The concept of hybridisation has been introduced to explain the shapes of molecules. It involves the intermixing of two or more atomic orbitals belonging to same atom but in or more atomic orbitals beloging to sasme atom but in different sub-shells so as to intermix and redistibute energies to from equivalent orbitals called hybrid orbitals.Depending upon toh enumber and nature of the orbitals involved, the hybridisation may be divided into sp (linear), `sp^(2)` (trigonal), `sp^(3)` (tetrahedral), `sp^(3)d` (trigonal bipyramidal), `sp^(3)d^(3)` (octahedral) and `sp^(3)d^(3)` (pentagonal bipyramidal) types. it may be noted that the orbitals of isolated atoms never hybridise and they do so at the time of bond formation.
The hybridisation of phosphorus in `POCl_(3)` is the same as:

Options

A. P in `PCl_(3)`
B. S in `SF_(4)`
C. Cl in `ClF_(3)`
D. B in `BCl_(3).`

Explanation

To determine the hybridization of phosphorus in \( POCl_3 \) and find a molecule with the same hybridization, we can follow these steps:

### Step 1: Identify the central atom and its valence electrons
- The central atom in \( POCl_3 \) is phosphorus (P).
- Phosphorus is in group 15 of the periodic table and has 5 valence electrons.

### Step 2: Count the number of monovalent atoms attached to the central atom
- In \( POCl_3 \), there are three chlorine (Cl) atoms, which are monovalent (each contributes one bond).

### Step 3: Use the hybridization formula
The formula for hybridization is:
\[ H = \frac{1}{2} \left( \text{Number of valence electrons on central atom} + \text{Number of monovalent atoms} - \text{Charge} \right) \]
For \( POCl_3 \):
- Valence electrons from phosphorus = 5
- Number of monovalent atoms (Cl) = 3
- Charge = 0 (neutral molecule)

Substituting these values into the formula:
\[ H = \frac{1}{2} \left( 5 + 3 + 0 \right) = \frac{1}{2} \times 8 = 4 \]

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

### Step 5: Find a molecule with the same hybridization
Now we need to check the hybridization of other molecules to find one that has \( sp^3 \) hybridization.

1. **For \( PCl_3 \)**:
- Valence electrons from phosphorus = 5
- Number of monovalent atoms (Cl) = 3
- Charge = 0
\[ H = \frac{1}{2} \left( 5 + 3 + 0 \right) = \frac{1}{2} \times 8 = 4 \quad \text{(sp}^3\text{ hybridization)} \]

2. **For \( SF_4 \)**:
- Valence electrons from sulfur = 6
- Number of monovalent atoms (F) = 4
- Charge = 0
\[ H = \frac{1}{2} \left( 6 + 4 + 0 \right) = \frac{1}{2} \times 10 = 5 \quad \text{(sp}^3d\text{ hybridization)} \]

3. **For \( ClF_3 \)**:
- Valence electrons from chlorine = 7
- Number of monovalent atoms (F) = 3
- Charge = 0
\[ H = \frac{1}{2} \left( 7 + 3 + 0 \right) = \frac{1}{2} \times 10 = 5 \quad \text{(sp}^3d\text{ hybridization)} \]

4. **For \( BCl_3 \)**:
- Valence electrons from boron = 3
- Number of monovalent atoms (Cl) = 3
- Charge = 0
\[ H = \frac{1}{2} \left( 3 + 3 + 0 \right) = \frac{1}{2} \times 6 = 3 \quad \text{(sp}^2\text{ hybridization)} \]

### Conclusion
The only molecule that has the same hybridization \( sp^3 \) as \( POCl_3 \) is \( PCl_3 \).

### Final Answer
The hybridization of phosphorus in \( POCl_3 \) is the same as in \( PCl_3 \).

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