Questions & explanations
1. The bond angle in NH3 is 107°, while in PH3 it is 93.5°. This decrease is best explained by:
- Hybridisation: NH3 is sp2 hybridised, while PH3 is sp3 hybridised.
- Electronegativity difference: N is more electronegative than P, so bond angle is larger.
- VSEPR theory: lone pair repulsion is stronger in PH3 due to larger size of P.
- Bent's rule: lone pair in NH3 has more s-character; in PH3, P uses nearly pure p orbitals.
Answer: Bent's rule: lone pair in NH3 has more s-character; in PH3, P uses nearly pure p orbitals.
According to Bent's rule, lone pairs occupy orbitals with more s-character. In NH3, N is sp3 hybridised, so the lone pair has 25% s-character. In PH3, the larger s-p energy gap makes hybridisation less favourable; P uses nearly pure p orbitals for bonding, and the lone pair stays in an s orbital. Bond angle then approaches 90° (pure p orbitals).
2. Which of the following compounds has the highest boiling point?
- H2Se
- H2S
- H2O
- H2Te
Answer: H2O
H2O exhibits strong intermolecular hydrogen bonding due to high electronegativity of oxygen and presence of lone pairs, resulting in a boiling point of 100°C. In contrast, H2S, H2Se, and H2Te lack hydrogen bonding and have boiling points that increase with molar mass but remain far below 100°C. Thus, H2O has the highest boiling point.
3. The bond angles of NH3, PH3, AsH3, and SbH3 are 107°, 93.5°, 92°, and 91° respectively. Which statement correctly explains this trend?
- Down the group, the number of lone pairs increases, so bond angles decrease.
- Down the group, atomic size increases, so bond angles increase due to reduced lone pair repulsion.
- Down the group, electronegativity of the central atom decreases, so bond angles increase.
- Down the group, the ns-np energy gap increases, so hybridisation becomes unfavourable; heavier hydrides use pure p orbitals, giving angles near 90°.
Answer: Down the group, the ns-np energy gap increases, so hybridisation becomes unfavourable; heavier hydrides use pure p orbitals, giving angles near 90°.
Drago's rule states that for heavier main-group hydrides, the ns-np energy gap is large, making hybridisation energetically unfavourable. The central atom uses pure p orbitals for bonding, with the lone pair in an s orbital, giving bond angles close to 90°. For NH3, the 2s-2p gap is small, so sp3 hybridisation occurs, giving 107°.
4. The dipole moment of NF3 (0.23 D) is much smaller than that of NH3 (1.47 D). Which statement explains this?
- N-F bond dipoles are smaller than N-H bond dipoles, and they oppose the lone pair dipole in NF3.
- NF3 is tetrahedral with no lone pair, so net dipole is zero.
- The lone pair on N in NF3 is in an s-orbital, contributing no dipole.
- N-F bonds are nonpolar because F is less electronegative than N.
Answer: N-F bond dipoles are smaller than N-H bond dipoles, and they oppose the lone pair dipole in NF3.
In NF3, the N-F bond dipoles point from N to F (away from N), opposing the lone pair dipole which points along the C3 axis from N outward. This partial cancellation results in a small net dipole (0.23 D). In NH3, the N-H bond dipoles point from H to N (toward N), adding to the lone pair dipole, giving a larger net dipole (1.47 D).
5. A substance has high melting point, is hard, and does not conduct electricity in solid or molten state. What type of bonding does it most likely have?
- Covalent network
- Ionic
- Metallic
- Molecular covalent
Answer: Covalent network
Covalent network solids (e.g., diamond, SiO2) have high melting points due to strong covalent bonds throughout the crystal, are hard, and do not conduct electricity because electrons are localized in bonds. Ionic solids conduct when molten; metallic solids conduct in solid; molecular covalent solids have low melting points.
6. In CH2F2, the H-C-H bond angle is 112° and the F-C-F bond angle is 108°. Which statement best explains this using Bent's rule?
- Carbon directs more s-character toward F than toward H, so F-C-F angle is larger.
- Carbon directs more s-character toward H than toward F, so H-C-H angle is larger.
