CBSE Class 11 Chemistry Chemical Bonding and Molecular Structure MCQs Set 07

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. In such type of compounds, participation of d-orbitals in bonding increases the number of valence electrons around the central atom to more than eight (10 around P in PF₅ and 12 around S in SF₆). Thus, it is termed as expanded octet.
2. P in PF₅ is sp³d and S in SF₆ is sp³d² hybridised, which is also true.
3. Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion (the octet is expanded because of the number of bonds formed using vacant d-orbitals).

Teacher's Note:
1. Expanded octets are possible only for elements of period 3 and beyond, which have vacant d-orbitals.
2. Hybridisation describes the geometry; it does not by itself explain why the octet expands.

Answer: (c) A is true, but R is false.

Explanation:
1. Octet theory does not explain the shape of molecules, so Assertion is correct.
2. Octet theory is totally silent about the energy of a molecule and about its relative stability, so Reason is incorrect.
3. ∴ Assertion is correct but Reason is incorrect.

Teacher's Note:
1. Limitations of the octet rule: no shapes, no relative stability or energy of molecules.
2. VSEPR and VBT were developed to explain shapes; MO theory explains stability.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Na loses one electron to form Na⁺ (2, 8) and Cl gains one electron to form Cl⁻ (2, 8, 8). Both ions acquire an octet (noble gas configuration), so NaCl is a stable compound.
2. Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Teacher's Note:
1. In the Kossel approach, ions are stable when they attain noble gas configuration.
2. The large lattice enthalpy released also makes NaCl formation energetically favourable.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Each ion, because of its uniformly distributed (spherically symmetrical) electric field, attracts oppositely charged ions equally in all directions, so the ionic bond is non-directional.
2. Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Teacher's Note:
1. Covalent bonds are directional because they depend on orbital overlap; ionic bonds are not.
2. This is why ionic compounds do not show structural (geometrical) isomerism.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Lattice energy ∝ \( \frac{\text{charge of ions}}{\text{size of ions}} \).
2. In LiF, NaF and KF the anion and the charges are the same; only the cation size changes. As the size of the alkali metal ion increases from Li⁺ to K⁺, the interionic distance increases and the lattice energy decreases.
3. So, LiF > NaF > KF. Both Assertion and Reason are correct, and Reason is the correct explanation of Assertion.

Teacher's Note:
1. With the same anion and charges, the smallest cation gives the highest lattice energy.
2. Lattice energy values: LiF ≈ 1037, NaF ≈ 926 and KF ≈ 821 kJ mol⁻¹.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Lone pair-lone pair repulsion > lone pair-bond pair repulsion > bond pair-bond pair repulsion.
2. In NH₃ there are three bond pairs and one lone pair. The three N—H bond pairs are pushed closer because of the lone pair-bond pair repulsion, and the H—N—H bond angle gets reduced from 109° 28′ (the tetrahedral angle) to 107°.
3. Both Assertion and Reason are correct and Reason is correct explanation of Assertion.

Teacher's Note:
1. lp–bp repulsion is stronger than bp–bp repulsion, so lone pairs always compress bond angles.
2. H₂O, with two lone pairs, has an even smaller angle of 104.5°.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Bond dissociation energy is the energy required to break one mole of bonds of a particular type so as to separate them into gaseous atoms. Bond energy is the average value of dissociation energies of the same type of bonds present in the molecule. For a diatomic molecule there is only one bond, so the two values are identical.
2. For polyatomic molecules, average bond energy is taken (e.g. bond energy of C—H bond in methane is 99.2 kcal/mol).
3. A bond with greater dissociation energy is stronger and less reactive.
4. Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

Teacher's Note:
1. Bond energy = average of successive dissociation energies; they coincide only when there is one bond.
2. Stronger bonds (higher dissociation energy) are harder to break, so they are less reactive.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. BF₃ is sp²⁻hybridised and trigonal planar. Dipole moment is a vector quantity. The three bond moments give a net sum of zero, as the resultant of any two is equal and opposite to the third.
2. Resultant dipole moment = 0 :
3. B—F bonds are polar because F is highly electronegative, which is also true.
4. Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion (zero dipole moment is due to the symmetrical shape).

