Q4 · CSIR NET Chemistry, December 2019
Paper: CSIR NET December 2019 · Subject: Inorganic Chemistry · Chapter: Chemical Bonding · Topic: Chemical Bonding – General · Marks: 2 · Difficulty: Easy
The ion having the highest bond order is
(a)$\mathrm{NO}^{+}$
(b)$\mathrm{O}_{2}^{+}$
(c)$\mathrm{N}_{2}^{+}$
(d)$\mathrm{C}_{2}^{+}$
Answer
SELF-PRACTICE — the source book printed no answer.
Nothing is invented here, so this question has no answer on record.
Explanation
The bond order can be calculated by molecular orbital theory Molecule Bond Order NO+ 3 O2+ 2.5 N2+ 2.5 C2+ 1.5 NO+ = σ21s < σ21s* < σ22s < σ22s* < π22px = π22py < σ22pz O2+ = σ21s < σ21s* < σ22s < σ22s* < π22px = π22py < σ22pz < π1*2pz = π1*2py N2+ = σ21s < σ21s* < σ22s < σ22s* < π22px = π22py < σ12pz C2+ = σ21s < σ21s* < σ22s < σ22s* < π22px = π12py 8-2 Bond order of NO+ = =3 2 8- 3 Bond order of O2+ = = 2.5 2 7- 2 Bond order of N2+ = = 2.5 2 5- 2 Bond order of C2+ = = 1.5 2
Explanation as extracted from the printed page; notation may be imperfect.
Study loop for Chemical Bonding