CBSE Class 11 Chemistry HOTs Chemical Bonding and Molecular Structure Set 02

Higher Order Thinking Questions

Question 1. Apart from the tetrahedral geometry, another possible geometry for CH₄ is square planar, with the four H atoms at the corners of a square and the C atom at the centre. Explain why CH₄ is not square planar.
Answer:
In CH₄ the carbon atom (2s² 2p²) promotes an electron to give 2s¹ 2p³ and mixes them to form four equivalent sp³ orbitals. These point to the corners of a tetrahedron, so CH₄ is tetrahedral (109.5°).
A square planar shape needs dsp² hybridisation, which requires an available d orbital of low energy. Carbon is in the second period and has no d orbitals.
VSEPR theory also rules out the square planar shape. The H–C–H angle would be only 90°, so the bond pairs would repel each other much more than in the tetrahedral arrangement (109.5°), and the molecule would be less stable.
In simple words: Carbon has no d orbitals, and the 90° angle of a square would crowd the bonds.

Teacher's Note:
1. Mention both the hybridisation argument and the VSEPR argument.
2. A four-bond arrangement with the maximum separation is the tetrahedron.

Question 2. Explain why the BeH₂ molecule has a zero dipole moment although the Be–H bonds are polar.
Answer:
Be has 2 valence electrons and forms two bonds, so BeH₂ has 2 bond pairs and no lone pair (AX₂). It is linear (H–Be–H, 180°, sp hybridisation).
Each Be–H bond is polar (H is more electronegative than Be, so H is δ⁻). The two bond dipoles are equal in magnitude and opposite in direction.
They cancel each other, so the net dipole moment is zero.
In simple words: Two equal and opposite bond dipoles cancel, so a linear molecule is non-polar.

Teacher's Note:
1. Dipole moment is a vector sum.
2. BeCl₂, CO₂ and CS₂ behave in the same way.

Question 3. Which has the higher dipole moment, NH₃ or NF₃? Explain.
Answer:
Both molecules are pyramidal, with one lone pair on N and three bond pairs. Fluorine is more electronegative than hydrogen, so one might expect NF₃ to have the larger moment. In fact NH₃ (1.47 D) has a larger dipole moment than NF₃ (0.23 D).
In NH₃ the N–H bond dipoles point towards N, in the same direction as the dipole due to the lone pair. They add up.
In NF₃ the N–F bond dipoles point towards F, opposite to the lone-pair dipole, so they partly cancel it. The resultant is small.
Hence the net dipole moment of NF₃ is less than that of NH₃.
In simple words: In NH₃ the dipoles add up and in NF₃ they oppose each other.

Teacher's Note:
1. Draw the lone pair and each bond dipole to see the direction.
2. The value depends on direction as well as on electronegativity.

Question 4. In the SF₄ molecule the lone pair of electrons occupies an equatorial position of the trigonal bipyramid, not an axial position. Why?
Answer:
S in SF₄ has 5 electron pairs (4 bond pairs and 1 lone pair), so the arrangement is trigonal bipyramidal (sp³d).
A lone pair in an equatorial position has only two neighbours at 90° (the axial bond pairs). A lone pair in an axial position would have three neighbours at 90° (the equatorial bond pairs).
Lone pair–bond pair repulsion is greater than bond pair–bond pair repulsion, so the structure with fewer 90° lone pair–bond pair repulsions is more stable. The lone pair therefore sits in the equatorial position, giving the see-saw shape.
In simple words: The equatorial position gives the lone pair more room, with fewer 90° neighbours.

Teacher's Note:
1. Count the 90° interactions for both positions.
2. Lone pairs prefer the roomier equatorial positions.

Question 5. Phosphorus forms PCl₅, but nitrogen does not form NCl₅. Explain.
Answer:
P has vacant 3d orbitals. One 3s electron is promoted to 3d, so P has five unpaired electrons (3s¹ 3p³ 3d¹) and forms five bonds by sp³d hybridisation. P can hold 10 electrons, an expanded octet.
N is in the second period. It has only 2s and 2p orbitals and no d orbitals, so it cannot promote an electron or expand its octet beyond 8. Its maximum covalency is 4. So NCl₅ does not exist.
In simple words: Elements of the second period have no d orbitals, so they cannot go beyond an octet.

Teacher's Note:
1. Give the maximum covalency of second-period elements: 4.
2. SF₆ and PF₅ are other expanded-octet examples.

Question 6. The I₃⁻ ion is known but F₃⁻ is not. Explain.
Answer:
The central iodine in I₃⁻ has 3 lone pairs and 2 bond pairs, so it holds 10 electrons (sp³d hybridisation, linear shape). Iodine is in the fifth period and has vacant d orbitals, so it can expand its octet.
Fluorine is in the second period. It has no d orbitals, so it cannot hold more than 8 electrons. Therefore F₃⁻ does not exist.
In simple words: Iodine can exceed an octet and fluorine cannot.

