What hybrid state of boron best explains the structures in `B_(2)H_(6)` ?

Options

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

Answer

To determine the hybrid state of boron that best explains the structure of \( B_2H_6 \) (diborane), we can follow these steps:

### Step 1: Understand the Structure of \( B_2H_6 \)
Diborane consists of two boron atoms and six hydrogen atoms. The structure can be visualized as two boron atoms connected by two hydrogen atoms, with the remaining hydrogen atoms bonded to each boron atom.

### Step 2: Determine the Valence Electrons
Each boron atom has 3 valence electrons. Therefore, in \( B_2H_6 \), the total number of valence electrons from the two boron atoms is \( 2 \times 3 = 6 \) electrons.

### Step 3: Identify Bonding in \( B_2H_6 \)
In \( B_2H_6 \), there are two types of bonds:
1. **B-H Bonds**: These are typical two-center, two-electron bonds.
2. **B-B Bonds**: The bond between the two boron atoms involves three centers sharing two electrons, often referred to as a "banana bond."

### Step 4: Analyze the Hybridization
To accommodate the bonding structure, we need to determine the hybridization of boron:
- In the ground state, boron has the electronic configuration of \( 1s^2 2s^2 2p^1 \).
- When boron is excited, one of the 2s electrons can be promoted to the 2p orbital, leading to the configuration \( 1s^2 2s^1 2p^2 \).

### Step 5: Hybridization Type
The hybridization that allows for the formation of the bonds in \( B_2H_6 \) is \( sp^3 \):
- One \( 2s \) orbital and three \( 2p \) orbitals combine to form four equivalent \( sp^3 \) hybrid orbitals.
- Two of these \( sp^3 \) hybrid orbitals from each boron atom form B-H bonds, while the third \( sp^3 \) orbital from one boron overlaps with the third \( sp^3 \) orbital of the other boron to form the banana bond.

### Conclusion
The hybrid state of boron that best explains the structures in \( B_2H_6 \) is **\( sp^3 \)** hybridization.

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