Question

Which of these is a reducing agent?
a.) $Cr{O_3}/{H^ + }$
b.) $KMn{O_4}$
c.) $LiAl{H_4}$
d.) ${O_3}$

Answer 1

Hint: To solve this question firstly we have to understand what oxidizing agents and reducing agents are and study their reactions. And with the help of an example, we will be studying reducing agents briefly.

Complete step-by-step answer:
An oxidizing agent is an agent that removes electrons from other reactants during a redox reaction. The oxidizing agent hence, takes electrons, thus gaining electrons and getting itself reduced.
Reducing agent is an element or compound that loses its electron to an electron recipient (oxidizing agent) in a redox chemical reaction.
Common examples of reducing agents are Lithium aluminium hydride $(LiAl{H_4})$, Nascent (atomic) hydrogen, Sodium amalgam (Na(Hg)), Sodium-lead alloy (Na + Pb), Sodium borohydride $\left( {NaB{H_4}} \right)$, etc.
Whenever the term reducing agent comes to mind, the first name that comes into the mind is Lithium aluminium hydride as it is one of the strongest reducing agents.

Hence, the correct answer is option C. $LiAl{H_4}$

$\left( {LiAl{H_4}} \right)$ is one of the strongest reducing agents. It will donate hydride to any $C = O$ containing a functional group. $LiAl{H_4}$ is a strong reducing agent. It will reduce almost any $C = O$ containing functional group to an alcohol.
The higher the electronegativity of the atom the less electron density will be on the hydrides, the less electron density of the hydrides, the less able they are to act as nucleophiles to reduce the carbonyl. These are the two factors combined to make $LiAl{H_4}$ a stronger reducing agent.

Note - $LiAl{H_4}$ reduces amides to amines. Mechanism depends slightly on whether amide has an $N - H$ or not. Sodium borohydride $\left( {NaB{H_4}} \right)$ being a mild reducing agent, is only having the ability of reducing aldehydes and ketones.