}{n_{2}^{2}}\)
\(\Rightarrow \ \frac{1}{n_{2}^{2}}=\frac{1}{4} \Rightarrow n_{2}^{2}=4 \Rightarrow n_{2}=2\)

Aluminium is oxidised because oxygen is added to it. Ferrous ferric oxide (Fe₃O₄) is reduced because oxygen has been removed from it.

10.

Isoelectronic species are those which have same number of electrons.
(i) Na has 10 electrons.Therefore, the species
N^3-, O^2-, F^-, Mg²⁺, Al³⁺ etc., each of which has also 10 electrons and hence, isoelectronic with it
(ii) Cl^- has 18 electrons. Therefore, the species P^3-, S^2-, Ar, K⁺ and Ca²⁺, each one of which contains 18 electrons and hence , isoelectronic with it.
(iii) Ca²⁺has 18 electrons. Therefore, the species P^3-, S^2-, Ar and K⁺, each of which also contains 18 electrons and hence, isoelectronic with it.
(iv) Rb⁺has 36 electrons. Therefore, the species Br^-, Kr or Sr²⁺ each of which also has 36 electrons and hence, isoelectronic with it.

Generic 5

2 x 5 = 10

11.

(a) An ionic or electrovalent bond is formed by the complete transferrence of one or more electrons from one atom to another.
Examples:
(i) Formation of (NaCl)
\(\longrightarrow\)\(\left[ \underset { 2,8 }{ Na } ^{ + } \right] \quad \left[ \overset { .. }{ :\underset { ..\\ 2,8,8 }{ CI } } : \right] \) or NaCI
Formation of (CaF2)
\(\longrightarrow\)\(\left[ \underset { 2,8,8 }{ Ca } \right] ^{ 2+ }\quad \left[ \overset { .. }{ :\underset { .. }{ F } : } \right] ^{ \\ - }\\ \quad \quad \quad \quad \quad \quad \left[ \overset { .. }{ :\underset { \quad ..\\ (2,8) }{ F } : } \right] ^{ - }\) or CaF₂
(b) Conditions favourable for the formation of ionic bond:
(i) Lesser the ionization enthalpy, easier will be the removal of an electron i.e., formation of a positive ion and hence greater the chances of formation of ionic bond.
(ii) Higher is the electron affinity, more is the energy released and stabler will be the negative ion produced. Consequently, the probability of formation of ionic bond will be enhanced

12.

Increase and decrease in oxidation number is already balanced. Add 4H⁺ towards LHS of the equation to balance charge.
\({ MnO }_{ 2 }+{ C }_{ 2 }{ O }_{ 4 }^{ 2- }+{ 4H }^{ + }\longrightarrow { Mn }^{ 2+ }+{ 2CO }_{ 2 }\)
Add 2 H₂O towards RHS of the equation to balance H-atoms
\({ MnO }_{ 2 }+{ C }_{ 2 }{ O }_{ 4 }^{ 2- }+{ 4H }^{ + }\longrightarrow { Mn }^{ 2+ }+{ 2CO }_{ 2 }+{ 2H }_{ 2 }O\)
This represents a balanced redox reaction.

Multiple Choice Question

10 x 1 = 10

13.

(a)
\( \left(\pi 2 p_{y}\right)<\left(\sigma 2 p_{z}\right)<\left(\pi^{*} 2 p_{x}\right) \approx\left(\pi^{*} 2 p_{y}\right) \)

14.

(a)
inverse proportion to the effective nuclear charge.

15.

(b)
84.3

16.

(a)
H⁺ ions

17.

(a)
Only I

18.

(a)
electron gain enthalpy

19.

(b)
\(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{9} 4 s^{2}\)

20.

(b)
4 λ

21.

22.

(a)
O₃

Assertion and reason

5 x 1 = 5

23.

24.

(c) Assertion is correct, reason is incorrect
Explanation:
He (1s2) should be placed along with s-block elements because of its electronic configuration but it has a completely filled valence shell and as a result it exhibits properties of noble gases, thus it is placed along with noble gases (ns2, np6).

25.

(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

26.

(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

27.

(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

Case Study Questions

5 x 4 = 20

28.

1.(C) 2
2.(C) 2 pi bonds and one sigma bond
3.(B) 2

29.

1. (b) Decreases
Explanation: Atomic size decreases across a period due to increased nuclear charge, pulling electrons closer.
2.(b) Cl
Explanation: Although F is more electronegative, electron-electron repulsion in its small size reduces its affinity. Cl has highest electron affinity among all elements.
3.(d) F
Explanation: Ionisation enthalpy increases across a period and decreases down a group. Fluorine has the smallest size and strongest nuclear hold on electrons.

30.

1. (c) They are negatively charged
Explanation:
Cathode rays (electrons) are deflected by electric and magnetic fields, indicating a negative charge.
2. (c) Protons
Explanation:
Canal rays (positive rays) were composed of positively charged particles → Protons.
3. (b) Thomson’s Plum Pudding Model
Explanation:
Thomson proposed a model with a positively charged sphere and embedded electrons (“plum pudding”).

31.

(a)

(b) Cs⁺ and \(\mathrm{I}_{3}^{-}\)
(c) \(\mathrm{CN}(14): \sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \pi^{2} p_{x}^{2} \pi^{2} p_{y}^{2} \sigma 2 p_{z}^{2}\)
\(\mathrm{BO}=\frac{1}{2}(10-4)=\frac{6}{2}=3\)
(d) lp - lp> lp - bp > bp - bp [bp is bond pair, lp is lone pair]
(e)

(f) \(\mathrm{B}_{2}(10): \sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \pi^{2} p_{x^{1}} \pi^{2} p_{y^{1}}\) is paramagnetic due to presence of two unpaired electrons.

32.

(a) 2.54 x 10^{- 3} m.
(b) Diamond.
(c) 3NO₂ + H₂O \(\rightarrow\) 2HNO₃ + NO
(d) \(\% \text { of } \mathrm{Na}=\frac{\text { Total mass of } \mathrm{Na}}{\text { Molar mass }} \times 100\)
\(=\frac{46}{106} \times 100=43.39 \%\)
(e) \(\frac{3 \times 35+1 \times 37}{4}=35.5\)
(f) CH₂O
(g) Mass of magnesium atoms \(=2 \times \frac{2.68}{100}\)
\(=\frac{5.36}{100}=0.0536 \mathrm{~g}\)