# How to Find Formal Charge | Lewis Structures

How to find formal charge? Before we find the formal charge of different atoms in molecules, we understand the term formal charge. The formal charge on an atom in a molecule or ion is defined as the charge on any atom in a molecule assuming that all electrons in bonds are equally shared.

## Formula to Calculate the Formal Charge

The formal charge on an atom in a molecule or ion is equal to the total number of valence electrons in the free atom minus the total number of electrons of lone pairs (non-bonding electrons) minus half of the total number of shared electrons bonding electrons.

F.C = Valance electrons in a free atom – lone pair electrons – bond pair electrons/2

Now we find the formal charge on an atom of different molecules.

## Formal Charge on all atoms in H2

H2 is a homodiatomic molecule consisting of two hydrogen atoms. Hydrogen is a half-filled valance orbital having configuration 1s1. It’s a universal rule that any atom wants to complete its duplet or octet and gain stability. Similarly, hydrogen wants to complete its duplet (2 electrons in valance shell) and become a stable molecule.

In a molecule of hydrogen, both hydrogen atoms combine together via a covalent bond. Now we need to find out if there is a formal charge on hydrogen h2. A single hydrogen atom has only one valance electron, no lone pair, and two bond pair electrons.

F.C(H) = Valance electrons in a free atom – lone pair electrons – bond pair electrons/2

F.C(H) = 1 – 0 – 2/2

F.C(H) = 1-1

F.C(H) = 0

Hydrogen has no formal charge.

### Lewis Dot Structure of H2

Do you know, How to draw the lewis structure of molecules? Click here To draw lewis’s structure of molecules, three terms are calculated, the Q (total elections), bond pair electrons, and lone pair electrons. Q (total electrons) which is equal to the valence electron of all-atom plus the number of negative charges minus the number of positive charges.

Q = Valance electron of all-atom + no of -ve charge – no of +ve charge

Then we find bond pair electrons which are equal to the number of bonds multiply by two.

B.P e = 2 × no of bonds

Then we find lone pair electrons by subtracting the bond pair electrons from total electrons Q.

L.P e = Q – B.P e

Lone pair electrons are used for both lone pairs as well as a double bond or triple bonds.

H2 lewis structure, first, we calculate Q.

Q = Valance electron of all atom + no of -ve charge – no of +ve charge

Q = (2 + 0 – 0)

Q = 2

B.P e = 2 × no of bonds

B.P e = 2 × 1

B.P e = 2 e

L.P e = Q – B.P e

L.P e = 2 – 2

L.P = 0

## Formal Charge on all atoms in NH3

Ammonia consists of three atoms of hydrogen and one atom of nitrogen in its molecule. Nitrogen has 5 valance electrons having configuration 1s2, 2s2, 2p3. Hydrogen has 1 valance electron having the configuration 1s1. Nitrogen is the central atom in an ammonia molecule and combines with three hydrogen atoms via a covalent bond. In an ammonia molecule, nitrogen has one lone pair and 6 bond pair electrons.

F.C(N) = Valance electrons in a free atom – lone pair electrons – bond pair electrons/2

F.C(N) = 5 – 2 – 6/2

F.C(N) = 3 – 3

F.C(N) = 0

Nitrogen has no formal charge.

A hydrogen atom in an ammonia molecule has no lone pair and 2 bond pair electrons.

F.C(H) = Valance electrons in a free atom – lone pair electrons – bond pair electrons/2

F.C(H) = 1 – 0 – 2/2

F.C(H) = 1-1

F.C(H) = 0

Hydrogen has no formal charge.

### Lewis Structure of NH3

NH3 lewis structure, first, we calculate Q.

