Chemistry · Chemical Bonding · NEET
sp3 hybridisation gives an ideal angle of 109.5°, but that ideal only holds when all four electron pairs are the SAME type (all bond pairs). In CH4 the central C has 4 bond pairs and 0 lone pairs, so it stays at 109.5°. In NH3 one of the four sp3 pairs is a lone pair, and in H2O two of them are lone pairs. Lone pairs repel more strongly, so they distort the shape and shrink the H–N–H and H–O–H angles below 109.5°.
A lone pair belongs to only ONE atom, so its electron cloud is fat and spread out close to the central atom. A bond pair is shared between two atoms, so it is pulled and pinned between the nuclei and stays thinner. Because the lone pair cloud is bigger and closer, it repels the neighbouring bond pairs harder and pushes them toward each other. That squeezing reduces the angle between the bonds.
It is about the NUMBER of lone pairs. NH3 has 1 lone pair pushing on 3 bond pairs, dropping the angle by about 2.5° (from 109.5° to 107°). H2O has 2 lone pairs pushing on only 2 bond pairs, so there is extra lone pair–lone pair and lone pair–bond pair repulsion. More lone pairs = more squeezing = smaller angle, so H2O falls further to 104.5°.
Lone pair–lone pair > lone pair–bond pair > bond pair–bond pair. This VSEPR order is directly asked (NEET 2016). It tells you that lone pairs are the strongest 'pushers'. So any time you add a lone pair to an sp3 centre, the bond angle drops. Use it to instantly rank: CH4 (no lp) > NH3 (1 lp) > H2O (2 lp) in bond angle.
Yes. Match them one-to-one: CH4 = 109.5°, NH3 = 107°, H2O = 104.5°. The order of bond angle is CH4 > NH3 > H2O, which is the same as the order of DECREASING lone pairs (0, 1, 2). If a question gives you these three molecules, this single line answers it.
No. Same sp3 hybridisation, but different shapes because lone pairs are invisible in the final shape. CH4 (0 lp) = tetrahedral. NH3 (1 lp) = trigonal pyramidal (bent down like a tripod). H2O (2 lp) = bent / angular (V-shape). The lone pairs still occupy sp3 positions, but we only 'see' the atoms, so the name changes.
Consider the molecules CH4, NH3 and H2O. Which of the given statements is false?
Predict the correct order of repulsion between electron pairs as per VSEPR theory:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
CH4 = 109.5°, NH3 = 107° (about 107.8°), and H2O = 104.5°. The angle decreases as the number of lone pairs on the central atom increases from 0 to 1 to 2.
Because lone pairs increase from 0 to 1 to 2. Lone pairs repel neighbouring bond pairs more strongly than bond pairs repel each other, so they push the bonds closer together and shrink the angle.
Yes, all three have an sp3 central atom with four electron pairs. The difference is only how many of those pairs are lone pairs, which changes the bond angle and the shape.
CH4 is tetrahedral, NH3 is trigonal pyramidal, and H2O is bent (angular). All come from an sp3 tetrahedral arrangement, but lone pairs are not counted in the visible shape.
H2O has the smallest bond angle (104.5°) because oxygen has two lone pairs, giving the maximum squeezing of the bond pairs.
