Q58 / 200 · NEET 2012 (Screening) · Chemistry

The pair of species with the same bond order is

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

Solution

Bond order of \(\mathrm{O}_2{ }^{2-}\) : Electronic : \(\sigma 1 \mathrm{~s}^2 \sigma^{\star}\left(\mathrm{s}^2 \sigma 2 \mathrm{~s}^2 \sigma^* 2 \mathrm{~s}^2 \sigma 2 \mathrm{P}_{\mathrm{z}}{ }^2\left(\pi 2 \mathrm{P}_{\mathrm{x}}{ }^2=\pi 2 \mathrm{P}_{\mathrm{y}}{ }^2\right)\right.\) \(\left(\pi 2 \mathrm{P}_{\mathrm{x}}^2=\pi 2^{\star} \mathrm{P}_{\mathrm{y}}^2\right) \sigma^{\star} 2 \mathrm{pz}{ }^{\circ}\) Bond order \(=\frac{\text { Bonding }- \text { Antibonding electrons }}{2}=\frac{10-8}{2}=1\) Bond order of \(B_2\) : \(\sigma 1 s^2 \sigma_{1 s}{ }^2 2 \sigma_{2 s} 2 \sigma^{\star} 2 s^2 \sigma 2 p_z^2\) Bond order \(=\frac{6-4}{2}=1\) \(\therefore \mathrm{O}_2{ }^{2-}\) and \(\mathrm{B}_2\) has same bond order 1.

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

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