Match List – I (species) and List – II (bond orders and select the correct answer :
List – I (species) List – II (bond orders)
A) $N_{2}$ I) 1.0
B) $O_{2}$ II) 2.0
C) $F_{2}$ III) 2.5
D) $O_{2}^{+}$ IV) 3.0

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A-IV, B-II, C-I, D-III

A-I, B-III, C-IV, D-II

A-I, B-II, C-IV, D-III

A-IV, B-III, C-I, D-II

answer is A.

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Detailed Solution

$N_{2} : σ {(2 s)}^{2} \overset{*}{σ} {(2 s)}^{2} π {(2 p_{x})}^{2} π {(2 p_{y})}^{2} σ {(2 p_{z})}^{2}$

Bond order $= \frac{1}{2} (8 - 2) = 3$
$O_{2} : {(σ 2 s)}^{2} {(\overset{*}{σ} 2 s)}^{2} {(σ 2 p_{z})}^{2} {(π_{y} 2 p_{y})}^{2} {(π_{z} 2 p_{x})}^{2} {(\overset{*}{π_{y}} 2 p_{y})}^{1} {(\overset{*}{π_{z}} 2 p_{x})}^{1}$
Bond order = $\frac{1}{2} (8 - 4) = 2$
$F_{2} : {(σ 2 s)}^{2} {(\overset{*}{σ} 2 s)}^{2} {(σ 2 p_{z})}^{2} {(π_{y} 2 p_{y})}^{2} {(π_{z} 2 p_{x})}^{2} {(\overset{*}{π_{y}} 2 p_{y})}^{2} {(\overset{*}{π_{z}} 2 p_{x})}^{2}$
Bond order = $\frac{1}{2} (8 - 6) = 1$
$O_{2}^{+} : {(σ 2 s)}^{2} {(\overset{*}{σ} 2 s)}^{2} {(σ 2 p_{z})}^{2} {(π_{y} 2 p_{y})}^{2} {(π_{z} 2 p_{x})}^{2} {(\overset{*}{π_{y}} 2 p_{y})}^{1}$
Bond order = $\frac{1}{2} (8 - 3) = 2.5$