ABC26IN0072 · Chemical Bonding

Subject: Inorganic Chemistry · Chapter: Chemical Bonding · Topic: Chemical Bonding – General · Exam: CSIR-NET DEC 2019 · 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
MoleculeBond Order
$\mathrm{NO^{+}}$3
$\mathrm{O_2^{+}}$2.5
$\mathrm{N_2^{+}}$2.5
$\mathrm{C_2^{+}}$1.5
\[ \mathrm{NO^{+}} = \sigma_{1s}^{2} < \sigma_{1s}^{*2} < \sigma_{2s}^{2} < \sigma_{2s}^{*2} < \pi_{2px}^{2}=\pi_{2py}^{2} < \sigma_{2pz}^{2} \]
\[ \mathrm{O_2^{+}} = \sigma_{1s}^{2} < \sigma_{1s}^{*2} < \sigma_{2s}^{2} < \sigma_{2s}^{*2} < \pi_{2px}^{2}=\pi_{2py}^{2} < \sigma_{2pz}^{2} < \pi_{2pz}^{*1}=\pi_{2py}^{*1} \]
\[ \mathrm{N_2^{+}} = \sigma_{1s}^{2} < \sigma_{1s}^{*2} < \sigma_{2s}^{2} < \sigma_{2s}^{*2} < \pi_{2px}^{2}=\pi_{2py}^{2} < \sigma_{2pz}^{1} \]
\[ \mathrm{C_2^{+}} = \sigma_{1s}^{2} < \sigma_{1s}^{*2} < \sigma_{2s}^{2} < \sigma_{2s}^{*2} < \pi_{2px}^{2}=\pi_{2py}^{1} \]
\[ \text{Bond order of } \mathrm{NO^{+}} = \frac{8-2}{2} = 3 \]
\[ \text{Bond order of } \mathrm{O_2^{+}} = \frac{8-3}{2} = 2.5 \]
\[ \text{Bond order of } \mathrm{N_2^{+}} = \frac{7-2}{2} = 2.5 \]
\[ \text{Bond order of } \mathrm{C_2^{+}} = \frac{5-2}{2} = 1.5 \]

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