Question

Compare Figs. 14.38 and 14.40. Why are they different? Because B₂ is known to be paramagnetic, the ${\mathrm{\pi }}_{2\mathrm{p}}$and ${\sigma }_{2\mathrm{p}}$ MOs must be switched from our first prediction. What is the rationale for this? Why might one expect the ${\sigma }_{2\mathrm{p}}$to be lower in energy than the ${\mathrm{\pi }}_{2\mathrm{p}}$? Why can't we use diatomic oxygen to help us decide whether the ${\sigma }_{2\mathrm{p}}$or ${\mathrm{\pi }}_{2\mathrm{p}}$is lower in energy?

Short answer

Figure 14.38 is formed without considering the s-p mixing of orbitals whereas figure 14.40 is created with consideration of s-p mixing.

Π_2p is low lying-in energy compared to σ_2p orbital in the case of molecules where s-p mixing of orbitals is considered.

σ_2p is expected to be lower in energy than Π_2p because σ_2p is formed by the direct overlap of orbitals with more electron density between the nucleus of the atoms whereasΠ_2p is formed by the lateral overlap of orbitals with less electron density between the nucleus of the atoms.

In case of diatomic oxygen, the s-p mixing is not considered due to its greater molecular weight.

Step-by-step solution

Step 1. Formation of sigma orbitals.

$\sigma$orbitals are formed by the direct overlap of $s-s$orbitals and $p_z-p_z$ orbitals.

The remaining degenerate $p_x,p_y$orbitals form a lateral overlap with $p_x-p_x$ and$p_y-p_y$ atomic orbitals forming degenerateΠ MOs.

Step 2. Explanation

These molecular orbitals afterward undergo s-p mixing for lighter elements increasing the energy of σ_2p overπ_2p

InB₂ molecule σ_2s and σ_2p orbitals have almost same energy that is why there is a s-p mixing and after mixing σ_2p orbital has higher energy and σ_2s orbital lowers its energy in B₂ molecule σ_2p orbital has higher energy then Π_2p orbital.

In case of diatomic oxygen, the s-p mixing is not considered due to its greater molecular weight