Question

Which reaction is not feasible?
A. $2KI\quad +\quad { Br }_{ 2 }\quad \longrightarrow \quad 3KBr\quad +\quad { I }_{ 2 }$
B. $2KBr\quad +\quad { I }_{ 2 }\quad \longrightarrow \quad 2KI\quad +\quad { Br }_{ 2 }$
C. $2KBr\quad +\quad { Cl }_{ 2 }\quad \longrightarrow \quad 2KCl\quad +\quad { Br }_{ 2 }$
D. $2{ H }_{ 2 }O\quad +\quad { 2F }_{ 2 }\quad \longrightarrow \quad 4HF\quad +\quad O_{ 2 }$

Answer 1

Hint: In the reactions given, they all involve halogens in them. In these reactions, halogens are replaced from their salts by other halogens. The reactions given are displacement reactions, in which one element displaces the other element from the corresponding compound.

Complete step by step answer:
We need to know which type of halogens replace the other halogens from their salts. The answer to this question is very simple, which is, the halogen with lower atomic number can only replace the halogen, which has greater atomic number, from their salts. The atomic number comes into play because the halogen with lower atomic number has higher electronegativity than the one with the higher atomic number. We know that down the group electronegativity decreases.

This can be illustrated as follows:
$2K{ X }\quad +\quad Y_{ 2 }\quad \longrightarrow \quad 2K{ Y }\quad +\quad { X }_{ 2 }$
In the above reaction X and Y are halogens. This reaction is feasible only if the atomic number of Y is less than that of X else it would not be feasible.
Now, let us look at the options one by one:

a) $2KI\quad +\quad { Br }_{ 2 }\quad \longrightarrow \quad 3KBr\quad +\quad { I }_{ 2 }$
In this reaction, the halogens involved are iodine and bromine. We know that the atomic number of bromine is less than the atomic number of iodine. Therefore, this reaction is feasible.

b) $2KBr\quad +\quad { I }_{ 2 }\quad \longrightarrow \quad 2KI\quad +\quad { Br }_{ 2 }$
In this reaction, the halogens involved are iodine and bromine. And we just saw that the atomic number of bromine is less than the atomic number of iodine. Therefore, this reaction is not feasible.

c) $2KBr\quad +\quad { Cl }_{ 2 }\quad \longrightarrow \quad 2KCl\quad +\quad { Br }_{ 2 }$
In this reaction, the halogens involved are bromine and chlorine. We know that the atomic number of chlorine is less than the atomic number of bromine. Therefore, this reaction is feasible.

d) $2{ H }_{ 2 }O\quad +\quad { 2F }_{ 2 }\quad \longrightarrow \quad 4HF\quad +\quad O_{ 2 }$
This reaction is also feasible. This due to the fact that fluorine is the most electronegative element in the periodic table. Also fluorine likes to exist in -1 oxidation state. So, if in this reaction, it is getting to shift from 0 oxidation in ${F}_{2}$ to -1 oxidation state in $HF$, why wouldn't it.
Therefore, the only reaction that is not feasible is the reaction in option (b).
Hence, the correct answer is option (b).

Note: The reactivity order of the elements in this question are $F>O>Cl>Br>I$. These types of reactions are known as single displacement reactions, in which the more reactive element replaces the less reactive one.