The shape and hybridisation of `PCI_(3)` molecule

Options

A. Tetahedral and `sp^(3)`
B. Pyramidal and `sp^(3)`
C. Angular and `sp^(3)`
D. Planar trigonal and `sp^(3)`

Answer

To determine the shape and hybridization of the \( PCl_3 \) molecule, 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. Each chlorine (Cl) atom is in group 17 and has 7 valence electrons. Since there are three chlorine atoms, we can calculate the total number of valence electrons in \( PCl_3 \).

- Valence electrons from P: 5
- Valence electrons from 3 Cl: \( 3 \times 7 = 21 \)

Total valence electrons = \( 5 + 21 = 26 \)

### Step 2: Determine the Hybridization
To find the hybridization, we need to consider the number of bonding pairs and lone pairs around the phosphorus atom.

- Phosphorus forms three single bonds with three chlorine atoms.
- This leaves one lone pair of electrons on the phosphorus atom.

Using the formula for hybridization:
\[ \text{Hybridization} = \frac{(N + M)}{2} \]
where \( N \) is the number of atoms bonded to the central atom (3 Cl) and \( M \) is the number of lone pairs (1).

So,
\[ \text{Hybridization} = \frac{(3 + 1)}{2} = \frac{4}{2} = 2 \]

This indicates that the hybridization is \( sp^3 \).

### Step 3: Determine the Shape
For \( sp^3 \) hybridization, the molecular geometry can be predicted. In the case of \( PCl_3 \), we have three bonding pairs and one lone pair. The presence of the lone pair affects the shape.

- The arrangement of the four electron pairs (3 bonding pairs + 1 lone pair) is tetrahedral.
- However, because of the lone pair, the actual shape of the molecule is trigonal pyramidal.

### Conclusion
Thus, the hybridization of \( PCl_3 \) is \( sp^3 \), and the shape of the molecule is trigonal pyramidal.

### Final Answer
- Hybridization: \( sp^3 \)
- Shape: Trigonal pyramidal

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