Shape of `H_3PO_4` is

Options

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

Answer

To determine the shape of H₃PO₄ (orthophosphoric acid), we can follow these steps:

### Step 1: Identify the central atom
The central atom in H₃PO₄ is phosphorus (P).

### Step 2: Determine the number of valence electrons
Phosphorus has 5 valence electrons, and each hydrogen (H) contributes 1 electron, while each oxygen (O) contributes 6 electrons. Therefore, the total number of valence electrons in H₃PO₄ is:
- P: 5 electrons
- H: 3 x 1 = 3 electrons
- O: 4 x 6 = 24 electrons
Total = 5 + 3 + 24 = 32 electrons.

### Step 3: Draw the Lewis structure
In the Lewis structure of H₃PO₄:
- Phosphorus is the central atom.
- It forms one double bond with one oxygen atom (P=O).
- It forms single bonds with three hydroxyl groups (P-OH).

### Step 4: Count the number of bonds and lone pairs
- There are 4 sigma bonds (1 P=O bond and 3 P-OH bonds).
- There are no lone pairs on the phosphorus atom.

### Step 5: Determine hybridization
The hybridization can be determined using the formula:
\[ \text{Hybridization index} = \text{Number of sigma bonds} + \text{Number of lone pairs} \]
Here, the number of sigma bonds is 4 and the number of lone pairs is 0:
\[ \text{Hybridization index} = 4 + 0 = 4 \]
This indicates that the hybridization of phosphorus in H₃PO₄ is sp³.

### Step 6: Determine the shape
For an sp³ hybridized atom with no lone pairs, the molecular geometry is tetrahedral.

### Conclusion
The shape of H₃PO₄ (orthophosphoric acid) is tetrahedral.

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