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QuestionChemistry

Draw the Lewis structure for I^3 -. Draw the molecule by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons. Show the formal charges of all atoms in the correct structure.
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Answer

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Step 1:
I'll solve this step-by-step, following the LaTeX formatting guidelines precisely:

Step 2:
: Determine the total number of valence electrons

- Total valence electrons: $$(3 \times 7) + 1 = 22$$ electrons
- Iodine (I) is in Group 17, so each I atom has 7 valence electrons - We have 3 I atoms, and the - 1 charge adds 1 additional electron

Step 3:
: Arrange the atoms

- Place the central iodine atom in the middle - Position the two other iodine atoms on either side

Step 4:
: Connect atoms with single bonds

- This uses $$2 \times 1 = 2$$ electron pairs
- Draw single bonds between the central iodine and the two terminal iodines

Step 5:
: Distribute remaining electrons as lone pairs

- Remaining electrons: $$22 - 4 = 18$$ electrons
- Place 3 lone pairs on each terminal iodine - Place 3 lone pairs on the central iodine

Step 6:
: Calculate formal charges

- Terminal I atoms: $$\frac{7}{1} - \frac{6}{2} - 3 = 0
- Central I atom: \frac{7}{1} - \frac{4}{2} - 4 = 0

Final Answer

The Lewis structure for I₃⁻ is a linear molecule with the central iodine single-bonded to two terminal iodines, with multiple lone pairs on each atom, maintaining a formal charge of zero on all atoms.