In $\text{KI}$ solution, ${{\text{I}}_{2}}$ readily dissolves and forms:
(A) ${{\text{I}}^{-}}$
(B) $\text{K}{{\text{I}}_{2}}$
(C) $\text{KI}_{2}^{-}$
(D) $\text{K}{{\text{I}}_{3}}$

Hint: Any molecule or element is soluble in any solvent when it can make a complex with the solvent or if it has any similar property the same as the solvent has. So, by observing the properties of iodine in potassium iodide solution, we can choose the correct answer.

Complete step by step solution:
- In the given question, we have to choose the correct product which is formed when iodine dissolves in the solution of potassium iodide.
- As we know that iodine is a halogen which has the atomic number 53 and valence electrons are 7. It is a non-polar atom because it does not have either a positive or negative charge.
- Whereas water which is formed of hydrogen and oxygen atoms is a polar molecule because there is a presence of partial negative charge on oxygen and a partial positive charge on the hydrogen atom.
- Also, we know that both solute and solvent should have the same property i.e. like dissolve like. It means for the solute to dissolve in the solvent both should be either polar or nonpolar.
- But here iodine is a nonpolar molecule whereas potassium iodide is also a nonpolar molecule so it will dissolve in it and forms a complex i.e.
 $\text{KI + }{{\text{I}}_{2}}\text{ }\to \text{ K}{{\text{I}}_{3}}$
- Here, $\text{K}{{\text{I}}_{3}}$ consist of poly halide ion i.e. $\text{I}_{3}^{-}$ due to which iodine is soluble in $\text{KI}$.

Therefore, option (D) is the correct answer.

Note: In case of water, iodine is very less soluble because as we know that iodine is nonpolar and water is a polar molecule so the hydrogen bond in water is much stronger and doesn't allow iodine to make bonds.