Teacher's Note:
1. A molecule with polar bonds can still be non-polar if its shape is symmetrical.
2. BF₃, CO₂, CH₄ and CCl₄ all have polar bonds but zero net dipole moment.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. A small cation has high polarising power and a large anion is easily polarised. Greater the polarisation of the anion by the cation, greater is the covalent character.
2. Assertion and Reason both are correct and Reason is the correct explanation of Assertion.

Teacher's Note:
1. Polarising power of cation ∝ charge/size; polarisability of anion increases with its size.
2. That is why LiI is more covalent than LiF, and AlCl₃ is more covalent than NaCl.

Answer: (d) A is false, but R is true.

Explanation:
1. NH₃ molecule is sp³⁻hybridised because out of four electron pairs, three are bonding pairs and one is a lone pair. Its electron pair geometry is tetrahedral but its shape is pyramidal (trigonal pyramidal).
2. Thus, Assertion is incorrect but Reason is correct.

Teacher's Note:
1. Shape describes only the positions of atoms; lone pairs are not counted in the shape name.
2. sp³ with one lone pair gives trigonal pyramidal; with two lone pairs gives bent (as in H₂O).

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Assertion and Reason both are correct and Reason is the correct explanation of Assertion.
2. In H₂O, two lone pairs on O cause greater repulsion (lp-lp > lp-bp), reducing the bond angle to 104.5°, whereas NH₃ with one lone pair has a bond angle of 107°.

Teacher's Note:
1. More lone pairs on the central atom mean a smaller bond angle for the same hybridisation.
2. CH₄ (0 lp) 109.5° > NH₃ (1 lp) 107° > H₂O (2 lp) 104.5°.

Answer: (d) A is false, but R is true.

Explanation:
1. Hybridisation of B in BCl₃ = \( \frac{1}{2}(V + Y - C + A) = \frac{1}{2}(3 + 3 - 0 + 0) = 3 \) ⇒ sp²
2. So, boron in BCl₃ is sp²⁻hybridised, not sp³. Assertion is incorrect.
3. An sp²⁻hybridised atom with no lone pair has trigonal planar geometry (bond angle 120°), so Reason is correct.
4. Thus, Assertion is incorrect but Reason is correct.

Teacher's Note:
1. Boron has only 3 valence electrons, forming 3 bond pairs and no lone pair in BCl₃.
2. BCl₃ becomes sp³ only when it accepts a lone pair, e.g. in BCl₃·NH₃ or BF₄⁻.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Electronic configuration of F₂ molecule is
2. σ1s² σ*1s² σ2s² σ*2s² σ2p_z² π2p_x² = π2p_y² π*2p_x² = π*2p_y²
3. Bond order = \( \frac{\text{No. of electrons in bonding MOs} - \text{No. of electrons in antibonding MOs}}{2} \) = \( \frac{10 - 8}{2} \) = 1
4. Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Teacher's Note:
1. A difference of 2 between bonding and antibonding electrons always gives bond order 1.
2. F₂ is diamagnetic since all its electrons are paired.

Answer: (c) A is true, but R is false.

Explanation:
1. The bond enthalpies of the two O—H bonds in H—O—H are not equal (502 kJ mol⁻¹ for the first and 427 kJ mol⁻¹ for the second). So, Assertion is correct.
2. This is because the electronic environment around O is not the same after breakage of one O—H bond (OH radical is left behind). So, Reason as stated is incorrect.
3. Thus, Assertion is correct but Reason is incorrect.

Teacher's Note:
1. Successive bond dissociation enthalpies differ because the remaining fragment has a different environment.
2. That is why mean bond enthalpy (464.5 kJ mol⁻¹ for O—H) is used for polyatomic molecules.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Water is an excellent solvent because it has a high value of dielectric constant.
2. Due to the high value of dielectric constant, the electrostatic forces of attraction between the ions decrease, and these ions get separated and ultimately get solvated by the solvent molecules.
3. H-bonding is also present in water molecules.
4. Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

Teacher's Note:
1. The dielectric constant of water (about 78) reduces interionic attraction about 78 times.
2. H-bonding explains the high boiling point of water, not its solvent power for ionic compounds.

Answer: (d) H₂²⁻

Explanation:
1. If the bond order of any species is zero, that cannot exist.
2. H₂²⁻ = σ1s², σ*1s²; bond order = \( \frac{2 - 2}{2} \) = 0
3. He₂²⁺ (σ1s²) has bond order 1, while He₂⁺ and H₂⁻ have bond order ½, so they can exist.
4. So, H₂²⁻ will not be an available molecule.

