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JEE Main 202622 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Chemical Bonding and Molecular Structure

JEE Main 2026 · 22 January · Morning Shift · Chemistry · Chemical Bonding and Molecular Structure

The formal charges on the atoms marked as (1) to (4) in the Lewis...

The formal charges on the atoms marked as (1) to (4) in the Lewis representation of HNO3\mathrm{HNO}_{3} molecule respectively are
The formal charges on the atoms marked as (1) to (4) in the Lewis representation of HNO_ 3 molecule respectively are

Options

  1. A0,0,−1,+10,0,-1,+1
  2. B0,−1,0,+10,-1,0,+1
  3. C+1,0,0,−1+1,0,0,-1
  4. D0,+1,0,−10,+1,0,-1

Correct answer

D. 0,+1,0,−10,+1,0,-1

Step-by-step solution

The formal charge (F.C.) of an atom in a Lewis structure is calculated using the formula:
F.C. = [Total valence electrons] - [Total non-bonding (lone pair) electrons] - 12\frac{1}{2}[Total bonding (shared) electrons]
For atom (1) Oxygen: Valence e−=6e^- = 6, Lone pair e−=4e^- = 4, Bonding e−=4e^- = 4.
F.C.=6−4−12(4)=0F.C. = 6 - 4 - \frac{1}{2}(4) = 0
For atom (2) Nitrogen: Valence e−=5e^- = 5, Lone pair e−=0e^- = 0, Bonding e−=8e^- = 8.
F.C.=5−0−12(8)=+1F.C. = 5 - 0 - \frac{1}{2}(8) = +1
For atom (3) Oxygen: Valence e−=6e^- = 6, Lone pair e−=4e^- = 4, Bonding e−=4e^- = 4.
F.C.=6−4−12(4)=0F.C. = 6 - 4 - \frac{1}{2}(4) = 0
For atom (4) Oxygen: Valence e−=6e^- = 6, Lone pair e−=6e^- = 6, Bonding e−=2e^- = 2.
F.C.=6−6−12(2)=−1F.C. = 6 - 6 - \frac{1}{2}(2) = -1
The formal charges are 0,+1,0,−10, +1, 0, -1 respectively.
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