Number of molecules having bond order 2 from the following molecules is _________.
C₂, O₂, Be₂, Li₂, Ne₂, N₂, He₂

Solution

Chemistry ma’am · 3 steps

Fill the MOs, count bonding and antibonding electrons, and halve the difference for each molecule.

1. Bond order formula. Bond order is half the difference between electrons in bonding MOs (N_b) and in antibonding MOs (Nₐ). Count all the electrons of the molecule, including the 1s ones.

   Bond order = (N_b - Nₐ)/2

2. Filling order. Up to N₂ (14 electrons), π 2p fills before σ 2p_z. For O₂ and Ne₂, σ 2p_z comes first. The order does not change N_b or Nₐ for these molecules, so the bond orders come out the same either way.

   Molecule	Total electrons	N_b	Nₐ	Bond order
   He₂	4	2	2	0
   Li₂	6	4	2	1
   Be₂	8	4	4	0
   C₂	12	8	4	2
   N₂	14	10	4	3
   O₂	16	10	6	2
   Ne₂	20	10	10	0

3. So. Only C₂ and O₂ have bond order 2, so the answer is 2.

The idea to keep. For homonuclear diatomics of the first two periods, remember the bond-order pattern by electron count: 12 and 16 give 2, 14 gives 3, and 4, 8 and 20 give 0.

Answer from the NTA final answer key. Solution by PrepWiser.