Question

which of these is the correct lewis structure for nitrogen tri-iodide, ni₃? options with lewis structure diagrams omitted as per instruction

Explanation:

Step1: Recall Lewis Structure Rules

Lewis structures show valence electrons, with atoms satisfying octet (or duet for H) and correct bonding. For \( \text{NI}_3 \), N (group 15) has 5 valence electrons, each I (group 17) has 7. Total valence electrons: \( 5 + 3\times7 = 26 \).

Step2: Analyze Each Option

- **First Option**: N has 2 lone electrons (incomplete octet for N? Wait, N should have 8 electrons. Wait, no—N is bonded to 3 I, so 3 single bonds (3×2=6) plus lone pairs. Wait, first option: N has 2 lone electrons? No, let's count. Wait, first option: N is in the center, bonded to 3 I (single bonds). N's valence: 3 bonds (6 e⁻) + 2 lone e⁻? No, N needs 8. Wait, no—wait, I has 7 valence, so each I should have 3 lone pairs (6 e⁻) plus 1 bond (2 e⁻) to satisfy octet (8 e⁻: 6 lone + 2 bond).
- **Second Option**: Linear structure \( \text{N-I-I-I} \) – incorrect, as \( \text{NI}_3 \) is trigonal pyramidal (N central, 3 I bonded, 1 lone pair on N). This structure has N bonded to 3 I in a chain, which is wrong.
- **Third Option**: I atoms have too many electrons? Wait, no—wait, the third option (blue circle) has I with 4 lone pairs? No, I should have 3 lone pairs (6 e⁻) and 1 bond. Wait, no, let's check the fourth option. Wait, the fourth option: N is central, bonded to 3 I (single bonds). Each I has 3 lone pairs (6 e⁻) + 1 bond (2 e⁻) → 8 e⁻ (octet). N: 3 bonds (6 e⁻) + 1 lone pair (2 e⁻) → 8 e⁻ (5 + 3 = 8? Wait, N has 5 valence, 3 bonds (3×1=3 electrons from N? No, bonding electrons are shared. Wait, total valence electrons: \( \text{N: } 5, \text{I: } 7\times3 = 21 \), total 26. Let's count electrons in each option:
- Fourth Option: N (central) – 3 single bonds to I. Each I: 3 lone pairs (6 e⁻) + 1 bond (2 e⁻) → 8 e⁻. N: 3 bonds (6 e⁻) + 1 lone pair (2 e⁻) → 8 e⁻. Total electrons: \( 3\times(6 + 2) + (6 + 2) = 3\times8 + 8 = 32 \)? No, wait, no—each bond is 2 e⁻ (shared). So for each I-N bond: 2 e⁻ (shared between N and I). So N's electrons: 3 bonds (3×2 e⁻, but N contributes 1 e⁻ per bond? No, Lewis structure counts total valence electrons. Let's calculate total electrons in fourth option: N has 1 lone pair (2 e⁻), 3 bonds (3×2 e⁻ = 6 e⁻) → N: 8 e⁻. Each I: 3 lone pairs (6 e⁻) + 1 bond (2 e⁻) → 8 e⁻. So total electrons: \( 2 (\text{N lone}) + 3\times2 (\text{bonds}) + 3\times6 (\text{I lone}) = 2 + 6 + 18 = 26 \), which matches. Wait, the third option (blue circle) – wait, no, the fourth option (last one) is the correct one? Wait, no, the third option (blue) has I with 4 lone pairs? Wait, no, let's look at the dots. Wait, the fourth option: each I has 3 lone pairs (6 dots) and 1 bond. N has 1 lone pair (2 dots) and 3 bonds. Let's count valence electrons:
N: 5 valence. Lone pair: 2 e⁻, bonds: 3×1 e⁻ (from N) + 3×1 e⁻ (from I) → no, bonding electrons are shared. Total valence electrons: N (5) + 3×I (7×3=21) = 26. In the fourth option:
- N: 1 lone pair (2 e⁻) + 3 bonds (3×2 e⁻ = 6 e⁻) → 8 e⁻ (5 + 3 = 8? Wait, no, valence electrons: N's 5 + 3×(I's 7) = 26. The fourth option:
- N: 2 lone e⁻ (wait, no, the fourth option's N has 2 lone dots? Wait, no, the fourth option: N is in the center, bonded to 3 I. Each I has 3 lone pairs (6 dots) and 1 bond. N has 1 lone pair (2 dots) and 3 bonds. So total electrons:
- N: 2 (lone) + 3×2 (bonds) = 8 e⁻ (correct, 5 + 3 = 8? Wait, no, valence electrons: N contributes 5, bonds: 3 bonds (each bond is 2 e⁻, but N and I each contribute 1 e⁻ per bond). So N's electrons: 5 (original) - 3 (used in bonds) + 3×1 (from I's bonds) + 2 (lone) = 5 - 3 + 3 + 2 = 7? No, I'm overcomplicating. The key is: \( \text{NI}_3 \) has N as central atom, 3 I atoms bonded (trigonal pyramidal), each I with 3 lone pairs, N with 1 lone pair. The fourth option (last one) has N central, 3 I bonded, each I with 3 lone pairs (6 dots), N with 1 lone pair (2 dots). The third option (blue) has I with 4 lone pairs? No, wait the third option's I: let's count dots. Third option: I has 4 lone pairs (8 e⁻) + 1 bond → 10 e⁻, which is incorrect (I should have 8). The fourth option: each I has 3 lone pairs (6 e⁻) + 1 bond (2 e⁻) → 8 e⁻. N has 1 lone pair (2 e⁻) + 3 bonds (6 e⁻) → 8 e⁻. Total valence electrons: \( 2 (\text{N lone}) + 3\times6 (\text{I lone}) + 3\times2 (\text{bonds}) = 2 + 18 + 6 = 26 \), which matches. The first option: N has 2 lone electrons (not a pair), so N has 6 e⁻ (3 bonds + 2 lone) → incomplete octet. The second option: linear, wrong structure. The third option: I has too many electrons. So the correct one is the fourth option (the last one, not the blue circle—wait, the blue circle is the third option? Wait, the image: first option (top), second (middle-top), third (middle-bottom, blue), fourth (bottom). Wait, no, the user's image: the third option (blue) has I with 4 lone pairs? No, let's re-express:

