Q5 / 160 · AP EAMCET 2020 (17 Sep Shift 1) · Chemistry

Which of the following sets of species have similar bond order?

  • A
    \(\mathrm{N}_2, \mathrm{O}_2^{2-}, \mathrm{NO}^{+}\)
  • B
    \(\mathrm{N}_2, \mathrm{~F}_2, \mathrm{O}_2^{2-}\)
  • C
    \(\mathrm{N}_2, \mathrm{~N}_2^{2-}, \mathrm{O}_2^{-}\)
  • Correct answer
    D
    \(\mathrm{N}_2, \mathrm{CO}, \mathrm{NO}^{+}\)

Solution

Bond order, \(\mathrm{BO}=\frac{N_b-N_a}{2}\) \(\begin{aligned} \Rightarrow \mathrm{N}_2\left(14 e^{-}\right): \mathrm{BO} & =\frac{10-4}{2}=3 \\ \mathrm{O}_2^{2-}\left(18 e^{-}\right): \mathrm{BO} & =\frac{10-8}{2}=1 \\ \mathrm{NO}^{+}\left(14 e^{-}\right): \mathrm{BO} & =\frac{10-4}{2}=3 \\ \mathrm{~F}_2\left(18 e^{-}\right): \mathrm{BO} & =\frac{10-8}{2}=1 \\ \mathrm{~N}_2^{2-}\left(16 e^{-}\right): \mathrm{BO} & =\frac{10-6}{2}=2 \\ \mathrm{O}_2^{-}\left(17 e^{-}\right): \mathrm{BO} & =\frac{10-7}{2}=1.5 \\ \mathrm{CO}\left(14 e^{-}\right): \mathrm{BO} & =\frac{10-4}{2}=3 \end{aligned}\) So, \(\left(\mathrm{N}_2, \mathrm{CO}, \mathrm{NO}^{+}\right)\)have similar bond order, 3 .

Hence, the correct answer is option: 'D'.

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