Chapter 9, Problem 9.91QP

(a)

Interpretation Introduction

Interpretation: The electronic configuration, bond order and the ionic species that can exist is to be predicted from given molecular ions.

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.

To determine: The orbital electronic configuration of the given ionic species.

Answer to Problem 9.91QP

Solution

The electronic configuration of molecular species is given as follows.

Explanation of Solution

Explanation

Nitrogen has five valence electrons.

Charge on ${\text{N}}_2^{+}$ is $+1$.

The total number of valence electrons in ${\text{N}}_{\text{2}}^{\text{+}}$ is calculated by formula,

$\begin{array}{c}\text{Total}\text{number}\text{of}\text{valence}\text{electrons}=\left(\begin{array}{c}\text{Number}\text{of}\text{N}\text{atoms}\times \text{Valence}\text{electrons in}\text{nitrogen}\\ \pm \text{Charge}\end{array}\right)\\ =2\times 5-1\\ =9\end{array}$

The total number of valence electrons in ${\text{N}}_{\text{2}}^{\text{+}}$ is $9$.

According to the molecular orbital theory the electronic configuration of ${\text{N}}_{\text{2}}^{\text{+}}$ is

${\text{N}}_{\text{2}}^{\text{+}}={{\text{(σ}}_{\text{2s}}\text{)}}^{\text{2}}\text{}{{\text{(σ}}^{\text{*}}{}_{\text{2s}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2p}}\text{)}}^{\text{4}}{{\text{(σ}}_{\text{2p}}\text{)}}^{\text{1}}$

Oxygen has six valence electrons.

Charge on ${\text{O}}_{\text{2}}^{\text{+}}$ is $+1$.

The total number of valence electrons in ${\text{O}}_{\text{2}}^{\text{+}}$ is calculated by formula,

$\begin{array}{c}\text{Total}\text{number}\text{of}\text{valence}\text{electrons}=\left(\begin{array}{c}\text{Number}\text{of}\text{O}\text{atoms}\times \text{Valence}\text{electrons}\text{of}\text{oxygen }\\ \pm \text{Charge}\end{array}\right)\\ =2\times 6-1\\ =11\end{array}$

The total number of valence electrons in ${\text{O}}_{\text{2}}^{\text{+}}$ is $11$.

According to the molecular orbital theory the electronic configuration of ${\text{O}}_{\text{2}}^{\text{+}}$ is

${\text{O}}_{\text{2}}^{\text{+}}={{\text{(σ}}_{\text{2s}}\text{)}}^{\text{2}}\text{}{{\text{(σ}}^{\text{*}}{}_{\text{2s}}\text{)}}^{\text{2}}{{\text{(σ}}_{\text{2p}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2p}}\text{)}}^{\text{4}}{{\text{(π}}^{\text{*}}{}_{\text{2p}}\text{)}}^{\text{1}}$

Carbon has four valence electrons.

Charge on ${\text{C}}_{\text{2}}^{\text{+}}$ is $+1$

The total number of valence electrons in ${\text{C}}_{\text{2}}^{\text{+}}$ is calculated by formula,

$\begin{array}{c}\text{Total}\text{number}\text{of}\text{valence}\text{electrons}=\left(\begin{array}{l}\text{Number}\text{of}\text{C}\text{atoms}\times \text{Valence}\text{electrons}\text{}\text{in}\text{carbon}\\ \pm \text{Charge}\end{array}\right)\\ =2\times 4-1\\ =7\end{array}$

The total number of valence electrons in ${\text{C}}_{\text{2}}^{\text{+}}$ is $7$.

According to the molecular orbital theory the electronic configuration of ${\text{C}}_{\text{2}}^{\text{+}}$ is

${\text{C}}_{\text{2}}^{\text{+}}={{\text{(σ}}_{\text{2s}}\text{)}}^{\text{2}}\text{}{{\text{(σ}}^{\text{*}}{}_{\text{2s}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2p}}\text{)}}^3$

Bromine has seven valence electrons.

Charge on ${\text{Br}}_{\text{2}}^{2-}$ is $-2$

The total number of valence electrons in ${\text{Br}}_{\text{2}}^{2-}$ is calculated by formula,

$\begin{array}{c}\text{Total}\text{number}\text{of}\text{valence}\text{electrons}=\left(\begin{array}{l}\text{Number}\text{of}\text{Br}\text{atoms}\times \text{Valence}\text{electrons}\text{}\text{in}\text{bromine}\\ \pm \text{Charge}\end{array}\right)\\ =2\times 7-(-2)\\ =16\end{array}$

The total number of valence electrons in ${\text{Br}}_{\text{2}}^{2-}$ is $16$.

According to the molecular orbital theory the electronic configuration of ${\text{Br}}_{\text{2}}^{\text{2-}}$ is

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

(b)

Interpretation Introduction

To determine: The bond order of given molecular ionic species.

Answer to Problem 9.91QP

Solution

All the given molecular ionic species except ${\text{Br}}_{\text{2}}^{2-}$ have nonzero bo9mnd order.

