Chapter 9, Problem 9.10P

Interpretation Introduction

Interpretation:

The ground state configuration, bond order and the bond length of the species has to be given.

Concept Introduction:

Bond order of a molecule or ion is calculated by the following formula,

Bond order $\text{=}\frac{\left({\text{N}}_{\text{b}}\right)\text{-}\left({\text{N}}_{\text{a}}\right)}{\text{2}}$

Where, ${\text{N}}_{\text{b}}$ is number of electrons in bonding orbital.

${\text{N}}_{\text{a}}$ is number of electrons in antibonding orbital.

Explanation of Solution

The ground state electronic configuration is the arrangement of electrons around the nucleus of an atom with lower energy levels.

Ground state electronic configurations of three species are as follows,

  $\text{NO}\text{=}{\left(\text{σ1s}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}\text{1s}\right)}^{\text{2}}{\left(\text{σ2s}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}\text{2s}\right)}^{\text{2}}{\left({\text{σ2p}}_{\text{z}}\right)}^{\text{2}}{\left({\text{π2p}}_{\text{x}}\right)}^{\text{2}}{\left({\text{π2p}}_{\text{y}}\right)}^{\text{2}}{\left({\text{π}}^{\text{*}}{\text{2p}}_{\text{x}}\right)}^{\text{1}}$

  ${\text{NO}}^{\text{+}}\text{=}{\left(\text{σ1s}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}\text{1s}\right)}^{\text{2}}{\left(\text{σ2s}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}\text{2s}\right)}^{\text{2}}{\left({\text{π2p}}_{\text{x}}\right)}^{\text{2}}{\left({\text{π2p}}_{\text{y}}\right)}^{\text{2}}{\left({\text{σ2p}}_{\text{z}}\right)}^{\text{2}}$

  ${\text{NO}}^{\text{-}}\text{=}{\left(\text{σ1s}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}\text{1s}\right)}^{\text{2}}{\left(\text{σ2s}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}\text{2s}\right)}^{\text{2}}{\left({\text{σ2p}}_{\text{z}}\right)}^{\text{2}}{\left({\text{π2p}}_{\text{x}}\right)}^{\text{2}}{\left({\text{π2p}}_{\text{y}}\right)}^{\text{2}}{\left({\text{π}}^{\text{*}}{\text{2p}}_{\text{x}}\right)}^{\text{1}}{\left({\text{π}}^{\text{*}}{\text{2p}}_{\text{y}}\right)}^{\text{1}}$

Bond order calculation:

Bond order of $\text{NO}$ can be calculated as,

  $\begin{array}{l}\text{Bond order}\text{=}\frac{\left(\text{10}\right)\text{-}\left(\text{5}\right)}{\text{2}}\\ \text{ =}\frac{\text{5}}{\text{2}}\\ \text{=2}\text{.5}\end{array}$

The bond order of $\text{NO}$ molecule is 2.5.

Bond order of ${\text{NO}}^{\text{+}}$ can be calculated as,

  $\begin{array}{l}\text{Bond order}\text{=}\frac{\left(\text{10}\right)\text{-}\left(\text{4}\right)}{\text{2}}\\ \text{ =}\frac{\text{6}}{\text{2}}\\ \text{=3}\end{array}$

The bond order of ${\text{NO}}^{\text{+}}$ is three.

Bond order of ${\text{NO}}^{\text{-}}$ can be calculated as

  $\begin{array}{l}\text{Bond order}\text{=}\frac{\left(\text{10}\right)\text{-}\left(\text{6}\right)}{\text{2}}\\ \text{ =}\frac{\text{4}}{\text{2}}\\ \text{=2}\end{array}$

The bond order of ${\text{NO}}^{\text{-}}$ is two.

Bond order of the species affects the bond length of the compound inversely.  If the bond order is low then the molecule has the longest bond length, if the bond order is high then its bond length will be shorter.

The increasing order of bond length is ${\text{NO}}^{\text{+}}{\text{,NO,NO}}^{\text{-}}\text{.}$

${\text{NO}}^{\text{-}}$ will have the longest bond length.