Chapter 9, Problem 48QP

Explain in molecular orbital terms the changes in $\text{H-H}$ internuclear distance that occur as the molecular ${\text{H}}_{\text{2}}$ is ionized first to ${\text{H}}_2^{+}$ and then to ${\text{H}}_2^{2+}$.

Interpretation Introduction

Interpretation: The changes in $\text{H-H}$ internuclear distance that occur as the molecule ${\text{H}}_{\text{2}}$ is ionized first to ${\text{H}}_{\text{2}}{}^{+}$

and then to ${\text{H}}_{\text{2}}{}^{2+}$ is to be explained.

Concept introduction:

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

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

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, divided by two.

Mathematically, bond order can be expressed as follows: $\text{B}\text{.O}=\frac{{\text{n}}_{\text{b}}-{\text{n}}_{\text{a}}}{2}$

Here, ${\text{n}}_{\text{b}}$ is the number of electrons in a bonding orbital and ${\text{n}}_{\text{a}}$ is the number of electrons in an antibonding orbital.

Answer to Problem 48QP

Solution: The $\text{H-H}$

internuclear distance increases as ${\text{H}}_{\text{2}}$ molecule ionizes to form ${\text{H}}_{\text{2}}{}^{+}$

and ${\text{H}}_{\text{2}}{}^{2+}$.

Explanation of Solution

The electron configuration of a hydrogen atom is ${\text{1s}}^{\text{1}}$.

The electronic configuration for ${\text{H}}_{\text{2}}$ molecule is ${\left({\sigma }_{1s}\right)}^2\left({\sigma }^{\ast }{}_{1s}\right)$.

The bond order of ${\text{H}}_{\text{2}}$ is given as follows:

$\text{B}\text{.O}=\frac{{\text{n}}_{\text{b}}-{\text{n}}_{\text{a}}}{2}$ …… (1)

Here, ${\text{n}}_{\text{b}}$ is the number of electrons in a bonding orbital and ${\text{n}}_{\text{a}}$ is the number of electrons in an antibonding orbital.

For ${\text{H}}_{\text{2}}$ molecule, there are 2 electrons in ${\sigma }_{1s}$ (bonding molecular orbital) and 0 electron in ${\sigma }^{\ast }{}_{1s}$ (antibonding molecular orbital).

Substitute 2 for ${\text{n}}_{\text{b}}$ and 0 for ${\text{n}}_{\text{a}}$ in equation (1) as follows:

$\begin{array}{c}\text{B}\text{.O}\left({\text{H}}_2\right)=\frac{\text{2}-\text{0}}{\text{2}}\\ =1\end{array}$

When ${\text{H}}_{\text{2}}$ loses one electron, it forms ${\text{H}}_{\text{2}}{}^{+}$. The electron configuration for ${\text{H}}_{\text{2}}{}^{+}$ is ${\left(\sigma 1s\right)}^1\left({\sigma }^{\ast }1s\right)$.

For ${\text{H}}_{\text{2}}{}^{+}$ molecule, there is 1 electron in ${\sigma }_{1s}$ (bonding molecular orbital) and 0 electron in ${\sigma }^{\ast }{}_{1s}$ (antibonding molecular orbital).

The bond order of ${\text{H}}_{\text{2}}{}^{+}$ is given as follows:

Substitute 1 for ${\text{n}}_{\text{b}}$ and 0 for ${\text{n}}_{\text{a}}$ in equation (1) as follows:

$\begin{array}{c}\text{B}\text{.O}\left({\text{H}}_{\text{2}}{}^{+}\right)=\frac{\text{1}-\text{0}}{\text{2}}\\ =0.5\end{array}$

When ${\text{H}}_{\text{2}}{}^{+}$ loses one electron, it forms ${\text{H}}_{\text{2}}{}^{2+}$. The electron configuration for ${\text{H}}_{\text{2}}{}^{2+}$ is $\left(\sigma 1s\right)\left({\sigma }^{\ast }1s\right)$.

For ${\text{H}}_{\text{2}}{}^{2+}$ molecule, there is 0 electron in ${\sigma }_{1s}$ (bonding molecular orbital) and 0 electron in ${\sigma }^{\ast }{}_{1s}$ (antibonding molecular orbital).

The bond order of ${\text{H}}_{\text{2}}{}^{2+}$ is given as follows:

Substitute 0 for ${\text{n}}_{\text{b}}$ and 0 for ${\text{n}}_{\text{a}}$ in equation (1) as follows:

$\begin{array}{c}\text{B}\text{.O}\left({\text{H}}_{\text{2}}{}^{2+}\right)=\frac{0-\text{0}}{\text{2}}\\ =0\end{array}$

Thus, as ${\text{H}}_{\text{2}}$ ionizes to form ${\text{H}}_{\text{2}}{}^{+}$, its bond order decreases and when it forms ${\text{H}}_{\text{2}}{}^{2+}$ its bond order becomes $0$. As bond order decreases the internuclear distance increases. This indicates that the $\text{H-H}$ internuclear distance increases as ${\text{H}}_{\text{2}}$ ionizes to form ${\text{H}}_{\text{2}}{}^{+}$, and as it ionizes to form ${\text{H}}_{\text{2}}{}^{2+}$, the $\text{H-H}$ internuclear distance becomes very long and no bond exists between the hydrogen atoms.

Conclusion

As ${\text{H}}_{\text{2}}$ ionizes to form ${\text{H}}_{\text{2}}{}^{+}$, its bond order decreases and when it forms ${\text{H}}_{\text{2}}{}^{2+}$ its bond order becomes $0$ and hence $\text{H-H}$ internuclear distance increases as ${\text{H}}_{\text{2}}$ molecule ionizes to form ${\text{H}}_{\text{2}}{}^{+}$

and ${\text{H}}_{\text{2}}{}^{2+}$.