Q = Valance electron of all atom + no of -ve charge – no of +ve charge

Q = (8 + 0 – 0)

Q = 8

B.P e = 2 × no of bonds

B.P e = 2 × 3

B.P e = 6 e

L.P e = Q – B.P e

L.P e = 8 – 6

L.P e = 2

## Formal Charge on all atoms in NH4+

When ammonia forms a coordinate covalent bond with H+, NH4+ is formed. As we discuss above, nitrogen has a lone pair in the ammonia molecule. It has a tendency to share its lone pair with the electron-deficient species and form a coordinate covalent bond. Normally, the valency of nitrogen is 3. Nitrogen can show a maximum of 4 co-valency (Nitrogen can never be pentavalent).

When a hydrogen atom loses its valence electron, H+ is formed. H+ has only one proton but no electron. By the sharing of lone pair electrons with H+, ammonia and hydrogen forms a coordinate covalent bond and complete their octet and duplet respectively, and get stability in the form of NH4+.

The nitrogen atom in the NH4+ molecule has 5 valance electrons, no lone pair, and 8 bond pair electrons.

F.C(N) = Valance electrons in a free atom – lone pair electrons – bond pair electrons/2

F.C(N) = 5 – 0 – 8/2

F.C(N) = 5-0-4

F.C(N) = 5-4

F.C(N) = +1

Nitrogen has a +1 formal charge.

F.C(H) = Valance electrons in a free atom – lone pair electrons – bond pair electrons/2

F.C(H) = 1 – 0 – 2/2

F.C(H) = 1-1

F.C(H) = 0

All Four hydrogen is similar to each other means only one valance electron, no lone pair, and 2 bond pair electrons. So, the hydrogen atom in NH4+ has no formal charge.

### Lewis Dot Structure of NH4+

NH4+ lewis structure, first, we calculate Q.

Q = Valance electron of all atom + no of -ve charge – no of +ve charge

Q = (9 + 0 – 1)

Q = 8

B.P e = 2 × no of bonds

B.P e = 2 × 4

B.P e = 8 e

L.P e = Q – B.P e

L.P e = 8 – 8

L.P e = 0

## Formal Charge on all atoms in NO3 -1

As we know, nitrogen is electron-rich species and has a tendency to share its lone pair of electrons. Nitrogen donates a lone pair to an oxygen atom and gets a positive charge. On the other hand, oxygen accepts the pair of electrons from nitrogen and will get a negative charge. Both the positive and negative charges cancel out with each other so, as a result, NO3 -1 is formed. For better understanding, first, we draw the lewis structure of NO3 -1.

• Nitrogen has 5 valance electrons.
• Oxygen has 6 valance electrons.

Q = Valance electron of all atom + no of -ve charge – no of +ve charge

Q = (23 + 1 – 0)

Q = 24

B.P e = 2 × no of bonds

B.P e = 2 × 3

B.P e = 6 e

L.P e = Q – B.P e

L.P e = 24 – 6

L.P e– = 18

By using lone pairs, we start to complete the octet of corner atoms.

As we see in the above structure, the octet of nitrogen is incomplete (6 electrons) so, 2 electrons of one oxygen shares with nitrogen and form a double bond.

One thing kept in mind is that the positive charge is always on the central atom and the negative charge is always on the corner atom. But there is three oxygen on the corner so, how can we decide which oxygen atom has a negative charge. To solve this problem we study how to find the formal charges.

### Formal Charge

F.C = Valance electrons – lone pair electrons – bond pair electrons/2

F.C(O1) = 6 – 4 – 4/2 = 6 – 4 – 2 = 0

F.C(O2) = 6 – 6 – 2/2 = 6 – 6 – 1 = -1

F.C(O3) = 6 – 6 – 2/2 = 6 – 6 – 1 = -1

F.C(N) = 5 – 0 – 8/2 = 5 – 0 – 4 = +1

So, the exact NO3-1 lewis dot structure is:

## Formal Charge on all atoms in NO2-1

Nitrite is the nitrogen oxoanion formed when a proton is lost from the nitrous acid.

NO2-1 lewis structure, first, we calculate Q.

• Nitrogen has 5 valance electrons.
• Oxygen has 6 valance electrons.

