How many of these molecules are diamagnetic and have bond order more than 2
`O_(2)^(2+), CO, overset(-)CN, NO, NO^(+), N_(2), O_(2)^(+), N_(2)^(2-), O_(2)^(2-)`

Answer

To determine how many of the given molecules are diamagnetic and have a bond order greater than 2, we will analyze each molecule step by step.

### Step 1: Identify the molecules and their electron configurations
The molecules we need to analyze are:
1. \( O_2^{2+} \)
2. \( CO \)
3. \( CN^- \)
4. \( NO \)
5. \( NO^+ \)
6. \( N_2 \)
7. \( O_2^+ \)
8. \( N_2^{2-} \)
9. \( O_2^{2-} \)

### Step 2: Calculate the number of electrons for each molecule
- **\( O_2^{2+} \)**: Oxygen has 8 electrons, so \( O_2 \) has 16 electrons. Removing 2 electrons gives us 14 electrons.
- **\( CO \)**: Carbon has 6 electrons and Oxygen has 8 electrons, so \( CO \) has 14 electrons.
- **\( CN^- \)**: Carbon has 6 electrons and Nitrogen has 7 electrons, plus 1 extra electron gives us 14 electrons.
- **\( NO \)**: Nitrogen has 7 electrons and Oxygen has 8 electrons, giving us 15 electrons.
- **\( NO^+ \)**: Removing 1 electron from \( NO \) gives us 14 electrons.
- **\( N_2 \)**: Each Nitrogen has 7 electrons, so \( N_2 \) has 14 electrons.
- **\( O_2^+ \)**: Removing 1 electron from \( O_2 \) gives us 15 electrons.
- **\( N_2^{2-} \)**: Adding 2 electrons to \( N_2 \) gives us 16 electrons.
- **\( O_2^{2-} \)**: Adding 2 electrons to \( O_2 \) gives us 18 electrons.

### Step 3: Determine the bond order for each molecule
The bond order is calculated using the formula:
\[ \text{Bond Order} = \frac{(\text{Number of bonding electrons}) - (\text{Number of antibonding electrons})}{2} \]

1. **\( O_2^{2+} \)**:
- Bonding: 6, Antibonding: 2 → Bond Order = \( \frac{6 - 2}{2} = 2 \)
- **Diamagnetic**: Yes (14 electrons, even)

2. **\( CO \)**:
- Bonding: 10, Antibonding: 4 → Bond Order = \( \frac{10 - 4}{2} = 3 \)
- **Diamagnetic**: Yes (14 electrons, even)

3. **\( CN^- \)**:
- Bonding: 10, Antibonding: 4 → Bond Order = \( \frac{10 - 4}{2} = 3 \)
- **Diamagnetic**: Yes (14 electrons, even)

4. **\( NO \)**:
- Bonding: 8, Antibonding: 7 → Bond Order = \( \frac{8 - 7}{2} = 0.5 \)
- **Diamagnetic**: No (15 electrons, odd)

5. **\( NO^+ \)**:
- Bonding: 8, Antibonding: 6 → Bond Order = \( \frac{8 - 6}{2} = 1 \)
- **Diamagnetic**: Yes (14 electrons, even)

6. **\( N_2 \)**:
- Bonding: 10, Antibonding: 4 → Bond Order = \( \frac{10 - 4}{2} = 3 \)
- **Diamagnetic**: Yes (14 electrons, even)

7. **\( O_2^+ \)**:
- Bonding: 10, Antibonding: 5 → Bond Order = \( \frac{10 - 5}{2} = 2.5 \)
- **Diamagnetic**: No (15 electrons, odd)

8. **\( N_2^{2-} \)**:
- Bonding: 10, Antibonding: 6 → Bond Order = \( \frac{10 - 6}{2} = 2 \)
- **Diamagnetic**: Yes (16 electrons, even)

9. **\( O_2^{2-} \)**:
- Bonding: 10, Antibonding: 8 → Bond Order = \( \frac{10 - 8}{2} = 1 \)
- **Diamagnetic**: Yes (18 electrons, even)

### Step 4: Count the molecules that are diamagnetic and have a bond order greater than 2
From our analysis:
- **Diamagnetic with bond order > 2**:
- \( CO \) (Bond Order = 3)
- \( CN^- \) (Bond Order = 3)
- \( N_2 \) (Bond Order = 3)

Thus, there are **3 molecules** that are diamagnetic and have a bond order greater than 2.

### Final Answer:
**3 molecules are diamagnetic and have a bond order greater than 2.**

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