Hybridisation of central sulphur in all oxo anions of sulphur is

Options

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

Answer

To determine the hybridization of the central sulfur atom in all oxo anions of sulfur, we can follow these steps:

### Step-by-Step Solution

1. **Identify the Oxo Anions of Sulfur**:
The common oxo anions of sulfur include sulfate (SO₄²⁻), sulfite (SO₃²⁻), thiosulfate (S₂O₃²⁻), and others. Here, we will analyze sulfate as an example.

2. **Determine the Valence Electrons of Sulfur**:
Sulfur is in group 16 of the periodic table and has 6 valence electrons.

3. **Count the Monovalent Atoms**:
In the case of sulfate (SO₄²⁻), there are 4 oxygen atoms. Each oxygen atom is bivalent, meaning it forms two bonds. Therefore, we do not count them as monovalent atoms.

4. **Calculate the Charge Contribution**:
The sulfate ion has a charge of -2. To find the effective number of electrons, we need to add the charge to the total number of valence electrons. Since there are no monovalent atoms, we only consider the charge:
- Total = Valence electrons (6) + Charge contribution (2) = 6 + 2 = 8.

5. **Calculate the Steric Number (SN)**:
The steric number is calculated using the formula:
\[ \text{Steric Number (SN)} = \text{Number of Valence Electrons} + \text{Number of Monovalent Atoms} + \text{Charge} \]
Here, since there are no monovalent atoms:
\[ SN = 6 + 0 + 2 = 8. \]

6. **Determine the Hybridization**:
The hybridization can be determined from the steric number:
- SN = 4 corresponds to sp³ hybridization.
- SN = 5 corresponds to sp³d hybridization.
- SN = 6 corresponds to sp³d² hybridization.
Since we calculated SN = 4 for sulfate, the hybridization of the central sulfur atom is **sp³**.

### Conclusion
Thus, the hybridization of the central sulfur in the oxo anions of sulfur, specifically in sulfate (SO₄²⁻), is **sp³**.

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