Chapter 10, Problem 10.94QE

(a)

Interpretation Introduction

Interpretation:

The species, which have stronger bonds between, ${\text{F}}_{\text{2}}\text{and}{\text{F}}_{\text{2}}{}^{\text{-}}$ 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 10.94QE

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

Explanation of Solution

The molecular electronic configuration of fluorine molecule is,

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

The number of electrons in the bonding orbital=10

The number of electrons in the antibonding orbital=8

Bond order of fluorine molecule can be calculated as,

Bond order $\begin{array}{l}\text{=}\frac{\left(\text{10}\right)\text{-}\left(8\right)}{\text{2}}\\ \text{=1}\end{array}$

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

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

The number of electrons in the bonding orbital is ten.

The number of electrons in the antibonding orbital is nine.

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

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

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

(b)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to part (a).

Answer to Problem 10.94QE

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

Explanation of Solution

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

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

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

Bond order of ${\text{O}}_{\text{2}}{}^{\text{-}}$ is 1.5.

${\text{O}}_{\text{2}}{}^{\text{+}}$ ion is formed when ${\text{O}}_{\text{2}}$ molecule losses its one electron from the outermost orbital.

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

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

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

Bond order of ${\text{O}}_{\text{2}}{}^{\text{+}}$ ion is 2.5.

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

(c)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to part (a).

Answer to Problem 10.94QE

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

Explanation of Solution

The atomic number of $\text{C}$ is six thus,

The electronic configuration of $\text{C}$=${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{2}}$

The molecular electronic configuration of ${\text{C}}_{\text{2}}$ molecule=${\text{σ1s}}^{\text{2}}{\text{σ}}^{\text{*}}{\text{1s}}^{\text{2}}{\text{σ2s}}^{\text{2}}{\text{σ}}^{\text{*}}{\text{2s}}^{\text{2}}{\text{σ2p}}_{\text{x}}{}^{\text{2}}{\text{σ2p}}_{\text{y}}{}^{\text{2}}$

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

  $\begin{array}{l}\text{Bond order}\text{=}\frac{\left(\text{8}\right)\text{-}\left(\text{4}\right)}{\text{2}}\\ \text{ =2}\end{array}$

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

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

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

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

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

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