Chapter 9, Problem 9C.4AE

Interpretation Introduction

Interpretation:

The bond order of each period 2 homonuclear diatomic molecule has to be calculated.

Concept introduction:

One of the most useful parameters to discuss the characteristics of bonds is bond order.  The bond order is directly related with the bond length and strength of the bond.  Higher the bond order, shorter will be the bond.  The bond dissociation energy determines the strength of a bond.

Answer to Problem 9C.4AE

The bond order of each period 2 homonuclear diatomic molecule is given below.

The bond order of ${\text{Li}}_{\text{2}}$ is $1$.

The bond order of ${\text{Be}}_{\text{2}}$ is $0$.

The bond order of ${\text{B}}_{\text{2}}$ is $1$.

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

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

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

The bond order of ${\text{F}}_{\text{2}}$ is $1$.

Explanation of Solution

The elements present in the second period of periodic table are $\text{Li, Be, B, C, N, O, F, Ne}$.  Except neon, all other elements can form homonuclear diatomic molecules. The homonuclear diatomic molecule are ${\text{Li}}_{\text{2}}{\text{, Be}}_{\text{2}}{\text{ ,B}}_{\text{2}}{\text{ ,C}}_{\text{2}}{\text{ ,N}}_{\text{2}}{\text{ ,O}}_{\text{2}}{\text{ ,F}}_{\text{2}}$.

The total number of valence electrons possessed by ${\text{Li}}_{\text{2}}$ is $\left(1+1\right)=2$.

The molecular orbitals are filled in the order shown below.

    ${\text{1σ}}_{\text{g}}{\text{1σ}}_{\text{u}}{\text{1π}}_{\text{u}}{\text{2σ}}_{\text{g}}{\text{1π}}_{\text{g}}{\text{2σ}}_{\text{u}}$

Therefore, the ground state electronic configuration of ${\text{Li}}_{\text{2}}$ molecule is shown below.

    ${\text{1σ}}_{\text{g}}{}^2$

The bond order ($b$) is calculated by the formula shown below.

    $b=\frac{1}{2}\left(N-N^{\ast }\right)$                                                                                              (1)

Where,

• $N$ is the number of electrons in bonding orbitals.
• $N^{\ast }$ is the number of electrons in anti-bonding orbitals.

The value of $N$ for ${\text{Li}}_{\text{2}}$ is $2$.

The value of $N^{\ast }$ for ${\text{Li}}_{\text{2}}$ is $0$.

Substitute the values of $N$ and $N^{\ast }$ in equation 1.

    $\begin{array}{c}b=\frac{1}{2}\left(N-N^{\ast }\right)\\ =\frac{1}{2}\left(2-0\right)\\ =\frac{2}{2}\\ =1\end{array}$

Therefore, the bond order of ${\text{Li}}_{\text{2}}$ is $1$.

The total number of valence electrons possessed by ${\text{Be}}_{\text{2}}$ is $\left(2+2\right)=4$.

The molecular orbitals are filled in the order shown below.

    ${\text{1σ}}_{\text{g}}{\text{1σ}}_{\text{u}}{\text{1π}}_{\text{u}}{\text{2σ}}_{\text{g}}{\text{1π}}_{\text{g}}{\text{2σ}}_{\text{u}}$

Therefore, the ground state electronic configuration of ${\text{Be}}_{\text{2}}$ molecule is shown below.

    ${\text{1σ}}_{\text{g}}{}^2{\text{1σ}}_{\text{u}}{}^2$

The bond order ($b$) is calculated by the formula shown below.

    $b=\frac{1}{2}\left(N-N^{\ast }\right)$                                                                                              (1)

Where,

• $N$ is the number of electrons in bonding orbitals.
• $N^{\ast }$ is the number of electrons in anti-bonding orbitals.

The value of $N$ for ${\text{Be}}_{\text{2}}$ is $2$.

The value of $N^{\ast }$ for ${\text{Be}}_{\text{2}}$ is $0$.

Substitute the values of $N$ and $N^{\ast }$ in equation 1.

