Chapter 9, Problem 9.12P

(a)

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:

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 carbon molecule ${\text{C}}_{\text{2}}{}^{\text{-}}$ ion is formed.

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

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

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

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

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

(b)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to part (a).

Explanation of Solution

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

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

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

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

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

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

(c)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to part (a).

Explanation of Solution

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

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

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

  $\begin{array}{l}{\text{N}}_{\text{b}}{\text{=2,N}}_{\text{a}}\text{=1}\\ \text{Bond}\text{order=}\frac{\left(2\right)\text{-}\left(1\right)}{\text{2}}\\ \text{=0}\text{.5}\end{array}$

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

(d)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to part (a).

Explanation of Solution

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

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

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

  $\begin{array}{l}\text{Bond order}\text{=}\frac{\left(10\right)\text{-}\left(9\right)}{\text{2}}\\ \text{ =}\frac{1}{\text{2}}\\ \text{=0}\text{.5}\end{array}$

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