JEE Advanced · Chemistry · 4. Chemical Bonding
The species having bond order different from that in \(\mathrm{CO}\) is
- A \(\mathrm{NO}^{-}\)
- B \(\mathrm{NO}^{+}\)
- C \(\mathrm{CN}^{-}\)
- D \(\mathrm{N}_2\)
Answer & Solution
Correct Answer
(A) \(\mathrm{NO}^{-}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{CO}\) : total \(e^{-s}=6+8=14\)
\(\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^{1+1}\) \(=\pi 2 p_y^{1+1}, \sigma 2 p_z^2 \)
\( \mathrm{BO}=\frac{10-4}{2}=3 \)
\( \mathrm{NO}^{-}=\text {Total } e^{-} s=7+8+1=16 \)
\( \sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^{1+1} \)
\( =\pi 2 p_y^{1+1}, \pi^* 2 p_x^1=\pi^* 2 p_y^1 \)
\( \mathrm{BO}=\frac{10-6}{2}=\frac{4}{2}=2\)
Total number of \(e^{-} s: \mathrm{CN}^{-}=6~+\) \(7+1=14\)
\(
\mathrm{N}_2=7+7=14
\)
\(\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^{1+1}\) \(=\pi 2 p_y^{1+1}, \sigma 2 p_z^2 \)
\( \mathrm{BO}=\frac{10-4}{2}=3 \)
\( \mathrm{NO}^{-}=\text {Total } e^{-} s=7+8+1=16 \)
\( \sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^{1+1} \)
\( =\pi 2 p_y^{1+1}, \pi^* 2 p_x^1=\pi^* 2 p_y^1 \)
\( \mathrm{BO}=\frac{10-6}{2}=\frac{4}{2}=2\)
Total number of \(e^{-} s: \mathrm{CN}^{-}=6~+\) \(7+1=14\)
\(
\mathrm{N}_2=7+7=14
\)
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