In sulphate ion the oxidation state of sulphur is +6 and the hybridization state of sulphur is 

Options

A. `sp`
B. `sp^(2)`
C. `sp^(3)`
D. `sp^(2) or sp^(3)d^(2)`

Answer

To determine the hybridization state of sulfur in the sulfate ion (SO₄²⁻), we can follow these steps:

### Step 1: Identify the sulfate ion and its components
The sulfate ion is represented as SO₄²⁻. It consists of one sulfur atom and four oxygen atoms.

**Hint:** Remember that the sulfate ion has a net charge of -2.

### Step 2: Calculate the total number of valence electrons
- Sulfur (S) has 6 valence electrons.
- Each oxygen (O) has 6 valence electrons, and there are four oxygen atoms.
- The ion has a -2 charge, which means we add 2 electrons.

Total valence electrons = (1 × 6) + (4 × 6) + 2 = 6 + 24 + 2 = 32 electrons.

**Hint:** Count the valence electrons carefully, considering the charge of the ion.

### Step 3: Draw the Lewis structure
- Place sulfur as the central atom and surround it with the four oxygen atoms.
- Form bonds between sulfur and each oxygen. Each bond uses 2 electrons, so 4 bonds will use 8 electrons.

Remaining electrons = 32 - 8 = 24 electrons.

**Hint:** Ensure that you are using the correct number of electrons for bonding.

### Step 4: Distribute the remaining electrons
- Each oxygen atom needs 6 electrons to complete its octet (since it already shares 2 with sulfur).
- Distributing the remaining 24 electrons gives 6 electrons (3 lone pairs) to each of the four oxygen atoms.

**Hint:** Remember that each oxygen atom needs to complete its octet.

### Step 5: Analyze the bonding and lone pairs
- In the final structure, sulfur has 4 bond pairs and 0 lone pairs.
- The geometry of the molecule is tetrahedral due to the four bonds.

**Hint:** The number of bond pairs and lone pairs will help determine the hybridization.

### Step 6: Determine the hybridization
- The tetrahedral geometry indicates that the hybridization is sp³.
- In the excited state, sulfur can promote one of its 3s electrons to the 3d orbital, allowing for hybridization.

**Hint:** Tetrahedral shapes typically correspond to sp³ hybridization.

### Conclusion
The hybridization state of sulfur in the sulfate ion (SO₄²⁻) is **sp³**.

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