Question

What's the lewis structure of ClI2-

Answer 1

The Lewis structure for
`\( \text{ClI}_2^- \)` features a central chlorine atom with two iodine atoms, each connected by a single bond. The chlorine atom carries a lone pair, and the ion has an overall negative charge.

To determine the Lewis structure of
`\( \text{ClI}_2^- \)` (chlorine diiodide ion), we need to consider the valence electrons of each element and the overall charge of the ion.

Chlorine (Cl) is in Group 17 of the periodic table and has 7 valence electrons. Iodine (I) is in Group 17 as well and also has 7 valence electrons. The negative charge
`(\(-\))` indicates an extra electron, so we add one more electron to the total.

Now, let's calculate the total number of valence electrons:

`\[ \text{Cl:} \, 7 \, \text{(from Cl)} \]`

`\[ \text{I:} \, 7 \, \text{(from I)} \]`

`\[ \text{Extra electron:} \, 1 \, \text{(from the negative charge)} \]`

Total valence electrons =
`\(7 + 7 + 1 = 15\)`

Now, we arrange the atoms with single bonds and distribute the remaining electrons as lone pairs to satisfy the octet rule for each atom. Since we have an extra electron, it will be placed on the central chlorine atom.

The Lewis structure for
`\( \text{ClI}_2^- \)` is as follows:

`\[ \text{ :Cl-I:}^- \]`

Here, the colon (:) represents a shared pair of electrons, and the lone pair on the chlorine atom is indicated by a single dot (:). The negative charge is placed on the entire ion to show that it has one extra electron.