Q. \( \text{i}2 \) lewis dot structure.

This question is asking for the Lewis dot structure of the ion (or species) that looks like “\( i2 \)”. However, “\( i2 \)” is not a standard chemical formula written in a way that uniquely determines the element(s) involved.

In Lewis structure problems, the symbol must specify the element. For example, common “diatomic” formulas include \( I_2 \) (iodine molecule), \( Br_2 \), \( Cl_2 \), etc. If you meant \( I_2 \), that would be iodine, and I can draw the correct Lewis dot structure. If you meant something else (like a charge state or an ion), the structure changes.

Based on the most likely interpretation: I will assume you meant \( I_2 \) (iodine molecule).

Step 1: Identify the valence electrons.

Iodine is in Group 17 (halogens). Each iodine atom has 7 valence electrons.

Total valence electrons in \( I_2 \):

\[
\text{Total} = 2 \times 7 = 14
\]

Step 2: Connect the atoms with a bond.

For a diatomic halogen molecule \( X_2 \), the atoms typically form a single covalent bond.

A single bond uses 2 electrons (one from each atom).

Step 3: Place remaining electrons as lone pairs.

After using 2 electrons for the bond, the remaining electrons are:

\[
14 – 2 = 12
\]

These 12 electrons become lone pairs on the two iodine atoms.

Each iodine atom has 3 lone pairs (that is \( 6 \) electrons per iodine).

So overall you place 3 lone pairs on the left iodine and 3 lone pairs on the right iodine.

Step 4: Final Lewis dot structure.

The Lewis structure of \( I_2 \) is a single bond between the two iodine atoms, with 3 lone pairs on each iodine.

Text version (showing lone pairs):

\[
\text{I} – \text{I}
\]

with:

\(\text{each I has 3 lone pairs}\).

How to draw it (exact placement rule):

1. Draw two iodine symbols \( \text{I} \) and \( \text{I} \).

2. Draw one bond line between them.

3. Add 3 pairs of dots around the first iodine (that is 6 dots total on that atom).

4. Add 3 pairs of dots around the second iodine (that is 6 dots total on that atom).