Chapter 14, Problem 39E

(a)

Interpretation Introduction

Interpretation: The reason of stability of ${\text{H}}_{\text{2}}$ whereas ${\mathrm{He}}_2$ is unstable needs to be explained.

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 39E

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

To draw the energy level and Molecular orbital diagram for ${\text{H}}_{\text{2}}$ and ${\mathrm{He}}_2$ molecule; write the electronic configuration of He:

  ${\text{He = 1s}}^{\text{2}}$

  ${\text{H = 1s}}^{\text{1}}$

  

  

Calculate bond order:

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

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.

(b)

Interpretation Introduction

Interpretation: The reason of paramagnetic nature of ${\text{B}}_{\text{2}}$ and ${\text{O}}_{\text{2}}$ whereas ${\text{C}}_{\text{2}}$ , ${\text{N}}_{\text{2}}$ and ${\text{F}}_{\text{2}}$ are diamagnetic needs to be explained.

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 39E

Only ${\text{B}}_{\text{2}}$ and ${\text{O}}_{\text{2}}$ molecules have unpaired electrons therefore they are paramagnetic.

Explanation of Solution

The molecular orbital electronic configuration of ${\text{B}}_{\text{2}}$ , ${\text{O}}_{\text{2}}$ , ${\text{C}}_{\text{2}}$ , ${\text{N}}_{\text{2}}$ and ${\text{F}}_{\text{2}}$ can be written as:

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

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

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

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

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

Only ${\text{B}}_{\text{2}}$ and ${\text{O}}_{\text{2}}$ molecules have unpaired electrons therefore they are paramagnetic.

(c)

Interpretation Introduction

Interpretation: The reason of large bond dissociation energy of ${\text{N}}_{\text{2}}$ molecule needs to be explained.

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 39E

Since the bond order for ${\text{N}}_{\text{2}}$ molecule is 3, therefore it has large bond dissociation energy.

Explanation of Solution

The molecular orbital electronic configuration of ${\text{N}}_{\text{2}}$ can be written as:

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

Calculate bond order:

  $\begin{array}{l}\text{Bond order = }\frac{{\text{bonding e}}^{\text{-}}{\text{- antibonding e}}^{\text{-}}}{2}\\ {\text{Bond order in N}}_{\text{2}}\text{ = }\frac{\text{10- 4}}{2}=3\end{array}$

Since the bond order for ${\text{N}}_{\text{2}}$ molecule is 3, therefore it has large bond dissociation energy.

(d)

Interpretation Introduction

Interpretation: The reason of more stability of ${\text{NO}}^{\text{+}}$ compare to ${\text{NO}}^{\text{-}}$ needs to be explained.

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 39E

Since there are more anti-bonding electrons in ${\text{NO}}^{\text{-}}$ ion therefore it is less stable than ${\text{NO}}^{\text{+}}$ which has lesser anti-bonding electrons.

Explanation of Solution

The molecular orbital electronic configuration of ${\text{NO}}^{\text{+}}$ and ${\text{NO}}^{\text{-}}$:

Number of electrons in N = 7

Number of electrons in O = 8

Total number of electrons in ${\text{NO}}^{\text{+}}$ = 7 + 8 -1 = 14

Total number of electrons in ${\text{NO}}^{\text{-}}$ = 7 + 8 +1 = 16

  ${\text{NO}}^{\text{+}}$ = ${{\text{(σ}}_{\text{1s}}\text{)}}^{\text{2}}{{\text{(σ}}_{{}_{\text{1s}}}^{\text{*}}\text{)}}^{\text{2}}{{\text{(σ}}_{\text{2s}}\text{)}}^{\text{2}}{{\text{(σ}}_{{}_{\text{2s}}}^{\text{*}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2py}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2pz}}\text{)}}^{\text{2}}{{\text{(σ}}_{\text{2px}}\text{)}}^{\text{2}}$

  ${\text{NO}}^{\text{-}}$ = ${{\text{(σ}}_{\text{1s}}\text{)}}^{\text{2}}{{\text{(σ}}_{{}_{\text{1s}}}^{\text{*}}\text{)}}^{\text{2}}{{\text{(σ}}_{\text{2s}}\text{)}}^{\text{2}}{{\text{(σ}}_{{}_{\text{2s}}}^{\text{*}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2py}}\text{)}}^{\text{2}}{{\text{(π}}_{\text{2pz}}\text{)}}^{\text{2}}{{\text{(σ}}_{\text{2px}}\text{)}}^{\text{2}}{{\text{(π}}_{{}_{\text{2py}}}^{*}\text{)}}^1{{\text{(π}}_{{}_{\text{2pz}}}^{*}\text{)}}^1$

Calculate bond order:

  $\begin{array}{l}\text{Bond order = }\frac{{\text{bonding e}}^{\text{-}}{\text{- antibonding e}}^{\text{-}}}{2}\\ {\text{Bond order in NO}}^{\text{+}}\text{ = }\frac{\text{8- 2}}{2}=3\\ {\text{Bond order in NO}}^{-}\text{ = }\frac{\text{8- 4}}{2}=2\end{array}$

Since there are more anti-bonding electrons in ${\text{NO}}^{\text{-}}$ ion therefore it is less stable than ${\text{NO}}^{\text{+}}$ which has lesser anti-bonding electrons. Also ${\text{NO}}^{\text{+}}$ has larger bond order compare to ${\text{NO}}^{\text{-}}$ that makes it more stable.