Chemical Bonding and Molecular Structure - Result Question 106
####108. The pair of species with the same bond order is :
(a)
(b)
(c)
(d)
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Solution:
- (a) Both
$B_2(10)=[\sigma 1 s^{2} \sigma^{} 1 s^{2} \sigma 2 s^{2} \sigma^{} 2 s^{2} \pi 2 p^{1} _{y} \pi 2 p_z{ }^{1}]$
Bond order
$ \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
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
Bond order
$(O_2^{-})=\sigma 1 s^{2} \sigma^{} 1 s^{2} \sigma 2 s^{2} \sigma^{} 2 s^{2} \sigma 2 P_z^{2}$
$\pi 2 p_x^{2}=\pi 2 p_y^{2} \pi^{} 2 p_x^{2} \pi^{} 2 p_y^{1}$
Bond order
$(O_2^{2-})=\sigma 1 s^{2} \sigma^{} 1 s^{2} \sigma 2 s^{2} \sigma^{} 2 s^{2} \sigma 2 p_z^{2} \pi 2 p_x^{2}$
$ =\pi 2 p_y^{2} \pi^{} 2 p_x^{2}=\pi^{} 2 p_y^{2} $
Bond order
