First slide

Molecular Orbital Theory

Question

Which of the following species is diamagnetic ?

Moderate

A

$N_{2}$
B

$N_{2}^{+}$
C

$N_{2}^{-}$
D

$N_{2}^{2 -}$

Solution

$N_{2} - {KK}^{'} {(σ_{2 s})}^{2} {(σ_{2 s}^{})}^{2} {(π_{2 p_{x}})}^{2} {(π_{2 p_{y}})}^{2} {({σ_{2 p}}_{z})}^{2}; N_{2}^{+} - {KK}^{'} {(σ_{2 s})}^{2} {(σ_{2 s}^{})}^{2} {(π_{2 p_{x}})}^{2} {(π_{2 p_{y}})}^{2} {({σ_{2 p}}_{z})}^{1}; N_{2}^{-} - {KK}^{'} {(σ_{2 s})}^{2} {(σ_{2 s}^{})}^{2} {(π_{2 p_{x}})}^{2} {(π_{2 p_{y}})}^{2} {({σ_{2 p}}_{z})}^{2} {(π_{2 p}^{})}^{1}; N_{2}^{2 -} - {KK}^{'} {(σ_{2 s})}^{2} {(σ_{2 s}^{})}^{2} {(π_{2 p_{x}})}^{2} {(π_{2 p_{y}})}^{2} {({σ_{2 p}}_{z})}^{2} {(π_{2 px}^{})}^{1} {(π_{2 py}^{*})}^{1};$
Bond order and magnetic moment :
$μ_{s} = \sqrt{n (n + 2)} B . M; N_{2}^{- 2} = \frac{10 - 6}{2} = 2; Paramagnetic . . . μ_{s} = 2.82 B . M; N_{2}^{-} = \frac{10 - 5}{2} = 2.5; Paramagnetic . . . μ_{s} = 1.732 B . M; N_{2} = \frac{10 - 4}{2} = 3; Diamagnetic . . . . . μ_{s} = 0; N_{2}^{+} = \frac{9 - 4}{2} = 2.5; Paramagnetic . . μ_{s} = 1.732 B . M;$
Bond order $\propto$ $\frac{1}{Bond length}$
Stability $\propto$ Bond order;
Stability----- $N_{2} (3) > N_{2}^{+} (2.5) > N_{2}^{-} (2.5) > N_{2}^{2 -} (2);$
Higher stability of $N_{2}^{+} over N_{2}^{-}$ is due to lesser number of ABMO electrons in $N_{2}^{+}$ .

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App