    $\begin{array}{c}b=\frac{1}{2}\left(N-N^{\ast }\right)\\ =\frac{1}{2}\left(2-2\right)\\ =0\end{array}$

Therefore, the bond order of ${\text{Be}}_{\text{2}}$ is $0$.

The total number of valence electrons possessed by ${\text{B}}_{\text{2}}$ is $\left(3+3\right)=6$.

The molecular orbitals are filled in the order shown below.

    ${\text{1σ}}_{\text{g}}{\text{1σ}}_{\text{u}}{\text{1π}}_{\text{u}}{\text{2σ}}_{\text{g}}{\text{1π}}_{\text{g}}{\text{2σ}}_{\text{u}}$

Therefore, the ground state electronic configuration of ${\text{B}}_{\text{2}}$ molecule is shown below.

    ${\text{1σ}}_{\text{g}}{}^2{\text{1σ}}_{\text{u}}{}^2{\text{1π}}_{\text{u}}{}^2$

The bond order ($b$) is calculated by the formula shown below.

    $b=\frac{1}{2}\left(N-N^{\ast }\right)$                                                                                              (1)

Where,

• $N$ is the number of electrons in bonding orbitals.
• $N^{\ast }$ is the number of electrons in anti-bonding orbitals.

The value of $N$ for ${\text{B}}_{\text{2}}$ is $4$.

The value of $N^{\ast }$ for ${\text{B}}_{\text{2}}$ is $2$.

Substitute the values of $N$ and $N^{\ast }$ in equation 1.

    $\begin{array}{c}b=\frac{1}{2}\left(N-N^{\ast }\right)\\ =\frac{1}{2}\left(4-2\right)\\ =1\end{array}$

Therefore, the bond order of ${\text{B}}_{\text{2}}$ is $1$.

The total number of valence electrons possessed by ${\text{C}}_{\text{2}}$ is $\left(4+4\right)=8$.

The molecular orbitals are filled in the order shown below.

    ${\text{1σ}}_{\text{g}}{\text{1σ}}_{\text{u}}{\text{1π}}_{\text{u}}{\text{2σ}}_{\text{g}}{\text{1π}}_{\text{g}}{\text{2σ}}_{\text{u}}$

Therefore, the ground state electronic configuration of ${\text{C}}_{\text{2}}$ molecule is shown below.

    ${\text{1σ}}_{\text{g}}{}^2{\text{1σ}}_{\text{u}}{}^2{\text{1π}}_{\text{u}}{}^4$

The bond order ($b$) is calculated by the formula shown below.

    $b=\frac{1}{2}\left(N-N^{\ast }\right)$                                                                                              (1)

Where,

• $N$ is the number of electrons in bonding orbitals.
• $N^{\ast }$ is the number of electrons in anti-bonding orbitals.

The value of $N$ for ${\text{C}}_{\text{2}}$ is $6$.

The value of $N^{\ast }$ for ${\text{C}}_{\text{2}}$ is $2$.

Substitute the values of $N$ and $N^{\ast }$ in equation 1.

    $\begin{array}{c}b=\frac{1}{2}\left(N-N^{\ast }\right)\\ =\frac{1}{2}\left(6-2\right)\\ =2\end{array}$

Therefore, the bond order of ${\text{C}}_{\text{2}}$ is $2$.

The total number of valence electrons possessed by ${\text{N}}_{\text{2}}$ is $\left(5+5\right)=10$.

The molecular orbitals are filled in the order shown below.

    ${\text{1σ}}_{\text{g}}{\text{1σ}}_{\text{u}}{\text{1π}}_{\text{u}}{\text{2σ}}_{\text{g}}{\text{1π}}_{\text{g}}{\text{2σ}}_{\text{u}}$

Therefore, the ground state electronic configuration of ${\text{N}}_{\text{2}}$ molecule is shown below.

    ${\text{1σ}}_{\text{g}}{}^2{\text{1σ}}_{\text{u}}{}^2{\text{1π}}_{\text{u}}{}^4{\text{2σ}}_{\text{g}}{}^2$

The bond order ($b$) is calculated by the formula shown below.

