Q.Write the Lewis structure of the nitrite ion, $NO_2^-$.

The key idea is to use Lewis Dot Structures to show how atoms share electrons while satisfying the octet rule and accounting for the ion’s negative charge.

Steps:

1. Count valence electrons: N has 5, each O has 6, plus 1 extra for the negative charge. Total = $5 + 6 + 6 + 1 = 18$ electrons.
2. Arrange atoms and draw single bonds: Place N in the centre, bonded to each O. This uses $2 \times 2 = 4$ electrons, leaving 14.
3. Complete octets on terminal oxygens: Give each O three lone pairs (6 electrons each). This uses 12 more electrons, leaving 2.
4. Place remaining electrons on N and form a double bond: Put the last 2 electrons on N as a lone pair. N now has only 6 electrons (2 from each single bond + 2 lone). To give N an octet, move one lone pair from an adjacent O to form a N=O double bond. The other O remains singly bonded with a negative charge. …

Why Lewis Structures Work for This

Lewis dot structures are our first tool for understanding how atoms bond. They show us which atoms are connected, how many bonds exist, and where lone pairs sit. For ions like $NO_2^-$, the key is counting valence electrons correctly — including the extra electron that gives the ion its negative charge.

The nitrite ion is a classic example of resonance. No single Lewis structure can capture the true electron distribution. Instead, we draw two equivalent structures and imagine the real ion as a blend of both.

Step-by-Step Construction

1. Count total valence electrons

Nitrogen has 5 valence electrons. Each oxygen has 6. The negative charge means we add 1 more electron.

This is our budget. Every bond and lone pair must account for exactly 18 electrons.

2. Identify the central atom

Nitrogen is less electronegative than oxygen, so it goes in the centre. The two oxygens attach to it.

3. Draw a skeleton and place a single bond between N and each O

Each single bond uses 2 electrons. Two bonds use 4 electrons.

4. Complete octets on the terminal oxygens first

Each oxygen needs 8 electrons total. Each already has 2 from the bond, so each needs 6 more — that's 3 lone pairs per oxygen.

Two oxygens × 3 lone pairs × 2 electrons per pair = 12 electrons used.

5. Place remaining electrons on the central nitrogen

Nitrogen currently has only 4 electrons (from the two single bonds). It needs 4 more to complete its octet. We have exactly 2 electrons left — that's one lone pair on nitrogen.

Now every atom has an octet. But we've used all 18 electrons.

6. Minimise formal charges by forming a double bond

Formal charge = (valence electrons) – (nonbonding electrons) – ½(bonding electrons).

For the current structure:

• N: $5 - 2 - \frac{1}{2}(4) = +1$
• Each O: $6 - 6 - \frac{1}{2}(2) = -1$

To reduce the positive charge on nitrogen, we can convert one lone pair from an oxygen into a bonding pair between N and that O. This creates a double bond.

Take one lone pair from either oxygen (2 electrons) and use them to form a second bond with nitrogen. Now:

• The double-bonded O has 2 lone pairs (4 nonbonding electrons) and shares 4 bonding electrons.
  • Formal charge: $6 - 4 - \frac{1}{2}(4) = 0$
• The single-bonded O still has 3 lone pairs (6 nonbonding electrons) and shares 2 bonding electrons.
  • Formal charge: $6 - 6 - \frac{1}{2}(2) = -1$
• N now has one lone pair (2 nonbonding electrons) and shares 6 bonding electrons (4 from the double bond, 2 from the single bond).
  • Formal charge: $5 - 2 - \frac{1}{2}(6) = 0$

The sum of formal charges is $0 + (-1) + 0 = -1$, which matches the ion's charge.

7. Recognise resonance

We could have chosen either oxygen to form the double bond. Both choices are equally valid. The two structures are:

$\text{Structure A: } \overset{\cdot\cdot}{\text{O}}=\overset{\cdot\cdot}{\text{N}}-\overset{\cdot\cdot\cdot\cdot}{\text{O}}^- \quad \longleftrightarrow \quad \text{Structure B: } ^-\overset{\cdot\cdot\cdot\cdot}{\text{O}}-\overset{\cdot\cdot}{\text{N}}=\overset{\cdot\cdot}{\text{O}}$ …