The hybridisation of the central atom
in the following species `NF_(3),BF_(3),PF_(5)` is `:`

Explanation

To determine the hybridization of the central atom in the species NF₃, BF₃, and PF₅, we can use the formula:

\[ H = \frac{1}{2} (V + M - C + A) \]

Where:
- \( H \) = hybridization
- \( V \) = number of valence electrons of the central atom
- \( M \) = number of monovalent atoms attached to the central atom
- \( C \) = charge on the cation
- \( A \) = charge on the anion

Let's analyze each compound step by step.

### Step 1: Hybridization of NF₃
1. **Identify the central atom**: Nitrogen (N).
2. **Determine the number of valence electrons (V)**: Nitrogen is in group 15, so it has 5 valence electrons.
3. **Count the number of monovalent atoms (M)**: There are 3 fluorine (F) atoms attached to nitrogen.
4. **Determine the charges (C and A)**: There are no charges on the molecule, so \( C = 0 \) and \( A = 0 \).

Now plug these values into the formula:
\[ H = \frac{1}{2} (5 + 3 - 0 + 0) \]
\[ H = \frac{1}{2} (8) = 4 \]

Since \( H = 4 \), the hybridization is \( sp^3 \).

### Step 2: Hybridization of BF₃
1. **Identify the central atom**: Boron (B).
2. **Determine the number of valence electrons (V)**: Boron is in group 13, so it has 3 valence electrons.
3. **Count the number of monovalent atoms (M)**: There are 3 fluorine (F) atoms attached to boron.
4. **Determine the charges (C and A)**: There are no charges on the molecule, so \( C = 0 \) and \( A = 0 \).

Now plug these values into the formula:
\[ H = \frac{1}{2} (3 + 3 - 0 + 0) \]
\[ H = \frac{1}{2} (6) = 3 \]

Since \( H = 3 \), the hybridization is \( sp^2 \).

### Step 3: Hybridization of PF₅
1. **Identify the central atom**: Phosphorus (P).
2. **Determine the number of valence electrons (V)**: Phosphorus is in group 15, so it has 5 valence electrons.
3. **Count the number of monovalent atoms (M)**: There are 5 fluorine (F) atoms attached to phosphorus.
4. **Determine the charges (C and A)**: There are no charges on the molecule, so \( C = 0 \) and \( A = 0 \).

Now plug these values into the formula:
\[ H = \frac{1}{2} (5 + 5 - 0 + 0) \]
\[ H = \frac{1}{2} (10) = 5 \]

Since \( H = 5 \), the hybridization is \( sp^3d \).

### Summary of Hybridizations
- **NF₃**: \( sp^3 \)
- **BF₃**: \( sp^2 \)
- **PF₅**: \( sp^3d \)