Wait, the correct Lewis structure for \( \text{NI}_3 \) is N in the center, bonded to 3 I (single bonds), each I has 3 lone pairs (6 electrons), and N has 1 lone pair (2 electrons). So let's check each:

1. Top option: N has 2 lone electrons (not a pair), I have no lone pairs? No, I should have 3 lone pairs. So incorrect.
2. Second option: N-I-I-I (chain) – incorrect, should be N central with 3 I bonded.
3. Third option (blue): I atoms have 4 lone pairs (8 electrons) + 1 bond → 10 electrons (exceeds octet for I, which is not allowed). Incorrect.
4. Fourth option: N central, 3 I bonded (single bonds), each I has 3 lone pairs (6 electrons) + 1 bond (2 electrons) → 8 electrons (octet). N has 1 lone pair (2 electrons) + 3 bonds (6 electrons) → 8 electrons (octet). Total valence electrons: \( 2 (N lone) + 3\times6 (I lone) + 3\times2 (bonds) = 2 + 18 + 6 = 26 \), which matches \( 5 + 3\times7 = 26 \). So the fourth option (bottom one) is correct. Wait, but the blue circle is on the third option. Wait, maybe I misread. Wait, the third option: let's count I's electrons. Third option: I has 4 lone pairs (8 dots) + 1 bond → 10 e⁻ (wrong). Fourth option: I has 3 lone pairs (6 dots) + 1 bond → 8 e⁻ (correct). N in fourth option: 1 lone pair (2 dots) + 3 bonds → 8 e⁻ (correct). So the correct structure is the fourth option (the bottom one, not the blue circle). Wait, but the user's image: the blue circle is on the third option, but that's incorrect. Wait, no—wait, maybe I made a mistake. Wait, N has 5 valence electrons. In \( \text{NI}_3 \), N forms 3 single bonds (3×2=6 e⁻) and has 1 lone pair (2 e⁻), so total 8 e⁻. Each I forms 1 single bond (2 e⁻) and has 3 lone pairs (6 e⁻), total 8 e⁻. So the correct structure is N central, 3 I bonded (trigonal pyramidal), each I with 3 lone pairs, N with 1 lone pair. So the fourth option (bottom) is correct.

Answer:

The bottom - most Lewis structure (the fourth option, with N in the center, bonded to three I atoms each having three lone pairs of electrons and N having one lone pair of electrons)