Question

The deep color produced when iodine is dissolved in a solution of potassium iodide is caused by the presence of:
(a)- $I_2$
(b)- $I^{-}$
(c)- ${I_3}^{-}$
(d)- ${I_2}^{-}$

Answer 1

Hint: Halogens are sparingly soluble in water but when they are dissolved in organic solvents they give colored solutions. When iodine is dissolved in a solution of potassium iodide there is a formation of potassium triiodide.

Complete step by step answer:
Iodine is an element of group 17. Some elements of group 17 give a colored solution when dissolved in organic solvents.
Fluorine and chlorine do not react with water. Bromine and iodine are very less soluble in water, but they become soluble in many organic solvents such as carbon disulfide, carbon tetrachloride, chloroform, and hydrocarbons such as benzene, hexane, etc to give a colored solution.
As the iodine is not very soluble in water, it is dissolved in the water in the presence of potassium iodide ($KI$). This leads to a formation of a potassium triiodide complex which is soluble and gives an intense blue-black color. The reaction is given below:
$KI+I_2\to K{I}_3$

The ionic form of potassium triiodide is:
$K{I}_3\rightleftharpoons K^{+}+{I_3}^{-}$

The formation of blue-black intense is used to check the starch in the compound. If starch is present then the color will be intense blue-black and if starch is not present then the color remains yellow or orange.
So, the intense color is due to the presence of triiodide ion (${I_3}^{-}$).
So, the correct answer is “Option C”.

Note: When iodine is dissolved in ethanol it gives a brown solution. When the iodine is dissolved in a carbon tetrachloride solution it gives the violet solution. So as we can see it gives different colours for different solutions.