Question

Iodine, \(\mathrm{I}_{2},\) is much more soluble in an aqueous solution of potassium iodide, KI, than it is in pure water. The anion found in solution is \(\mathrm{I}_{3}^{-}\). (a) Draw an electron dot structure for \(\mathrm{I}_{3}^{-}\). (b) Write an equation for this reaction, indicating the Lewis acid and the Lewis base.

Short answer

(a) I3- is linear, central I with 3 lone pairs and both outer I's with 3 lone pairs. (b) Reaction: I2 + I- -> I3-; Lewis base is I-, Lewis acid is I2.

Step-by-step solution

Determine Valence Electrons for I3-

First, identify the total number of valence electrons in \( \mathrm{I}_{3}^{-} \). Each iodine atom contributes 7 electrons, and the negative charge adds one extra electron, making it a total of \( 3 \times 7 + 1 = 22 \) electrons.

Distribute Electrons in I3-

Arrange the iodine atoms in a linear formation: I — I — I. Place a pair of electrons between each iodine to form bonds, using 4 electrons. This leaves us with 18 electrons.

Complete Octets

Fill the remaining electrons around the outer iodine atoms to complete their octets. Each outer iodine already has 2 electrons from bonding, needing 6 more: placing 6 electrons around each of the outer iodines uses up 12 electrons (3 pairs around each iodine), leaving 6 electrons.

Assign Lone Pairs to Central Iodine

Place the remaining 6 electrons as lone pairs around the central iodine atom, completing its octet with 10 electrons in total due to sharing.

Evaluate Formal Charges

Check formal charges to ensure stability. The central iodine will have a formal charge of -1, which matches the overall charge of \( \mathrm{I}_{3}^{-} \). Each outer iodine has a formal charge of 0.

Write Reaction Equation

Write the equation describing the formation of \( \mathrm{I}_{3}^{-} \): \( \mathrm{I}_{2} + \mathrm{I}^{-} \rightarrow \mathrm{I}_{3}^{-} \). Here, \( \mathrm{I}^{-} \) acts as the Lewis base by donating an electron pair, and \( \mathrm{I}_{2} \) acts as the Lewis acid, accepting the electron pair.

Key concepts

Valence Electrons

Valence electrons are the outermost electrons of an atom. They play a key role in chemical bonding and reactions. For iodine (\(\mathrm{I}\)), there are 7 valence electrons as it is in group 17 of the periodic table. In the \(\mathrm{I}_{3}^{-}\) ion, we need to account for an additional electron due to the negative charge. This leads to a total of 22 valence electrons:

• 3 iodine atoms each contributing 7 electrons = 21 electrons


• 1 extra electron due to the negative charge = 1 electron

Understanding valence electrons is crucial because it helps us predict how atoms will bond and form the structure of molecules such as \(\mathrm{I}_{3}^{-}\).

Octet Rule

The octet rule is a guiding principle in chemistry that atoms tend to surround themselves with eight electrons in their valence shell, much like the noble gases. In the case of \(\mathrm{I}_{3}^{-}\), each outer iodine seeks 8 electrons to fulfill its octet:

• Each bond shares 2 electrons, so bonded atoms can count these shared electrons toward their octet.


• For the outer iodine atoms, 6 additional electrons must be placed as lone pairs to fulfill their octet.

Interestingly, the central iodine in \(\mathrm{I}_{3}^{-}\) has more than 8 electrons due to expanded octet capability, accommodating 10 electrons through bonding and lone pairs. This helps iodine form stable compounds by using d sublevel orbitals.

Formal Charge

Formal charge is a concept that helps in verifying the stability of a Lewis structure by accounting for the electron distribution among atoms:

• The formula is: \( \text{Formal charge} = \text{Valence electrons} - (\text{Non-bonding electrons} + \frac{1}{2}\text{Bonding electrons}) \).


• For \(\mathrm{I}_{3}^{-}\), the central iodine has a formal charge of -1, which aligns with the overall charge of the ion.


• Both outer iodine atoms have a formal charge of 0, indicating they are stable as part of the structure.

Calculating formal charges ensures the justification of the molecule's configuration, ensuring it reflects the actual electron distribution.

Lewis Acid-Base Reaction

A Lewis acid-base reaction involves the transfer of an electron pair. In the reaction to form \(\mathrm{I}_{3}^{-}\):

• The \(\mathrm{I}^{-}\) ion donates an electron pair, making it the Lewis base.


• The \(\mathrm{I}_{2}\) molecule accepts this pair, acting as the Lewis acid.

This interaction stabilizes the iodine chain by the formation of \(\mathrm{I}_{3}^{-}\), illustrating how electron sharing enables new compound formation. Lewis acid-base theories help us understand how reactions proceed at a molecular level, where molecules actively exchange electron pairs.