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JEE Advanced 2007Paper 1
Exam
JEE Advanced
Year
2007
Paper
Paper 1
Subject
Chemistry
Chapter
Chemical Bonding and Molecular Structure

JEE Advanced 2007 · Paper 1 · Chemistry · Chemical Bonding and Molecular Structure

The species having bond order different from that in CO is

The species having bond order different from that in CO\mathrm{CO} is

Options

  1. ANO−\mathrm{NO}^{-}
  2. BNO+\mathrm{NO}^{+}
  3. CCN−\mathrm{CN}^{-}
  4. DN2\mathrm{N}_2

Correct answer

A. NO−\mathrm{NO}^{-}

Step-by-step solution

CO\mathrm{CO} : total e−s=6+8=14σ1s2,σ∗1s2,σ2s2,σ∗2s2,π2px1+1=π2py1+1,σ2pz2BO=10−42=3NO−=Total e−s=7+8+1=16σ1s2,σ∗1s2,σ2s2,σ∗2s2,σ2pz2,π2px1+1=π2py1+1,π∗2px1=π∗2py1BO=10−62=42=2Totalnumberofe^{-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 ofe^{-} s: CN−=\mathrm{CN}^{-}=6+7+1=14 N2=7+7=14\mathrm{N}_2=7+7=14
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