Chapter 9, Problem 9.11P

(a)

Interpretation Introduction

Interpretation:

The ground state electronic configuration and the bond order of ${\text{O}}_{\text{2}}{}^{\text{-}}$ ion have to be given.

Concept Introduction:

The bond order is the number of electrons pairs shared between two atoms in the formation of the bond.

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

If one electron is added to the oxygen molecule ${\text{O}}_{\text{2}}{}^{\text{-}}$ ion is formed.

The atomic number of ${\text{O}}_{\text{2}}{}^{\text{-}}$ ion is 17.

The ground state electronic configuration of an ${\text{O}}_{\text{2}}{}^{\text{-}}$ ion is,

  ${\text{O}}_{\text{2}}{}^{\text{-}}$=${\left(\text{σ1s}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}\text{1s}\right)}^{\text{2}}{\left(\text{σ2s}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}\text{2s}\right)}^{\text{2}}{\left({\text{σ2p}}_{\text{z}}\right)}^{\text{2}}{\left({\text{π2p}}_{\text{x}}\right)}^{\text{2}}{\left({\text{π2p}}_{\text{y}}\right)}^{\text{2}}{\left({\text{π}}^{\text{*}}{\text{2p}}_{\text{x}}\right)}^2{\left({\text{π}}^{\text{*}}{\text{2p}}_{\text{y}}\right)}^1$

Bond order of ${\text{O}}_{\text{2}}{}^{\text{-}}$ ion can be calculated as,

  $\begin{array}{l}{\text{N}}_{\text{b}}\text{=10}\text{}{\text{,N}}_{\text{a}}\text{=7}\\ \text{Bond}\text{order=}\frac{\left(\text{10}\right)\text{-}\left(7\right)}{\text{2}}\\ \text{=1}\text{.5}\end{array}$

The bond order of ${\text{O}}_{\text{2}}{}^{\text{-}}$ ion is 1.5.

(b)

Interpretation Introduction

Interpretation:

The ground state electronic configuration and the bond order of ${\text{C}}_{\text{2}}{}^{\text{+}}$ ion have to be given

Concept Introduction:

Refer to part(a).

Explanation of Solution

If one electron is loosed by the carbon molecule ${\text{C}}_{\text{2}}{}^{\text{+}}$ ion is formed.

The atomic number of ${\text{C}}_{\text{2}}{}^{\text{+}}$ ion is 11.

The ground state electronic configuration of an ${\text{C}}_{\text{2}}{}^{\text{+}}$ ion is,

  ${\text{C}}_{\text{2}}{}^{\text{+}}$=${\left(\text{σ1s}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}\text{1s}\right)}^{\text{2}}{\left(\text{σ2s}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}\text{2s}\right)}^{\text{2}}{\left({\text{σ2p}}_{\text{x}}\right)}^{\text{2}}{\left({\text{σ2p}}_{\text{y}}\right)}^{\text{1}}$

Bond order of ${\text{C}}_{\text{2}}{}^{\text{+}}$ ion can be calculated as,

  $\begin{array}{l}{\text{N}}_{\text{b}}\text{=7}\text{}{\text{,N}}_{\text{a}}\text{=4}\\ \text{Bond}\text{order=}\frac{\left(7\right)\text{-}\left(4\right)}{\text{2}}\\ \text{=1}\text{.5}\end{array}$

The bond order of ${\text{C}}_{\text{2}}{}^{\text{+}}$ ion is 1.5.

(c)

Interpretation Introduction

Interpretation:

The ground state electronic configuration and the bond order of ${\text{Be}}_{\text{2}}{}^{\text{+}}$ ion have to be given

Concept Introduction:

Refer to part (a).

Explanation of Solution

The atomic number of ${\text{Be}}_{\text{2}}{}^{\text{+}}$ ion is 7.

The ground state electronic configuration of an ${\text{Be}}_{\text{2}}{}^{\text{+}}$ ion is,

  ${\text{Be}}_{\text{2}}{}^{\text{+}}$=${\left(\text{σ1s}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}\text{1s}\right)}^{\text{2}}{\left(\text{σ2s}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}\text{2s}\right)}^{\text{1}}$

Bond order of ${\text{C}}_{\text{2}}{}^{\text{+}}$ ion can be calculated as,

  $\begin{array}{l}{\text{N}}_{\text{b}}{\text{=4,N}}_{\text{a}}\text{=3}\\ \text{Bond}\text{order=}\frac{\left(\text{4}\right)\text{-}\left(\text{3}\right)}{\text{2}}\\ \text{=0}\text{.5}\end{array}$

The bond order of ${\text{Be}}_{\text{2}}{}^{\text{+}}$ ion is 0.5.

(d)

Interpretation Introduction

Interpretation:

The ground state electronic configuration and the bond order of ${\text{N}}_{\text{2}}{}^{\text{+}}$ ion have to be given

Concept Introduction:

Refer to part (a).

Explanation of Solution

The ground state electronic configuration of an ${\text{N}}_{\text{2}}{}^{\text{+}}$ ion is,

  ${\text{N}}_{\text{2}}{}^{\text{+}}$=${\left(\text{σ1s}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}\text{1s}\right)}^{\text{2}}{\left(\text{σ2s}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}\text{2s}\right)}^{\text{2}}{\left({\text{π2p}}_{\text{x}}\right)}^{\text{2}}{\left({\text{π2p}}_{\text{y}}\right)}^{\text{2}}{\left({\text{σ2p}}_{\text{z}}\right)}^{\text{1}}$

Bond order of ${\text{N}}_{\text{2}}{}^{\text{+}}$ ion can be calculated as,

  $\begin{array}{l}\text{Bond order}\text{=}\frac{\left(\text{9}\right)\text{-}\left(\text{4}\right)}{\text{2}}\\ \text{ =}\frac{\text{5}}{\text{2}}\\ \text{=2}\text{.5}\end{array}$

Bond order of nitronium ion is 2.5

The bond order of ${\text{N}}_{\text{2}}{}^{\text{+}}$ ion is 2.5.