Chapter 14, Problem 42E

(a)

Interpretation Introduction

Interpretation: The most stable species from the given options needs to be identified.

  ${\text{H}}_{{}_{\text{2}}}^{+}{\text{,H}}_{{}_{\text{2}}}^{},{\text{H}}_{{}_{\text{2}}}^{-},{\text{H}}_{{}_{\text{2}}}^{2-}$

Concept Introduction: Molecular orbital theory explained the bonding, magnetic and spectral properties of molecule. It is based on the formation of molecular orbitals by the combination of atomic orbitals. On the basis of energy and stability these molecular orbitals can be further classified in three types:

• Bonding molecular orbitals (BMO): They have lesser energy than atomic orbital therefore more stable compare to atomic orbital.
• Antibonding molecular orbitals (ABMO): They have higher energy than atomic orbital therefore less stable compare to atomic orbital.
• Non-bonding molecular orbitals (NBMO): They have same energy as atomic orbital.

Molecular orbital diagrams represents the distribution of electrons in different molecular orbitals in increasing order of their energy. Hence lower energy molecular orbitals occupy first then only electron moves in higher energy orbitals.

Answer to Problem 42E

Since the bond order of ${\mathrm{He}}_2$ is zero hence it is unstable whereas ${\text{H}}_{\text{2}}$ is stable as it has bond order 1.

Explanation of Solution

The molecular orbital configuration of ${\text{H}}_{{}_{\text{2}}}^{+}{\text{,H}}_{{}_{\text{2}}}^{},{\text{H}}_{{}_{\text{2}}}^{-},{\text{H}}_{{}_{\text{2}}}^{2-}$ are:

  ${\text{H}}_{{}_{\text{2}}}^{+}$ = ${{\text{(σ}}_{\text{1s}}\text{)}}^1$

  ${\text{H}}_{{}_{\text{2}}}^{}$ = ${{\text{(σ}}_{\text{1s}}\text{)}}^{\text{2}}$

  ${\text{H}}_{{}_{\text{2}}}^{-}$ = ${{\text{(σ}}_{\text{1s}}\text{)}}^{\text{2}}{{\text{(σ}}_{{}_{\text{1s}}}^{\text{*}}\text{)}}^1$

  ${\text{H}}_{{}_{\text{2}}}^{2-}$ = ${{\text{(σ}}_{\text{1s}}\text{)}}^{\text{2}}{{\text{(σ}}_{{}_{\text{1s}}}^{\text{*}}\text{)}}^{\text{2}}$

  ${\text{H}}_{{}_{\text{2}}}^{}$ must be most stable one here because it does not have any anti-bonding electron and also the bond order of 1 whereas ${\text{H}}_{{}_{\text{2}}}^{+}$ also does not have any anti-bonding electron yet the bond order is 0.5 that is not stable.

  $\begin{array}{l}\text{Bond order = }\frac{{\text{bonding e}}^{\text{-}}{\text{- antibonding e}}^{\text{-}}}{2}\\ {\text{Bond order in H}}_{\text{2}}\text{ = }\frac{\text{2- 0}}{2}=1\\ {\text{Bond order in H}}_{{}_{\text{2}}}^{+}\text{ = }\frac{\text{1- 0}}{2}=0.5\end{array}$

(b)

Interpretation Introduction

Interpretation: The most stable species from the given options needs to be identified.

  ${\text{N}}_{{}_{\text{2}}}^{\text{2-}}{\text{,O}}_{{}_{\text{2}}}^{\text{2-}}{\text{,F}}_{{}_{\text{2}}}^{\text{2-}}$

Concept Introduction: Molecular orbital theory explained the bonding, magnetic and spectral properties of molecule. It is based on the formation of molecular orbitals by the combination of atomic orbitals. On the basis of energy and stability these molecular orbitals can be further classified in three types:

• Bonding molecular orbitals (BMO): They have lesser energy than atomic orbital therefore more stable compare to atomic orbital.
• Antibonding molecular orbitals (ABMO): They have higher energy than atomic orbital therefore less stable compare to atomic orbital.
• Non-bonding molecular orbitals (NBMO): They have same energy as atomic orbital.

