Animated Solution for Chemistry - s and p-Block Elements: The reaction of zinc with dilute and concentrated nitric acid, respectively, produce

When we think of acids reacting with metals, our minds immediately jump to the classic production of hydrogen gas. However, nitric acid ($ext{HNO}_3$) is a rebel. It is a remarkably strong oxidizing agent. When a metal like zinc is dropped into nitric acid, the acid doesn't just sit back and let hydrogen escape; it actively oxidizes the hydrogen into water ($ext{H}_{2}extO$) while getting reduced itself.

The fascinating part of this chemistry is that the reduction product of nitric acid is not fixed. It is highly sensitive to two main factors: the concentration of the acid and the reactivity of the metal.

Pushing the Limits

Let's first examine what happens when zinc reacts with dilute nitric acid. Zinc is a fairly reactive metal (it sits comfortably above hydrogen in the electrochemical series). Because the acid is dilute, there are fewer $ext{HNO}_3$ molecules available per zinc atom. The highly reactive zinc forces the nitrogen in nitric acid, which starts at a $+5$ oxidation state, to undergo a deep reduction all the way down to a $+1$ oxidation state.

The resulting gas is nitrous oxide ($ext{N}_{2}extO$), famously known as laughing gas. The balanced chemical equation for this vigorous reaction is:

A Shallow Reduction

Now, imagine pouring concentrated nitric acid over zinc granules. The environment is now swarming with $ext{HNO}_3$ molecules. The oxidizing power of the solution is immense. Because there is such an overwhelming excess of the oxidizing agent, the reduction of nitrogen doesn't go as deep. It only drops from $+5$ to $+4$.

The product here is nitrogen dioxide ($ext{NO}_2$), a highly toxic gas characterized by its dense, reddish-brown fumes. The balanced equation is:

The question specifically asks for the products formed with dilute and concentrated nitric acid, respectively. Based on our chemical analysis, dilute acid yields $ext{N}_{2}extO$ and concentrated acid yields $ext{NO}_2$. Therefore, the correct sequence is $ext{N}_{2}extO$ and $ext{NO}_2$, which perfectly matches option (d).