Chapter 6, Problem 6.83QP

Draw Lewis structures for the following four isoelectronic species: (a) CO, (b) NO⁺, (c) CN−, (d) N₂. Show formal charges. (See Problem 6.69.)

Interpretation Introduction

Interpretation: The Lewis structures for the given isoelectronic species should be shown.

Concept Introduction:

• Lewis structures are diagrams that represent the chemical bonding of covalently bonded molecules and coordination compounds.
• It is also known as Lewis dot structures which represents the bonding between atoms of a molecule and the lone pairs of electrons that may exist in the molecule.
• The Lewis structure is based on the concept of the octet rule so that the electrons shared in each atom should have 8 electrons in its outer shell.

To find: The Lewis structure for the given set of isoelectronic species.

Answer to Problem 6.83QP

Explanation of Solution

Given isoelectronic species is below.

$\text{CO}$

Lewis structure of the above isoelectronic species is drawn below.

The total number of valence electrons is found to be 10, where carbon and oxygen has 4 and 6 valence electrons respectively.

The 8 electrons getting after reducing two electrons for each bond from the total valence electron are distributed on nitrogen atom to complete the octet.

Since the octets of carbon atoms are not filled, a triple bond was made between carbon and oxygen atoms in expense of two electrons where the remaining four electrons are distributed equally over two atoms present in the given species.

Interpretation Introduction

Interpretation: The Lewis structures and the formal charges for the given isoelectronic species should be shown.

Concept Introduction:

• Lewis structures are diagrams that represent the chemical bonding of covalently bonded molecules and coordination compounds.
• It is also known as Lewis dot structures which represents the bonding between atoms of a molecule and the lone pairs of electrons that may exist in the molecule.
• The Lewis structure is based on the concept of the octet rule so that the electrons shared in each atom should have 8 electrons in its outer shell.

To find: The Lewis structure for the given set of isoelectronic species.

Answer to Problem 6.83QP

Explanation of Solution

Given isoelectronic species is below.

${\text{NO}}^{+}$

Lewis structure of the above isoelectronic species is drawn below.

The total number of valence electrons is found to be 11, where nitrogen and oxygen contains 5 and 6 valence electrons respectively. The whole charge of the molecule is +1 that results in the total number of valence electrons as 10.

The 8 electrons getting after reducing two electrons for each bond from the total valence electron are distributed on nitrogen atom to complete the octet.

Since the octets of nitrogen atoms are not filled, a triple bond is made between nitrogen and oxygen atoms in expense of two electrons where the remaining four electrons are distributed over the 2 atoms present in the given molecule.

Interpretation Introduction

Interpretation: The Lewis structures and the formal charges for the given isoelectronic species should be shown.

Concept Introduction:

• Lewis structures are diagrams that represent the chemical bonding of covalently bonded molecules and coordination compounds.
• It is also known as Lewis dot structures which represents the bonding between atoms of a molecule and the lone pairs of electrons that may exist in the molecule.
• The Lewis structure is based on the concept of the octet rule so that the electrons shared in each atom should have 8 electrons in its outer shell.

To find: The Lewis structure for the given set of isoelectronic species.

Answer to Problem 6.83QP

Explanation of Solution

Given isoelectronic molecule is below.

${\text{CN}}^{-}$

Lewis structure of the above isoelectronic species is drawn below.

The total number of valence electrons is found to be 9, where nitrogen and carbon contributes 5 and 4 electrons respectively. The whole charge of the molecule is -1 making the total number of valence electrons 10.

The 8 electrons getting after reducing two electrons for each bond from the total valence electron are distributed on nitrogen atom to complete the octet.

Since the octets of carbon atoms are not filled, a triple bond is made between carbon and nitrogen atoms in expense of two electrons where the remaining four electrons are distributed over the atoms present in the given molecule.

Interpretation Introduction

Interpretation: The Lewis structures and the formal charges for the given isoelectronic species should be shown.

Concept Introduction:

• Lewis structures are diagrams that represent the chemical bonding of covalently bonded molecules and coordination compounds.
• It is also known as Lewis dot structures which represents the bonding between atoms of a molecule and the lone pairs of electrons that may exist in the molecule.
• The Lewis structure is based on the concept of the octet rule so that the electrons shared in each atom should have 8 electrons in its outer shell.

To find: The Lewis structure for the given set of isoelectronic species.

Answer to Problem 6.83QP

Explanation of Solution

Given isoelectronic species is below.

${\text{N}}_{\text{2}}$

Lewis structure of above isoelectronic species is drawn below.

The total number of valence electrons is found to be 10, where both nitrogen atoms contribute 5 electrons.

The 8 electrons getting after reducing two electrons for each bond from the total valence electron are distributed on nitrogen atom to complete the octet.

Since the octets of nitrogen atoms are not filled, a triple bond is made between both nitrogen atoms.

Interpretation Introduction

Interpretation: The formal charges for the given isoelectronic species should be shown.

Concept Introduction

A formal charge (FC) is the charge assigned to an atom in a molecule, irrespective of relative electronegativity by thinking that electrons in all chemical bonds are shared equally among atoms.

This method is used to identify the most probable Lewis structures if more than one possibility exists for a compound.

The Lewis structure with formal charge on each of the atoms close to zero is taken as the most plausible structure.

Formal charge of an atom can be determined by the given formula.

