Chemical Bonding and Molecular Structure - Result Question 95
####97. 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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Solution:
- (a) Oxygen molecule $(O_2)$ - Total number of electrons $=16$ and electronic configuration is
$\sigma 1 s^{2}<\sigma^{} 1 s^{2}<\sigma 2 s^{2}<\sigma^{} 2 s^{2}<\sigma 2 p_x^{2}$
$ <\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
$ \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
$ \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^{2}=\pi^{} 2 p_z^{1} \end{aligned} $
Bond order $=\frac{(N_b-N_a)}{2}=\frac{10-7}{2}=\frac{3}{2}=1 \frac{1}{2}$
