Chapter 2, Problem 2G.4E

(a)

Interpretation Introduction

Interpretation:

Valence shell electronic configuration of ${\text{N}}_2^{+}$, ${\text{N}}_2^{2+}$ and ${\text{N}}_2^{2-}$ species 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,

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

Explanation of Solution

For ${\text{N}}_2^{+}$ molecule

The symbol for nitrogen is $\text{N}$ with atomic number 7. Its electronic configuration is $\left[\text{He}\right]{\text{2s}}^{2}2{\text{p}}^3$. Thus, it possesses 5 valence electrons.

One positive charge is added up in the total valence count.

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

  $\begin{array}{c}\text{Total valence electrons}=5\left(2\right)-1\\ =9\end{array}$

Hence, 9 electrons are to be arranged in each molecular orbital to obtain an electronic configuration. Since number of electrons in ${\text{N}}_2^{+}$ is less than $14$ so electronic configuration is ${\left({\text{σ}}_{\text{2s}}\right)}^2{\left({\text{σ}}_{\text{2s}}^{\text{*}}\right)}^2{\left({\text{π}}_{\text{2p}}\right)}^4{\left({\text{σ}}_{\text{2p}}\right)}^1$.

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

For ${\text{N}}_2^{2+}$ molecule

The symbol for nitrogen is $\text{N}$ with atomic number 7. Its electronic configuration is $\left[\text{He}\right]{\text{2s}}^{2}2{\text{p}}^3$. Thus, it possesses 5 valence electrons.

Two positive charges are subtracted from the total valence count.

Thus total valence electrons are sum of the valence electrons for each atom in ${\text{N}}_2^{2+}$. It is calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=5\left(2\right)-2\\ =8\end{array}$

Hence, 8 electrons are to be arranged in each molecular orbital to obtain an electronic configuration. Since number of electrons in ${\text{N}}_2^{2+}$ is less than $14$ so electronic configuration is ${\left({\text{σ}}_{{\text{2s}}^{\text{2}}}\right)}^2{\left({\text{σ}}_{\text{2s}}^{\text{*}}\right)}^2{\left({\text{π}}_{\text{2p}}\right)}^4$.

Hence, the electronic configuration of ${\text{N}}_2^{2+}$ is ${\left({\text{σ}}_{{\text{2s}}^{\text{2}}}\right)}^2{\left({\text{σ}}_{\text{2s}}^{\text{*}}\right)}^2{\left({\text{π}}_{\text{2p}}\right)}^4$.

For ${\text{N}}_2^{2-}$ molecule

The symbol for nitrogen is $\text{N}$ with atomic number 7. Its electronic configuration is $\left[\text{He}\right]{\text{2s}}^{2}2{\text{p}}^3$. Thus, it possesses 5 valence electrons.

Two negative charges are added up in total valence count.

Thus total valence electrons are sum of the valence electrons for each atom in ${\text{N}}_2^{2-}$. It is calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=5\left(2\right)+2\\ =12\end{array}$

Hence, 12 electrons are to be arranged in each molecular orbital to obtain an electronic configuration. Since number of electrons in ${\text{N}}_2^{2-}$ is more than $14$ so electronic configuration is ${\left({\text{σ}}_{\text{2s}}\right)}^2{\left({\text{σ}}_{\text{2s}}^{\text{*}}\right)}^2{\left({\text{σ}}_{\text{2p}}\right)}^2{\left({\text{π}}_{\text{2p}}\right)}^4{\left({\text{π}}_{\text{2p}}^{\text{*}}\right)}^2$.

Hence, the electronic configuration of ${\text{N}}_2^{2-}$ is ${\left({\text{σ}}_{\text{2s}}\right)}^2{\left({\text{σ}}_{\text{2s}}^{\text{*}}\right)}^2{\left({\text{σ}}_{\text{2p}}\right)}^2{\left({\text{π}}_{\text{2p}}\right)}^4{\left({\text{π}}_{\text{2p}}^{\text{*}}\right)}^2$.

(b)

Interpretation Introduction

Interpretation:

The bond order of ${\text{N}}_2^{+}$ ${\text{N}}_2^{2+}$ and ${\text{N}}_2^{2-}$ species has to be determined.

Concept Introduction:

Bond order $b$ 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 ${\text{N}}_2^{+}$ molecule

The electronic configuration of ${\text{N}}_2^{+}$ is ${\left({\text{σ}}_{\text{2s}}\right)}^2{\left({\text{σ}}_{\text{2s}}^{\text{*}}\right)}^2{\left({\text{π}}_{\text{2p}}\right)}^4{\left({\text{σ}}_{\text{2p}}\right)}^1$.

Substitute 7 for number of electrons in bonding orbitals and 2 for number of electrons in antibonding orbitals in equation (1) to calculate bond order.

  $\begin{array}{c}b=\frac{1}{2}\left(7-2\right)\\ =\frac{5}{2}\\ =2.5\end{array}$

Hence, the bond order of the molecule ${\text{N}}_2^{+}$ is 2.5.

For ${\text{N}}_2^{2+}$ molecule

The electronic configuration of ${\text{N}}_2^{2+}$ is ${\left({\text{σ}}_{{\text{2s}}^{\text{2}}}\right)}^2{\left({\text{σ}}_{\text{2s}}^{\text{*}}\right)}^2{\left({\text{π}}_{\text{2p}}\right)}^4$.

