Chapter 10, Problem 10.89QE

(a)

Interpretation Introduction

Interpretation:

The ground state electronic configuration, the bond order of ${\text{C}}_{\text{2}}{}^{\text{+}}$ ion and the number of unpaired electrons 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 lost 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,

  ${\text{C}}_{\text{2}}{}^{\text{+}}$=${\text{σ1s}}^{\text{2}}{\text{σ}}^{\text{*}}{\text{1s}}^{\text{2}}{\text{σ2s}}^{\text{2}}{\text{σ}}^{\text{*}}{\text{2s}}^{\text{2}}{\text{π2p}}_{\text{x}}{}^{\text{2}}{\text{π2p}}_{\text{y}}{}^{\text{1}}$

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.  It has one lone pair of electrons in bonding orbital and based on the bond order value it is a stable species.

(b)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to part (a).

Explanation of Solution

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

  ${\text{N}}_{\text{2}}{}^{\text{-}}$=${\text{σ1s}}^{\text{2}}{\text{σ}}^{\text{*}}{\text{1s}}^{\text{2}}{\text{σ2s}}^{\text{2}}{\text{σ}}^{\text{*}}{\text{2s}}^{\text{2}}{\text{π2p}}_{\text{x}}{}^{\text{2}}{\text{π2p}}_{\text{y}}{}^{\text{2}}{\text{σ2p}}_{\text{z}}{}^{\text{2}}{\text{π}}^{\text{*}}{\text{2p}}_{\text{x}}{}^{\text{1}}$

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

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

The bond order of ${\text{N}}_{\text{2}}{}^{\text{-}}$ ion is $\text{2}\text{.5}$ .  There is no lone pair of electrons in ${\text{N}}_{\text{2}}{}^{\text{-}}$ ion and based on the bond order it is a stable species.

(c)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to part (a).

Explanation of Solution

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

  ${\text{Be}}_{\text{2}}{}^{-}$=${\text{σ1s}}^{\text{2}}{\text{σ}}^{\text{*}}{\text{1s}}^{\text{2}}{\text{σ2s}}^{\text{2}}{\text{σ}}^{\text{*}}{\text{2s}}^{\text{2}}$

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

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

The bond order of ${\text{Be}}_{\text{2}}{}^{-}$ ion is zero.  Based on the bond order it is an unstable species.