Chapter 9, Problem 73AP

Interpretation Introduction

Interpretation:

The electronic configuration, bond order, and magnetic properties of ${\text{P}}_{\text{2}}$ are to be determined.

Concept introduction:

Electronic configuration of a molecule is the arrangement of electrons in the molecular orbitals of the molecule.

Two atomic orbitals combine to form a bonding molecular orbital and an antibonding molecular orbital. Orbitals that lie on internuclear axis combine to form $\sigma$

(sigma) molecular orbitals, and orbitals parallel to each other combine to form $\pi$

molecular orbitals.

The molecular orbitals formed by the combination of $\text{1s}$

orbital are a bonding molecular orbital, designated by ${\sigma }_{1s}$, and an antibonding molecular orbital, designated by ${\sigma }^{\ast }{}_{1s}$. The $\text{2s}$

orbital forms corresponding molecular orbitals.

The molecular orbitals formed by the combination of ${\text{2p}}_{\text{x}}$

orbitals, are a bonding molecular orbital, designated by ${\sigma }_{2{\text{p}}_{\text{x}}}$, and an antibonding molecular orbital, designated by ${\sigma }^{\ast }{}_{2{\text{p}}_{\text{x}}}$.

The molecular orbitals formed by combining ${\text{2p}}_{\text{y}}$ and ${\text{2p}}_{\text{z}}$

orbitals are bonding molecular orbitals, designated by ${\text{π}}_{{\text{2p}}_{\text{y}}}$

and ${\text{π}}_{{\text{2p}}_{\text{z}}}$, and antibonding molecular orbitals, designated by ${\text{π}}^{\ast }{}_{{\text{2p}}_{\text{y}}}$

and ${\text{π}}^{\ast }{}_{{\text{2p}}_{\text{z}}}$.

Electrons are filled in the molecular orbitals in increasing order of energy.

Bond order is determined by subtracting the number of electrons in antibonding orbitals from the number of electrons in bonding orbitals, and dividing by two.

A molecule in which no unpaired electron is present is diamagnetic, whereas one in which all electrons are paired is paramagnetic.

Answer to Problem 73AP

Solution:

a) $\left[{\text{Ne}}_{\text{2}}\right]{\left({\sigma }_{3s}\right)}^2{\left({\sigma }^{\ast }{}_{3s}\right)}^2{\left({\text{π}}_{{\text{3p}}_{\text{y}}}\right)}^2{\left({\text{π}}_{{\text{3p}}_{\text{z}}}\right)}^2{\left({\sigma }_{{\text{3p}}_{\text{x}}}\right)}^2$

b) $3$

c) Diamagnetic

Explanation of Solution

a) The electronic configuration of ${\text{P}}_{\text{2}}$

The electronic configuration of a phosphorus atom is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}3{\text{s}}^2{\text{3p}}^{\text{3}}$. The electronic configuration of $\text{Ne}$ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}$. Thus, the electronic configuration of phosphorus atom can be written as $\left[\text{Ne}\right]3{\text{s}}^2{\text{3p}}^{\text{3}}$.

The electronic configuration of a nitrogen atom is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{3}}$

and that of a nitrogen molecule is as follows:

${\left({\sigma }_{1s}\right)}^2{\left({\sigma }^{\ast }{}_{1s}\right)}^2{\left({\sigma }_{2s}\right)}^2{\left({\sigma }^{\ast }{}_{2s}\right)}^2{\left({\text{π}}_{{\text{2p}}_{\text{y}}}\right)}^2{\left({\text{π}}_{{\text{2p}}_{\text{z}}}\right)}^2{\left({\sigma }_{{\text{2p}}_{\text{x}}}\right)}^2$

Based on nitrogen molecule, and considering only the valence electrons, we can write the electronic configuration of ${\text{P}}_{\text{2}}$

as follows:

$\left[{\text{Ne}}_{\text{2}}\right]{\left({\sigma }_{3s}\right)}^2{\left({\sigma }^{\ast }{}_{3s}\right)}^2{\left({\text{π}}_{{\text{3p}}_{\text{y}}}\right)}^2{\left({\text{π}}_{{\text{3p}}_{\text{z}}}\right)}^2{\left({\sigma }_{{\text{3p}}_{\text{x}}}\right)}^2$

b) The bond order of ${\text{P}}_{\text{2}}$

Bond order is determined by subtracting the number of electrons in antibonding orbitals from the number of electrons in bonding orbitals, and dividing by two.

In ${\text{P}}_{\text{2}}$, after the configuration of $\text{Ne}$, there are eight electrons in bonding molecular orbital and two in antibonding orbital. Thus, the bond order is calculated as follows:

$\begin{array}{c}\text{b}\text{.o}=\frac{8-2}{2}\\ =3\end{array}$

c) Magnetic properties of ${\text{P}}_{\text{2}}$

The electronic configuration of ${\text{P}}_{\text{2}}$

molecule shows that all electrons are paired. Thus, the molecule is diamagnetic.