The hybridisation of `BeF_3-` is ________

Options

A. `sp^3`
B. `sp^2`
C. sp
D.

Answer

To determine the hybridization of the ion \( \text{BeF}_3^- \), we can follow these steps:

### Step 1: Determine the Valence Electrons
Beryllium (Be) has 2 valence electrons, and each fluorine (F) atom has 7 valence electrons. Since there are 3 fluorine atoms, we calculate the total valence electrons as follows:

- Valence electrons from Be: 2
- Valence electrons from 3 F atoms: \( 3 \times 7 = 21 \)
- Total valence electrons without considering the negative charge: \( 2 + 21 = 23 \)
- The negative charge indicates that there is an additional electron, so we add 1 more electron: \( 23 + 1 = 24 \)

### Step 2: Calculate the Steric Number
The steric number is calculated by adding the number of bonded atoms and the number of lone pairs on the central atom (beryllium in this case).

- Beryllium is bonded to 3 fluorine atoms (3 bonds).
- The negative charge indicates there is 1 lone pair on beryllium.

Thus, the steric number is:
\[ \text{Steric Number} = \text{Number of Bonds} + \text{Number of Lone Pairs} = 3 + 1 = 4 \]

### Step 3: Determine the Hybridization
The hybridization can be determined based on the steric number:
- A steric number of 4 corresponds to \( \text{sp}^3 \) hybridization.
- However, since we have 3 bonds and 1 lone pair, the geometry is trigonal planar, which corresponds to \( \text{sp}^2 \) hybridization.

### Conclusion
The hybridization of \( \text{BeF}_3^- \) is \( \text{sp}^2 \).

### Final Answer
The hybridization of \( \text{BeF}_3^- \) is **sp²**.

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