Two pi and half sigma bonds are present in :

Options

A. NO
B. CN
C. CO
D. `N_2`

Explanation

To determine which compound has 2 pi and half a sigma bond, we will analyze the molecular structure and bonding of each option provided in the question. We will focus on the number of electrons and the types of bonds formed in each molecule.

### Step-by-Step Solution:

1. **Identify the Electron Count for Each Compound**:
- For the first compound, **NO (Nitric Oxide)**, it has a total of 15 electrons.
- For the second compound, **Cyanide (CN⁻)**, it has a total of 13 electrons.
- For the third compound, **CO (Carbon Monoxide)**, it has a total of 14 electrons.
- For the fourth compound, **N₂ (Nitrogen gas)**, it also has a total of 14 electrons.

2. **Construct the Molecular Orbital Diagram for Each Compound**:
- For **NO**:
- The molecular orbital configuration is:
- σ(1s)², σ*(1s)², σ(2s)², σ*(2s)², π(2p_x)², π(2p_y)², σ(2p_z)¹
- This results in 1 sigma bond and 1 half pi bond (due to the unpaired electron in σ(2p_z)).

- For **Cyanide (CN⁻)**:
- The molecular orbital configuration is:
- σ(1s)², σ*(1s)², σ(2s)², σ*(2s)², π(2p_x)², π(2p_y)², σ(2p_z)¹
- This results in 2 pi bonds (from π(2p_x) and π(2p_y)) and 1 half sigma bond (from σ(2p_z)).

- For **CO**:
- The molecular orbital configuration is:
- σ(1s)², σ*(1s)², σ(2s)², σ*(2s)², π(2p_x)², π(2p_y)², σ(2p_z)²
- This results in 2 pi bonds and 1 sigma bond.

- For **N₂**:
- The molecular orbital configuration is:
- σ(1s)², σ*(1s)², σ(2s)², σ*(2s)², π(2p_x)², π(2p_y)²
- This results in 2 pi bonds and 1 sigma bond.

3. **Count the Bonds**:
- **NO**: 1 sigma + 0.5 pi = 1.5 bonds (not our answer).
- **Cyanide (CN⁻)**: 2 pi + 0.5 sigma = 2.5 bonds (this is our answer).
- **CO**: 2 pi + 1 sigma = 3 bonds (not our answer).
- **N₂**: 2 pi + 1 sigma = 3 bonds (not our answer).

4. **Conclusion**:
- The only compound that has 2 pi bonds and half a sigma bond is **Cyanide (CN⁻)**.

### Final Answer:
The compound with 2 pi and half a sigma bond is **Cyanide (CN⁻)**.