Chapter 9, Problem 9.110QP

(a)

Interpretation Introduction

Interpretation: The diatomic molecule whose bond order increases with the loss of two electrons is to be found from the given molecules.

Concept introduction: When two atomic orbitals come close to each other they lose their identity and form new pair of orbitals knows as molecular orbitals. Among the two molecular orbitals formed one has energy lower than the atomic orbitals is known as bonding molecular orbital and the other has energy higher than the atomic orbitals and is known as antibonding molecular orbital. The filling electrons in molecular orbitals follow Aufbau’s principle and Hund’s rule.

To determine: If the bond order of diatomic molecule ${\text{B}}_{\text{2}}$ increases after the loss of two electrons.

Answer to Problem 9.110QP

Solution

The bond order of ${\text{B}}_{\text{2}}$ decreases with the loss of two electrons.

Explanation of Solution

Explanation

The electronic configuration of ${\text{B}}_{\text{2}}$ is,

${{\text{(σ}}_{\text{2s}}\text{)}}^{\text{2}}\text{}{{\text{(σ}}^{\text{*}}{}_{\text{2s}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2p}}\text{)}}^2$

The bond order for ${\text{B}}_{\text{2}}$ is calculated by using formula,

$\text{Bond}\text{order}\text{of}{\text{B}}_{\text{2}}=\frac{1}{2}\left(\begin{array}{l}\text{Number}\text{of}\text{}\text{bonding}\text{electrons}\text{in}{\text{B}}_{\text{2}}\\ -\text{Numberof}\text{of}\text{antibondong}\text{elctrons}\text{in}{\text{B}}_{\text{2}}\end{array}\right)$

The number of bonding electrons in ${\text{B}}_{\text{2}}=4$.

The number of antibonding electrons in ${\text{B}}_{\text{2}}=2$

Substitute the value of number of electrons in bonding and antibonding orbitals in ${\text{B}}_{\text{2}}$ in the above equation.

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

The bond order of ${\text{B}}_2=1$

The ${\text{B}}_{\text{2}}^{\text{2+}}$ ion is formed by the loss of two electrons from ${\text{B}}_{\text{2}}$. The electronic configuration of ${\text{B}}_{\text{2}}^{\text{2+}}$ is,

${{\text{(σ}}_{\text{2s}}\text{)}}^{\text{2}}\text{}{{\text{(σ}}^{\text{*}}{}_{\text{2s}}\text{)}}^{\text{2}}$

The bond order of ${\text{B}}_{\text{2}}^{\text{2+}}$ is calculated by formula,

$\text{Bond}\text{order}\text{of}{\text{B}}_{\text{2}}^{\text{2+}}\frac{1}{2}\left(\begin{array}{l}\text{Number}\text{of}\text{}\text{bonding}\text{electrons}\text{in}{\text{B}}_{\text{2}}^{\text{2+}}\\ -\text{Numberof}\text{of}\text{antibondong}\text{elctrons}\text{in}{\text{B}}_2^{2+}\end{array}\right)$

The number of bonding electrons in ${\text{B}}_2^{2+}=2$

The number of antibonding electrons in ${\text{B}}_2^{2+}=2$

Substitute the value of number of electrons in bonding and antibonding orbitals in ${\text{B}}_{\text{2}}^{\text{2+}}$ in the above equation.

$\begin{array}{c}\text{Bond}\text{order}=\frac{1}{2}\left(2-2\right)\\ =0\end{array}$

The bond order of ${\text{B}}_{\text{2}}^{\text{2+}}=0$.

Hence, the bond order of ${\text{B}}_{\text{2}}$ molecule decreases with the loss of two electrons.

(b)

Interpretation Introduction

To determine: If the bond order of diatomic molecule ${\text{C}}_{\text{2}}$ increases after the loss of two electrons.

Answer to Problem 9.110QP

Solution

The bond order of ${\text{C}}_{\text{2}}$ decreases with the loss of two electrons.

