Sulphate ion has ____geometry

Options

A. Pyramidal
B. Tetrahedral
C. Square planar
D. See-saw

Answer

To determine the geometry of the sulfate ion (SO₄²⁻), we can follow these steps:

### Step 1: Identify the structure of the sulfate ion
The sulfate ion consists of one sulfur atom (S) and four oxygen atoms (O). The sulfate ion has a charge of -2, which means it has gained two extra electrons.

### Step 2: Determine the number of valence electrons
Sulfur has 6 valence electrons, and each oxygen has 6 valence electrons. Therefore, the total number of valence electrons in the sulfate ion is:
- Sulfur: 6 electrons
- Oxygen: 4 × 6 = 24 electrons
- Total: 6 + 24 - 2 (for the two negative charges) = 28 valence electrons

### Step 3: Draw the Lewis structure
In the Lewis structure, sulfur is the central atom surrounded by four oxygen atoms. The structure can be represented as:
- One sulfur atom is double bonded to two oxygen atoms and single bonded to two other oxygen atoms, which carry a negative charge (O⁻).

### Step 4: Count the sigma bonds and lone pairs
In the sulfate ion:
- There are 4 sigma bonds (one for each S-O bond).
- There are no lone pairs on the sulfur atom.

### Step 5: Calculate the hybridization
The hybridization can be determined using the formula:
\[ \text{Hybridization Index} = \text{Number of Sigma Bonds} + \text{Number of Lone Pairs} \]
Here:
- Number of sigma bonds = 4
- Number of lone pairs = 0

Thus, the hybridization index = 4 + 0 = 4.

### Step 6: Determine the hybridization type
A hybridization index of 4 corresponds to sp³ hybridization.

### Step 7: Identify the geometry
For a molecule with sp³ hybridization and no lone pairs on the central atom, the geometry is tetrahedral.

### Conclusion
The geometry of the sulfate ion (SO₄²⁻) is tetrahedral.

### Final Answer
Sulphate ion has tetrahedral geometry.

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