Chemical Bonding and Molecular StructureNEET Chemistry · Class 11 · NCERT Chapter 4

Calculate the steric number = (number of atoms bonded to the central atom) + (number of lone pairs on the central atom). Steric number 2 = sp, 3 = sp², 4 = sp³, 5 = sp³d, 6 = sp³d². For example, in H₂O, O has 2 bonds + 2 lone pairs = steric number 4 = sp³. In SO₂, S has 2 bonds + 1 lone pair = steric number 3 = sp².

Lewis structures cannot capture this. In MO theory, the 16 electrons of O₂ fill up to the π*2p level. There are two degenerate π* orbitals, and by Hund's rule, each gets one electron with parallel spins. Two unpaired electrons means O₂ is paramagnetic. This was experimentally confirmed: liquid O₂ is attracted to magnets. It was one of the early triumphs of MO theory over VB theory.

Bond order is the number of bonds between two atoms (can be fractional in resonance/MO theory). Bond length is the physical distance between the nuclei. Higher bond order means stronger, shorter bond. For example: C-C (order 1, length 154 pm), C=C (order 2, length 134 pm), C≡C (order 3, length 120 pm). In benzene, the C-C bond order is 1.5 and the bond length is 139 pm, between single and double.

Both N and O in NH₃ and H₂O are sp³ hybridised. NH₃ has 3 bonding pairs and 1 lone pair on N. Bond angle = 107° (lone pair pushes bonding pairs slightly closer). H₂O has 2 bonding pairs and 2 lone pairs on O. The 2 lone pairs push the bonding pairs further together, giving bond angle = 104.5°.

H-C≡N: H-C is a single bond = 1 sigma. C≡N is a triple bond = 1 sigma + 2 pi. Total: 2 sigma bonds + 2 pi bonds. The C is sp hybridised (1 sigma bond to H, 1 sigma + 2 pi bonds to N, using all 2 sp orbitals for sigma bonds and 2 unhybridised p orbitals for the 2 pi bonds).

In ortho-nitrophenol, the OH group and the NO₂ group are close enough that an intramolecular hydrogen bond forms (within the same molecule), satisfying the H-bond donor and acceptor within one molecule. This prevents intermolecular hydrogen bonding between molecules. With fewer intermolecular H-bonds, less energy is needed to vaporise the compound, so its boiling point is lower. Para-nitrophenol cannot form intramolecular H-bonds (groups too far apart), so it relies on intermolecular H-bonding, raising its boiling point.