# Formal Charge

## What is a Formal Charge ^{[1-8]}

A formal charge is a charge assigned to an atom in a molecule, assuming that all electrons in the chemical bonds are shared equally between the atoms. This assumption excludes the electronegativity difference between the atoms. The sum of formal charges of all the atoms is equal to the compound’s actual charge. If the compound is neutral, the net formal charge is zero.

## How to Calculate the Formal Charge ^{[1-7]}

In order to calculate the formal charge of a molecule or an ion, one has to know about **Lewis structure **and **resonance**. A Lewis dot structure is a straightforward representation of valence shell electrons in an atom, ion, or molecule. It shows how electrons are positioned around the atoms of the compound. Dots represent paired electrons, and dash lines show bonded electrons.

Sometimes more than one Lewis structure can be drawn to represent the same compound. The difference among these structures is the position of the electrons and bond formation. The free electrons can move throughout the compound and are said to be delocalized. This phenomenon of shifting electrons among the atoms is known as resonance. These equivalent structures are known as resonance structures. Depending on the compound, the shifting of electrons may cause a change in the atoms’ formal charges. However, the net formal charge in all the resonance structures remains the same.

From the Lewis structure representation of the compound, the formal charge can be calculated by two methods.

### 1. Using Equation

The following formula gives the formal charge of an atom in a molecule.

*q _{f} = V – N – B/2*

where,

*V*: number of valence electrons of the neutral atom

*N*: number of non-bonding valence electrons of the neutral atom (also known as lone pair electrons)

*B*: total number of electrons shared in bonds with other atoms

This formula explicitly gives the relationship between the number of bonding and non-bonding electrons. It states how many electrons are formally “owned” by the atom. The number of valence electrons for an atom is equal to its group number.

#### Examples

**1. Carbon dioxide (CO _{2})**

Carbon (C) has a double bond with each of the oxygen (O) atom. The structure of CO_{2} is O=C=O.

For carbon,

V = 4, N = 0, B = 8

Therefore, formal charge on carbon is given by,

q_{f} = 4 – 0 – 8/2 = 0

For oxygen,

V = 6, N = 4, B = 4

Therefore, formal charge on carbon is given by,

q_{f} = 6 – 4 – 4/2 = 0

The net formal charge on carbon dioxide is,

0 + 0 + 0 = 0

**2. Carbon monoxide (CO)**

Carbon has a double bond with oxygen. The structure of CO is C=O.

For carbon,

V = 4, N = 2, B = 6

Therefore,

q_{f} = 4 – 2 – 6/2 = -1

For oxygen,

V = 6, N = 2, B = 6

Therefore,

q_{f} = 6 – 2 – 6/2 = 1

The net formal charge is

-1 + 1 = 0

**3. Nitrate (NO _{3}^{–})**

Nitrogen has a double bond with one oxygen atom and two single bonds with two other oxygen atoms. The three bonds in NO_{3}^{–} are represented by O=N^{+}, N^{+}-O^{–}, and N^{+}-O^{–}.

For nitrogen,

V = 5, N = 0, B = 8

Therefore,

q_{f} = 5 – 0 – 8/2 = 1

For oxygen 1,

V = 6, N = 4, B = 4

Therefore,

q_{f} = 6 – 4 – 4/2 = 0

For oxygen 2 and oxygen 3,

V = 6, N = 6, B = 2

Therefore,

q_{f} = 6 – 6 – 2/2 = -1

The net formal charge on nitrate ion is,

1 – 0 – 1 – 1 = – 1

**4. Nitrite (NO _{2}^{–})**

Nitrogen (N) has a double bond with each oxygen (O) atom. The structure of NO_{2}^{–} is O=N-O^{–}.

For oxygen 1,

V = 6, N = 4, B = 4

Therefore,

q_{f} = 6 – 4 – 4/2 = 0

For nitrogen,

V = 5, N = 2, B = 6

Therefore,

q_{f} = 5 – 2 – 6/2 = 0

For oxygen 2,

V = 6, N = 6, B = 2

Therefore,

q_{f} = 6 – 6 – 2/2 = -1

The net formal charge of nitrite ion is

0 + 0 – 1 =-1

**5. Nitrous oxide (N _{2}O)**

Nitrogen (N) bonds with oxygen (O) and with itself to form two stable resonance structures.

