The number of sigma and pi bonds present in inorganic benzene are respectively

Options

A. `3sigma,12pi`
B. `12sigma,3pi`
C. `3sigma,3pi`
D. `12sigma,12pi`

Answer

To determine the number of sigma and pi bonds in inorganic benzene (B3N3H6), we can follow these steps:

### Step 1: Write the molecular formula
The molecular formula of inorganic benzene is B3N3H6.

### Step 2: Draw the structure of inorganic benzene
Inorganic benzene has a cyclic structure similar to benzene, where boron and nitrogen atoms alternate. The structure can be represented as follows:
- Each boron (B) atom is bonded to two nitrogen (N) atoms and one hydrogen (H) atom.
- Each nitrogen atom is bonded to one boron atom and one hydrogen atom.
- The alternating B and N atoms form double bonds between some B and N pairs.

### Step 3: Identify the types of bonds
- **Sigma (σ) bonds**: These are formed by the head-on overlap of atomic orbitals. In the structure, each single bond (B-H, N-H, B-N) counts as one sigma bond.
- **Pi (π) bonds**: These are formed by the side-to-side overlap of p-orbitals and are present in double bonds. Each double bond consists of one sigma bond and one pi bond.

### Step 4: Count the sigma bonds
1. Count the B-H bonds: There are 3 B-H bonds (1 for each boron).
2. Count the N-H bonds: There are 3 N-H bonds (1 for each nitrogen).
3. Count the B-N bonds: There are 3 B-N bonds (1 for each boron bonded to nitrogen).
4. Count the double bonds between B and N: There are 3 double bonds (each contributing 1 sigma bond).

Total sigma bonds:
- B-H: 3
- N-H: 3
- B-N: 3
- B=N (double bonds): 3 (1 sigma bond each)

Total sigma bonds = 3 (B-H) + 3 (N-H) + 3 (B-N) + 3 (B=N) = 12 sigma bonds.

### Step 5: Count the pi bonds
Each double bond contributes 1 pi bond. Since there are 3 double bonds between boron and nitrogen, we have:
Total pi bonds = 3 (from the 3 double bonds).

### Final Answer
The number of sigma and pi bonds present in inorganic benzene is:
- Sigma bonds: 12
- Pi bonds: 3