$\text{Formal}\text{charge}\left(\text{F}\text{C}\right)\text{=}\left(\begin{array}{l}\text{no}\text{.of}\text{valence}\\ \text{electron}\text{in}\text{atom}\end{array}\right)-\frac{\text{1}}{\text{2}}\left(\begin{array}{l}\text{no}\text{.of}\text{bonding}\\ \text{electrons}\end{array}\right)-\left(\begin{array}{l}\text{no}\text{.of}\text{non-bonding}\\ \text{electrons}\end{array}\right)$

Answer to Problem 6.83QP

Explanation of Solution

Formal charge of the given isoelectronic species is given below

The formal charge of the given isoelectronic species is calculated,

• Carbon atom

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=4}\\ \text{Number}\text{of}\text{bonding}\text{electron=}6\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}2\end{array}$

Substituting these values to the equation,

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

• Oxygen atom

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=6}\\ \text{Number}\text{of}\text{bonding}\text{electron=}6\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}2\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}6-\left(\frac{\text{1}}{\text{2}}\text{×6}\right)-2\\ \text{=}\text{+1}\end{array}$

Interpretation Introduction

Interpretation: The formal charges for the given isoelectronic species should be shown.

Concept Introduction

A formal charge (FC) is the charge assigned to an atom in a molecule, irrespective of relative electronegativity by thinking that electrons in all chemical bonds are shared equally among atoms.

This method is used to identify the most probable Lewis structures if more than one possibility exists for a compound.

The Lewis structure with formal charge on each of the atoms close to zero is taken as the most plausible structure.

Formal charge of an atom can be determined by the given formula.

$\text{Formal}\text{charge}\left(\text{F}\text{C}\right)\text{=}\left(\begin{array}{l}\text{no}\text{.of}\text{valence}\\ \text{electron}\text{in}\text{atom}\end{array}\right)-\frac{\text{1}}{\text{2}}\left(\begin{array}{l}\text{no}\text{.of}\text{bonding}\\ \text{electrons}\end{array}\right)-\left(\begin{array}{l}\text{no}\text{.of}\text{non-bonding}\\ \text{electrons}\end{array}\right)$

Answer to Problem 6.83QP

Explanation of Solution

Formal charge of the given isoelectronic species is given below

The formal charge of the given isoelectronic species is calculated,

• Nitrogen atom

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron= 5}\\ \text{Number}\text{of}\text{bonding}\text{electron=}6\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}2\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=5}-\left(\frac{\text{1}}{\text{2}}\text{×6}\right)-2\\ \text{=}0\end{array}$

• Oxygen atom

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=6}\\ \text{Number}\text{of}\text{bonding}\text{electron=}6\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}2\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{=}6-\left(\frac{\text{1}}{\text{2}}\text{×6}\right)-2\\ \text{=}\text{+1}\end{array}$

Interpretation Introduction

Interpretation: The formal charges for the given isoelectronic species should be shown.

Concept Introduction

A formal charge (FC) is the charge assigned to an atom in a molecule, irrespective of relative electronegativity by thinking that electrons in all chemical bonds are shared equally among atoms.

This method is used to identify the most probable Lewis structures if more than one possibility exists for a compound.

The Lewis structure with formal charge on each of the atoms close to zero is taken as the most plausible structure.

Formal charge of an atom can be determined by the given formula.

$\text{Formal}\text{charge}\left(\text{F}\text{C}\right)\text{=}\left(\begin{array}{l}\text{no}\text{.of}\text{valence}\\ \text{electron}\text{in}\text{atom}\end{array}\right)-\frac{\text{1}}{\text{2}}\left(\begin{array}{l}\text{no}\text{.of}\text{bonding}\\ \text{electrons}\end{array}\right)-\left(\begin{array}{l}\text{no}\text{.of}\text{non-bonding}\\ \text{electrons}\end{array}\right)$

Answer to Problem 6.83QP

Explanation of Solution

Formal charge of the given isoelectronic species is given below

The formal charge of the given isoelectronic species is calculated,

• Carbon atom

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron=4}\\ \text{Number}\text{of}\text{bonding}\text{electron=}6\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}2\end{array}$

Substituting these values to the equation,

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

• Nitrogen atom

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron= 5}\\ \text{Number}\text{of}\text{bonding}\text{electron=}6\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}2\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{= 5}-\left(\frac{\text{1}}{\text{2}}\text{×6}\right)-2\\ \text{=}0\end{array}$

Interpretation Introduction

Interpretation: The formal charges for the given isoelectronic species should be shown.

Concept Introduction

A formal charge (FC) is the charge assigned to an atom in a molecule, irrespective of relative electronegativity by thinking that electrons in all chemical bonds are shared equally among atoms.

This method is used to identify the most probable Lewis structures if more than one possibility exists for a compound.

The Lewis structure with formal charge on each of the atoms close to zero is taken as the most plausible structure.

Formal charge of an atom can be determined by the given formula.

$\text{Formal}\text{charge}\left(\text{F}\text{C}\right)\text{=}\left(\begin{array}{l}\text{no}\text{.of}\text{valence}\\ \text{electron}\text{in}\text{atom}\end{array}\right)-\frac{\text{1}}{\text{2}}\left(\begin{array}{l}\text{no}\text{.of}\text{bonding}\\ \text{electrons}\end{array}\right)-\left(\begin{array}{l}\text{no}\text{.of}\text{non-bonding}\\ \text{electrons}\end{array}\right)$

Answer to Problem 6.83QP

Explanation of Solution

Formal charge of the given isoelectronic species is given below

The formal charge of the given isoelectronic species is calculated,

• Nitrogen atom

$\begin{array}{c}\text{Number}\text{of}\text{valence}\text{electron= 5}\\ \text{Number}\text{of}\text{bonding}\text{electron=}6\\ \text{Number}\text{of}\text{non-bonding}\text{electron}\text{=}2\end{array}$

Substituting these values to the equation,

$\begin{array}{c}\text{FC}\text{= 5}-\left(\frac{\text{1}}{\text{2}}\text{×6}\right)-2\\ \text{=}0\end{array}$

• Since both the nitrogen atoms are similar, the formal charge of the nitrogen atoms is zero.