N ≡ N^{+}–O^{–} ↔ N^{–}=N^{+}=O

**A.** The most stable resonance structure of N_{2}O is N ≡ N^{+}–O^{–}

For nitrogen 1,

V = 5, N = 2, B = 6

Therefore,

q_{f} = 5 – 2 – 6/2 = 0

For nitrogen 2,

V = 5, N = 0, B = 8

Therefore,

q_{f} = 5 – 0 – 8/2 = 1

For oxygen,

V = 6, N = 6, B = 2

Therefore,

q_{f} = 6 – 6 – 2/2 = -1

Total formal charge on nitrous oxide (N_{2}O) is,

0 + 1 – 1 = 0

**B.** Another stable resonance structure of N_{2}O is N^{–}=N^{+}=O.

For nitrogen 1,

V = 5, N = 4, B = 4

Therefore,

q_{f} = 5 – 4 – 4/2 = -1

For nitrogen 2,

V = 5, N = 0, B = 8

Therefore,

q_{f} = 5 – 0 – 8/2 = 1

For oxygen,

V = 6, N = 4, B = 4

Therefore,

q_{f} = 6 – 4 – 4/2 = 0

The net formal charge on nitrous oxide is,

-1 + 1 + 0 = 0

Therefore, for both the resonance structure, the formal charge of nitrous oxide is 0.

**6. Ozone (O _{3})**

Ozone has two resonance structures given by,

O=O^{+}-O^{–} ↔ O^{–}-O^{+}=O

The formal charge on the first structure can be found as follows.

For oxygen 1,

V = 6, N = 4, B = 4

Therefore,

q_{f} = 6 – 4 – 4/2 = 0

For oxygen 2,

V = 6, N = 2, B = 6

Therefore,

q_{f} = 6 – 2 – 6/2 = 1

For oxygen 3,

V = 6, N = 6, B = 2

Therefore,

q_{f} = 6 – 6 – 2/2 = -1

The net formal charge on ozone is,

0 + 1 – 1 = 0

**7. Ammonia (NH _{3})**

Nitrogen (N) atom is bonded to three hydrogen (H) atoms like N–H.

For nitrogen,

V = 5, N = 2, B = 6

Therefore,

q_{f} = 5 – 2 – 6/2 = 0

For hydrogen 1, hydrogen 2, and hydrogen 3,

V = 1, N = 0, B = 2

Therefore,

q_{f} = 1 – 0 – 2/2 = 0

The net force on ammonia is,

0 + 0 + 0 = 0

**8. Sulfur dioxide (SO _{2})**

Sulfur (S) has a double bond with each oxygen (O) atom and SO_{2} undergoes the following resonance.

O=S=O ↔ O=S^{+}–O^{–}

**A.** Resonance structure: O=S=O

For sulfur,

V = 6, N = 2, B = 8

Therefore,

q_{f} = 6 – 2 – 8/2 = 0

For oxygen 1 and oxygen 2,

V = 6, N = 4, B = 4

Therefore,

q_{f} = 6 – 4 – 4/2 = 0

The net formal charge on sulfur dioxide is,

0 + 0 + 0 = 0

**B.** Resonance structure O=S^{+}–O^{–}.

For sulfur,

V = 6, N = 2, B = 6

Therefore,

q_{f} = 6 – 2 – 6/2 = 1

For oxygen 1, q_{f} = 0

For oxygen 2,

V = 6, N = 6, B = 2

Therefore,

q_{f} = 6 – 6 – 2/2 = -1

The net formal charge on sulfur dioxide is,

1 + 0 – 1 = 0

**9. Sulfur trioxide (SO _{3})**

Sulfur (S) has a double bond with each of the oxygen (O) atoms. There will be three S=O bonds.

For sulfur,

V = 6, N = 0, B = 12

Therefore,

q_{f} = 6 – 0 – 12/2 = 0

For oxygen 1, oxygen 2, and oxygen 3,

V = 6, N = 4, B = 4

Therefore,

q_{f} = 6 – 4 – 4/2 = 0

The net formal charge on sulfur trioxide is,

0 + 3 x 0 = 0

**10. Sulfate (SO _{4}^{2-})**

Sulfur (S) has a single bonded with two oxygen (O) atoms and a double bond with two other oxygen atoms. The bonds are S=O and S-O^{–}_{.}

For sulfur,

V = 6, N = 0, B = 12

Therefore,

q_{f} = 6 – 0 – 12/2 = 0

For oxygen 1 and oxygen 2,

V = 6, N = 4, B = 4

Therefore,

q_{f} = 6 – 4 – 4/2 = 0

For oxygen 3 and oxygen 4,

V = 6, N = 6, B = 2

Therefore,

q_{f} = 6 – 6 – 2/2 = -1

The net formal charge on sulfate is,

0 + 2 x 0 + 2 x -1 = -2

**11. Perchlorate (ClO _{4}^{–})**

Chlorine (Cl) has a double bond with three of the oxygen (O) atoms and a single bond with the fourth oxygen atom.

