Question

What if the${\mathit{\pi }}_{\mathbf{2}\mathbf{p}}$orbitals were lower in energy than the${\mathit{\sigma }}_{\mathbf{2}\mathbf{p}}$orbitals? What would you expect the${\mathit{B}}_{\mathbf{2}}$molecular orbital energy level diagram to look like (without considering p-s mixing)? Compare your expected diagram to Figs. 14.37 and 14.38 and state the differences for each.

Short answer

The new MO diagram looks like this;

The calculated MO is paramagnetic, and the one in the book is diamagnetic.

Step-by-step solution

The Molecular Orbital (MO) Diagram.

A molecular orbital diagram helps explain the chemical bonding in differentmolecules through thelinear combination of atomic orbitals (LCAO).

Here,atomic orbitals are attached to formmolecular orbitals of the same number and are used for diatomic molecules only.

The diagram also helps in predicting bond strength and the differentelectronic transitions taking place.

The expected molecular orbital energy level diagram.

The molecule of boron contains 10 electrons in total.

$$\begin{gathered} \text{B}=1s^{2}2s^{2}2{\text{p}}^1 \\ {\text{B}}_{\text{2}}=1s^{2}1\text{s}{*}^{2}2s^{2}2{\text{s*}}^{2}2{\text{p}}^{1}2{\text{p*}}^1 \end{gathered}$$

If the ${\pi }_{2p}$orbitals were lower in energy than the ${\sigma }_{2p}$orbitals.

The new ${\text{B}}_{\text{2}}$molecular orbital energy level diagram will look like;

Here, themolecule is Paramagnetic because of the presence of unpaired electron.

The electronic configuration for 14 electrons and below goes as;

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

The filling of ${\pi }_{2p_y},{\pi }_{2p_z}$is done together because they are present on same energy level.

In Paramagnetism, the substances are attracted toward the inducing magnetic field.

The molecular orbital energy level diagram as per book.

The molecular orbital diagram in the book looks like;

The electronic configuration of electrons more than 14 goes as;

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

Here, the molecule is Diamagnetic because of the absence of unpaired electron.

In Diamagnetism, the substances are repelled from the inducing magnetic field.