Question

$NO^{+}$ has bond order

• A. 2
• B. 2.5
• C. 3
• D. 3.5

Answer 1

Answer: (C)

3

Answer 2

Bond order of $NO ^{+}$ can be calculated by following formula.

Bond order $=\frac{N_{b}-N_{a}}{2}$

The increasing order of energies of various molecular orbitals for $NO ^{+}$ is given below

$NO ^{+}(14)=\sigma l s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2},\left(\pi 2 p_{x}+\pi 2 p_{y}\right)$
$\sigma 2 p_{z^{\prime}}^{2}\left(\pi^{*} 2 p_{x}^{0}=\pi^{*} 2 p_{y}^{0}\right), \sigma^{*} 2 p_{z}^{0}$

The bond order can be calculated as

Bond order $=\frac{N_{b}-N_{a}}{2}$
$=\frac{10-4}{2}=\frac{6}{2}=3$