Chemical Bonding and Molecular Structure - Result Question 106
####108. The pair of species with the same bond order is :
(a) $O_2^{2-}, B_2$
(b) $O_2^{+}, NO^{+}[2012]$
(c) $NO, CO$
(d) $N_2, O_2$
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Solution:
- (a) Both $O_2^{2-}$ and $B_2$ has bond order equal to 1 .
$B_2(10)=[\sigma 1 s^{2} \sigma^{} 1 s^{2} \sigma 2 s^{2} \sigma^{} 2 s^{2} \pi 2 p^{1} _{y} \pi 2 p_z{ }^{1}]$
Bond order $=\frac{N_b-N_a}{2}=\frac{6-4}{2}=\frac{2}{2}=1$
$B_2$ is known to be in the gas phase.
$ \begin{aligned} & O_2{ }^{2-}=\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} \end{aligned} $
Bond order $=\frac{1}{2}(10-8)=1$
109
(b) $(O_2)=\sigma 1 s^{2} \sigma^{} 1 s^{2} \sigma 2 s^{2} \sigma^{} 2 s^{2} \sigma 2 p_z^{2} \pi 2 p_x^{2}$
$ =\pi 2 p_y^{2} \pi^{} 2 p_x^{1}=\pi^{} 2 p_y^{1} $
Bond order $=\frac{N_b-N_a}{2}=\frac{10-6}{2}=\frac{4}{2}=2$
$(O_2^{+}.$ion $)=\sigma 1 s^{2} \sigma^{} 1 s^{2} \sigma 2 s^{2} \sigma^{} 2 s^{2}$
$ \sigma 2 p_z^{2} \pi 2 p_x^{2}=\pi 2 p_y^{2} \pi^{*} 2 p_x^{1} $
Bond order $=\frac{N_b-N_a}{2}=\frac{10-5}{2}=\frac{5}{2}=2 \frac{1}{2}$
$(O_2^{-})=\sigma 1 s^{2} \sigma^{} 1 s^{2} \sigma 2 s^{2} \sigma^{} 2 s^{2} \sigma 2 P_z^{2}$
$\pi 2 p_x^{2}=\pi 2 p_y^{2} \pi^{} 2 p_x^{2} \pi^{} 2 p_y^{1}$
Bond order $=\frac{(N_b-N_a)}{2}=\frac{10-7}{2}=\frac{3}{2}=1 \frac{1}{2}$
$(O_2^{2-})=\sigma 1 s^{2} \sigma^{} 1 s^{2} \sigma 2 s^{2} \sigma^{} 2 s^{2} \sigma 2 p_z^{2} \pi 2 p_x^{2}$
$ =\pi 2 p_y^{2} \pi^{} 2 p_x^{2}=\pi^{} 2 p_y^{2} $
Bond order $\frac{N_b-N_a}{2}=\frac{10-8}{2}=\frac{2}{2}=1$
