Question

What are the formal charges on each atom in sulphite, $S{O}_3^{2-}$ and chlorite, $Cl{O}_2^{-}$ ions?

Answer 1

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:
 $F=V-N-{\displaystyle \frac{B}{2}}-(i)$
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:
  $F=6-2-{\displaystyle \frac{8}{2}}$
 $\Rightarrow F=0$
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:
  $F=6-4-{\displaystyle \frac{4}{2}}$
 $\Rightarrow F=0$
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:
  $F=6-6-{\displaystyle \frac{2}{2}}$
 $\Rightarrow F=-1$
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:
  $F=7-4-{\displaystyle \frac{6}{2}}$
 $\Rightarrow F=0$
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:
  $F=6-4-{\displaystyle \frac{4}{2}}$
 $\Rightarrow F=0$
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:
  $F=6-6-{\displaystyle \frac{2}{2}}$
 $\Rightarrow F=-1$
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.