Question

Without consulting tables of data, predict which species has the greater bond length, ${\mathbf{H}}_{\mathbf{2}}^{\mathbf{+}}$or${\mathbf{H}}_{\mathbf{2}}$.

Short answer

${{\mathrm{H}}_2}^{+}$has greater bond length.

Step-by-step solution

Concept of bonding and antibonding orbital

Two or more two atomic orbitals overlap to form bonds; these orbitals are called molecular orbitals. The count of molecular orbitals obtained is the same as that of atomic orbitals mixed.

Two forms of molecular orbital are obtained. These are bonding and antibonding orbital.

Bonding orbitals are those in that electrons are in between nuclei of two atoms.

Antibonding orbitals are those in which electrons are away from the nucleus of the two-atom. Also, electrons in the antibonding orbital have higher energy than bonding orbital.

In the sigma $\mathbf{\left(}\mathit{\sigma }\mathbf{\right)}$bonding orbital, the electron density is shared directly between bonding atoms, along the bonding axis.

In the sigma$\mathbf{\left(}{\mathit{\sigma }}^{\mathbf{*}}\mathbf{\right)}$antibonding orbital, the orbital is empty. These orbital electrons are along the nuclear axis.

In the pi$\left(\pi \right)$bonding orbital, the bonding electron lies above and below the bonding axis and has no electron on the bonding axis.

In the pi$\mathbf{\left(}{\mathbf{\pi }}^{\mathbf{*}}\mathbf{\right)}$antibonding orbital, the orbital is empty. In these orbital, electrons are perpendicular to the nuclear axis.

Relation between bond order and bond length

Bond order is calculated as half the difference between the number of electrons present in the bonding orbital and the antibonding orbital.

$\mathrm{Bond} \mathrm{order} =\frac{\mathrm{Bonding} \mathrm{electrons}- \mathrm{Antibonding} \mathrm{electrons}}{2}$

The bond order is inversely proportional to bond energy. The higher the bond order, the larger will be the bond energy and the smaller will be the bond length. It is because molecules having higher bond energy has stronger bonds and smaller bond lengths.

Calculating bond order of H2

The atomic number of H is 1.

Therefore${\mathrm{H}}_2$ is made of two H atoms and has 2 electrons.

The ground state electronic configuration of${\mathrm{H}}_2$ is ${\left({\sigma }_{g1s}\right)}^2$.

Also, bonding electrons will be 2, and antibonding electrons will be 0.

The number of bonding electrons is 2.

The number of antibonding electrons is 0.

Substitute the values in the above formula.

Bond order of ${\mathrm{H}}_2=\frac{1}{2}$(bonding electrons - antibonding electrons)

$\begin{array}{l}=\frac{1}{2}(2-0)\\ =1\end{array}$

${\mathrm{H}}_2$is stable and has a bond order 1.

Therefore, it has a smaller bond length; the higher the bond order, the stronger the pull between two atoms, and the shorter the bond length.

Calculating bond order of H2+

The atomic number of H is 1.

Therefore ${\mathrm{H}}_2$is made of two H atoms and has 2 electrons.

${\mathrm{H}}_2^{+}$has a +1 positive charge.

Therefore, it has 1 electron less than the original ${\mathrm{H}}_2$molecule and has only 1 electron.

The ground state electronic configuration of ${\mathrm{H}}_2^{+}$is ${\left({\sigma }_{g1s}\right)}^1$.

Also, bonding electrons will be 1, and antibonding electrons will be 0.

The number of bonding electrons is 2.

The number of antibonding electrons is 1.

Substitute the values in the above formula.

Bond order of ${\mathrm{H}}_2^{+}=\frac{1}{2}$(bonding electrons - antibonding electrons)

$\begin{array}{l}=\frac{1}{2}(1-0)\\ =0.5\end{array}$

Conclusion

Since, the bond order of${{\mathrm{H}}_2}^{+}$ is less than${\mathrm{H}}_2$ so,${{\mathrm{H}}_2}^{+}$ will have a greater bond length.

Therefore, ${{\mathrm{H}}_2}^{+}$has a greater bond length.