Hybridisation of 'P' in `PO_4^(3-)` is same as that of : -

Options

A. I in `Icl_4^(-)`
B. S in `SO_3`
C. N is `NO_3^(-)`
D. S in `SO_4^(2-)`

Answer

To determine the hybridization of phosphorus (P) in the phosphate ion \( PO_4^{3-} \) and compare it with other species, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure of \( PO_4^{3-} \)**:
- The phosphate ion \( PO_4^{3-} \) consists of one phosphorus atom and four oxygen atoms. The phosphorus atom is the central atom.

2. **Count the Valence Electrons**:
- Phosphorus has 5 valence electrons, and each oxygen has 6 valence electrons. Since there are four oxygen atoms, the total number of valence electrons contributed by oxygen is \( 4 \times 6 = 24 \).
- The overall charge of \( -3 \) means we add 3 more electrons.
- Total valence electrons = \( 5 + 24 + 3 = 32 \).

3. **Draw the Lewis Structure**:
- In the Lewis structure, phosphorus forms single bonds with each of the four oxygen atoms. To satisfy the octet rule for oxygen, we can form double bonds with some of the oxygen atoms.
- The final structure will show phosphorus with four single bonds to oxygen and a formal charge distribution that satisfies the overall charge of \( -3 \).

4. **Determine the Hybridization**:
- The phosphorus atom in \( PO_4^{3-} \) is surrounded by four regions of electron density (the four P-O bonds).
- According to VSEPR theory, four regions of electron density correspond to \( sp^3 \) hybridization.

5. **Compare with Other Options**:
- Now we need to find the hybridization of the other options provided:
- **A. \( ICl_4^{-} \)**: The iodine atom has 4 bonding pairs and 2 lone pairs, leading to \( sp^3d^2 \) hybridization.
- **B. \( NO_3^{-} \)**: The nitrogen atom has 3 bonding pairs and no lone pairs, leading to \( sp^2 \) hybridization.
- **C. \( SO_4^{2-} \)**: The sulfur atom has 4 bonding pairs and no lone pairs, leading to \( sp^3 \) hybridization.

6. **Conclusion**:
- The hybridization of phosphorus in \( PO_4^{3-} \) is \( sp^3 \), which is the same as that of sulfur in \( SO_4^{2-} \).
- Therefore, the correct answer is option D.

### Final Answer:
The hybridization of 'P' in \( PO_4^{3-} \) is the same as that of sulfur in \( SO_4^{2-} \).

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