Chapter 2, Problem 2G.17E

(a)

Interpretation Introduction

Interpretation:

Bond order of ${\text{F}}_2$ and ${\text{F}}_2^{-}$ and species with stronger bond among ${\text{F}}_2$ and ${\text{F}}_2^{-}$ has to be determined.

Concept Introduction:

Molecular orbital diagram is a linear combination of atomic orbitals of similar energy and similar symmetry. It is formed by the proper overlap of the atomic orbitals.

There are 3 types of molecular orbitals as follows:

1. Bonding molecular orbital: They are formed by the constructive interference of atomic orbitals and electrons in it stabilize the molecule and are of lesser in energy.

2. Antibonding molecular orbital: This type of orbitals increases the energy of molecule and destabilizes it and weakens the bond between the atoms.

3. Non-bonding molecular orbital: These types of orbitals have energy similar to atomic orbitals that is addition or removal of electron does not change the energy of molecule.

The order of energy in molecular orbital follows two rules as follows:

1. For atomic number less than or equal to 14 order of energy is,

  ${\sigma }_{1\text{s}}<{\sigma }_{1\text{s}}^{\ast }<{\sigma }_{\text{2s}}<{\sigma }_{\text{2s}}^{\ast }<\left({\text{π}}_{2{\text{p}}_{\text{x}}}={\text{π}}_{2{\text{p}}_{\text{y}}}\right)<{\sigma }_{{\text{2p}}_{\text{z}}}<\left({\text{π}}_{2{\text{p}}_{\text{x}}}^{\ast }={\text{π}}_{2{\text{p}}_{\text{y}}}^{\ast }\right)<{\sigma }_{{}_{{\text{2p}}_{\text{z}}}}^{\ast }$

2. For atomic number more than 14 order of energy is,

  ${\sigma }_{1\text{s}}<{\sigma }_{1\text{s}}^{\ast }<{\sigma }_{\text{2s}}<{\sigma }_{\text{2s}}^{\ast }<{\sigma }_{{\text{2p}}_{\text{z}}}<\left({\text{π}}_{2{\text{p}}_{\text{x}}}={\text{π}}_{2{\text{p}}_{\text{y}}}\right)<\left({\text{π}}_{2{\text{p}}_{\text{x}}}^{\ast }={\text{π}}_{2{\text{p}}_{\text{y}}}^{\ast }\right)<{\sigma }_{{}_{{\text{2p}}_{\text{z}}}}^{\ast }$

Bond order $\left(b\right)$ is defined as total number of bonds between the atoms. It is calculated as follows:

  $b=\frac{1}{2}\left[\begin{array}{l}\text{number of elctrons in bonding orbitals}-\\ \text{number of electrons in antibonding orbitals}\end{array}\right]$        (1)

Explanation of Solution

For ${\text{F}}_2$ molecule:

The symbol for fluorine is $\text{F}$ with atomic number 9. Its electronic configuration is $\left[\text{He}\right]{\text{2s}}^2{\text{2p}}^5$. Thus, it possesses 7 valence electrons.

Thus total valence electrons are sum of the valence electrons for each atom in ${\text{F}}_2$. It is calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=7\left(2\right)\\ =14\end{array}$

Hence, 14 electrons are to be arranged in each molecular orbital to obtain an electronic configuration. Since, number of electrons in ${\text{F}}_2$ is more than $14$ so electronic configuration is ${\left({\text{σ}}_{\text{2s}}\right)}^2{\left({\text{ σ}}_{\text{2s}}^{\text{*}}\right)}^2{\left({\text{σ}}_{\text{2p}}\right)}^2{\left({\text{π}}_{\text{2p}}\right)}^4{\left({\text{ π}}_{\text{2p}}^{\text{*}}\right)}^4$.

Substitute 8 for number of electrons in bonding orbitals and 6 for number of electrons in antibonding orbitals in equation (1) to calculate bond order.

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

Hence, the bond order of the molecule ${\text{F}}_2$ is 1 and electronic configuration is ${\left({\text{σ}}_{\text{2s}}\right)}^2{\left({\text{ σ}}_{\text{2s}}^{\text{*}}\right)}^2{\left({\text{σ}}_{\text{2p}}\right)}^2{\left({\text{π}}_{\text{2p}}\right)}^4{\left({\text{ π}}_{\text{2p}}^{\text{*}}\right)}^4$.

For ${\text{F}}_2^{-}$ molecule:

The symbol for fluorine is $\text{F}$ with atomic number 9. Its electronic configuration is $\left[\text{He}\right]{\text{2s}}^2{\text{2p}}^5$. Thus, it possesses 7 valence electrons.

One negative charge is added up in total valence count.

