Question

Find the no. of molecules which will not exhibit sp mixing.
$NO$, $N_2{B}_2,,C_2,O_2,O{{}_2}^{2^{+}}O{{}_2}^{2^{-}},N{{}_2}^{2^{+}},N{{}_2}^{+}$

Answer 1

5
Moving across the period as the electronegativity of atoms increases, the nucleus more strongly pulls the nearer '2s' than for the '2p'. Due to the difference between attraction of
the nucleus for 2s and 2p, the energy gap between 2s and 2p orbitals increases from boron to fluorine.
Example: $O_{2,}{F}_{2,}{O}_2^{2+},NO,O_2^{2+}etc.$ (Diatomic molecules having more than 14 electrons).
In atoms like Li, Be, B, C, N the energy difference between 2s and 2p orbitals quite close as expected and low gap between energy levels of these orbitals causes $\sigma \ast 2s$ orbitals and $\sigma \ast 2p_z$ orbital the diatomic molecule formed by mixing these atoms to overlap. This is called $"2s-2p\ast "$ mixing or s-p mixing. Due to this, the energy of $\sigma 2p_z$ increases and it's energy gets higher than $\pi ,2p_{x}and\pi 2p_{y^{\prime }}$
Example: Diatomic molecules having less than or equal to 14 electrons in all show s-p mixing.
$N_2^{+},N_2,CO,C_2,B_2,N_2^{2+}etc.$
$B_2\to 10e^{-5},N_2\to 14e^{-5},N_2^{+}\to 13e^{-5},C_2\to 13e^{-5}$
So, correct option is 5