Question

what is the oxidation state of sulphur in so32-​

Answer 1

Explanation:

The oxidation state of the sulfur is +6 (work it out!); therefore, the ion is more properly named the sulfate(VI) ion. The sulfite ion is SO32-. The oxidation state of the sulfur is +4. This ion is more properly named the sulfate(IV) ion.

Answer 2

The oxidation state of sulphur in $SO_3^{2-}$  is +4

Rules for Oxidation Numbers :

1. The oxidation number of a free element is always zero.
2. The oxidation number of a monatomic ion equals the charge of the ion.
3. The oxidation number of   Hydrogen (H)  is +1, but it is -1 in when combined with less electronegative elements.
4. The oxidation number of  oxygen (O)  in compounds is usually -2, but it is -1 in peroxides.
5. The oxidation number of a Group 1 element in a compound is +1.
6. The oxidation number of a Group 2 element in a compound is +2.
7. The oxidation number of a Group 17 element in a binary compound is -1.
8. The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.
9. The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

The given compound is, $SO_3^{2-}$

Let the oxidation state of S be, 'x'

$x+3(-2)=-2\\\\x=+4$

Hence, the oxidation state of S is, (+4)

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