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 (boragene, B3N3H3), we can follow these steps:

### Step 1: Understand the Structure of Inorganic Benzene
Inorganic benzene, also known as boragene, has a cyclic structure similar to benzene but consists of alternating boron and nitrogen atoms. The chemical formula is B3N3H3, indicating three boron atoms and three nitrogen atoms in a hexagonal arrangement.

**Hint:** Remember that the structure of boragene is similar to benzene but includes boron and nitrogen atoms.

### Step 2: Identify the Hybridization of Boron
Boron in boragene is bonded to one hydrogen atom and two nitrogen atoms. To determine the hybridization:
- Count the number of regions of electron density around the boron atom. Boron has three bonds (one with hydrogen and two with nitrogen).
- The hybridization can be determined using the formula:
\[ \text{Hybridization} = \text{Number of bonds} + \text{Lone pairs} \]
- Here, boron has 3 regions of electron density, which corresponds to \( sp^2 \) hybridization.

**Hint:** For boron, count the bonds it forms and remember that three bonds indicate \( sp^2 \) hybridization.

### Step 3: Identify the Hybridization of Nitrogen
Each nitrogen atom in boragene is bonded to one boron atom and one hydrogen atom. To determine the hybridization:
- Each nitrogen atom also has one lone pair.
- Count the regions of electron density around the nitrogen atom. Each nitrogen has two bonds (one with boron and one with hydrogen) and one lone pair.
- Using the same formula:
\[ \text{Hybridization} = \text{Number of bonds} + \text{Lone pairs} \]
- Here, nitrogen has 3 regions of electron density, which corresponds to \( sp^2 \) hybridization.

**Hint:** For nitrogen, consider both the bonds and the lone pairs to determine its hybridization state.

### Final Answer
- The hybridization state of boron (B) in inorganic benzene is \( sp^2 \).
- The hybridization state of nitrogen (N) in inorganic benzene is also \( sp^2 \).