Teacher's Note:
1. Any species with equal bonding and antibonding electrons has bond order zero and does not exist.
2. H₂²⁻ is isoelectronic with He₂ (4 electrons), which also does not exist.

Answer: (b) NaClO₄

Explanation:
1. NaClO₄ contains two parts: it has ionic bonding between Na⁺ and ClO₄⁻ ions.
2. In ClO₄⁻ (perchlorate ion), Cl is sp³⁻hybridised and the Cl—O bonds are covalent.
3. NaCl is purely ionic, while PCl₃ and POCl₃ are purely covalent.

Teacher's Note:
1. Salts of polyatomic ions (NaClO₄, KNO₃, NH₄Cl) contain both ionic and covalent bonds.
2. The bonds inside the polyatomic ion are covalent; the attraction between the ions is ionic.

Answer: (a) −0.75, 1.25

Explanation:
1. The formal charge on each O-atom in PO₄³⁻ = \( -\frac{3}{4} \) = −0.75
2. P—O bond order in PO₄³⁻ = \( \frac{\text{total number of bonds between atoms}}{\text{total number of resonating structures}} \) = \( \frac{5}{4} \) = 1.25
3. (One P=O double bond and three P—O single bonds give 5 bonds shared by 4 equivalent P—O links.)

Teacher's Note:
1. Average formal charge = ionic charge ÷ equivalent atoms; average bond order = total bonds ÷ equivalent links.
2. Compare with NO₃⁻ : formal charge −1/3 on each O and N—O bond order 4/3.

Answer: (b) The lowering of potential energy of the system as the two H-atoms come near to each other

Explanation:
1. On the basis of VB theory, the formation of H₂ molecule from two H-atoms involves lowering of potential energy of the system as the two H-atoms come near to each other.
2. This energy will be minimum at the equilibrium internuclear distance. If the nuclei are brought still closer, repulsion increases and the energy rises.

Teacher's Note:
1. The 1s orbitals that overlap are half-filled, not vacant.
2. Energy is minimum (not maximum) at the equilibrium bond length.

Answer: (c) acetophenone

Explanation:
1. trans-2-Butene is symmetrical (μ = 0) and 1,3-dimethylbenzene has a very small dipole moment.
2. Ethanol has a dipole moment of about 1.7 D.
3. The dipole moment of acetophenone is about 3 Debye, the highest amongst the given compounds, as it has a strong electron withdrawing carbonyl (C=O) group.

Teacher's Note:
1. The polar C=O group gives carbonyl compounds large dipole moments.
2. trans isomers of symmetrical alkenes have zero dipole moment.

Answer: (c) BF₃

Explanation:
1. In BF₃, the dipole moment is zero although the B—F bonds are oriented at an angle of 120° to one another; the three bond moments give a net sum of zero as the resultant of any two is equal and opposite to the third.
2. Structure of BF₃ molecule is as follows :
3. HF, H₂O (bent) and CHCl₃ have net dipole moments.

Teacher's Note:
1. Symmetrical shapes (linear AB₂, trigonal planar AB₃, tetrahedral AB₄) cancel bond dipoles.
2. CHCl₃ is polar because one C—H bond differs from the three C—Cl bonds.

Answer: (d) PCl₃

Explanation:
1. PCl₃ contains three bond pairs and one lone pair around the central atom.
2. H₂O and NH₂⁻ have 2 bp + 2 lp, while BF₃ has 3 bp + 0 lp.

Teacher's Note:
1. Group 15 central atoms with three single bonds (NH₃, PCl₃, PH₃) have 3 bp + 1 lp.
2. NH₂⁻ is isoelectronic with H₂O and has two lone pairs.

Answer: (c) Repulsive effect

Explanation:
1. Repulsive effects result in deviations from idealised shapes and alterations in bond angles in the molecules. Presence of lone pair of electrons causes repulsion between lone pair-bond pair.
2. In H—O—H, two lone pairs of oxygen repel the two bond pairs of —OH.
3. In NH₃, one lone pair of nitrogen repels the three bond pairs of N—H.