Q = Valance electron of all atom + no of -ve charge – no of +ve charge

Q = (17 + 1 – 0)

Q = 18

B.P e = 2 × no of bonds

B.P e = 2 × 2

B.P e = 4 e

L.P e = Q – B.P e

L.P e = 18 – 4

L.P e = 14

The octet of corner atoms is complete by using the lone pair electrons. The corner atom has 8 electrons (octet complete) but the central atom nitrogen has an incomplete octet. So, a lone pair of one oxygen is shared with nitrogen and forms a double bond.

### Formal Charge

F.C = Valance electrons – lone pair electrons – bond pair electrons/2

F.C(O1) = 6 – 4 – 4/2 = 6 – 4 – 2 = 0

F.C(O2) = 6 – 6 – 2/2 = 6 – 6 – 1 = -1

So, the exact NO2-1 lewis dot structure is:

## Formal Charge on all atoms in SO3 2-

H2SO3 is known as sulfurous acid. H2SO3 is the oxy acid of sulfur. It is formed by dissolving SO2 in H2O.

SO2 + H2O → H2SO3

H2SO3 is dibasic acid and ionizes as:

H2SO3 → 2H+ + SO32-

SO3 2- lewis structure, first, we calculate Q.

Q = Valance electron of all-atom + no of -ve charge – no of +ve charge

Q = (24 + 2 – 0)

Q = 26

B.P e = 2 × no of bonds

B.P e = 2 × 3

B.P e = 6 e

L.P e = Q – B.P e

L.P e = 26 – 6

L.P e = 20

By using lone pairs, first, we complete the octet of corner atoms.

Sulfur has one lone pair. All oxygens have 8 electrons in their valance shell. The octet of sulfur is also complete but in the above structure, sulfur shows 5 co-valency instead of 6. So, a lone pair of one oxygen is shared with sulfur and forms a double bond.

Out of three oxygens, which two oxygens have a formal charge? Lets find out

### Formal Charge

F.C = Valance electrons – lone pair electrons – bond pair electrons/2

F.C(O1) = 6 – 4 – 4/2 = 6 – 4 – 2 = 0

F.C(O2) = 6 – 6 – 2/2 = 6 – 6 – 1 = -1

F.C(O3) = 6 – 6 – 2/2 = 6 – 6 – 1 = -1

F.C(S) = 6 – 2 – 8/2 = 6 – 2 – 4 = 0

So, the exact SO3 2- lewis dot structure is:

## Formal Charge on all atoms in SO3

SO3 exists in liquid form at room temperature which strongly fumes in the air. Sulfuric acid is formed when SO3 is combined with water. For gases, it acts as a drying agent. It acts as a solvent for the manufacture of H2SO4.

• Sulfur has 6 valance electrons.
• Oxygen has 6 valance electrons.

SO3 lewis structure, first, we calculate Q.

Q = Valance electron of all-atom + no of -ve charge – no of +ve charge

Q = (24 + 0 – 0)

Q = 24

B.P e = 2 × no of bonds

B.P e = 2 × 3

B.P e = 6 e

L.P e = Q – B.P e

L.P e = 24 – 6

L.P e = 18

The sulfur has an incomplete octet so, the lone pair of one oxygen is shared with sulfur and makes a double bond.

The octet of sulfur is complete but sulfur shows 4 co-valency rather than 6. So, the two lone pairs of two oxygens are shared with sulfur and make double bonds.

### Formal Charge

F.C = Valance electrons – lone pair electrons – bond pair electrons/2

F.C(O1) = 6 – 4 – 4/2 = 6 – 4 – 2 = 0

F.C(O2) = 6 – 4 – 4/2 = 6 – 4 – 2 = 0

F.C(O3) = 6 – 6 – 4/2 = 6 – 6 – 2 = 0

F.C(S) = 6 – 0 – 12/2 = 6 – 0 – 6 = 0

SO3 has no formal charge.

## Can You Solve These Questions: Test Yourself

#### Find the formal charges on all atoms in PO4 3-?

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