Question

which of the following is the correct lewis structure for n₂?
n - n with no lone pairs
n = n with two lone pairs on each nitrogen
n ≡ n with one lone pair on each nitrogen
n - n = n with two lone pairs on each nitrogen

Explanation:

Step1: Recall Lewis Structure Rules

Nitrogen has 5 valence electrons. For \( \text{N}_2 \), we need to satisfy octet (or duet for H, but N needs 8). Each N contributes 5, so total valence electrons: \( 5 + 5 = 10 \).

Step2: Analyze Bonding and Lone Pairs

• A triple bond (\( \equiv \)) uses 6 electrons (3 pairs), leaving \( 10 - 6 = 4 \) electrons, which is 2 lone pairs? Wait, no: Wait, each N in \( \text{N}_2 \) has a triple bond (\( \text{N} \equiv \text{N} \)) and one lone pair (2 electrons) on each N. Let's check: Triple bond (3 bonds, 6 electrons) + 2 lone electrons per N (total 4) = 10. Wait, no: Each N in \( \text{N}_2 \) has 1 lone pair (2 electrons) because \( 5 - 3 \) (bonds, each bond is 1 electron from each N) = 2? Wait, Lewis structure for \( \text{N}_2 \) is \( :\text{N} \equiv \text{N}: \)? No, wait, no: Wait, nitrogen's valence: 5. To form a triple bond, each N shares 3 electrons with the other, so each N has \( 5 - 3 = 2 \) non - bonding electrons, which is 1 lone pair (since a lone pair is 2 electrons). So the correct structure is \( \text{N} \equiv \text{N} \) with one lone pair on each nitrogen.

Step3: Evaluate Options

• Option 1 (N - N with no lone pairs? No, valence electrons don't add up.
• Option 2 (N = N with two lone pairs? Bonding electrons: 4, lone pairs: 4 per N? No, total electrons would be too much.
• Option 3 (N \( \equiv \) N with one lone pair on each N): This matches the electron count (triple bond: 6 electrons, 2 lone electrons (1 pair) per N: 2*2 = 4; total 10).
• Option 4 (N - N with two lone pairs? Bonding electrons: 2, lone pairs: 4 per N? Total electrons too high.

Answer:

\( \text{N} \equiv \text{N} \) with one lone pair on each nitrogen (the second option in the list, assuming the options are ordered as: 1. N - N no lone pairs, 2. N = N two lone pairs, 3. N \( \equiv \) N one lone pair, 4. N - N two lone pairs? Wait, from the image, the correct option is the one with \( \text{N} \equiv \text{N} \) and one lone pair on each nitrogen.