Chemical Bonding And Molecular Structure Ques 124
124. The correct bond order in the following species is:
[2015]
(a) $O_2^{2+}<O_2^{-}<O_2^{+}$
(b) $O_2^{+}<O_2^{-}<O_2^{2+}$
(c) $O_2^{-}<O_2^{+}<O_2^{2+}$
(d) $O_2^{2+}<O_2^{+}<O_2^{-}$
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Answer:
Correct Answer: 124.(c)
Solution:
(c) $O_2^{+}$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 $
$\hspace{30mm} <\pi 2 p_y^{2} =\pi 2 p_z^{2}<\pi^{*} 2 p_y^{1}$
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
$\sigma 1 s^2 < \sigma^* 1 s^2 < \sigma 2 s^2 < \sigma^* 2 s^2 < \sigma 2 p_x^2$
$\hspace{30mm} < \pi 2 p_y^2=\pi 2 p_z^2 < \pi^* 2 p_y^2=\pi^* 2 p_z^1$
Bond order $=\frac{(N_b-N_a)}{2}=\frac{10-7}{2}=\frac{3}{2}=1 \frac{1}{2}$
$O_2^{+2}$ ion: Total number of electrons
$=(16-2)=14$ Electronic configuration
$\sigma 1 s^2 < \sigma^* 1 s^2 < \sigma 2 s^2 < \sigma^* 2 s^2 < \sigma 2 p_x^2 < \pi 2 p_y{ }^2$
$=\pi 2 p_z^{2}$
Bond order $=\frac{(N_b-N_a)}{2}=\frac{10-4}{2}=\frac{6}{2}=3$
So bond order: $O_2{ }^{-}<O_2{ }^{+}<O_2{ }^{2+}$
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