Question

Write the electronic configuration of Lithium $\left( {{\text{L}}{{\text{i}}_{\text{2}}}} \right)$ molecule. What is its bond order?

Answer 1

The number of molecular orbitals formed is equal to the number of interacting atomic orbitals. You can use the following formula for the bond order.
$BO = \dfrac{{{N_b} - {N_a}}}{2}$
Here, $BO,{N_b}$ and ${N_a}$ represents the bond order, the number of electrons in bonding molecular orbitals and the number of electrons in antibonding molecular orbitals respectively.

Complete step by step answer:
The atomic number of lithium is 3. Thus, one lithium atom has 3 electrons. Out of these, two electrons are core electrons and one electron is valence electron. Write the electronic configuration of one atom of lithium.
${\text{1}}{{\text{s}}^2}2{{\text{s}}^1}$
Here 1s subshell is core shell and 2s subshell is the valence shell.
Two lithium atoms combine to form Lithium $\left( {{\text{L}}{{\text{i}}_{\text{2}}}} \right)$ molecule. Each lithium atom contributes 3 electrons. So, one Lithium $\left( {{\text{L}}{{\text{i}}_{\text{2}}}} \right)$ molecule will have $3 + 3 = 6$ electrons. Each lithium atom contributes 1 valence electron. So, one Lithium $\left( {{\text{L}}{{\text{i}}_{\text{2}}}} \right)$ molecule will have $1 + 1 = 2$ valence electrons.
Write the electronic configuration of Lithium $\left( {{\text{L}}{{\text{i}}_{\text{2}}}} \right)$ molecule.
${\left( {\sigma {\text{1s}}} \right)^2}{\left( {{\sigma ^*}{\text{1s}}} \right)^2}{\left( {\sigma 2{\text{s}}} \right)^2}$
Here, $\sigma$ represents bonding molecular orbital and ${\sigma ^*}$ represents an antibonding molecular orbital. When two 1s atomic orbitals of two lithium atoms interact, $\sigma {\text{1s}}$ bonding molecular orbital and ${\sigma ^*}{\text{1s}}$ antibonding molecular orbital are formed. Similarly, when two 2s atomic orbitals of two lithium atoms interact, $\sigma {\text{2s}}$ bonding molecular orbital and ${\sigma ^*}{\text{2s}}$ antibonding molecular orbital are formed.
Calculate the bond order:
$BO = \dfrac{{{N_b} - {N_a}}}{2} = \dfrac{{4 - 2}}{2} = \dfrac{2}{2} = 1$
Thus, a single bond is present between two lithium atom in Lithium $\left( {{\text{L}}{{\text{i}}_{\text{2}}}} \right)$ molecule.

Note: When 2 orbitals of each lithium interact, total 4 atomic orbitals interact and 4 molecular orbitals are formed. Out of these, two are bonding molecular orbitals and two are antibonding molecular orbitals are formed. There are 6 electrons, these electrons occupy two bonding and one antibonding molecular orbitals. One antibonding molecular orbital remains empty.