Which charge for the `N_2` molecule would give a bond order of 2.5 ?

Options

A. `+1`
B. `+2`
C. `-1`
D. `-2`

Explanation

To determine which charge on the \( N_2 \) molecule would give a bond order of 2.5, we need to analyze the molecular orbital (MO) configuration of \( N_2 \) and apply the bond order formula.

### Step-by-Step Solution:

1. **Understanding Bond Order**:
The bond order can be calculated using the formula:
\[ \text{Bond Order} = \frac{N_b - N_a}{2} \]
where \( N_b \) is the number of electrons in bonding molecular orbitals and \( N_a \) is the number of electrons in anti-bonding molecular orbitals.

2. **Molecular Orbital Configuration of \( N_2 \)**:
The molecular orbital configuration for \( N_2 \) (which has 14 electrons) is:
- Bonding orbitals: \( \sigma_{1s}^2, \sigma^*_{1s}^2, \sigma_{2s}^2, \sigma^*_{2s}^2, \pi_{2p_y}^2, \pi_{2p_z}^2, \sigma_{2p_x}^2 \)
- Anti-bonding orbitals: \( \pi^*_{2p_y}^0, \pi^*_{2p_z}^0, \sigma^*_{2p_x}^0 \)

Therefore, we have:
- \( N_b = 10 \) (from bonding orbitals)
- \( N_a = 4 \) (from anti-bonding orbitals)

The bond order for \( N_2 \) is:
\[ \text{Bond Order} = \frac{10 - 4}{2} = 3 \]

3. **Finding the Charge for Bond Order of 2.5**:
To decrease the bond order from 3 to 2.5, we can either:
- Add electrons to the anti-bonding orbitals (which would decrease the bond order).
- Remove electrons from the bonding orbitals (which would also decrease the bond order).

Let's analyze both cases:

**Case 1: Adding Electrons to Anti-bonding Orbitals**:
- Let \( x \) be the number of electrons added to the anti-bonding orbitals.
- The new bond order equation becomes:
\[ \frac{10 - (4 + x)}{2} = 2.5 \]
Simplifying gives:
\[ 10 - 4 - x = 5 \implies 6 - x = 5 \implies x = 1 \]
Thus, adding 1 electron to the anti-bonding orbitals results in a charge of -1 (i.e., \( N_2^- \)).

**Case 2: Removing Electrons from Bonding Orbitals**:
- Let \( y \) be the number of electrons removed from the bonding orbitals.
- The new bond order equation becomes:
\[ \frac{(10 - y) - 4}{2} = 2.5 \]
Simplifying gives:
\[ 10 - y - 4 = 5 \implies 6 - y = 5 \implies y = 1 \]
Thus, removing 1 electron from the bonding orbitals results in a charge of +1 (i.e., \( N_2^+ \)).

4. **Final Answer**:
The charges that would give a bond order of 2.5 for the \( N_2 \) molecule are:
- \( N_2^- \) (charge of -1)
- \( N_2^+ \) (charge of +1)