When `N_(2)` goes to `N_(2)^(+)`, the `N-N` bond distance …………

Explanation

To determine how the bond distance changes when \( N_2 \) transitions to \( N_2^+ \), we can follow these steps:

### Step 1: Understand the Bond Order
The bond order is a measure of the number of chemical bonds between a pair of atoms. For \( N_2 \), the bond order is 3, indicating a triple bond between the two nitrogen atoms.

**Hint:** Bond order can be calculated using the formula:
\[ \text{Bond Order} = \frac{(\text{Number of bonding electrons} - \text{Number of antibonding electrons})}{2} \]

### Step 2: Determine the Effect of Ionization
When \( N_2 \) loses an electron to form \( N_2^+ \), one electron is removed from the bonding molecular orbital. This decreases the bond order from 3 to 2.5.

**Hint:** Removing an electron from a bonding orbital decreases the bond order, which can affect bond strength and length.

### Step 3: Analyze the Relationship Between Bond Order and Bond Length
As bond order decreases, bond strength also decreases. A weaker bond means that the atoms can be further apart, resulting in an increase in bond length.

**Hint:** Remember that higher bond order typically corresponds to shorter bond lengths due to stronger interactions between the bonded atoms.

### Step 4: Conclusion
Since the bond order decreases from 3 to 2.5 when \( N_2 \) becomes \( N_2^+ \), the bond strength decreases, leading to an increase in the \( N-N \) bond distance.

**Final Answer:** The \( N-N \) bond distance increases when \( N_2 \) goes to \( N_2^+ \).