Explanation of Solution

Explanation

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

${\text{N}}_{\text{2}}^{\text{+}}={{\text{(σ}}_{\text{2s}}\text{)}}^{\text{2}}\text{}{{\text{(σ}}^{\text{*}}{}_{\text{2s}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2p}}\text{)}}^{\text{4}}{{\text{(σ}}_{\text{2p}}\text{)}}^{\text{1}}$

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

$\text{Bond}\text{order}=\frac{1}{2}\left(\begin{array}{l}\text{Number}\text{of}\text{electrons}\text{in}\text{bonding}\text{molecular}\text{orbitals}-\\ \text{Number}\text{of}\text{electron}\text{in}\text{antibonding}\text{molecular}\text{orbitals}\end{array}\right)$

The number of bonding electrons in ${\text{N}}_{\text{2}}^{\text{+}}=7$.

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

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

$\begin{array}{c}\text{Bond}\text{order}=\frac{1}{2}(7-2)\\ =2.5\end{array}$

The bond order of ${\text{N}}_{\text{2}}^{\text{+}}$ is $2.5$.

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

${\text{O}}_{\text{2}}^{\text{+}}={{\text{(σ}}_{\text{2s}}\text{)}}^{\text{2}}\text{}{{\text{(σ}}^{\text{*}}{}_{\text{2s}}\text{)}}^{\text{2}}{{\text{(σ}}_{\text{2p}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2p}}\text{)}}^{\text{4}}{{\text{(π}}^{\text{*}}{}_{\text{2p}}\text{)}}^{\text{1}}$

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

$\text{Bond}\text{order}=\frac{1}{2}\left(\begin{array}{l}\text{Number}\text{of}\text{electrons}\text{in}\text{bonding}\text{molecular}\text{orbitals}-\\ \text{Number}\text{of}\text{electron}\text{in}\text{antibonding}\text{molecular}\text{orbitals}\end{array}\right)$

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

The number of antibonding electrons in ${\text{O}}_{\text{2}}^{\text{+}}=3$.

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

$\begin{array}{c}\text{Bond}\text{order}=\frac{1}{2}(8-3)\\ =2.5\end{array}$

The bond order of ${\text{O}}_{\text{2}}^{\text{+}}$ is $2.5$.

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

${\text{C}}_{\text{2}}^{\text{+}}={{\text{(σ}}_{\text{2s}}\text{)}}^{\text{2}}\text{}{{\text{(σ}}^{\text{*}}{}_{\text{2s}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2p}}\text{)}}^3$

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

$\text{Bond}\text{order}=\frac{1}{2}\left(\begin{array}{l}\text{Number}\text{of}\text{electrons}\text{in}\text{bonding}\text{molecular}\text{orbitals}-\\ \text{Number}\text{of}\text{electron}\text{in}\text{antibonding}\text{molecular}\text{orbitals}\end{array}\right)$

The number of bonding electrons in ${\text{C}}_{\text{2}}^{\text{+}}=5$.

The number of antibonding electrons in ${\text{C}}_{\text{2}}^{\text{+}}=3$.

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

$\begin{array}{c}\text{Bond}\text{order}=\frac{1}{2}(5-3)\\ =1.5\end{array}$

The bond order of ${\text{C}}_{\text{2}}^{\text{+}}$ is $1.5$.

The electronic configuration of ${\text{Br}}_{\text{2}}^{\text{2-}}$ is,

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

The bond order of ${\text{Br}}_{\text{2}}^{\text{2-}}$ is calculate by formula,

$\text{Bond}\text{order}=\frac{1}{2}\left(\begin{array}{l}\text{Number}\text{of}\text{electrons}\text{in}\text{bonding}\text{molecular}\text{orbitals}-\\ \text{Number}\text{of}\text{electron}\text{in}\text{antibonding}\text{molecular}\text{orbitals}\end{array}\right)$

The number of bonding electrons in ${\text{Br}}_{\text{2}}^{\text{2-}}=8$.

The number of antibonding electrons in ${\text{Br}}_{\text{2}}^{\text{2-}}=8$.

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

$\begin{array}{c}\text{Bond}\text{order}=\frac{1}{2}(8-8)\\ =0\end{array}$

The bond order of ${\text{Br}}_{\text{2}}^{\text{2-}}$ is $0$.

(c)

Interpretation Introduction

To determine: The molecular ion species that are expected to exist

Answer to Problem 9.91QP

Solution

All the molecular species except ${\text{Br}}_{\text{2}}^{\text{2-}}$ are expected to exist.

Explanation of Solution

Explanation

The bond order of a molecule is in direct proportion with its stability. Higher the bond order higher is the stability of the molecule.

Molecular species with nonzero bond order are expected to exist. All the given molecular ions except ${\text{Br}}_{\text{2}}^{\text{2-}}$ have nonzero bond order. Therefore, all the molecular species except ${\text{Br}}_{\text{2}}^{\text{2-}}$ are expected to exist.

Conclusion

The molecular species with nonzero bond order are expected to exist.