    $b=\frac{1}{2}\left(N-N^{\ast }\right)$                                                                                              (1)

Where,

• $N$ is the number of electrons in bonding orbitals.
• $N^{\ast }$ is the number of electrons in anti-bonding orbitals.

The value of $N$ for ${\text{N}}_{\text{2}}$ is $8$.

The value of $N^{\ast }$ for ${\text{N}}_{\text{2}}$ is $2$.

Substitute the values of $N$ and $N^{\ast }$ in equation 1.

    $\begin{array}{c}b=\frac{1}{2}\left(N-N^{\ast }\right)\\ =\frac{1}{2}\left(8-2\right)\\ =3\end{array}$

Therefore, the bond order of ${\text{N}}_{\text{2}}$ is $3$.

The total number of valence electrons possessed by ${\text{O}}_{\text{2}}$ is $\left(6+6\right)=12$.

The molecular orbitals are filled in the order shown below.

    ${\text{1σ}}_{\text{g}}{\text{1σ}}_{\text{u}}{\text{1π}}_{\text{u}}{\text{2σ}}_{\text{g}}{\text{1π}}_{\text{g}}{\text{2σ}}_{\text{u}}$

Therefore, the ground state electronic configuration of ${\text{O}}_{\text{2}}$ molecule is shown below.

    ${\text{1σ}}_{\text{g}}{}^2{\text{1σ}}_{\text{u}}{}^2{\text{1π}}_{\text{u}}{}^4{\text{2σ}}_{\text{g}}{}^2{\text{1π}}_{\text{g}}{}^2$

The bond order ($b$) is calculated by the formula shown below.

    $b=\frac{1}{2}\left(N-N^{\ast }\right)$                                                                                              (1)

Where,

• $N$ is the number of electrons in bonding orbitals.
• $N^{\ast }$ is the number of electrons in anti-bonding orbitals.

The value of $N$ for ${\text{O}}_{\text{2}}$ is $8$.

The value of $N^{\ast }$ for ${\text{O}}_{\text{2}}$ is $4$.

Substitute the values of $N$ and $N^{\ast }$ in equation 1.

    $\begin{array}{c}b=\frac{1}{2}\left(N-N^{\ast }\right)\\ =\frac{1}{2}\left(8-4\right)\\ =2\end{array}$

Therefore, the bond order of ${\text{O}}_{\text{2}}$ is $2$.

The total number of valence electrons possessed by ${\text{F}}_{\text{2}}$ is $\left(7+7\right)=14$.

The molecular orbitals are filled in the order shown below.

    ${\text{1σ}}_{\text{g}}{\text{1σ}}_{\text{u}}{\text{1π}}_{\text{u}}{\text{2σ}}_{\text{g}}{\text{1π}}_{\text{g}}{\text{2σ}}_{\text{u}}$

Therefore, the ground state electronic configuration of ${\text{F}}_{\text{2}}$ molecule is shown below.

    ${\text{1σ}}_{\text{g}}{}^2{\text{1σ}}_{\text{u}}{}^2{\text{1π}}_{\text{u}}{}^4{\text{2σ}}_{\text{g}}{}^2{\text{1π}}_{\text{g}}{}^4$

The bond order ($b$) is calculated by the formula shown below.

    $b=\frac{1}{2}\left(N-N^{\ast }\right)$                                                                                              (1)

Where,

• $N$ is the number of electrons in bonding orbitals.
• $N^{\ast }$ is the number of electrons in anti-bonding orbitals.

The value of $N$ for ${\text{F}}_{\text{2}}$ is $8$.

The value of $N^{\ast }$ for ${\text{F}}_{\text{2}}$ is $6$.

Substitute the values of $N$ and $N^{\ast }$ in equation 1.

    $\begin{array}{c}b=\frac{1}{2}\left(N-N^{\ast }\right)\\ =\frac{1}{2}\left(8-6\right)\\ =1\end{array}$

Therefore, the bond order of ${\text{F}}_{\text{2}}$ is $1$.