Among the following, the molecule expected to be stabilized by anion formation is : `C_(2), O_(2), NO, F_(2)`

Options

A. `C_(2)`
B. NO
C. `O_(2)`
D. `F_(2)`

Explanation

To determine which molecule is expected to be stabilized by anion formation among C2, O2, NO, and F2, we need to analyze the bond order of their respective anions. Here’s a step-by-step solution:

### Step 1: Understand the concept of anions
An anion is a negatively charged ion that has gained one or more electrons. For each molecule, we will consider its anion and the total number of electrons it contains.

### Step 2: Count the total number of electrons for each molecule
- **C2**: Each carbon atom contributes 4 valence electrons, so C2 has a total of 8 electrons. The anion C2^- will have 9 electrons.
- **O2**: Each oxygen atom contributes 6 valence electrons, so O2 has a total of 12 electrons. The anion O2^- will have 13 electrons.
- **NO**: Nitrogen contributes 5 electrons and oxygen contributes 6 electrons, so NO has a total of 11 electrons. The anion NO^- will have 12 electrons.
- **F2**: Each fluorine atom contributes 7 valence electrons, so F2 has a total of 14 electrons. The anion F2^- will have 15 electrons.

### Step 3: Write the molecular orbital configuration
We will use the general molecular orbital configuration for diatomic molecules to determine the bond order.

1. **C2^- (9 electrons)**:
- Configuration: σ1s² σ*1s² σ2s² σ*2s² σ2p_z² π2p_x² π2p_y¹
- Bonding electrons = 9, Antibonding electrons = 4
- Bond order = (Bonding - Antibonding) / 2 = (9 - 4) / 2 = 2.5

2. **O2^- (13 electrons)**:
- Configuration: σ1s² σ*1s² σ2s² σ*2s² σ2p_z² π2p_x² π2p_y²
- Bonding electrons = 10, Antibonding electrons = 3
- Bond order = (10 - 3) / 2 = 3.5

3. **NO^- (12 electrons)**:
- Configuration: σ1s² σ*1s² σ2s² σ*2s² σ2p_z² π2p_x²
- Bonding electrons = 8, Antibonding electrons = 4
- Bond order = (8 - 4) / 2 = 2.0

4. **F2^- (15 electrons)**:
- Configuration: σ1s² σ*1s² σ2s² σ*2s² σ2p_z² π2p_x² π2p_y²
- Bonding electrons = 12, Antibonding electrons = 3
- Bond order = (12 - 3) / 2 = 4.5

### Step 4: Compare bond orders
From the calculations:
- C2^- has a bond order of 2.5
- O2^- has a bond order of 3.5
- NO^- has a bond order of 2.0
- F2^- has a bond order of 4.5

### Step 5: Conclusion
The anion with the highest bond order is **F2^-** with a bond order of 4.5, indicating that it is the most stable anion among the given options.

### Final Answer
The molecule expected to be stabilized by anion formation is **F2**.