- Fluorine's lone pairs repel each other more strongly, compressing the F-C-F angle.
- Hydrogen's small size allows closer approach, increasing the H-C-H angle.
Answer: Carbon directs more s-character toward H than toward F, so H-C-H angle is larger.
Bent's rule states that a central atom directs hybrid orbitals with greater s-character toward more electropositive substituents. Hydrogen is less electronegative than fluorine, so carbon puts more s-character into C-H bonds. Higher s-character leads to larger bond angles, so H-C-H (112°) is larger than F-C-F (108°).
7. What is the formal charge on the central carbon atom in the most stable Lewis structure of CO2?
- +2
- +1
- -1
- 0
Answer: 0
In the most stable Lewis structure O=C=O, each oxygen has 6 valence electrons, 4 lone-pair electrons, and 4 bonding electrons, giving FC = 6 - 4 - 4/2 = 0. The central carbon has 4 valence electrons, 0 lone-pair electrons, and 8 bonding electrons, so FC = 4 - 0 - 8/2 = 0. Thus the formal charge on carbon is 0.
8. What is the hybridisation and shape of XeF4?
- sp3d2, octahedral
- sp3d, tetrahedral
- sp3d2, square planar
- sp3d, square pyramidal
Answer: sp3d2, square planar
XeF4 has 8 valence electrons from Xe and 4 from F atoms, total 12 electrons. Xe forms 4 sigma bonds (4 bond pairs) and has 2 lone pairs. Steric number = 4 + 2 = 6, so hybridisation is sp3d2. According to VSEPR, two lone pairs occupy trans positions to minimise repulsion, giving a square planar shape.
9. Which property of metals is NOT explained by the electron-sea model?
- Directional bonding
- Malleability
- High electrical conductivity
- Metallic lustre
Answer: Directional bonding
The electron-sea model describes delocalised electrons that are free to move, explaining conductivity, malleability, and lustre. However, metallic bonds are non-directional; the model does not predict directional bonding. Directional bonding is characteristic of covalent bonds, not metallic bonds.
10. Arrange F2, Cl2, Br2, I2 in increasing order of boiling point.
- Cl2 < F2 < Br2 < I2
- I2 < Br2 < Cl2 < F2
- F2 < I2 < Br2 < Cl2
- F2 < Cl2 < Br2 < I2
Answer: F2 < Cl2 < Br2 < I2
All are nonpolar molecules, so only London dispersion forces exist. London forces increase with molecular size and polarisability. As we go down the group, molar mass increases, so boiling point increases: F2 (gas), Cl2 (gas), Br2 (liquid), I2 (solid). Thus correct order is F2 < Cl2 < Br2 < I2.
11. Trisilylamine, N(SiH3)3, is planar while trimethylamine, N(CH3)3, is pyramidal. What is the reason?
- In N(CH3)3, nitrogen uses sp2 hybridisation due to hyperconjugation from methyl groups.
- In N(SiH3)3, nitrogen uses sp3 hybridisation but steric hindrance forces planarity.
- In N(SiH3)3, nitrogen uses sp2 hybridisation due to pπ-dπ back bonding from N to Si.
- In N(SiH3)3, silicon atoms are larger, so the molecule adopts a planar geometry.
Answer: In N(SiH3)3, nitrogen uses sp2 hybridisation due to pπ-dπ back bonding from N to Si.
In N(SiH3)3, nitrogen's lone pair is donated into empty 3d orbitals of silicon via pπ-dπ back bonding, so nitrogen uses sp2 hybridisation (trigonal planar). In N(CH3)3, carbon has no empty d orbitals, so back bonding is absent; nitrogen remains sp3 hybridised, giving a pyramidal shape.
12. Which of the following has the highest magnetic moment (spin-only)?
- O2^2-
- O2+
- O2-
- O2
Answer: O2
Spin-only magnetic moment μ = √(n(n+2)) BM, where n = number of unpaired electrons. Using molecular orbital theory, O2 has 2 unpaired electrons in π* orbitals (n=2, μ=2.83 BM). O2+ and O2- have 1 unpaired electron (μ=1.73 BM). O2^2- has 0 unpaired (μ=0). Hence O2 has the highest μ.