First slide

Molecular Orbital Theory

Question

Bond order in $O_{2}^{+}$ is ‘X’. Bond order in $O_{2}^{2 -}$ is ‘Y’. Difference between X and Y is

Moderate

A

1
B

0.5
C

Zero
D

1.5

Solution

$O_{2}^{- 2} - {KK}^{'} {(σ_{2 s})}^{2} {(σ_{2 s}^{})}^{2} {({σ_{2 p}}_{z})}^{2} {(π_{2 p_{x}})}^{2} {(π_{2 p_{y}})}^{2} {(π_{2 p_{x}}^{})}^{2} {(π_{2 p_{y}}^{})}^{2}$
$O_{2}^{-} - {KK}^{'} {(σ_{2 s})}^{2} {(σ \overset{}{_{2 s}})}^{2} {({σ_{2 p}}_{z})}^{2} {(π_{2 p_{y}})}^{2} {(π_{2 p_{y}})}^{2} {(π_{2 p_{x}}^{})}^{2} {(π_{2 p_{y}}^{})}^{1}$
$O_{2}^{} - {KK}^{'} {(σ_{2 s})}^{2} {(σ_{2 s}^{})}^{2} {({σ_{2 p}}_{z})}^{2} {(π_{2 p_{x}})}^{2} {(π_{2 p_{y}})}^{2} {(π_{2 p_{x}}^{})}^{1} {(π_{2 p_{y}}^{})}^{1}$
$O_{2}^{+} - {KK}^{'} {(σ_{2 s})}^{2} {(σ_{2 s}^{})}^{2} {(σ_{2 p_{z}})}^{2} {(π_{2 p_{x}})}^{2} {(π_{2 p_{y}})}^{2} {(π_{2 p_{x}}^{})}^{1} {(π_{2 p_{y}}^{})}^{0}$
Bond order and magnetic moment :
$μ_{s} = \sqrt{n (n + 2)} B . M; O_{2}^{- 2} = \frac{10 - 8}{2} = 1; Diamagnetic . . . μ_{s} = 0; O_{2}^{-} = \frac{10 - 7}{2} = 1.5; Paramagnetic . . μ_{s} = 1.732 B . M; O_{2} = \frac{10 - 6}{2} = 2; Paramagnetic . . . . . . . μ_{s} = 2.82 B . M; O_{2}^{+} = \frac{10 - 5}{2} = 2.5; Paramagnetic . . . . . μ_{s} = 1.732 B . M;$
O $—$ O bond length in
$O_{2}^{- 2} > O_{2}^{-} > O_{2} > O_{2}^{+}$

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