Animated Solution for Chemistry - d and f-Block Elements: Which of the following combination will produce ${\text{H}}_2$ gas?

Welcome to a fascinating exploration of inorganic chemistry! Today, we are embarking on a quest to find which combination of metal and reagent will successfully produce hydrogen gas. This isn't just a simple memory test; it's a beautiful journey through the diverse chemical personalities of d-block elements. Let's dive in and analyze each option meticulously.

Our mission is to identify the reaction that yields ${\text{H}}_2$ gas. To do this, we must understand the unique properties of each metal involved—Zinc, Gold, Copper, and Iron—and how they interact with their respective reagents. Some metals are amphoteric, some are noble, and some fall victim to the overwhelming oxidizing power of certain acids.

The Amphoteric Nature of Zinc

Let's begin with Zinc ($\text{Zn}$) reacting with aqueous Sodium Hydroxide ($\text{NaOH}$). Zinc is a fascinating element because it is amphoteric. This means it has the remarkable ability to react with both strong acids and strong bases.

When Zinc is dropped into a strong base like $\text{NaOH}$, it doesn't just sit there. It actively displaces hydrogen from the hydroxide. The reaction proceeds as follows:

Here, a soluble complex called Sodium Zincate (${\text{Na}}_2{\text{ZnO}}_2$) is formed, and glorious bubbles of hydrogen gas are released. So, right off the bat, we have found our hydrogen producer! But as rigorous chemists, we must investigate the other options to understand why they fail to produce ${\text{H}}_2$.

The Macarthur-Forrest Process

Next, we look at Gold ($\text{Au}$) reacting with Sodium Cyanide ($\text{NaCN}$) in the presence of air. Gold is a noble metal, notoriously unreactive. However, it can be coaxed into reacting through the famous Macarthur-Forrest cyanide process, which is extensively used in gold metallurgy.

In this process, the oxygen from the air acts as a crucial oxidizing agent. The reaction is:

Notice what happens here. The gold is oxidized to form a soluble complex, Sodium dicyanoaurate(I). The oxygen is reduced to hydroxide ions. There is absolutely no hydrogen gas evolved in this elegant extraction process.

The Oxidizing Power of Nitric Acid

Moving on to Copper ($\text{Cu}$) and concentrated Nitric Acid (${\text{HNO}}_3$). Copper sits below hydrogen in the electrochemical reactivity series. This fundamental positioning means Copper simply lacks the driving force to displace hydrogen from any acid.

Furthermore, concentrated ${\text{HNO}}_3$ is not your average acid; it is a powerful oxidizing agent. Instead of releasing hydrogen, the nitric acid attacks the copper, oxidizing it while reducing itself. The reaction unfolds as:

The result is Copper(II) nitrate, water, and the evolution of Nitrogen Dioxide (${\text{NO}}_2$), a toxic, reddish-brown gas. Once again, no hydrogen is produced.

The Phenomenon of Passivation

Finally, we examine Iron ($\text{Fe}$) with concentrated Nitric Acid. Iron is an active metal and easily displaces hydrogen from dilute, non-oxidizing acids like $\text{HCl}$ or dilute ${\text{H}}_2{\text{SO}}_4$. However, concentrated ${\text{HNO}}_3$ changes the rules of the game entirely.

Because concentrated ${\text{HNO}}_3$ is such a strong oxidizing agent, it rapidly oxidizes the surface of the iron, forming a microscopic, invisible, and highly protective layer of Iron(III) oxide (${\text{Fe}}_2{\text{O}}_3$).

This phenomenon is known as passivation. The oxide layer acts as an impenetrable shield, stopping any further reaction dead in its tracks. While a tiny amount of ${\text{NO}}_2$ gas might evolve in the first fraction of a second, the reaction halts completely, and no hydrogen gas is ever produced.

After a thorough chemical interrogation of all four options, the truth is undeniable. Only the amphoteric Zinc possesses the specific chemical nature required to liberate hydrogen gas from a strong base like Sodium Hydroxide. The correct answer is unequivocally Option A.