Draw the lewis structure.

Draw the lewis structure.
Draw the Lewis structure for I3- in the window below and then answer the questions that follow. What is the electron-pair geometry for I in I3-? What is the shape (molecular geometry) of I3-?

The Correct Answer and Explanation is:

Here’s the Lewis structure in ASCII form, showing one of the equivalent resonance contributors for I₃⁻ (the full resonance involves relocating the “–” to either terminal iodine):

   :Ö:         :Ö:
    |            |
 ..I — I — I..
    |            |
   :Ö:         :Ö:

• Each “I” symbol has lone pairs (dots) around it: the central I has three lone pairs (6 e⁻), and each terminal I has three lone pairs (6 e⁻).
• A formal negative charge (shown as “–”) is delocalized over the three-atom chain.

Answers

• Electron‐pair geometry around the central iodine: trigonal bipyramidal
• Molecular geometry (shape) of I₃⁻: linear

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Explanation (≈300 words)

To build the Lewis structure for I₃⁻, start by counting valence electrons:

1. Each iodine atom brings 7 valence electrons: 3 × 7 = 21 e⁻
2. The extra negative charge adds 1 more electron: +1 e⁻
3. Total = 22 valence electrons.

Place the three iodines in a row, with the central iodine bonded to each terminal iodine by a single bond (2 bonds × 2 e⁻ = 4 e⁻). Distribute the remaining 18 electrons as lone pairs to satisfy the octet (or expanded octet) rule:

• Each terminal I needs 6 more electrons for an octet (2 from bond + 6 lone = 8). That consumes 2 × 6 = 12 e⁻.
• The central I then has 22 – 4 (bonds) – 12 (terminal lone pairs) = 6 e⁻, or three lone pairs.

This gives five total electron‐pair domains around the central iodine: two bonding pairs and three nonbonding pairs.

According to VSEPR (Valence Shell Electron Pair Repulsion) theory, five electron domains arrange themselves in a trigonal bipyramidal geometry to minimize repulsion. However, when naming the molecular shape, we consider only the positions of the atoms (not lone pairs). Here, the three lone pairs preferentially occupy the three equatorial positions (further apart), leaving the two bonding pairs in the axial positions. That leaves the three iodine nuclei collinear, giving a linear molecular geometry.

Finally, resonance must be considered: the negative charge is delocalized over the three iodines, so there are three equivalent resonance structures (each with the “–” on one terminal iodine). This delocalization further stabilizes the ion but does not alter the geometry.

Thus, central I has trigonal bipyramidal electron‐pair geometry, and the observed shape of I₃⁻ is linear.