Question

The bond order of the underlined species $N{{O}^{+}}HSO_{4}^{-}$ $NOHS{{O}_{4}}$ is

Answer 1

Hint: Bond order is the number of bonding pairs of electrons between two atoms . In a covalent bond between two elements the number of bonds it forms with each other is called the bond order . Firstly it needs to be taken in consideration that in which ions it dissociates in the water. Now the ions to which it resembles in molecular electronic configuration will help in the calculation of bond order.

Complete step by step solution:The compound $NOHS{{O}_{4}}$ exists as two ions which are $N{{O}^{+}}HSO_{4}^{-}$. Now here in these ions we know that the $N{{O}^{+}}$ion is a derivative of oxygen and so resembles the molecular orbital configuration with the oxygen. Bond order is the number of bonding pairs of electrons between two atoms. It helps us to know the number of participating electrons in the formation of bonds .. Higher bond order confers more stability. Bond order also helps us to understand the bond length. Formula for calculating the bond order of any specie can be represented as :
Bond order=$\frac{10-4}{2}$ $\frac{{{\sigma }_{n}}-{{\sigma }_{n}}^{*}}{2}$
Where ${{\sigma }_{n}}$=no. of bonding electrons
And ${{\sigma }_{n}}^{*}$=no. of antibonding electrons
Here, the molecular orbital configuration of the ion $N{{O}^{+}}$ is given as-
$N{{O}^{+}}$=${{\sigma }_{1s}}^{2}\sigma _{1s}^{*2}\sigma _{2s}^{2}\sigma _{2s}^{*2}\sigma _{2{{p}_{z}}}^{2}\pi _{2{{p}_{x}}}^{2}\pi _{2{{p}_{y}}}^{2}$
Thus, here Bond order=$\frac{{{\sigma }_{n}}-{{\sigma }_{n}}^{*}}{2}$=$\frac{10-4}{2}$=3

Thus , the bond order of the species $NOHS{{O}_{4}}$is 3.

Note: The bond order of the NO and $N{{O}^{+}}$ differs as in $N{{O}^{+}}$there is lack of 1 electron from the antibonding orbital. And thus there should not be confusion in calculating the bond order of $N{{O}^{+}}$as 3 rather than 2.5.