Hybridisation of Boron in `B_(2)H_(6)` molecule is :

Options

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

Explanation

To determine the hybridization of boron in the molecule \( B_2H_6 \), we can follow these steps:

### Step 1: Determine the Electronic Configuration of Boron
Boron has an atomic number of 5. Its electronic configuration is:
\[ 1s^2 \, 2s^2 \, 2p^1 \]

### Step 2: Identify the Ground State and Excited State
In the ground state, the boron atom has:
- 2 electrons in the \( 1s \) orbital
- 2 electrons in the \( 2s \) orbital
- 1 electron in the \( 2p \) orbital

To undergo hybridization, one of the \( 2s \) electrons can be promoted to the \( 2p \) orbital, resulting in the excited state:
- \( 1s^2 \, 2s^1 \, 2p^2 \)

### Step 3: Determine the Hybridization
In the excited state, boron has:
- 1 electron in the \( 2s \) orbital
- 2 electrons in the \( 2p \) orbitals

This configuration allows for the formation of four equivalent hybrid orbitals. The hybridization can be determined as follows:
- The mixing of one \( 2s \) and three \( 2p \) orbitals results in four \( sp^3 \) hybrid orbitals.

### Step 4: Bond Formation in \( B_2H_6 \)
In \( B_2H_6 \):
- Each boron atom forms bonds with hydrogen atoms using the \( sp^3 \) hybrid orbitals.
- The bonding involves three centered two-electron bonds (also known as banana bonds) between the boron atoms and hydrogen atoms.

### Conclusion
Thus, the hybridization of boron in \( B_2H_6 \) is \( sp^3 \).

### Final Answer
The hybridization of boron in \( B_2H_6 \) is \( sp^3 \).

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