So, the Oxidation state of S in SO₃²⁻ is +4.
Sulfite ion is the ion depicted.
The oxygen atoms in the ion are currently each in the oxidation state of 2.
The charge of the same entity to which they belong is determined by the sum of the oxidation states of all the atoms in a molecule, ion, or other entity.
Let's say that the sulfur is in the "x" oxidation state.
So,
x + [3×(−2)] = −2.
∴ x + (−6) = −2.
∴x = +4.
S thus has an oxidation state of +4 in SO₃²⁻.
<h3>
How do you determine an ion's oxidation number?</h3>
For simple ions, the charge of the ion is equal to its oxidation number. For instance, the sodium ion, Na+, has an oxidation number of 1, while the chlorine ion, Cl-, has an oxidation number of -1. In combinations with nonmetals like CH4, NH3, H2O, and HCl, hydrogen has an oxidation number of one (+1).
<h3>
What is an oxidation number?</h3>
An oxidation number, which reflects the general distribution of electrons among the bonded atoms in a molecular compound or molecular ion, is assigned to each atom in a molecular compound or molecular ion.
<h3>
What is the general rule for assigning oxidation numbers? </h3>
The connected atom's oxidation number is a representation of the charge that would exist if its electrons were transferred from the connection to the atom of a more electronegative element.
Learn more about oxidation number: brainly.com/question/10079361
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