The correct order of hybridisation of the central atom in the following species ` NH_3 ,[PtCl_4 ]^(2-) , PCl_5` and ` BCl_3` is :

Options

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

Explanation

To determine the correct order of hybridization of the central atom in the species \( NH_3 \), \( [PtCl_4]^{2-} \), \( PCl_5 \), and \( BCl_3 \), we will analyze each compound step by step.

### Step 1: Hybridization of \( NH_3 \)
1. **Count the valence electrons**: Nitrogen (N) has 5 valence electrons and each hydrogen (H) has 1. Therefore, for \( NH_3 \):
\[ \text{Total valence electrons} = 5 + 3 \times 1 = 8 \]
2. **Determine the number of bond pairs and lone pairs**:
- \( NH_3 \) has 3 bond pairs (N-H bonds) and 1 lone pair.
3. **Determine hybridization**:
- The hybridization can be determined using the formula \( \text{Hybridization} = \text{Number of bond pairs} + \text{Number of lone pairs} \).
- Here, \( 3 + 1 = 4 \), which corresponds to \( sp^3 \) hybridization.

### Step 2: Hybridization of \( [PtCl_4]^{2-} \)
1. **Determine the oxidation state of platinum (Pt)**:
- Let the oxidation state of Pt be \( x \). The equation is:
\[ x + 4(-1) = -2 \implies x = +2 \]
2. **Determine the hybridization**:
- \( Pt^{2+} \) typically forms square planar complexes. For square planar geometry, the hybridization is \( dsp^2 \).

### Step 3: Hybridization of \( PCl_5 \)
1. **Count the valence electrons**: Phosphorus (P) has 5 valence electrons and each chlorine (Cl) has 7. Therefore, for \( PCl_5 \):
\[ \text{Total valence electrons} = 5 + 5 \times 7 = 40 \]
2. **Determine the number of bond pairs and lone pairs**:
- Using the formula \( V/8 \) (where \( V \) is total valence electrons), we get:
\[ \frac{40}{8} = 5 \quad (\text{number of bond pairs}) \quad \text{and remainder} = 0 \quad (\text{lone pairs} = 0) \]
3. **Determine hybridization**:
- Since there are 5 bond pairs and 0 lone pairs, the hybridization is \( sp^3d \).

### Step 4: Hybridization of \( BCl_3 \)
1. **Count the valence electrons**: Boron (B) has 3 valence electrons and each chlorine (Cl) has 7. Therefore, for \( BCl_3 \):
\[ \text{Total valence electrons} = 3 + 3 \times 7 = 24 \]
2. **Determine the number of bond pairs and lone pairs**:
- Using the formula \( V/8 \):
\[ \frac{24}{8} = 3 \quad (\text{number of bond pairs}) \quad \text{and remainder} = 0 \quad (\text{lone pairs} = 0) \]
3. **Determine hybridization**:
- Since there are 3 bond pairs and 0 lone pairs, the hybridization is \( sp^2 \).

### Summary of Hybridizations
- \( NH_3 \): \( sp^3 \)
- \( [PtCl_4]^{2-} \): \( dsp^2 \)
- \( PCl_5 \): \( sp^3d \)
- \( BCl_3 \): \( sp^2 \)

### Final Order of Hybridization
The correct order of hybridization from the species is:
1. \( NH_3 \) - \( sp^3 \)
2. \( BCl_3 \) - \( sp^2 \)
3. \( [PtCl_4]^{2-} \) - \( dsp^2 \)
4. \( PCl_5 \) - \( sp^3d \)

### Answer
The correct order of hybridization is:
\[ sp^3 < sp^2 < dsp^2 < sp^3d \]