Chemical Bonding And Molecular Structure Ques 11
11. Bond order of $1.5$ is shown by:
[2012]
(a) $\mathrm{O}_2^{+}$
(b) $\mathrm{O}_2^{-}$
(c) $\mathrm{O}_2^{2-}$
(d) $\mathrm{O}_2$
Show Answer
Answer:
Correct Answer: 11.(b)
Solution: (b)
$\left(O_2\right)=\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^2 $
$ =\pi 2 p_y^2 \pi^* 2 p_x^1=\pi^* 2 p_y^1 $
$ \text { Bond order }=\frac{N_b-N_a}{2}=\frac{10-6}{2}=\frac{4}{2}=2 $
$ \left(\mathrm{O}_2^{+} \text {ion }\right)=\sigma \mid s^2 \sigma^4 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 $
$ \sigma 2 p_z^2 \pi 2 p_x^2=\pi 2 p_y^2 \pi^* 2 p_x^1 $
$ \text { Bond order }=\frac{N_b-N_a}{2}=\frac{10-5}{2}=\frac{5}{2}=2 \frac{1}{2} $
$ \left(\mathrm{O}_2^{-}\right)=\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^2=\pi 2 p_y^2 \pi^* 2 p_x^2 \pi^* 2 p_y^1 $
$ \text { Bond order }=\frac{\left(N_b-N_a\right)}{2}=\frac{10-7}{2}=\frac{3}{2}=1 \frac{1}{2} $
$ \left(\mathrm{O}_2^{2-}\right)=\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^2 $
$ =\pi 2 p_y^2 \pi^* 2 p_x^2=\pi^* 2 p_y^2 $
$ \text { Bond order } \frac{N_b-N_a}{2}=\frac{10-8}{2}=\frac{2}{2}=1 $
BETA