Teacher's Note:
1. Both H₂O and NH₃ are sp³; the deviation from 109.5° comes from lone pair repulsion, not hybridisation.
2. Greater number of lone pairs means greater deviation (H₂O 104.5°, NH₃ 107°).

Answer: (d) BCl₃

Explanation:
1. BCl₃ is sp²⁻hybridised and the central atom does not have any lone pair of electrons. So, it is trigonal planar with bond angles of 120°.
2. Structure of BCl₃ :
3. PH₃ and NCl₃ are pyramidal (angles less than 109.5°) and ClF₃ is T-shaped (angles close to 90°).

Teacher's Note:
1. 120° bond angles belong to sp² central atoms with no lone pair (BF₃, BCl₃, SO₃).
2. PH₃ has a bond angle of only about 93.5°.

Answer: (a) Structure (a)

Explanation:
1. Structure (a) is the most stable geometry because both the lone pairs are present at equatorial positions.
2. Due to this, repulsion is minimum in the molecule as compared to the other structures, where a lone pair is at an axial position.
3. This gives ClF₃ its T-shape.

Teacher's Note:
1. In a trigonal bipyramid, lone pairs always prefer the roomier equatorial positions.
2. The same rule explains the see-saw shape of SF₄ and the linear shape of XeF₂.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Formal charge on atom = [total valence electrons in free atom] − [total non-bonding (lone pair) electrons] − \( \frac{1}{2} \) [total shared electrons]
2. The net charge belongs to the polyatomic ion as a whole; formal charges are only used to keep track of how the valence electrons are assigned to individual atoms.
3. Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Teacher's Note:
1. The sum of formal charges of all atoms equals the net charge of the ion.
2. Formal charge is a bookkeeping tool; real charge is spread over the whole ion.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Lewis structure of O₃ :
2. Formal charge on O₁ = 6 − 2 − \( \frac{1}{2} \)(6) = +1
3. Formal charge on O₂ = 6 − 4 − \( \frac{1}{2} \)(4) = 0
4. Formal charge on O₃ = 6 − 6 − \( \frac{1}{2} \)(2) = −1
5. Hence, correct representation of O₃ is
6. Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Teacher's Note:
1. In O₃, the central O carries +1 and one terminal O carries −1, so the molecule is overall neutral.
2. The O—O bond order in O₃ is 1.5 due to resonance between two such structures.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Formal charge does not indicate real charge separation within the molecule; it is based on a pure covalent view with electrons shared equally.
2. Formal charge helps in the selection of the lowest energy structure from a number of possible structures for a given species (generally the one with the smallest formal charges).
3. Hence, both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.

Teacher's Note:
1. The lowest energy Lewis structure is usually the one with the smallest formal charges on the atoms.
2. Real charge distribution depends on electronegativity, which formal charge ignores.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Due to the presence of vacant d-orbitals on S-atom, it can expand its octet as in the case of H₂SO₄ (12 electrons around S).
2. Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Teacher's Note:
1. Period 3 elements (P, S, Cl) can show expanded octets using vacant 3d-orbitals.
2. Period 2 elements such as N and O can never exceed an octet because they have no d-orbitals.

Answer: (c) A is true, but R is false.

Explanation:
1. Sulphur forms many compounds which obey the octet rule (e.g. H₂S, SCl₂).
2. However, sulphur has vacant 3d-orbitals, so it can also expand its octet (as in SF₆ and H₂SO₄). Hence, the Reason is incorrect.
3. Hence, Assertion is correct but Reason is incorrect.

Teacher's Note:
1. Having d-orbitals allows an expanded octet but does not force it.
2. H₂S and SCl₂ obey the octet rule, while SF₄ and SF₆ do not.

Answer: (b) μ_NH3 > μ_NF3

Explanation:
1. The lone pair of nitrogen opposes the resultant dipole moment of the N—F bonds in NF₃, while it adds to the dipole moment of the N—H bonds in NH₃.
2. So, μ_NH3 > μ_NF3 :
3. Dipole moment of NH₃ = 4.90 × 10⁻³⁰ C m (1.47 D) and of NF₃ = 0.80 × 10⁻³⁰ C m (0.24 D).

Teacher's Note:
1. In NH₃ the bond dipoles point towards N, the same direction as the lone pair moment.
2. In NF₃ the bond dipoles point towards F, opposite to the lone pair moment.

