To draw the Lewis structure for HNO₃ (nitric acid), follow these steps:
- Count the total number of valence electrons:
- Hydrogen (H) has 1 valence electron.
- Nitrogen (N) has 5 valence electrons.
- Each Oxygen (O) has 6 valence electrons, and there are 3 Oxygen atoms, so 6 × 3 = 18 valence electrons.
- Total valence electrons = 1 + 5 + 18 = 24.
- Determine the central atom:
- Nitrogen (N) is the central atom because it is less electronegative than Oxygen and can form multiple bonds.
- Draw the skeletal structure:
- Place Nitrogen in the center and surround it with the three Oxygen atoms and one Hydrogen atom.
- Distribute the electrons:
- Start by forming single bonds between Nitrogen and each Oxygen atom, and between Nitrogen and Hydrogen.
- This uses 8 electrons (4 single bonds × 2 electrons each).
- Remaining electrons = 24 – 8 = 16.
- Complete the octets:
- Add lone pairs to the Oxygen atoms to complete their octets. Each Oxygen needs 6 more electrons (3 lone pairs).
- This uses 18 electrons (3 Oxygen atoms × 6 electrons each), but we only have 16 electrons left. This indicates the need for resonance structures.
- Draw resonance structures:
- Move lone pairs to form double bonds between Nitrogen and Oxygen atoms, ensuring that the formal charges are minimized.
- There are three possible resonance structures, each with a double bond between Nitrogen and a different Oxygen atom.
- Calculate formal charges:
- Formal charge = (Valence electrons) – (Non-bonding electrons) – (Bonding electrons / 2).
- For Nitrogen: 5 – 0 – (8 / 2) = +1.
- For Oxygen with a double bond: 6 – 4 – (4 / 2) = 0.
- For Oxygen with a single bond: 6 – 6 – (2 / 2) = -1.
- For Hydrogen: 1 – 0 – (2 / 2) = 0.
Here is the final Lewis structure with resonance structures and formal charges:
O || H-O-N=O | O
Remember that the actual structure is a hybrid of the resonance forms, with partial double bond character between Nitrogen and each Oxygen atom.