Question

Which of the following species has bond angle $\geq 120^\circ$, w.r.t. underlined atom?

• A. $\underline{\text{O}}(\text{CH}_3)_2$
• B. $\underline{\text{N}}(\text{CH}_3)_3$
• C. $\underline{\text{N}}(\text{SiH}_3)_3$
• D. $\underline{\text{P}}(\text{SiH}_3)_3$

Answer 1

Answer: (C)

N(SiH3)3

Answer 2

Let's analyze the structure and hybridization of the central atom in each species to determine the bond angle.

(A) $\underline{\text{O}}(\text{CH}_3)_2$: The central atom is oxygen ($\underline{\text{O}}$). Oxygen has 6 valence electrons. It forms two single bonds with two methyl ($\text{CH}_3$) groups. The number of valence electrons remaining is $6 - 2 = 4$, which form two lone pairs. The number of electron domains around oxygen is 2 bonding pairs + 2 lone pairs = 4. According to VSEPR theory, this corresponds to $sp^3$ hybridization. The electron geometry is tetrahedral, and the molecular geometry is bent (V-shaped). The ideal bond angle for $sp^3$ hybridization is $109.5^\circ$. The presence of two lone pairs repels the bonding pairs, reducing the bond angle. The $\angle \text{C-O-C}$ bond angle in dimethyl ether is approximately $111^\circ$, which is less than $120^\circ$.

(B) $\underline{\text{N}}(\text{CH}_3)_3$: The central atom is nitrogen ($\underline{\text{N}}$). Nitrogen has 5 valence electrons. It forms three single bonds with three methyl ($\text{CH}_3$) groups. The number of valence electrons remaining is $5 - 3 = 2$, which form one lone pair. The number of electron domains around nitrogen is 3 bonding pairs + 1 lone pair = 4. According to VSEPR theory, this corresponds to $sp^3$ hybridization. The electron geometry is tetrahedral, and the molecular geometry is trigonal pyramidal. The ideal bond angle for $sp^3$ hybridization is $109.5^\circ$. The presence of one lone pair repels the bonding pairs, reducing the bond angle. The $\angle \text{C-N-C}$ bond angle in trimethylamine is approximately $108^\circ$, which is less than $120^\circ$.

(C) $\underline{\text{N}}(\text{SiH}_3)_3$: The central atom is nitrogen ($\underline{\text{N}}$). Nitrogen has 5 valence electrons. It forms three single bonds with three silyl ($\text{SiH}_3$) groups. The number of valence electrons remaining is $5 - 3 = 2$, which form one lone pair. Based on the number of electron domains (4), we might initially predict $sp^3$ hybridization and trigonal pyramidal geometry. However, this molecule exhibits planarity around the nitrogen atom. This is explained by the delocalization of the lone pair on nitrogen into the empty d-orbitals of the silicon atoms (p$\pi$-d$\pi$ backbonding). For effective p$\pi$-d$\pi$ bonding, the nitrogen atom undergoes $sp^2$ hybridization. Three $sp^2$ hybrid orbitals form sigma bonds with the three silicon atoms, and the remaining unhybridized p-orbital contains the lone pair, which overlaps with the empty d-orbitals on silicon. This leads to a planar arrangement of the three silicon atoms and the nitrogen atom. In a trigonal planar geometry, the ideal bond angle is $120^\circ$. The $\angle \text{Si-N-Si}$ bond angle in $\text{N}(\text{SiH}_3)_3$ is found experimentally to be $120^\circ$.

(D) $\underline{\text{P}}(\text{SiH}_3)_3$: The central atom is phosphorus ($\underline{\text{P}}$). Phosphorus has 5 valence electrons. It forms three single bonds with three silyl ($\text{SiH}_3$) groups. The number of valence electrons remaining is $5 - 3 = 2$, which form one lone pair. The number of electron domains around phosphorus is 3 bonding pairs + 1 lone pair = 4. According to VSEPR theory, this corresponds to $sp^3$ hybridization. The electron geometry is tetrahedral, and the molecular geometry is trigonal pyramidal. Although p$\pi$-d$\pi$ backbonding between P (3p lone pair) and Si (empty 3d orbitals) is possible, it is significantly weaker than in the case of N (2p lone pair) and Si (empty 3d orbitals). The lone pair on phosphorus remains largely localized. Therefore, the molecule retains its trigonal pyramidal structure. The ideal bond angle for $sp^3$ hybridization is $109.5^\circ$. The presence of one lone pair reduces the bond angle. The $\angle \text{Si-P-Si}$ bond angle in $\text{P}(\text{SiH}_3)_3$ is approximately $96^\circ$, which is less than $120^\circ$.

Comparing the bond angles: (A) $\text{O}(\text{CH}_3)_2$: $\approx 111^\circ < 120^\circ$ (B) $\text{N}(\text{CH}_3)_3$: $\approx 108^\circ < 120^\circ$ (C) $\text{N}(\text{SiH}_3)_3$: $\approx 120^\circ \geq 120^\circ$ (D) $\text{P}(\text{SiH}_3)_3$: $\approx 96^\circ < 120^\circ$

Only $\text{N}(\text{SiH}_3)_3$ has a bond angle equal to $120^\circ$, which satisfies the condition $\geq 120^\circ$.