Chemical Bonding And Molecular Structure Ques 125
125. Which of the following options represents the correct bond order?
[2015]
(a) $O_2^{-}<O_2<O_2^{+}$
(b) $O_2^{-}>O_2<O_2^{+}$
(c) $O_2^{-}<O_2>O_2^{+}$
(d) $O_2^{-}>O_2>O_2^{+}$
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Answer:
Correct Answer: 125.(a)
Solution:
(a) Oxygen molecule $(O_2)$ - Total number of electrons $=16$ and electronic configuration is
$\sigma 1 s^2 < \sigma^* 1 s^2 < \sigma 2 s^2 < \sigma^* 2 s^2 < \sigma 2 p_x^2$
$\hspace{20mm} <\pi 2 p_y^2=\pi 2 p_z^2 < \pi^* 2 p_y^1=\pi^* 2 p_z^1 $
Bond order $=\frac{N_b-N_a}{2}=\frac{10-6}{2}=\frac{4}{2}=2$
$O_2^{+}$ion - Total number of electrons $(16-1)=15$.
Electronic configuration
$\sigma 1 s^2 < \sigma^* 1 s^2 < \sigma 2 s^2 < \sigma^* 2 s^2 < \sigma 2 p_x^2 $
$\hspace{20mm} <\pi 2 p_y^2 =\pi 2 p_z^2<\pi^* 2 p_y^1$
Bond order $=\frac{N_b-N_a}{2}=\frac{10-5}{2}=\frac{5}{2}=2 \frac{1}{2}$
$O_2^{-}$(Super oxide ion) Total number of electrons $(16+1)=17$. Electronic configuration
$\sigma 1 s^2 < \sigma^* 1 s^2 < \sigma 2 s^2 < \sigma^* 2 s^2 < \sigma 2 p_x^2 $
$\hspace{20mm} <\pi 2 p_y^2=\pi 2 p_z^2 < \pi^* 2 p_y^2=\pi^* 2 p_z^1$
Bond order $=\frac{(N_b-N_a)}{2}=\frac{10-7}{2}=\frac{3}{2}=1 \frac{1}{2}$
BETA