Explanation of Solution

Explanation

The electronic configuration of ${\text{C}}_{\text{2}}$ is,

${{\text{(σ}}_{\text{2s}}\text{)}}^{\text{2}}\text{}{{\text{(σ}}^{\text{*}}{}_{\text{2s}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2p}}\text{)}}^4$

The bond order for ${\text{C}}_{\text{2}}$ is calculated by using formula,

$\text{Bond}\text{order}\text{of}{\text{C}}_{\text{2}}=\frac{1}{2}\left(\begin{array}{l}\text{Number}\text{of}\text{}\text{bonding}\text{electrons}\text{in}{\text{C}}_{\text{2}}\\ -\text{Numberof}\text{of}\text{antibondong}\text{elctrons}\text{in}{\text{C}}_{\text{2}}\end{array}\right)$

The number of bonding electrons in ${\text{C}}_{\text{2}}=6$

The number of antibonding electrons in ${\text{C}}_{\text{2}}=2$

Substitute the value of number of electrons in bonding and antibonding orbitals in ${\text{C}}_{\text{2}}$ in the above equation.

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

The bond order of ${\text{C}}_{\text{2}}=2$

The ${\text{C}}_2^{2+}$ ion is formed by the loss of two electrons from ${\text{C}}_{\text{2}}$. The electronic configuration of ${\text{C}}_2^{2+}$ is,

${{\text{(σ}}_{\text{2s}}\text{)}}^{\text{2}}\text{}{{\text{(σ}}^{\text{*}}{}_{\text{2s}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2p}}\text{)}}^2$

The bond order for ${\text{C}}_2^{2+}$ is calculated by using formula,

$\text{Bond}\text{order}\text{of}{\text{C}}_2^{2+}=\frac{1}{2}\left(\begin{array}{l}\text{Number}\text{of}\text{}\text{bonding}\text{electrons}\text{in}{\text{C}}_2^{2+}\\ -\text{Numberof}\text{of}\text{antibondong}\text{elctrons}\text{in}{\text{C}}_{\text{2}}^{\text{2+}}\end{array}\right)$

The number of bonding electrons in ${\text{C}}_2^{2+}=4$

The number of antibonding electrons in ${\text{C}}_2^{2+}=2$

Substitute the value of number of electrons in bonding and antibonding orbitals in ${\text{C}}_2^{2-}$ in the above equation.

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

The bond order of ${\text{C}}_2^{2+}=1$

Hence, the bond order of ${\text{C}}_{\text{2}}$ molecule decreases with the loss of two electrons.

(c)

Interpretation Introduction

To determine: If the bond order of diatomic molecule ${\text{N}}_{\text{2}}$ after the loss of two electrons.

Answer to Problem 9.110QP

Solution

The bond order of ${\text{N}}_{\text{2}}$ decreases with the loss of two electrons.

Explanation of Solution

Explanation

The electronic configuration of ${\text{N}}_{\text{2}}$ is,

${{\text{(σ}}_{\text{2s}}\text{)}}^{\text{2}}\text{}{{\text{(σ}}^{\text{*}}{}_{\text{2s}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2p}}\text{)}}^4{{\text{(σ}}_{\text{2p}}\text{)}}^{\text{2}}$

The bond order for ${\text{N}}_{\text{2}}$ is calculated by using formula,

$\text{Bond}\text{order}\text{of}{\text{N}}_{\text{2}}=\frac{1}{2}\left(\begin{array}{l}\text{Number}\text{of}\text{}\text{bonding}\text{electrons}\text{in}{\text{N}}_{\text{2}}\\ -\text{Numberof}\text{of}\text{antibondong}\text{elctrons}\text{in}{\text{N}}_{\text{2}}\end{array}\right)$

The number of bonding electrons in ${\text{N}}_{\text{2}}=8$

The number of antibonding electrons in ${\text{N}}_{\text{2}}=2$

Substitute the value of number of electrons in bonding and antibonding orbitals in ${\text{N}}_{\text{2}}$ in the above equation.

$\begin{array}{c}\text{Bond}\text{order}=\frac{1}{2}\left(8-2\right)\\ =3\end{array}$

The bond order of ${\text{N}}_{\text{2}}=\text{3}$

The ${\text{N}}_2^{2+}$ ion is formed by the loss of two electrons from ${\text{N}}_{\text{2}}$. The electronic configuration of ${\text{N}}_2^{2+}$ is,

${{\text{(σ}}_{\text{2s}}\text{)}}^{\text{2}}\text{}{{\text{(σ}}^{\text{*}}{}_{\text{2s}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2p}}\text{)}}^4$