Answer: (a) 0.25

Explanation:
1. Partial charge = \( \frac{\text{Dipole moment}}{\text{Bond distance}} \) = \( \frac{1.2 \times 10^{-18}\ \text{esu cm}}{1 \times 10^{-8}\ \text{cm}} \) = 1.2 × 10⁻¹⁰ esu
2. Fractional charge = \( \frac{1.2 \times 10^{-10}}{4.8 \times 10^{-10}} \) = 0.25
3. (4.8 × 10⁻¹⁰ esu is the charge of one electron.)

Teacher's Note:
1. Convert debye to esu cm first: 1 D = 10⁻¹⁸ esu cm.
2. Fractional charge = partial charge ÷ electronic charge (4.8 × 10⁻¹⁰ esu).

Answer: (c) H₂O

Explanation:
1. Dipole moments : BeF₂ : μ = 0; BF₃ : μ = 0; H₂O : μ = 1.84 D
2. In linear BeF₂ and trigonal planar BF₃, the bond moments cancel. In bent H₂O, the O—H bond moments do not cancel, so H₂O has the highest dipole moment.

Teacher's Note:
1. Bond angles of 180° (AB₂) and 120° (AB₃) with identical atoms always give zero dipole moment.
2. A bent shape (104.5°) leaves a resultant dipole.

Answer: (d) Both (b) and (c)

Explanation:
1. O=C=O is a linear molecule.
2. Its two C=O bond dipoles are equal and opposite, so its dipole moment is zero and hence it is non-polar.

Teacher's Note:
1. CO₂ has polar bonds but is a non-polar molecule because of its linear shape.
2. Compare with SO₂, which is bent and therefore polar.

Answer: (a) μ = q × d

Explanation:
1. Dipole moment = charge (q) × distance of separation (d).

Teacher's Note:
1. Dipole moment is a vector pointing (in chemistry) from the positive to the negative end.
2. Units: C m in SI and debye (D) in practice.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. According to the assumptions of MOT, two atomic orbitals combine to form two molecular orbitals; the bonding MO is lowered and the antibonding MO is raised in energy with respect to the parent atomic orbitals, so that the total energy remains the same.
2. Thus, both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

Teacher's Note:
1. The number of molecular orbitals formed always equals the number of atomic orbitals combined.
2. The antibonding MO is raised by about as much as the bonding MO is lowered.

Answer: (c) A is true, but R is false.

Explanation:
1. The linear combination of atomic orbitals to form molecular orbitals takes place only if the combining atomic orbitals have the same or nearly the same energy.
2. A 1s-orbital cannot combine with a 2s-orbital but can combine with another 1s-orbital, because the energy of the 2s-orbital is appreciably higher than that of the 1s-orbital.
3. Thus, Assertion is correct but Reason is incorrect.

Teacher's Note:
1. Conditions for LCAO: comparable energy, proper symmetry about the molecular axis and maximum overlap.
2. 1s combines with 1s and 2s with 2s in homonuclear diatomic molecules.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Bond dissociation energy is the energy required to break one mole of bonds of a particular type so as to separate them into gaseous atoms. Bond energy is the average value of dissociation energies of the same type of bond present in the molecule. In a diatomic molecule there is only one bond, so both are identical.
2. Bond energy of C—H bond in methane is 99.2 kcal/mol (average value). Thus, for polyatomic molecules, average bond energy is taken as the dissociation bond energy.
3. Hence, both Assertion and Reason are correct and Reason is not the correct explanation of Assertion.

Teacher's Note:
1. For polyatomic molecules successive dissociation energies differ, so an average bond energy is used.
2. A higher bond dissociation energy means a stronger, less reactive bond.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. With increase in bond order, bond length decreases and hence bond energy increases.
2. Bond orders are C—C (1) < C=C (2) < C≡C (3), so bond energies follow the same order.
3. So, both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Teacher's Note:
1. Approximate values: C—C 348, C=C 614 and C≡C 839 kJ mol⁻¹.
2. Bond energy does not triple for a triple bond, because π bonds are weaker than σ bonds.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. He₂ = σ1s² σ*1s²; bond order = \( \frac{1}{2}(2 - 2) = 0 \)
2. Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Teacher's Note:
1. Equal bonding and antibonding electrons cancel each other’s effect, giving bond order zero.
2. That is why He₂ does not exist as a stable molecule.