Molecular orbital diagrams represents the distribution of electrons in different molecular orbitals in increasing order of their energy. Hence lower energy molecular orbitals occupy first then only electron moves in higher energy orbitals.

Answer to Problem 42E

  ${\text{N}}_{{}_{\text{2}}}^{\text{2-}}$ must be most stable one as it has less number of antibonding electrons that makes the bond order 2 for this ion.

Explanation of Solution

The molecular orbital configuration of ${\text{N}}_{{}_{\text{2}}}^{\text{2-}}{\text{,O}}_{{}_{\text{2}}}^{\text{2-}}{\text{,F}}_{{}_{\text{2}}}^{\text{2-}}$ are:

  ${\text{N}}_{{}_{\text{2}}}^{\text{2-}}$ = ${{\text{(σ}}_{\text{1s}}\text{)}}^{\text{2}}{{\text{(σ}}_{{}_{\text{1s}}}^{\text{*}}\text{)}}^{\text{2}}{{\text{(σ}}_{\text{2s}}\text{)}}^{\text{2}}{{\text{(σ}}_{{}_{\text{2s}}}^{\text{*}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2py}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2pz}}\text{)}}^{\text{2}}{{\text{(σ}}_{\text{2px}}\text{)}}^{\text{2}}{{\text{(π}}_{{}_{\text{2py}}}^{*}\text{)}}^{\text{1}}{{\text{(π}}_{{}_{\text{2pz}}}^{*}\text{)}}^{\text{1}}$

  ${\text{O}}_{{}_{\text{2}}}^{\text{2-}}$ = ${{\text{(σ}}_{\text{1s}}\text{)}}^{\text{2}}{{\text{(σ}}_{{}_{\text{1s}}}^{\text{*}}\text{)}}^{\text{2}}{{\text{(σ}}_{\text{2s}}\text{)}}^{\text{2}}{{\text{(σ}}_{{}_{\text{2s}}}^{\text{*}}\text{)}}^{\text{2}}{{\text{(σ}}_{\text{2px}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2py}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2pz}}\text{)}}^{\text{2}}{{\text{(π}}_{{}_{\text{2py}}}^{*}\text{)}}^{\text{2}}{{\text{(π}}_{{}_{\text{2pz}}}^{*}\text{)}}^2$

  ${\text{F}}_{{}_{\text{2}}}^{\text{2-}}$ = ${{\text{(σ}}_{\text{1s}}\text{)}}^{\text{2}}{{\text{(σ}}_{{}_{\text{1s}}}^{\text{*}}\text{)}}^{\text{2}}{{\text{(σ}}_{\text{2s}}\text{)}}^{\text{2}}{{\text{(σ}}_{{}_{\text{2s}}}^{\text{*}}\text{)}}^{\text{2}}{{\text{(σ}}_{\text{2px}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2py}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2pz}}\text{)}}^{\text{2}}{{\text{(π}}_{{}_{\text{2py}}}^{*}\text{)}}^2{{\text{(π}}_{{}_{\text{2pz}}}^{*}\text{)}}^2{{\text{(σ}}_{{}_{\text{2px}}}^{*}\text{)}}^{\text{2}}$

Calculate bond order:

  $\begin{array}{l}\text{Bond order = }\frac{{\text{bonding e}}^{\text{-}}{\text{- antibonding e}}^{\text{-}}}{2}\\ {\text{Bond order in N}}_{{}_{\text{2}}}^{\text{2-}}\text{ = }\frac{\text{10- 6}}{2}=2\\ {\text{Bond order in O}}_{{}_{\text{2}}}^{\text{2-}}\text{ = }\frac{\text{10- 8}}{2}=1\\ {\text{Bond order in F}}_{{}_{\text{2}}}^{\text{2-}}\text{ = }\frac{\text{10- 10}}{2}=0\end{array}$

  ${\text{N}}_{{}_{\text{2}}}^{\text{2-}}$ must be most stable one as it has less number of antibonding electrons that makes the bond order 2 for this ion.

