When ${{I}_{2}}$ is passed through $KCL,\,KF,\,KBr$ solution:
A. $C{{l}_{2}}\,\,and\,\,B{{r}_{2}}$are evolved
B. $C{{l}_{2}}$ is evolved
C. $C{{l}_{2}},B{{r}_{2}},{{F}_{2}}$ are evolved
D. None of these

Hint: In this question, we have to answer the gas which evolves when iodine is passed through $KCL,\,KF,\,KBr$. To answer this question, first we know the reactivity of the halogens then we are able to answer the question and choose the correct option.

Complete Step by Step Solution:
We answer this question based on their reactivity.
As we know fluorine, chlorine, bromine and iodine belongs to the halogens family, first we discuss about the reactivity of halogens:
As we know when we go down the group, atomic radius goes on increasing and due to this, the attraction faced by valence electrons goes on decreasing and thus, electronegativity decreases. Smaller negative ion will have more capability to attract a proton towards itself. Thus on moving down the group, reactivity decreases.

As the fluorine lies on the top followed by chlorine, bromine and iodine. So fluorine is the most reactive and iodine is the least reactive.
As ${{I}_{2}}$ is less reactive than Cl, Br and F, so it does not displace any halogen.
So ${{I}_{2}}$ cannot displace $C{{l}_{2}},B{{r}_{2}},{{F}_{2}}$from $KCL,\,KF,\,KBr$.
Thus, Option (D) is correct.

Note: Chlorine, bromine are used as disinfectants. Atomic radius of halogens increases down the group whereas ionisation energy decreases down the group and electronegativity and the electron affinity also decreases down the group. Students must know why they increase or decrease when we go down the group.