Question

Why does the molecular orbital model do a better job in explaining the bonding in $\mathit{N}{\mathit{O}}^{\mathbf{-}}$and NO then the hybrid orbital model?

Short answer

The molecular orbital model does a better job in explaining the bonding in $N{O}^{-}$and NO because it can predict the accurate bond order and bonding pattern of the molecule whereas the hybridization model fails to do so.

Step-by-step solution

Determine the electronic configuration

Electronic configuration N atom: $1s^{2}2s^{2}2p^3$

Electronic configuration O atom:$1s^{2}2s^{2}2p^4$

Molecular orbital configuration of NO is

${\left(\sigma 1s\right)}^2{\left(\sigma 1s^{*}\right)}^2{\left(\sigma 2s\right)}^2{\left(\sigma 2s^{*}\right)}^2{\left(\pi 2p_x\right)}^2{\left(\pi 2p_y\right)}^2{\left(\sigma 2p_x\right)}^2{\left(\pi 2{p_x}^{*}\right)}^1$

Molecular orbital configuration of $N{O}^{-}$

${\left(\sigma 1s\right)}^2{\left(\sigma 1s^{*}\right)}^2{\left(\sigma 2s\right)}^2{\left(\sigma 2s^{*}\right)}^2{\left(\pi 2p_x\right)}^2{\left(\pi 2p_y\right)}^2{\left(\sigma 2p_x\right)}^2{\left(\pi 2{p_x}^{*}\right)}^2$

Determine the bond order 

Bond order of $N{O}^{-}$:$\frac{1}{2}\frac{2}{a}\left(10-6\right)=2$

Bond order of NO:${\frac{1}{2}}^2\left(10-5\right)=25$

Explanation

Electrons are present in antibonding π orbitals which helps to predict the bonding pattern whereas the hybrid model fails to do so.

The molecular orbital model does a better job in explaining the bonding in and NO because it can predict the accurate bond order and bonding pattern of the molecule whereas the hybridization model fails to do so.