(c)

Interpretation Introduction

Interpretation: The most stable species from the given options needs to be identified.

  ${\text{Be}}_{{}_{\text{2}}}^{}{\text{,B}}_{{}_{\text{2}}}^{}{\text{,Ne}}_{{}_{\text{2}}}^{}$

Concept Introduction: Molecular orbital theory explained the bonding, magnetic and spectral properties of molecule. It is based on the formation of molecular orbitals by the combination of atomic orbitals. On the basis of energy and stability these molecular orbitals can be further classified in three types:

• Bonding molecular orbitals (BMO): They have lesser energy than atomic orbital therefore more stable compare to atomic orbital.
• Antibonding molecular orbitals (ABMO): They have higher energy than atomic orbital therefore less stable compare to atomic orbital.
• Non-bonding molecular orbitals (NBMO): They have same energy as atomic orbital.

Molecular orbital diagrams represents the distribution of electrons in different molecular orbitals in increasing order of their energy. Hence lower energy molecular orbitals occupy first then only electron moves in higher energy orbitals.

Answer to Problem 42E

  ${\text{B}}_{{}_{\text{2}}}^{}$ must be most stable one as it has bond order 1 whereas other two molecules do not exist.

Explanation of Solution

The molecular orbital configuration of ${\text{Be}}_{{}_{\text{2}}}^{}{\text{,B}}_{{}_{\text{2}}}^{}{\text{,Ne}}_{{}_{\text{2}}}^{}$ are:

  ${\text{Be}}_{{}_{\text{2}}}^{}$ = ${{\text{(σ}}_{\text{1s}}\text{)}}^{\text{2}}{{\text{(σ}}_{{}_{\text{1s}}}^{\text{*}}\text{)}}^{\text{2}}{{\text{(σ}}_{\text{2s}}\text{)}}^{\text{2}}{{\text{(σ}}_{{}_{\text{2s}}}^{\text{*}}\text{)}}^{\text{2}}$

  ${\text{B}}_{{}_{\text{2}}}^{}$ = ${{\text{(σ}}_{\text{1s}}\text{)}}^{\text{2}}{{\text{(σ}}_{{}_{\text{1s}}}^{\text{*}}\text{)}}^{\text{2}}{{\text{(σ}}_{\text{2s}}\text{)}}^{\text{2}}{{\text{(σ}}_{{}_{\text{2s}}}^{\text{*}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2py}}\text{)}}^1{{\text{(π}}_{\text{2pz}}\text{)}}^1$

  ${\text{Ne}}_{{}_{\text{2}}}^{}$ = ${{\text{(σ}}_{\text{1s}}\text{)}}^{\text{2}}{{\text{(σ}}_{{}_{\text{1s}}}^{\text{*}}\text{)}}^{\text{2}}{{\text{(σ}}_{\text{2s}}\text{)}}^{\text{2}}{{\text{(σ}}_{{}_{\text{2s}}}^{\text{*}}\text{)}}^{\text{2}}{{\text{(σ}}_{\text{2px}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2py}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2pz}}\text{)}}^{\text{2}}{{\text{(π}}_{{}_{\text{2py}}}^{*}\text{)}}^2{{\text{(π}}_{{}_{\text{2pz}}}^{*}\text{)}}^2{{\text{(σ}}_{{}_{\text{2px}}}^{*}\text{)}}^{\text{2}}$

Calculate bond order:

  $\begin{array}{l}\text{Bond order = }\frac{{\text{bonding e}}^{\text{-}}{\text{- antibonding e}}^{\text{-}}}{2}\\ {\text{Bond order in Be}}_{{}_{\text{2}}}^{}\text{ = }\frac{\text{4- 4}}{2}=0\\ {\text{Bond order in B}}_{{}_{\text{2}}}^{}\text{ = }\frac{\text{6- 4}}{2}=1\\ {\text{Bond order in Ne}}_{{}_{\text{2}}}^{}\text{ = }\frac{\text{10- 10}}{2}=0\end{array}$

  ${\text{B}}_{{}_{\text{2}}}^{}$ must be most stable one as it has bond order 1 whereas other two molecules do not exist.