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**Hint:**Formal charge is a theoretical value assigned to an atom in a molecule which reflects the equal sharing of electrons in a chemical bond, neglecting the electronegativity difference between the atoms. Sketch the Lewis diagram for the given ions to find the value of formal charge of each atom.

**Complete answer:**

The formal charge can be assigned to an atom with the help of following formula:

$ F = V - N - \dfrac{B}{2}\,\,\,\,\,\,\, - (i) $

Where, $ F $ is the formal charge on the atom, $ V $ is the number of electrons of atom in its ground state, $ N $ is the number of lone pair of electrons present on the atom and $ B $ is the number of bonding electrons in that atom.

The Lewis structure for sulphite ions is as follows:

Formal charge for sulphur atom:

Number of electrons of sulphur in its ground state $ = 6 $

Count of nonbonded electrons present on sulphur atom $ = 2 $

Number of bonding electrons $ = 8 $

Substituting values in equation $ (i) $ , formal charge of sulphur atom will be as follows:

$ F = 6 - 2 - \dfrac{8}{2} $

$ \Rightarrow F = 0 $

Formal charge for doubly bonded oxygen atom:

Number of electrons of oxygen in its ground state $ = 6 $

Count of nonbonded electrons or number of lone pair of electrons present on oxygen atom $ = 4 $

Number of bonding electrons $ = 4 $

Substituting values in equation $ (i) $ , formal charge of sulphur atom will be as follows:

$ F = 6 - 4 - \dfrac{4}{2} $

$ \Rightarrow F = 0 $

Formal charge for single bonded oxygen atoms:

Number of electrons of oxygen in its ground state $ = 6 $

Count of nonbonded electrons or number of lone pair of electrons present on oxygen atom $ = 6 $

Number of bonding electrons $ = 2 $

Substituting values in equation $ (i) $ , formal charge of sulphur atom will be as follows:

$ F = 6 - 6 - \dfrac{2}{2} $

$ \Rightarrow F = - 1 $

Hence, in sulphite ions, the formal charge of sulphur atom, doubly bonded oxygen atom and singly bonded oxygen atoms is $ 0,\,0 $ and $ - 1 $ respectively.

The Lewis structure for chlorite ions is as follows:

Formal charge for chlorine atom:

Number of electrons of chlorine in its ground state $ = 7 $

Count of nonbonded electrons present on chlorine atom $ = 4 $

Number of bonding electrons $ = 6 $

Substituting values in equation $ (i) $ , formal charge of sulphur atom will be as follows:

$ F = 7 - 4 - \dfrac{6}{2} $

$ \Rightarrow F = 0 $

Formal charge for doubly bonded oxygen atom:

Number of electrons of oxygen in its ground state $ = 6 $

Count of nonbonded electrons or number of lone pair of electrons present on oxygen atom $ = 4 $

Number of bonding electrons $ = 4 $

Substituting values in equation $ (i) $ , formal charge of sulphur atom will be as follows:

$ F = 6 - 4 - \dfrac{4}{2} $

$ \Rightarrow F = 0 $

Formal charge for single bonded oxygen atom:

Number of electrons of oxygen in its ground state $ = 6 $

Count of nonbonded electrons or number of lone pair of electrons present on oxygen atom $ = 6 $

Number of bonding electrons $ = 2 $

Substituting values in equation $ (i) $ , formal charge of sulphur atom will be as follows:

$ F = 6 - 6 - \dfrac{2}{2} $

$ \Rightarrow F = - 1 $

Hence, in chlorite ions, the formal charge of chlorine atom, doubly bonded oxygen atom and singly bonded oxygen atom is $ 0,\,0 $ and $ - 1 $ respectively.

**Note:**

Lewis structures in which the formal charges are zero for most atoms in the compound, are more preferably considered than the one with non-zero formal charges. Moreover, the negative formal charge should be present on the most electronegative element in the compound.

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