In sulphite ion, the oxidation state of sulphur is `+4` and `S` is `sp^3` hybridised.

Answer

To determine whether the statement about the sulfite ion (SO₃²⁻) is true or false, we will analyze the oxidation state of sulfur and its hybridization step by step.

### Step 1: Determine the oxidation state of sulfur in sulfite ion (SO₃²⁻)
1. The sulfite ion has the formula SO₃²⁻.
2. Oxygen typically has an oxidation state of -2.
3. In the sulfite ion, there are three oxygen atoms, contributing a total of -6 (3 × -2).
4. Let the oxidation state of sulfur be x.
5. The overall charge of the sulfite ion is -2. Therefore, we can set up the equation:
\[ x + (-6) = -2 \]
6. Solving for x gives:
\[ x - 6 = -2 \implies x = +4 \]
7. Thus, the oxidation state of sulfur in the sulfite ion is +4.

### Step 2: Determine the hybridization of sulfur in the sulfite ion
1. The structure of the sulfite ion can be represented as follows:
- Sulfur is bonded to three oxygen atoms: two with single bonds and one with a double bond.
2. To visualize the bonding, we can draw the Lewis structure:
- Sulfur has 6 valence electrons.
- It forms 3 sigma bonds (one with each oxygen) and has one lone pair of electrons.
3. The steric number (the number of bonded atoms plus lone pairs) for sulfur in the sulfite ion is:
\[ \text{Steric number} = 3 \text{ (bonds)} + 1 \text{ (lone pair)} = 4 \]
4. A steric number of 4 corresponds to an sp³ hybridization.
5. Therefore, sulfur in the sulfite ion is indeed sp³ hybridized.

### Conclusion
Both parts of the statement are true:
- The oxidation state of sulfur in the sulfite ion is +4.
- Sulfur is sp³ hybridized.

Thus, the statement is **true**.

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