What is the hybridisation state of B and N in inorganic benzene respectively ?

Options

A. `sp^(2)` and `sp^(3)`
B. `sp^(3)` and `sp^(2)`
C. Both have `sp^(2)`
D. Both have `sp^(3)`

Answer

To determine the hybridization state of boron (B) and nitrogen (N) in inorganic benzene (B3N3H6), we can follow these steps:

### Step 1: Understand the Structure of Inorganic Benzene
Inorganic benzene has the formula B3N3H6. It consists of alternating boron and nitrogen atoms in a cyclic structure, similar to benzene. The structure can be visualized as follows:

- The arrangement is B-N-B-N-B-N, with alternating boron and nitrogen atoms.
- Each boron atom is bonded to one nitrogen atom and has one hydrogen atom attached.
- Each nitrogen atom is bonded to two boron atoms.

### Step 2: Analyze the Bonding
For each boron atom:
- Each boron forms three sigma (σ) bonds: one with nitrogen and two with adjacent boron atoms.
- Additionally, boron will participate in one pi (π) bond with the nitrogen atoms.

For each nitrogen atom:
- Each nitrogen forms three sigma (σ) bonds: one with boron and one with another boron atom.
- Nitrogen also forms one pi (π) bond with the adjacent boron atom.

### Step 3: Determine Hybridization for Boron
- Boron has three sigma bonds and one pi bond.
- To form three sigma bonds, boron requires three hybrid orbitals.
- Therefore, the hybridization state of boron is **sp²**.

### Step 4: Determine Hybridization for Nitrogen
- Nitrogen also has three sigma bonds and one pi bond.
- Similarly, nitrogen requires three hybrid orbitals to form these sigma bonds.
- Thus, the hybridization state of nitrogen is also **sp²**.

### Conclusion
Both boron and nitrogen in inorganic benzene are **sp² hybridized**.

### Final Answer
The hybridization state of B and N in inorganic benzene is **sp²** for both.

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