Q3 / 180 · KCET 2011 · Chemistry

When $\mathrm{O}_{2}$ is converted into $\mathrm{O}_{2}^{+}$

  • A
    both paramagnetic character and bond order increase
  • B
    bond order decreases
  • C
    paramagnetic character increases
  • Correct answer
    D
    paramagnetic character decreases and the bond order increases

Solution

(i) $\mathrm{O}_{2}-(\sigma 1 s)^{2}(\sigma * 1 s)^{2}(\sigma 2 s)^{2}\left(\sigma^{*} 2 s\right)^{2}$ $$ \begin{aligned} &\left(\sigma 2 p_{z}\right)^{2}\left(\pi 2 p_{x}\right)^{2}\left(\pi 2 p_{y}\right)^{2}\left(\pi * 2 p_{x}\right)^{1} \\ &\left(\pi^{*} 2 p_{y}\right)^{1} \end{aligned} $$ Bond order $=\frac{N_{b}-N_{a}}{2}$ $$ =\frac{8-4}{2} $$ $$ =2 $$ $\mathrm{O}_{2}$ molecule having 2 unpaired electron. (ii) $\mathrm{O}_{2}^{+}-(\sigma 1 s)^{2}\left(\sigma^{*} 1 s\right)^{2}(\sigma 2 s)^{2}\left(\sigma^{*} 2 s\right)^{2}$ $$ \begin{aligned} &\left(\sigma 2 p_{z}\right)^{2}\left(\pi 2 p_{x}\right)^{2}\left(\pi 2 p_{y}\right)^{2}\left(\pi^{*} 2 p_{x}\right)^{1} \\ &\left(\pi^{*} 2 p_{y}\right)^{0} \end{aligned} $$ Bond order $=\frac{8-3}{2}$ $$ =2.5 $$ $\mathrm{O}_{2}^{+}$ion having only 1 unpaired electron. Hence, when $\mathrm{O}_{2}$ is converted into $\mathrm{O}_{2}^{+}$ paramagnetic character decrease and the bond order increases.

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

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