Chapter 9, Problem 92AP

Write the ground-state electron configuration for ${\text{B}}_{\text{2}}\text{.}$ Is the molecule diamagnetic or paramagnetic?

Interpretation Introduction

Interpretation:

The electron configuration of ${\text{B}}_{\text{2}}$

in ground state and magnetic properties of ${\text{B}}_{\text{2}}$

are to be determined.

Concept introduction:

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

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

molecular orbital, and orbitals parallel to each other combine to form $\pi$

molecular orbitals.

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

orbital forms bonding molecular orbital designated as ${\sigma }_{1s}$ and antibonding molecular orbital ${\sigma }^{\ast }{}_{1s}$. The $\text{2s}$ orbital forms corresponding molecular orbitals.

Molecular orbital formed by the combination of ${\text{2p}}_{\text{x}}$

orbital forms a bonding molecular orbital designated as ${\sigma }_{2{\text{p}}_{\text{x}}}$

and an antibonding molecular orbital designated as ${\sigma }^{\ast }{}_{2{\text{p}}_{\text{x}}}$.

Molecular orbitals formed by combining ${\text{2p}}_{\text{y}}$ and ${\text{2p}}_{\text{z}}$ orbital forms bonding molecular orbitals designated as ${\text{π}}_{{\text{2p}}_{\text{y}}}$

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

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

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

Answer to Problem 92AP

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

Explanation of Solution

The electron configuration of a boron atom is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{1}}$. In a ${\text{B}}_{\text{2}}$ molecule, two boron atoms are present. Thus, there are total 10 electrons to be accommodated in the molecular orbitals of ${\text{B}}_{\text{2}}$.

The electron configuration for ${\text{B}}_{\text{2}}$

in ground state is as follows:

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

There are two unpaired electrons in ${\text{π}}_{{\text{2p}}_{\text{y}}}$

and ${\text{π}}_{{\text{2p}}_{\text{z}}}$

orbitals (because p_y and p_x orbitals have same energy. Hence, pairing of electrons will be done only when each element is singly occupied). Hence, ${\text{B}}_{\text{2}}$

molecule is paramagnetic (due to the presence of unpaired electron in the molecular orbital).

Conclusion

The ${\text{B}}_{\text{2}}$

molecule is paramagnetic and its electron configuration in ground state is as follows:

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