NBrS (N is central) Number of total electrons?Lewis structure?Bond order for N-BrBond order for N-S
The correct answer and explanation is:
To analyze the molecule NBrS (with nitrogen as the central atom), we will follow these steps:
🔢 1. Total Number of Valence Electrons
- Nitrogen (N): 5 valence electrons
- Bromine (Br): 7 valence electrons
- Sulfur (S): 6 valence electrons
Total = 5 + 7 + 6 = 18 valence electrons
✏️ 2. Lewis Structure of NBrS
- Place N in the center (less electronegative than Br and S).
- Connect Br and S to N with single bonds initially.
- Subtract bonding electrons (2 per bond × 2 bonds = 4 electrons).
- Remaining = 18 – 4 = 14 electrons
Distribute remaining 14 electrons to complete octets on terminal atoms (Br and S):
- Br gets 6 electrons (3 lone pairs)
- S gets 6 electrons (3 lone pairs)
Now N has only 4 electrons (from single bonds), so we add a lone pair (2 electrons) to nitrogen.
Total used: 6 (Br) + 6 (S) + 2 (N) + 4 (2 bonds) = 18 electrons
However, this gives nitrogen only 6 electrons total. Nitrogen tends to form multiple bonds to complete its octet.
Alternative structure: Add a double bond between N and S:
- N–Br: single bond
- N=S: double bond
- This way, nitrogen has 8 electrons (1 single + 1 double + 1 lone pair = 8e⁻)
- Formal charges are minimized:
- N: 5 valence – (2 lone + 3 bonding) = 0
- Br: 7 – (6 lone + 1 bond) = 0
- S: 6 – (4 lone + 2 bonds) = 0
✅ Correct Lewis Structure:
Br – N = S
:
With lone pairs on Br, S, and one on N.
⚖️ 3. Bond Orders
- Bond order = number of bonding electron pairs / number of bonds
- N–Br: single bond → bond order = 1
- N=S: double bond → bond order = 2
🧠 Summary
- Total valence electrons: 18
- Lewis structure: Br–N=S, with lone pairs to complete octets
- Bond order N–Br: 1
- Bond order N–S: 2
This structure satisfies the octet rule, minimizes formal charges, and correctly accounts for bonding.