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# Question 88946

Jun 17, 2015

The answer is d) none of these

#### Explanation:

In a redox reaction, a reducing agent loses electrons and an oxidizing agent gains electrons.

This means that stronger reducing agents will lose electrons more easily than weaker reducing agents.

In the case of hydracids, oxidation can only remove an electron from the halogen ion, which means that the stability of the halogen atom will determine how strong the reducing ability of the hydracid will be.

In other words, the easier it is to remove an electron from a halogen ion, the stronger reducing agent the hydracid will be.

Because the size of the ions increases down a group, the ions with the largest ionic size will tend to lose an electron more easily.

This implies that hydriodic acid, $H I$, is the strongest reducing agent of the group, because the iodide ion has the largest ionic size of the group. A larger ioniz size implies that the outermost electrons are further away from the nucleus, and can thus be removed more easily.

Hydrobromic acid, $H B r$, comes after $H I$ in terms of reducing strength, followed by hydrochloric acid, $H C l$.

Hydrofluoric acid, $H F$, is not only the weakest reducing agent of the group, but also a poor reducing agent overall. That happens because the fluoride ion, ${F}^{-}$, is a very small ion, which implies that its outermost electrons are actually very close to the nucleus.

As a result, electrons are more difficult to remove from ${F}^{-}$, which is why $H F$ is such a weak reducing agent, and therefore not capable of reducing any of the compounds listed. The take-home message thus is

"easily removable electrons" -> color(green)("strong reducing agent")

"not so easily removable electrons" -> color(red)("weak reducing agent")#

Final answer, d) none of these.