Q4. How many unpaired electrons do N22− and O22+ have? Which neutral molecule is each isoelectronic with?
For this formal occupancy exercise (not a claim of a stable free gas-phase dianion), N22− has 16 electrons, like O2: one in each π*2p, so 2 unpaired (bond order 2). O22+ has 14, like N2: all paired (bond order 3).
Q5. Which of N2+, O2+, NO+ and CN have the same bond order?
N2+ (13 electrons) 2.5, O2+ (15) 2.5, NO+ (14) 3, CN (13) 2.5. So N2+, O2+ and CN share bond order 2.5.
Q6. Removing one electron lengthens the bond in N2 but shortens it in O2. Explain.
The electron leaves the highest occupied orbital. In N2 that is the bonding σ2pz: Nb falls from 10 to 9 and the bond order from 3 to 2.5. In O2 it is an antibonding π*2p: Na falls from 6 to 5 and the bond order rises from 2 to 2.5.
Q7. The elementary diagram gives Be2 and Ne2 zero bond order. What does it predict for Be2+ and Ne2+?
Be2+ (7 electrons): Nb = 4, Na = 3, bond order 0.5. Ne2+ (19 electrons) loses a σ*2pz electron: Nb = 10, Na = 9, bond order 0.5. Both are positive, so both can exist (weakly bound), each with one unpaired electron.
Q8. B2 and F2 both have bond order 1. Use NCERT Fig. 4.21 to test whether equal bond order means equal bond strength and length.
It does not: B2 (290 kJ mol−1, 159 pm) has a bond both stronger and longer than F2 (159 kJ mol−1, 143 pm). Bond order compares species of the same element, such as O2+, O2, O2− and O22−; between different elements, atom size and lone-pair repulsion (large in small F2) also matter.