The correct order of hybridization 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^(2),dsp^(3),sp^(2)`
C. `dsp^(2),sp^(2),sp^(3),dsp^(3)`
D. `dsp^(2),sp^(3),sp^(2),dsp^(3)`

Explanation

To determine the correct order of hybridization of the central atom in the species NH₃, [PtCl₄]²⁻, PCl₅, and BCl₃, we will analyze each compound step by step.

### Step 1: Analyze NH₃ (Ammonia)
- **Valence Electrons**: Nitrogen (N) has 5 valence electrons, and each Hydrogen (H) has 1 valence electron.
- **Bonding**: Nitrogen forms 3 bonds with 3 Hydrogen atoms, using 3 of its valence electrons. This leaves 2 electrons as a lone pair.
- **Steric Number**: The steric number is calculated as the number of bond pairs plus the number of lone pairs. Here, it is 3 (bond pairs) + 1 (lone pair) = 4.
- **Hybridization**: With a steric number of 4, the hybridization is **sp³**.

### Step 2: Analyze [PtCl₄]²⁻ (Platinum Tetrachloride Ion)
- **Oxidation State**: The oxidation state of Platinum (Pt) can be determined by considering the charges of Chlorine (Cl) which is -1. Thus, for 4 Cl atoms, the total charge is -4. The overall charge of the complex is -2, so Pt must be +2.
- **Electron Configuration**: The electron configuration of Pt in the +2 state is [Xe] 4f¹⁴ 5d⁸.
- **Bonding**: Pt forms 4 bonds with Cl atoms. Since Cl is a weak field ligand, it does not cause pairing of d electrons.
- **Steric Number**: The steric number is 4 (bond pairs) + 0 (lone pairs) = 4.
- **Hybridization**: The hybridization is **dsp²**.

### Step 3: Analyze PCl₅ (Phosphorus Pentachloride)
- **Valence Electrons**: Phosphorus (P) has 5 valence electrons, and each Chlorine (Cl) has 1 valence electron.
- **Bonding**: Phosphorus forms 5 bonds with 5 Chlorine atoms, using all 5 of its valence electrons.
- **Steric Number**: The steric number is 5 (bond pairs) + 0 (lone pairs) = 5.
- **Hybridization**: The hybridization is **sp³d**.

### Step 4: Analyze BCl₃ (Boron Trichloride)
- **Valence Electrons**: Boron (B) has 3 valence electrons, and each Chlorine (Cl) has 1 valence electron.
- **Bonding**: Boron forms 3 bonds with 3 Chlorine atoms, using all 3 of its valence electrons.
- **Steric Number**: The steric number is 3 (bond pairs) + 0 (lone pairs) = 3.
- **Hybridization**: The hybridization is **sp²**.

### Summary of Hybridizations
- NH₃: sp³
- [PtCl₄]²⁻: dsp²
- PCl₅: sp³d
- BCl₃: sp²

### Correct Order of Hybridization
Now, we can arrange the hybridizations in order:
1. BCl₃: sp²
2. NH₃: sp³
3. [PtCl₄]²⁻: dsp²
4. PCl₅: sp³d

Thus, the correct order of hybridization of the central atom in the species is:
**sp² < sp³ < dsp² < sp³d**