Q1 / 66 · JEE Advanced 2007 (Paper 1) · Chemistry
The species having bond order different from that in $\mathrm{CO}$ is
- Correct answerA$\mathrm{NO}^{-}$
- B$\mathrm{NO}^{+}$
- C$\mathrm{CN}^{-}$
- D$\mathrm{N}_2$
Solution
$\mathrm{CO}$ : total $e^{-s}=6+8=14$
$$
\begin{gathered}
\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^{1+1}=\pi 2 p_y^{1+1}, \sigma 2 p_z^2 \\
\mathrm{BO}=\frac{10-4}{2}=3 \\
\mathrm{NO}^{-}=\text {Total } e^{-} s=7+8+1=16 \\
\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^{1+1} \\
=\pi 2 p_y^{1+1}, \pi^* 2 p_x^1=\pi^* 2 p_y^1 \\
\mathrm{BO}=\frac{10-6}{2}=\frac{4}{2}=2
\end{gathered}
$$
Total number of $e^{-} s: \mathrm{CN}^{-}=6+7+1=14$
$$
\mathrm{N}_2=7+7=14
$$
Hence, the correct answer is option: 'A'.