The hybridisation of phosphorous in`PO_(4)^(3-) ` is

Options

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

Explanation

To determine the hybridization of phosphorus in the phosphate ion \( PO_4^{3-} \), we can follow these steps:

### Step 1: Determine the Valence Electrons
Phosphorus (P) is in group 15 of the periodic table and has 5 valence electrons. Oxygen (O) is in group 16 and has 6 valence electrons.

### Step 2: Count the Total Electrons in the Ion
In \( PO_4^{3-} \):
- Phosphorus contributes 5 electrons.
- Each of the 4 oxygen atoms contributes 6 electrons, giving a total of \( 4 \times 6 = 24 \) electrons.
- The \( 3- \) charge means we add 3 more electrons.

Total electrons = \( 5 + 24 + 3 = 32 \) electrons.

### Step 3: Draw the Lewis Structure
In the phosphate ion, phosphorus is the central atom bonded to four oxygen atoms. Each oxygen atom forms a single bond with phosphorus, and to satisfy the octet rule, one of the oxygen atoms forms a double bond with phosphorus.

### Step 4: Determine the Steric Number
The steric number is calculated by adding the number of sigma bonds and lone pairs on the central atom (phosphorus):
- Phosphorus forms 4 sigma bonds (one with each oxygen).
- There are no lone pairs on phosphorus in this structure.

Steric number = Number of sigma bonds + Number of lone pairs = \( 4 + 0 = 4 \).

### Step 5: Determine the Hybridization
Based on the steric number:
- A steric number of 4 corresponds to \( sp^3 \) hybridization.

### Step 6: Determine the Geometry
With \( sp^3 \) hybridization, the geometry of the phosphate ion is tetrahedral.

### Final Answer
The hybridization of phosphorus in \( PO_4^{3-} \) is \( sp^3 \).

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