Chemical Bonding and Molecular Structure - Result Question 94
####96. 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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Solution:
- (c) $O_2^{+}$ion - Total number of electrons $(16-1)=15$.
Electronic configuration
$ \begin{aligned} \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}<\pi^{*} 2 p_y^{1} \end{aligned} $
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
$\sigma 1 s^{2}<\sigma^{} 1 s^{2}<\sigma 2 s^{2}<\sigma^{} 2 s^{2}<\sigma 2 p_x^{2}$ $<\pi 2 p_y^{2}=\pi 2 p_z^{2}<\pi^{} 2 p_y^{2}=\pi^{} 2 p_z^{1}$
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: $O_2{ }^{-}<O_2{ }^{+}<O_2{ }^{2+}$
