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# What are the formal charges on each atom in sulphite, $SO_3^{2 - }$ and chlorite, $ClO_2^ -$ ions?

Last updated date: 13th Jul 2024
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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.

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.