Describe the shapes of `BF_(3)` and `BH_(4)^(-)`. Assign the hybridisation of boron atom in these species.

Options

A. `BF_(3)` - Trigonal, `sp^(2)` hybridisation,
`BH_(4)^(-)` - square planar, `sp^(3)` hybridisation
B. `BF_(3)`-Triangle, `sp^(3)` hybridisation,
`BH_(4)^(-)` - Hexagonal, `sp^(2)` d hybridisation
C. `BF_(3)` - Trigonal, `sp^(2)` hybridisation,
`BH_(4)^(-)` - Tetrahedral, `sp^(3)` hybridisation
D. `BF_(3)` - Tetrahedral, `sp^(3)` hybridisation
`BH_(4)^(-)`- Tetrahedral, `sp^(3)` hybridisation

Explanation

To describe the shapes of \( BF_3 \) and \( BH_4^- \) and assign the hybridization of the boron atom in these species, we can follow these steps:

### Step 1: Determine the hybridization of \( BF_3 \)

1. **Identify the central atom**: In \( BF_3 \), boron (B) is the central atom.
2. **Count the valence electrons of boron**: Boron has 3 valence electrons (electronic configuration: \( 1s^2 2s^2 2p^1 \)).
3. **Count the number of monovalent atoms attached**: There are 3 fluorine (F) atoms, which are monovalent (each forms one bond).
4. **Check for any charge**: There are no anions or cations present in \( BF_3 \).
5. **Use the hybridization formula**:
\[ \text{Hybridization} = \frac{(V + M + A - C)}{2} \]
where \( V \) = valence electrons of B (3), \( M \) = number of monovalent atoms (3), \( A \) = number of anions (0), \( C \) = number of cations (0).
\[ \text{Hybridization} = \frac{(3 + 3 + 0 - 0)}{2} = \frac{6}{2} = 3 \]
This indicates \( sp^2 \) hybridization.

### Step 2: Determine the shape of \( BF_3 \)

1. **Identify the hybridization**: Since the hybridization is \( sp^2 \), the shape is trigonal planar.
2. **Draw the shape**: In a trigonal planar arrangement, the three fluorine atoms are positioned at the corners of an equilateral triangle around the boron atom.

### Step 3: Determine the hybridization of \( BH_4^- \)

1. **Identify the central atom**: In \( BH_4^- \), boron (B) is again the central atom.
2. **Count the valence electrons of boron**: Boron has 3 valence electrons.
3. **Count the number of monovalent atoms attached**: There are 4 hydrogen (H) atoms, which are monovalent.
4. **Check for any charge**: There is a negative charge (-1) on the species.
5. **Use the hybridization formula**:
\[ \text{Hybridization} = \frac{(V + M + A - C)}{2} \]
where \( V \) = 3, \( M \) = 4, \( A \) = 1 (for the negative charge), \( C \) = 0.
\[ \text{Hybridization} = \frac{(3 + 4 + 1 - 0)}{2} = \frac{8}{2} = 4 \]
This indicates \( sp^3 \) hybridization.

### Step 4: Determine the shape of \( BH_4^- \)

1. **Identify the hybridization**: Since the hybridization is \( sp^3 \), the shape is tetrahedral.
2. **Draw the shape**: In a tetrahedral arrangement, the four hydrogen atoms are positioned at the corners of a tetrahedron around the boron atom.

### Summary of Results

- **Shape of \( BF_3 \)**: Trigonal planar
- **Hybridization of boron in \( BF_3 \)**: \( sp^2 \)
- **Shape of \( BH_4^- \)**: Tetrahedral
- **Hybridization of boron in \( BH_4^- \)**: \( sp^3 \)