Question

Arrange the following in the decreasing order of their reducing character.
$HF,{\text{ }}HCl,{\text{ }}HBr,{\text{ }}HI$

Answer 1

Hint : We must know that the reducing character is the tendency of an atom to donate an electron. The easier it is for an atom the more reducing character it has.

Complete step by step solution :
Let’s start with basic understanding of reducing character, it is the tendency of an element to donate hydrogen atom or to accept oxygen atom. To be clearer it is the tendency of a species to donate electrons during a redox reaction. The donation of this electron leads to the generation of positive charge on the species and the species with reducing character oxidizes itself reducing the other species.
In the periodic table, As we move down the group the distance of the valence shell form the nuclei increases, as the distance increases the forces between the two weaken, due to the weakening of forces the electron and nuclei are not strongly bonded so it is easy for the atom to lose an electron. Such behaviour shows that as we move down the group the reducing characteristic of elements increases.
So, in the question we are given $HF,{\text{ }}HCl,{\text{ }}HBr,{\text{ }}HI$ out of these the fluorine (F) is present in 2nd period, Chlorine present in 3rd period, Bromine present in 4th period and Iodine present in 5th period. So $HI$will have the highest reducing characteristic followed by$\;HBr,{\text{ }}HCl{\text{ }}and{\text{ }}HF$. $HF$is having the least reducing characteristic due to its compact size.
So, the answer will be$HI > HBr > HCl > HF$.

Note : We must understand that the main reason behind the corrosion of metals is the reduction and oxidation reactions. Corrosion requires an anode and cathode, anode being the element which acts as a reducing agent and cathode is the element which acts as oxidizing agent. Whenever oxidative potential difference is generated the anode starts oxidizing and hence it corrodes.