For chlorine,

V = 7, N = 0, B = 14

Therefore,

q_{f} = 7 – 0 – 14/2 = 0

For oxygen 1, oxygen 2, and oxygen 3

V = 6, N = 4, B = 4

Therefore,

q_{f} = 6 – 4 – 4/2 = 0

For oxygen 4,

V = 6, N = 6, B = 2

Therefore,

q_{f} = 6 – 6 – 2/2 = -1

The net formal charge of perchlorate is

0 + 3 x 0 – 1 = -1

**12. Cyanate (NCO ^{-1})**

Carbon is double bonded to nitrogen (N) and oxygen (O). The resonance structures are,

N^{-1}=C=O ↔ C^{-2}=N^{+}=O ↔ C^{-2}=O^{+2}=N^{-1}

**A.** Resonance structure: N^{-1}=C=O

For nitrogen,

V = 5, N = 4, B = 4

Therefore,

q_{f} = 5 – 4 – 4/2 = -1

For carbon,

V = 4, N = 0, B = 8

Therefore,

q_{f} = 4 – 0 – 8/2 = 0

For oxygen,

V = 6, N = 4, B = 4

Therefore,

q_{f} = 6 – 4 – 4/2 = 0

The net formal charge is,

-1 + 0 + 0 = -1

**B.** Resonance structure: C^{-2}=N^{+}=O

For carbon,

V = 4, N = 4, B = 4

Therefore,

q_{f} = 4 – 4 – 4/2 = -2

For nitrogen,

V = 5, N = 0, B = 8

Therefore,

q_{f} = 5 – 0 – 8/2 = 1

For oxygen,

V = 6, N = 4, B = 4

Therefore,

q_{f} = 6 – 4 – 4/2 = 0

The net formal charge is,

-2 + 1 + 0 = -1

**C.** Resonance structure: C^{-2}=O^{+2}=N^{-1}

For carbon,

V = 4, N = 4, B = 4

Therefore,

q_{f} = 4 – 4 – 4/2 = -2

For oxygen,

V = 6, N = 2, B = 8

Therefore,

q_{f} = 6 – 2 – 8/2 = 0

For nitrogen,

V = 5, N = 4, B = 4

Therefore,

q_{f }= 5 – 4 – 4/2 = -1

The net formal charge is,

-2 + 2 – 1 = -1

### 2. Visual Method

Apart from the formula, the formal charge can be calculated pictorially using the diagrammatic method. Here are the steps to calculate the formal charge.

**Step 1**: Draw a circle around the atom for which formal charge is being calculated

**Step 2**: Count the number of electrons in the concerned atom’s circle, including electrons shared in covalent bonds.

**Step 3**: The formal charge is given by,

Formal charge = group number (old IUPAC) of the atom – electrons in the atom’s circle

## Oxidation Number vs. Formal Charge ^{[8]}

The oxidation number is the number of electrons a particular atom can lose, gain, or share with another atom. This term applies to any atom in a molecule. However, it is commonly used for the central metal atoms of coordination complexes. The oxidation number gives the degree of oxidation of an atom in a compound and is a whole number.

The critical difference between formal charge and oxidation number is that the formal charge is the charge of an atom in a molecule that is calculated assuming that electrons in chemical bonds are shared equally between atoms. On the other hand, the oxidation number is the number of electrons an atom loses, gains, or shares with another atom. For example, the formal charge of the nitrogen atom in the ammonia molecule is 0. However, the oxidation number is +3.

**References**

- Chem.libretexts.org
- Opentextbc.ca
- Blog.cambridgecoaching.com
- Guweb2.gonzaga.edu
- Leah4sci.com
- w5wsh.com
- Chem.ucalgary.ca
- Chem.libretexts.org