The shape of `SO_4^(2-)` ion is

Options

A. square planar
B. square pyramid
C. tetrahedral
D. none of these

Explanation

To determine the shape of the sulfate ion \( \text{SO}_4^{2-} \), we can follow these steps:

### Step 1: Determine the Valence Electrons
First, we need to find the total number of valence electrons in the sulfate ion. Sulfur (S) has 6 valence electrons, and each oxygen (O) has 6 valence electrons. Since there are 4 oxygen atoms, we have:

\[ \text{Total valence electrons} = 6 + (4 \times 6) + 2 = 6 + 24 + 2 = 32 \]

### Step 2: Calculate the Hybridization
Next, we will use the formula for hybridization:

\[ \text{Hybridization} = \frac{\text{Valence electrons of central atom} + \text{Number of monovalent atoms} + \text{Charge}}{2} \]

In this case:
- Valence electrons of sulfur = 6
- Number of monovalent atoms = 0 (since all are divalent)
- Charge = +2 (because the ion has a -2 charge)

Thus, we have:

\[ \text{Hybridization} = \frac{6 + 0 + 2}{2} = \frac{8}{2} = 4 \]

### Step 3: Identify the Hybridization Type
A hybridization value of 4 corresponds to \( \text{sp}^3 \) hybridization.

### Step 4: Determine the Shape
The \( \text{sp}^3 \) hybridization leads to a tetrahedral geometry. In a tetrahedral shape, the bond angles are approximately \( 109.5^\circ \).

### Step 5: Draw the Structure
In the sulfate ion, sulfur is the central atom bonded to four oxygen atoms. Two of these bonds are double bonds (due to resonance), but the overall geometry remains tetrahedral.

### Conclusion
The shape of the sulfate ion \( \text{SO}_4^{2-} \) is tetrahedral.

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