Question

3. Bond order of NO +is x and bond order of NO- is y. x + y is equal to

Answer 1

Answer:

C−O: σ1s

2

σ

∗

1s

2

σ2s

2

σ

∗

2s

2

π2P

x

2

=π2p

y

2

σ2P

y

2

Bondorder=

2

No.ofbondingelectrons(10)−No.ofelectronsinABMO(4)

=3

N−O: σ1s

2

σ

∗

1s

2

σ2s

2

σ

∗

2s

2

π2P

x

2

=π2P

z

2

σ2P

z

2

π

∗

2P

x

1

Bondorder=

2

10−5

=2.5

Answer 2

Answer:

The Bond order of  NO⁺ (x) is 3 and bond order of NO⁻ (y) is equal to 2.

The value of (x+y) is equal to 5.

Explanation:  

From the energy level diagram of molecular orbitals of NO, the electron distribution will be as:

$N-O: \sigma 1s^{2}\; \sigma^{*}1s^{2}\; \sigma2s^{2}\; \sigma ^{*} 2s^{2}\; \pi 2p_{x}^{2}=\pi 2p_{z}^{2}\ \sigma2p_{z}^{2}\ \pi ^{*}2p_{x}^{1}$

During the formation of NO⁺, one electron is removed from NO:

$N-O^{+}: \sigma 1s^{2}\; \sigma^{*}1s^{2}\; \sigma2s^{2}\; \sigma ^{*} 2s^{2}\; \pi 2p_{x}^{2}=\pi 2p_{z}^{2}\ \sigma2p_{z}^{2}\ \pi ^{*}2p_{x}^{0}$

Bond order $=[\frac{(Number \ of \: Bonding \; electron)- (Number \ of \ Antibondig \ electrons)}{2}]$

Therefore, Bond order of  NO⁺ $=\frac{10-4}{2}=\frac{6}{2} =3$

Now, During the formation of NO⁻, one electron added in NO:

$N-O^{-}: \sigma 1s^{2}\; \sigma^{*}1s^{2}\; \sigma2s^{2}\; \sigma ^{*} 2s^{2}\; \pi 2p_{x}^{2}=\pi 2p_{z}^{2}\ \sigma2p_{z}^{2}\ \pi ^{*}2p_{x}^{2}$

Bond order of  NO⁻ $=\frac{10-6}{2}=\frac{4}{2} =2$

Now, bond order of  NO⁺, $x=3$

Bond order of  NO⁻ , $y=2$

So, $x+y=3+2=5$

Therefore, the value of (x + y) is equal to 5.