Chapter 10, Problem 10.93QE

(a)

Interpretation Introduction

Interpretation:

The species, which have stronger bonds between, ${\text{B}}_{\text{2}}\text{and}{\text{B}}_{\text{2}}{}^{\text{-}}$ has to be predicted.

Concept Introduction:

Bond energy:

The bond order is the number of electrons pairs shared between two atoms in the formation of the bond.  The amount of energy required to break a bond is called bond dissociation energy.  Thus greater the bond order greater is the bon energy.

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

Answer to Problem 10.93QE

The bond in ${\text{B}}_{\text{2}}$ molecule is stronger than the ${\text{B}}_{\text{2}}{}^{-}$ ion.

Explanation of Solution

The molecular electronic configuration of boron molecule is,

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

The number of electrons in the bonding orbital=6

The number of electrons in the antibonding orbital=4

The bond order of boron molecule can be calculated as,

Bond order $\begin{array}{l}\text{=}\frac{\left(6\right)\text{-}\left(4\right)}{\text{2}}\\ \text{=1}\end{array}$

The molecular electronic configurations of ${\text{B}}_{\text{2}}{}^{-}$ ion is,

  ${\text{B}}_{\text{2}}{}^{-}$=${\text{σ1s}}^{\text{2}}{\text{σ}}^{\text{*}}{\text{1s}}^{\text{2}}{\text{σ2s}}^{\text{2}}{\text{σ}}^{\text{*}}{\text{2s}}^{\text{2}}{\text{π2p}}_{\text{x}}{}^1$

The number of electrons in the bonding orbital is five.

The number of electrons in the antibonding orbital is four.

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

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

The bond order of ${\text{B}}_{\text{2}}$ is greater ${\text{B}}_{\text{2}}{}^{-}$ ion which means that the bond in ${\text{B}}_{\text{2}}$ molecule is stronger than the ${\text{B}}_{\text{2}}{}^{-}$ ion.

(b)

Interpretation Introduction

Interpretation:

The species, which have stronger bonds between, ${\text{C}}_{\text{2}}{}^{-}\text{and}{\text{C}}_{\text{2}}{}^{\text{+}}$ has to be predicted.

Concept Introduction:

Refer to part (a).

Answer to Problem 10.93QE

The bond in ${\text{C}}_{\text{2}}{}^{-}$ ion molecule is stronger than the ${\text{C}}_{\text{2}}{}^{\text{+}}$ ion.

Explanation of Solution

The molecular electronic configuration of ${\text{C}}_{\text{2}}{}^{-}$ molecule is,

  ${\text{σ1s}}^{\text{2}}{\text{σ}}^{\text{*}}{\text{1s}}^{\text{2}}{\text{σ2s}}^{\text{2}}{\text{σ}}^{\text{*}}{\text{2s}}^{\text{2}}{\text{π2P}}_{\text{x}}{}^{\text{2}}{\text{π2p}}_{\text{y}}{}^{\text{2}}{\text{σ2p}}_{\text{z}}{}^{\text{1}}$

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

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

There are six electrons in bonding and four electrons in antibonding orbitals.

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

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

  ${\text{σ1s}}^{\text{2}}{\text{σ}}^{\text{*}}{\text{1s}}^{\text{2}}{\text{σ2s}}^{\text{2}}{\text{σ}}^{\text{*}}{\text{2s}}^{\text{2}}{\text{σ2p}}_{\text{x}}{}^1$

The number of electrons in the bonding orbital is five and the number of electrons in antibonding orbital is 4.

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

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

The bond order of the ${\text{C}}_{\text{2}}{}^{-}$ is greater ${\text{C}}_{\text{2}}{}^{\text{+}}$ ion which means that the bonds in ${\text{C}}_{\text{2}}{}^{-}$ is stronger than the ${\text{C}}_{\text{2}}{}^{\text{+}}$ ion.

(c)

Interpretation Introduction

Interpretation:

The species that have stronger bonds between ${\text{O}}_{\text{2}}\text{and}{\text{O}}_{\text{2}}{}^{\text{2+}}$ has to be predicted.

Concept Introduction:

Refer to part (a).

Answer to Problem 10.93QE

The bond in ${\text{O}}_{\text{2}}{}^{\text{2+}}$ ion molecule is stronger than the ${\text{O}}_{\text{2}}$ ion.

Explanation of Solution

The molecular configuration of ${\text{O}}_{\text{2}}$=${\text{σ1s}}^{\text{2}}{\text{σ}}^{\text{*}}{\text{1s}}^{\text{2}}{\text{σ2s}}^{\text{2}}{\text{σ}}^{\text{*}}{\text{2s}}^{\text{2}}{\text{σ2p}}_{\text{z}}{}^{\text{2}}{\text{π2p}}_{\text{x}}{}^{\text{2}}{\text{π2p}}_{\text{y}}{}^{\text{2}}{\text{π}}^{\text{*}}{\text{2p}}_{\text{x}}{}^{\text{1}}{\text{π}}^{\text{*}}{\text{2p}}_{\text{y}}{}^{\text{1}}$

There are ten electrons in bonding orbital and six electrons antibonding orbitals.

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

Bond order of ${\text{O}}_{\text{2}}$ is two.

The molecular configuration of ${\text{O}}_{\text{2}}{}^{\text{2+}}$=${\text{σ1s}}^{\text{2}}{\text{σ}}^{\text{*}}{\text{1s}}^{\text{2}}{\text{σ2s}}^{\text{2}}{\text{σ}}^{\text{*}}{\text{2s}}^{\text{2}}{\text{σ2p}}_{\text{z}}{}^{\text{2}}{\text{π2p}}_{\text{x}}{}^{\text{2}}{\text{π2p}}_{\text{y}}{}^{\text{2}}$

Bond order of ${\text{O}}_{\text{2}}{}^{\text{2+}}$ ion is,

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

Bond order of ${\text{O}}_{\text{2}}{}^{\text{2+}}$ ion is three.

The bond order of the ${\text{O}}_{\text{2}}{}^{\text{2+}}$ ion is greater than the oxygen molecule that means that bonds in ${\text{O}}_{\text{2}}{}^{\text{2+}}$ ion are stronger than the oxygen molecule.