Question

explain the observation that the bond length in N-2 is 0.02Å
greater
than in N2, while the bond length in
NO+ is 0.09Å less than in No​

Answer 1

AnSwer :-

$\Large\star$The electronic configuration should be written for the 4 molecules according to the buildup principle

$\sf N_2 \: \: \: \: \: \: \: \: \: \: \: \: (K \: electrons) \sigma^{2}_s \sigma^{*2}_s\pi^{4}_{x,y} \sigma^{2}_{p_z}$

$\sf N ^{ + } _2 \: \: \: \: \: \: \: \: \: \: \: \: (K electrons) \sigma^{2}_s \sigma^{*2}_s\pi^{4}_{x,y} \sigma^{1}_{p_z}$

$\sf NO\: \: \: \: \: \: \: \: \: \: \: \: \: (K electrons) \sigma^{2}_s \sigma^{*2}_s\pi^{4}_{x,y} \sigma^{2}_{p_z} \pi^{*1}_{x,y}$

$\sf NO^{ + } \: \: \: \: \: \: \: \: \: (K electrons) \sigma^{2}_s \sigma^{*2}_s\pi^{4}_{x,y} \sigma^{2}_{p_z}$

The computed bond orders for N$_2$ and⠀ N $_{2}^{+}$ are 3 and $\sf 2 \dfrac{1}{2}$, respectively.

H$_{2}$ therefore has the stronger bond and should have the shorter bond length. The computed bond orders for NO and NO+ are $\sf 2 \dfrac{1}{2}$ and 3, respectively.

NO+ has the stronger bond and should have the shorter bond length. As opposed to ionization of N$_2$, which involves the loss of an electron in a bonding orbital, ionization of NO involves the loss of an electron in an antibonding orbital.