The bond order for ${\text{N}}_2^{2+}$  is calculated by using formula,

$\text{Bond}\text{order}\text{of}{\text{N}}_2^{2+}=\frac{1}{2}\left(\begin{array}{l}\text{Number}\text{of}\text{}\text{bonding}\text{electrons}\text{of}{\text{N}}_2^{2+}\\ -\text{Numberof}\text{of}\text{antibondong}\text{elctrons}\text{in}{\text{N}}_2^{2+}\end{array}\right)$

The number of bonding electrons in ${\text{N}}_2^{2+}=6$

The number of antibonding electrons in ${\text{N}}_2^{2+}=2$

Substitute the value of number of electrons in bonding and antibonding orbitals in ${\text{N}}_2^{2-}$ in the above equation.

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

The bond order of ${\text{N}}_2^{2+}=2$

Hence, the bond order of ${\text{N}}_{\text{2}}$ molecule decreases with the loss of two electrons.

(d)

Interpretation Introduction

To determine: The bond order of diatomic molecule ${\text{O}}_{\text{2}}$ after the loss of two electrons.

Answer to Problem 9.110QP

Solution

The bond order of ${\text{O}}_{\text{2}}$ increases with the loss of two electrons.

Explanation of Solution

Explanation

The electronic configuration of ${\text{O}}_{\text{2}}$ is,

${{\text{(σ}}_{\text{2s}}\text{)}}^{\text{2}}\text{}{{\text{(σ}}^{\text{*}}{}_{\text{2s}}\text{)}}^{\text{2}}{{\text{(σ}}_{\text{2p}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2p}}\text{)}}^4{{\text{(π}}_{2\text{p}}^{*}\text{)}}^2$

The bond order for ${\text{O}}_{\text{2}}$ is calculated by using formula,

$\text{Bond}\text{order}\text{of}{\text{O}}_{\text{2}}=\frac{1}{2}\left(\begin{array}{l}\text{Number}\text{of}\text{}\text{bonding}\text{electrons}\text{in}{\text{O}}_{\text{2}}\\ -\text{Numberof}\text{of}\text{antibondong}\text{elctrons}\text{in}{\text{O}}_{\text{2}}\end{array}\right)$

The number of bonding electrons in ${\text{O}}_{\text{2}}=8$

The number of antibonding electrons in ${\text{O}}_{\text{2}}=4$

Substitute the value of number of electrons in bonding and antibonding orbitals in ${\text{O}}_{\text{2}}$ in the above equation.

$\begin{array}{c}\text{Bond}\text{order=}\frac{1}{2}\left(8-4\right)\\ =2\end{array}$

The bond order of ${\text{O}}_{\text{2}}=2$

he ${\text{O}}_2^{2+}$ ion is formed by the loss of two electrons from ${\text{N}}_{\text{2}}$. The electronic configuration of ${\text{O}}_2^{2-}$ is,

${{\text{(σ}}_{\text{2s}}\text{)}}^{\text{2}}\text{}{{\text{(σ}}^{\text{*}}{}_{\text{2s}}\text{)}}^{\text{2}}{{\text{(σ}}_{\text{2p}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2p}}\text{)}}^4$

The bond order for ${\text{O}}_2^{2+}$ is calculated by using formula,

$\text{Bond}\text{order}{\text{of O}}_2^{2+}=\frac{1}{2}\left(\begin{array}{l}\text{Number}\text{of}\text{}\text{bonding}\text{electrons}\text{of}{\text{O}}_2^{2+}\\ -\text{Numberof}\text{of}\text{antibondong}\text{elctrons}\text{in}{\text{O}}_2^{2+}\end{array}\right)$

The number of bonding electrons in ${\text{O}}_2^{2+}=8$.

The number of antibonding electrons in ${\text{O}}_2^{2+}=2$.

Substitute the value of number of electrons in bonding and antibonding orbitals in ${\text{O}}_2^{2+}$ in the above equation.

$\begin{array}{c}\text{Bond}\text{order=}\frac{1}{2}\left(8-2\right)\\ =3\end{array}$

The bond order of ${\text{O}}_2^{2+}=3$

Hence, the bond order of ${\text{O}}_{\text{2}}$ molecule increases with the loss of two electrons.

Conclusion

Higher the bond order of a molecule higher will be its bond strength. The bond order of oxygen increases while as the bond order of boron, carbon and nitrogen molecules decreases with the loss of two electrons.