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QuestionChemistry

Draw the Lewis structure for the azide $\left(\mathrm{N}_{3}^{-}\right)$ion.
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Step 1:
I'll solve this Lewis structure problem step by step:

Step 2:
: Determine the total number of valence electrons

- Total valence electrons: $$(3 \times 5) + 1 = 16$$ electrons
- Nitrogen (N) has 5 valence electrons - We have 3 nitrogen atoms - The ion has a - 1 charge, which adds 1 more electron

Step 3:
: Arrange the atoms

- Place the three nitrogen atoms in a linear arrangement - The central nitrogen will connect to the two outer nitrogens

Step 4:
: Create initial single bonds

- Remaining electrons: $$16 - 6 = 10$$ electrons
- Draw single bonds between the central nitrogen and the two outer nitrogens

Step 5:
: Add remaining electrons as lone pairs

- Distribute the remaining $$10$$ electrons as lone pairs
- Prioritize outer atoms first - Each outer nitrogen will have 3 lone pairs - The central nitrogen will have 1 lone pair

Step 6:
: Check formal charges

\mathrm{N} \equiv \mathrm{N}^{-} \equiv \mathrm{N}
- Verify that the structure minimizes formal charges - The structure will have a resonance form with electrons delocalized Final Lewis Structure: (Triple bond between outer nitrogens, with a negative charge)

Final Answer

The Lewis structure for the azide ion \left(\mathrm{N}_{3}^{-}\right) is a linear structure with triple bonds and a negative charge.