Question

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

Short answer

${\mathrm{H}}_2$has larger bond energy.

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

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

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

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

Relation between bond order and bond energy

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 directly proportional to bond energy. The higher the bond order, the larger will be the bond energy.

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 formula to calculate bond order of ${\mathrm{H}}_2$is as follows:

Bond order of localid="1663664444133" $H_2=\frac{1}{2}$(bonding electrons - antibonding electrons)

The number of bonding electrons is 2.

The number of antibonding electrons is 0.

Substitute the values in the above formula.

Bond order of $H_2=\frac{1}{2}$

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

$H_2$is stable and has a bond order 1.

$H_2^{+}$ has a +1 positive charge.

Therefore, it has 1 electron less than the original $H_2^{+}$molecule and has only 1 electron.

The ground state electronic configuration of $H_2^{+}$is ${\left({\sigma }_{g1s}\right)}^1$.

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

The formula to calculate bond order of ${\mathrm{H}}_2^{+}$is as follows:

Bond order of (bonding electrons - antibonding electrons)

The number of bonding electrons is 1.

The number of antibonding electrons is 0.

Substitute the values in the above formula.

Bond order of $H_2^{+}=\frac{1}{2}$(bonding electrons - antibonding electrons)

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

$H_2^{+}$is unstable and has a bond order of 0.5

${\mathrm{H}}_2$ has greater bond order than $H_2^{+}$. Hence, role="math" localid="1663664824317" $H_2$has lesser bond energy than $H_2^{+}$as greater the bond length, the weaker bond will be, and less energy will be required to break the bond.

Therefore, $H_2$has larger bond energy.