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When $\mathrm{NO}$ is converted to $\mathrm{NO}^{+}$ ,bond order of $\mathrm{N}-\mathrm{O}$ bond changes from
Option: 1
3 to 2

Option: 2
2 to 3

Option: 3
2.5 to 3

Option: 4
remains same.

Answers (1)

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Part 1: Bond order of NO

The total electron of $\mathrm{NO}$ molecules is $7+8=15$

The molecular orbital electronic configuration can be written as

$\mathrm{\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z{ }^2, \pi 2 \mathrm{p}_{\mathrm{x}}{ }^2=\pi 2 \mathrm{py}^2, \pi^* 2 \mathrm{p}_{\mathrm{x}}{ }^1=\pi^* 2 \mathrm{py}_{\mathrm{y}}}$

That is, the number of electrons in bonding $\left(N_b\right)$ is 10

The number of electrons in antibonding $\mathrm{\left(N_b\right)}$ is 5

5 So the bond order of $\mathrm{\mathrm{NO}}$ is;

$\mathrm{B. \mathrm{O}=\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2} }$

$\mathrm{B. \mathrm{O}=\frac{10-5}{2}}$

$\mathrm{\text { B. } O=\frac{5}{2}}$

$\mathrm{\text { B. } \mathrm{O}=2.5}$

Part 2: Bond order of $\mathrm{NO}^{+}$

The total electron of $\mathrm{NO}$ molecule is $7+8=15$

The positive charge indicates the loss of one electron

So the total electron of $\mathrm{NO}^{+}$ molecules is $15-1=14$

The molecular orbital electronic configuration can be written as

$\mathrm{\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z{ }^2, \pi 2 \mathrm{px}^2=\pi 2 \mathrm{py}^2}$

That is, the number of electrons in bonding $\left(N_b\right)$ is 10

The number of electrons in antibonding $\left(N_b\right)$ is 4

So the bond order of $\mathrm{NO}^{+}$ is;

$\mathrm{B. \mathrm{O}=\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}}$
$\mathrm{B. \mathrm{O}=\frac{10-4}{2}}$
$\mathrm{B. \mathrm{O}=\frac{6}{2}}$
$\mathrm{B. \mathrm{O}=3}$

Posted by

Gautam harsolia

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