Teacher's Note:
1. Count the electrons on the central atom: 3 lone pairs + 2 bond pairs = 10.
2. Compare XeF₂, which has the same linear shape.

Question 7. Chlorine has almost the same electronegativity as nitrogen, but HCl does not form hydrogen bonds. Explain.
Answer:
For hydrogen bonding the atom bonded to H must be both highly electronegative and small, so that its electron density is high enough to attract the H atom of another molecule.
Cl is as electronegative as N, but it is much larger, so its electron density is spread over a large volume. The attraction for H is weak, and no significant hydrogen bond forms. N, O and F are small and do form hydrogen bonds.
In simple words: Chlorine is too big to attract a hydrogen atom strongly.

Teacher's Note:
1. Mention both electronegativity and atomic size.
2. That is why HCl is a gas while H₂O is a liquid.

Question 8. Assertion (A): O₂ is paramagnetic.
Reason (R): There are two unpaired electrons in the π* molecular orbitals of O₂.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (a) The configuration of O₂ is KK (σ2s)² (σ*2s)² (σ2pz)² (π2px)² (π2py)² (π*2px)¹ (π*2py)¹. The two unpaired electrons make O₂ paramagnetic. So both statements are true and R explains A.
In simple words: Unpaired electrons make a substance paramagnetic.

Teacher's Note:
1. Write the molecular orbital configuration to find unpaired electrons.

Question 9. Assertion (A): CO₂ is a linear molecule, whereas SO₂ is a bent molecule.
Reason (R): The central carbon atom of CO₂ has no lone pair, while the central sulphur atom of SO₂ has one lone pair.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (a) C in CO₂ has two bond pairs (two double bonds) and no lone pair (AX₂), so it is linear. S in SO₂ has two bond pairs and one lone pair (AX₂E), so the three electron pairs are trigonal planar and the molecule is bent. R explains A.
In simple words: The lone pair on S bends the SO₂ molecule.

Teacher's Note:
1. Compare AX₂ with AX₂E.

Question 10. Assertion (A): The N₂⁺ ion is paramagnetic, whereas the N₂ molecule is diamagnetic.
Reason (R): N₂⁺ has an unpaired electron in the σ2pz molecular orbital.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (a) N₂ has all its electrons paired, (σ2pz)². Removing one electron gives (σ2pz)¹ in N₂⁺, which leaves one unpaired electron, so N₂⁺ is paramagnetic. R correctly explains A.
In simple words: Taking one electron out of a filled orbital leaves an unpaired electron.

Teacher's Note:
1. N₂⁺ has a bond order of 2.5.

Question 11. Assertion (A): The boiling point of H₂O is higher than that of HF.
Reason (R): The hydrogen bond F–H···F is stronger than the hydrogen bond O–H···O.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (b) Water boils at 373 K and HF at 293 K, so A is true. F–H···F is indeed the strongest hydrogen bond, so R is true.
But R is not the explanation. Each H₂O molecule can form up to four hydrogen bonds (two through its H atoms and two through the lone pairs on O), whereas an HF molecule can form only two. The larger number of hydrogen bonds per molecule raises the boiling point of water.
In simple words: Water forms more hydrogen bonds per molecule than HF does.

Teacher's Note:
1. Strength of one bond is different from the total effect of many bonds.

Question 12. Which of the following species are isoelectronic (same number of valence electrons) and isostructural? (more than one may be correct)
(a) NO₃⁻ and CO₃²⁻
(b) SO₃ and NO₃⁻
(c) ClO₃⁻ and CO₃²⁻
(d) SO₃ and CO₃²⁻
Answer:
Ans. (a), (b) and (d)
NO₃⁻ (5 + 18 + 1 = 24), CO₃²⁻ (4 + 18 + 2 = 24) and SO₃ (6 + 18 = 24) each have 24 valence electrons and three bond pairs on the central atom with no lone pair, so each is trigonal planar.
ClO₃⁻ has 7 + 18 + 1 = 26 electrons and one lone pair on Cl, so it is pyramidal and not isostructural with CO₃²⁻.
In simple words: Count the valence electrons and the lone pairs of each species.

Teacher's Note:
1. Isoelectronic species with the same pair count have the same shape.

Question 13. Arrange NO₂⁺, NO₂ and NO₂⁻ in the decreasing order of their bond angles and give the reason.
Answer:
NO₂⁺ (180°) > NO₂ (about 134°) > NO₂⁻ (about 115°).
NO₂⁺ has no lone pair on N (two bond pairs only), so it is linear.
NO₂ has one unpaired electron on N, which repels the two bond pairs less than a lone pair would, so the angle is large (134°).
NO₂⁻ has one lone pair on N, which repels the bond pairs strongly and pushes the angle down to about 115°.
In simple words: The more electron density sits on nitrogen outside the bonds, the smaller the angle.

Teacher's Note:
1. Compare an empty position, a single electron and a full lone pair.
2. NO₂⁺ is isoelectronic with CO₂.