Thus total valence electrons are sum of the valence electrons for each atom in ${\text{F}}_2^{-}$. It is calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=7\left(2\right)+1\\ =15\end{array}$

Hence, 15 electrons are to be arranged in each molecular orbital to obtain an electronic configuration. Since, number of electrons in ${\text{F}}_2^{-}$ is more than $14$ so electronic configuration is ${\left({\text{σ}}_{\text{2s}}\right)}^2{\left({\text{σ}}_{\text{2s}}^{\text{*}}\right)}^2{\left({\text{σ}}_{\text{2p}}\right)}^2{\left({\text{π}}_{\text{2p}}\right)}^4{\left({\text{π}}_{\text{2p}}^{\text{*}}\right)}^4{\left({\text{σ}}_{2\text{p}}^{\ast }\right)}^1$.

Substitute 8 for number of electrons in bonding orbitals and 7 for number of electrons in antibonding orbitals in equation (1) to calculate bond order.

  $\begin{array}{c}b=\frac{1}{2}\left(8-7\right)\\ =\frac{1}{2}\\ =0.5\end{array}$

Hence, the bond order of the molecule ${\text{F}}_2^{-}$ is 0.5 and electronic configuration is ${\left({\text{σ}}_{\text{2s}}\right)}^2{\left({\text{σ}}_{\text{2s}}^{\text{*}}\right)}^2{\left({\text{σ}}_{\text{2p}}\right)}^2{\left({\text{π}}_{\text{2p}}\right)}^4{\left({\text{π}}_{\text{2p}}^{\text{*}}\right)}^4{\left({\text{σ}}_{2\text{p}}^{\ast }\right)}^1$.

Since the bond order of ${\text{F}}_2$ is higher than ${\text{F}}_2^{-}$ that means ${\text{F}}_2$ is bonded by more number of bonds between the atoms hence difficult to break and therefore, has higher bond strength and thus ${\text{F}}_2$ has strong bonds.

(b)

Interpretation Introduction

Interpretation:

Bond order of ${\text{B}}_2$ and ${\text{B}}_2^{+}$ and species with stronger bond has to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

For ${\text{B}}_2$ molecule:

The symbol for boron is $\text{B}$ with atomic number 5. Its electronic configuration is $\left[\text{He}\right]{\text{2s}}^2{\text{2p}}^1$. Thus, it possesses 3 valence electrons.

Thus total valence electrons are sum of the valence electrons for each atom in ${\text{B}}_2$. It is calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=3\left(2\right)\\ =6\end{array}$

Hence, 6 electrons are to be arranged in each molecular orbital to obtain an electronic configuration. Since, number of electrons in ${\text{B}}_2$ is less than $14$ so electronic configuration is ${\left({\text{σ}}_{2\text{s}}\right)}^2{\left({\text{σ}}_{2\text{s}}^{\ast }\right)}^2{\left({\text{π}}_{2\text{p}}\right)}^2$.

Substitute 4 for number of electrons in bonding orbitals and 2 for number of electrons in antibonding orbitals in equation (1) to calculate bond order.

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

Hence, the bond order of the molecule ${\text{B}}_2$ is 1 and electronic configuration is ${\left({\text{σ}}_{2\text{s}}\right)}^2{\left({\text{σ}}_{2\text{s}}^{\ast }\right)}^2{\left({\text{π}}_{2\text{p}}\right)}^2$.

For ${\text{B}}_2^{+}$ molecule:

The symbol for boron is $\text{B}$ with atomic number 5. Its electronic configuration is $\left[\text{He}\right]{\text{2s}}^2{\text{2p}}^1$. Thus, it possesses 3 valence electrons.

Thus total valence electrons are sum of the valence electrons for each atom in ${\text{B}}_2$. It is calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=3\left(2\right)-1\\ =5\end{array}$

Hence, 5 electrons are to be arranged in each molecular orbital to obtain an electronic configuration. Since, number of electrons in ${\text{B}}_2^{+}$ is less than $14$ so electronic configuration is ${\left({\text{σ}}_{2\text{s}}\right)}^2{\left({\text{σ}}_{2\text{s}}^{\ast }\right)}^2{\left({\text{ π}}_{2\text{p}}\right)}^1$.

Substitute 3 for number of electrons in bonding orbitals and 2 for number of electrons in antibonding orbitals in equation (1) to calculate bond order.

  $\begin{array}{c}b=\frac{1}{2}\left(3-2\right)\\ =\frac{1}{2}\\ =0.5\end{array}$

Hence, the bond order of the molecule ${\text{B}}_2^{+}$ is 0.5 and electronic configuration is ${\left({\text{σ}}_{2\text{s}}\right)}^2{\left({\text{σ}}_{2\text{s}}^{\ast }\right)}^2{\left({\text{ π}}_{2\text{p}}\right)}^1$.

Since the bond order of ${\text{B}}_2$ is higher than ${\text{B}}_2^{+}$ that means ${\text{B}}_2$ is bonded by more number of bonds between the atoms hence difficult to break and therefore, has higher bond strength. Thus ${\text{B}}_2$ has strong bonds.