Question

Calculate the bond order for an ion with this configuration: \({\left( {{{\bf{\sigma }}_{{\bf{2s}}}}} \right)^{\bf{2}}}{\left( {{\bf{\sigma }}_{{\bf{2s}}}^{\bf{*}}} \right)^{\bf{2}}}{\left( {{{\bf{\sigma }}_{{\bf{2px}}}}} \right)^{\bf{2}}}{\left( {{{\bf{\pi }}_{{\bf{2py}}}}{\bf{,}}{{\bf{\pi }}_{{\bf{2pz}}}}} \right)^{\bf{4}}}{\left( {{\bf{\pi }}_{{\bf{2py}}}^{\bf{*}}{\bf{,\pi }}_{{\bf{2pz}}}^{\bf{*}}} \right)^{\bf{3}}}\).

Short answer

The bond order is 1.5.

Step-by-step solution

Definition of ion

An ion is a net electrically charged atom or molecule. The charge of an electron is traditionally thought to be negative, and it is equal to and opposite the charge of a proton, which is traditionally thought to be positive.

Explanation

Let us calculate the bond order for an ion with this configuration,

\({\left( {{\sigma _{2s}}} \right)^2}{\left( {\sigma _{2s}^*} \right)^2}{\left( {{\sigma _{2px}}} \right)^2}{\left( {{\pi _{2py}},{\pi _{2pz}}} \right)^4}{\left( {\pi _{2py}^*,\pi _{2pz}^*} \right)^3}\)\(\begin{array}{c}{\rm{bond order }} = \frac{{{\rm{ (number of bonding electrons }}) - {\rm{ (number of antibonding electrons) }}}}{2}\\ = \frac{{8 - 5}}{2}\\ = 1.5\end{array}\)

Therefore, the bond amount is 1.5.