Question
Why does have a smaller bond angle () than (), even though both have hybridization on the central atom?
Solution — Step by Step
Both N (in NH) and C (in CH) are hybridized. Each has four electron pairs around the central atom, giving an idealized tetrahedral arrangement with angles.
In CH: all four pairs are bond pairs (with H atoms). In NH: three are bond pairs (with H atoms) and one is a lone pair on N.
The order of electron-pair repulsion strength is:
A lone pair occupies more space than a bond pair because it’s held by only one nucleus, not two.
The lone pair on N pushes the three N–H bond pairs closer together, compressing the H–N–H angle from the ideal down to . CH has no lone pairs, so all repulsions are equal and the angle stays at .
Final Answer: Lone-pair repulsion in NH compresses bond angles below the tetrahedral value. CH, with no lone pairs, retains the ideal angle.
Why This Works
VSEPR theory says molecular shape is determined by minimizing repulsion among all electron pairs (bonding + lone). A lone pair “occupies more angular space” because its electron density is concentrated near one atom, so it bullies the bond pairs into a tighter arrangement.
Continuing the trend: HO has two lone pairs on O, and the H–O–H angle drops further to . The pattern is consistent — each additional lone pair compresses the bond angle by roughly .
Alternative Method
Argue from electron-density orientation: lone pair density is mostly close to N, while bond pair density extends out toward the H. So the lone pair “pushes” each H closer to its neighbours. Same conclusion, more visual.
“Both NH and CH are tetrahedral, so they must have the same bond angle.” This conflates electron geometry (tetrahedral in both) with molecular geometry (tetrahedral CH vs trigonal pyramidal NH). Always distinguish the two.
Quick angle-prediction recipe: count lone pairs on the central atom. Subtract roughly per lone pair from . NH → , HO → . NEET often asks “rank HO, NH, CH by bond angle” — answer: CH > NH > HO.