Substitute 6 for number of electrons in bonding orbitals and 2 for number of electrons in antibonding orbitals in equation (1) to calculate bond order.

  $\begin{array}{c}b=\frac{1}{2}\left(6-2\right)\\ =\frac{4}{2}\\ =2\end{array}$

Hence, the bond order of the molecule ${\text{N}}_2^{2+}$ is 2.

For ${\text{N}}_2^{2-}$ molecule

The electronic configuration of ${\text{N}}_2^{2-}$ is ${\left({\text{σ}}_{\text{2s}}\right)}^2{\left({\text{σ}}_{\text{2s}}^{\text{*}}\right)}^2{\left({\text{σ}}_{\text{2p}}\right)}^2{\left({\text{π}}_{\text{2p}}\right)}^4{\left({\text{π}}_{\text{2p}}^{\text{*}}\right)}^2$.

Substitute 8 for number of electrons in bonding orbitals and 4 for number of electrons in antibonding orbitals in equation (1) to calculate bond order.

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

Hence, the bond order of the molecule ${\text{N}}_2^{2-}$ is 2.

(c)

Interpretation Introduction

Interpretation:

Molecular orbital diagram of ${\text{N}}_2^{+}$, ${\text{N}}_2^{2+}$ and ${\text{N}}_2^{2-}$ species have to be drawn.

Concept Introduction:

Refer to (a)

Explanation of Solution

For ${\text{N}}_2^{+}$ molecule

The electronic configuration of ${\text{N}}_2^{+}$ is ${\left({\text{σ}}_{\text{2s}}\right)}^2{\left({\text{σ}}_{\text{2s}}^{\text{*}}\right)}^2{\left({\text{π}}_{\text{2p}}\right)}^4{\left({\text{σ}}_{\text{2p}}\right)}^1$. The molecular orbital diagram of ${\text{N}}_2^{+}$ is as follows:

Since ${\text{N}}_2^{+}$ has one unpaired electron, therefore it is paramagnetic.

For ${\text{N}}_2^{2+}$ molecule

The electronic configuration of ${\text{N}}_2^{2+}$ is ${\left({\text{σ}}_{{\text{2s}}^{\text{2}}}\right)}^2{\left({\text{σ}}_{\text{2s}}^{\text{*}}\right)}^2{\left({\text{π}}_{\text{2p}}\right)}^4$. The molecular orbital diagram of ${\text{N}}_2^{2+}$ is as follows:

Since ${\text{N}}_2^{2+}$ molecule has no unpaired electron, therefore it is diamagnetic.

For ${\text{N}}_2^{2-}$ molecule

The electronic configuration of ${\text{N}}_2^{2-}$ is ${\left({\text{σ}}_{\text{2s}}\right)}^2{\left({\text{σ}}_{\text{2s}}^{\text{*}}\right)}^2{\left({\text{σ}}_{\text{2p}}\right)}^2{\left({\text{π}}_{\text{2p}}\right)}^4{\left({\text{π}}_{\text{2p}}^{\text{*}}\right)}^2$. The molecular orbital diagram of ${\text{N}}_2^{2-}$ molecule is as follows:

Since ${\text{N}}_2^{2-}$ has two unpaired electrons, therefore it is paramagnetic.

(d)

Interpretation Introduction

Interpretation:

Character of highest energy orbital of ${\text{N}}_2^{+}$ , ${\text{N}}_2^{2+}$ and ${\text{N}}_2^{2-}$ species has to be determined.

Concept Introduction:

Refer to (a)

Explanation of Solution

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

The electronic configuration of ${\text{N}}_2^{+}$ is ${\left({\text{σ}}_{\text{2s}}\right)}^2{\left({\text{σ}}_{\text{2s}}^{\text{*}}\right)}^2{\left({\text{π}}_{\text{2p}}\right)}^4{\left({\text{σ}}_{\text{2p}}\right)}^1$. The highest energy orbital electron in ${\text{N}}_2^{+}$ is present in ${\left({\text{σ}}_{\text{2p}}\right)}^1$. Therefore, it contains $\text{σ}$ character.

For ${\text{N}}_2^{2+}$ molecule

The electronic configuration of ${\text{N}}_2^{2+}$ is ${\left({\text{σ}}_{{\text{2s}}^{\text{2}}}\right)}^2{\left({\text{σ}}_{\text{2s}}^{\text{*}}\right)}^2{\left({\text{π}}_{\text{2p}}\right)}^4$. The highest energy orbital electron in ${\text{N}}_2^{2+}$ is present in ${\text{ π}}_{2{\text{p}}^4}$. Therefore, it contains $\text{π}$ character.

For ${\text{N}}_2^{2-}$ molecule:

The electronic configuration of ${\text{N}}_2^{2-}$ is ${\left({\text{σ}}_{\text{2s}}\right)}^2{\left({\text{σ}}_{\text{2s}}^{\text{*}}\right)}^2{\left({\text{σ}}_{\text{2p}}\right)}^2{\left({\text{π}}_{\text{2p}}\right)}^4{\left({\text{π}}_{\text{2p}}^{\text{*}}\right)}^2$. The highest energy orbital electron in ${\text{N}}_2^{2-}$ is present in ${\text{ π}}_{{\text{2p}}^2}^{\text{*}}$. Therefore, it contains $\text{π}$ character.