Which one of the following pairs of species have the same bond order? [2017]

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Solution

Q.
Which one of the following pairs of species have the same bond order [2017]

1 $O_{2}, {NO}^{+}$

2 ${CN}^{-}, CO$

3 $N_{2}, O_{2}^{-}$

4 $CO, NO$

Ans.
(2)

Molecular orbital electronic configurations and bond order values are:

$O_{2} (16) : σ 1 s^{2}, σ^{} 1 s^{2}, σ 2 s^{2}, σ^{} 2 s^{2}, σ 2 p_{z}^{2}, π 2 p_{x}^{2} = π 2 p_{y}^{2}, π^{} 2 p_{x}^{1} = π^{} 2 p_{y}^{1}$

$B . O . = \frac{1}{2} (N_{b} - N_{a}) = \frac{1}{2} (10 - 6) = 2$

$N O^{+} (14) : σ 1 s^{2}, σ^{} 1 s^{2}, σ 2 s^{2}, σ^{} 2 s^{2}, σ 2 p_{z}^{2}, π 2 p_{x}^{2} = π 2 p_{y}^{2}$

$B . O . = \frac{1}{2} (10 - 4) = 3$

$C N^{-} (14) : σ 1 s^{2}, σ^{} 1 s^{2}, σ 2 s^{2}, σ^{} 2 s^{2}, π 2 p_{x}^{2} = π 2 p_{y}^{2}, σ 2 p_{z}^{2}$

$B . O . = \frac{1}{2} (10 - 4) = 3$

$C O (14) : σ 1 s^{2}, σ^{} 1 s^{2}, σ 2 s^{2}, σ^{} 2 s^{2}, π 2 p_{x}^{2} = π 2 p_{y}^{2}, σ 2 p_{z}^{2}$

$B . O . = \frac{1}{2} (10 - 4) = 3$

$N_{2} (14) : σ 1 s^{2}, σ^{} 1 s^{2}, σ 2 s^{2}, σ^{} 2 s^{2}, π 2 p_{x}^{2} = π 2 p_{y}^{2}, σ 2 p_{z}^{2}$

$B . O . = \frac{1}{2} (10 - 4) = 3$

$O_{2}^{-} (17) : σ 1 s^{2}, σ^{} 1 s^{2}, σ 2 s^{2}, σ^{} 2 s^{2}, σ 2 p_{z}^{2}, π 2 p_{x}^{2} = π 2 p_{y}^{2}, π^{} 2 p_{x}^{2} = π^{} 2 p_{y}^{1}$

$B . O . = \frac{1}{2} (10 - 7) = 1.5$

$N O (15) : σ 1 s^{2}, σ^{} 1 s^{2}, σ 2 s^{2}, σ^{} 2 s^{2}, σ 2 p_{z}^{2}, π 2 p_{x}^{2} = π 2 p_{y}^{2}, π^{*} 2 p_{x}^{1}$

$B . O . = \frac{1}{2} (10 - 5) = 2.5$

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