A metal on combustion in excess air forms X. X upon hydrolysis with water yields H2O2 and O2 along w

Q: Why doesn't Lithium form a superoxide?

Lithium is a very small cation. It has a high positive charge density, which strongly polarizes the electron cloud of large anions like superoxide or peroxide, making them unstable. It prefers to form a simple oxide ($Li_2O$) where the small $O^{2-}$ anion matches its size, resulting in high lattice energy.

Q: What would be the products if the metal was Sodium?

Sodium reacts with excess oxygen to form Sodium peroxide ($Na_2O_2$). When hydrolyzed, it produces Sodium hydroxide ($NaOH$) and hydrogen peroxide ($H_2O_2$), but no oxygen gas is evolved.

Q: Is Magnesium an alkali metal?

No, Magnesium is an alkaline earth metal (Group 2). It burns in oxygen to form a simple oxide, Magnesium oxide ($MgO$), which upon hydrolysis yields only Magnesium hydroxide ($Mg(OH)_2$).

Q: What exactly is a superoxide ion?

A superoxide ion is a diatomic oxygen anion with a charge of -1, written as $O_2^-$. It has an unpaired electron, making it a radical and paramagnetic.

Q: Why is lattice energy important here?

Lattice energy is the energy released when gaseous ions bind to form a solid crystal. A stable crystal requires a good 'fit' between cations and anions. Large cations (like $Rb^+$) pack efficiently with large anions (like $O_2^-$), maximizing the lattice energy and thus the stability of the compound.

Question

A metal on combustion in excess air forms X. X upon hydrolysis with water yields $H_2O_2$ and $O_2$ along with another product. The metal is

Options:

(A) Li

(B) Mg

(C) Rb

(D) Na

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are in a chemistry lab. You take a piece of an unknown metal, let's call it $M$, and you burn it in a flask filled with an excess of air. The metal reacts vigorously, forming a new compound which we will call $X$.

Next, you take this compound $X$ and drop it into water—a process known as hydrolysis. The reaction bubbles and fizzes, and upon analyzing the products, you find hydrogen peroxide ($H_2O_2$) and oxygen gas ($O_2$), along with another product (likely a metal hydroxide). Our mission is to use these clues to unmask the identity of the mystery metal $M$.

### The Clue in the Bubbles
The most critical piece of evidence lies in the products of the hydrolysis. We didn't just get one product; we got a specific combination: $H_2O_2$ and $O_2$. This is a massive, flashing neon sign in inorganic chemistry!

Let's break down why. If compound $X$ were a normal oxide (containing the $O^{2-}$ ion), reacting it with water would simply yield a metal hydroxide. If $X$ were a peroxide (containing the $O_2^{2-}$ ion), hydrolysis would produce a metal hydroxide and hydrogen peroxide ($H_2O_2$).

However, the simultaneous evolution of **both** hydrogen peroxide and oxygen gas is the signature characteristic of a **superoxide**. A superoxide contains the $O_2^-$ ion. Therefore, we can confidently conclude that compound $X$ is a metal superoxide, with the formula $MO_2$.

### The Lineup of Suspects
Now that we know we are looking for a metal that forms a superoxide when burned in excess oxygen, let's examine our suspects: Lithium (Li), Magnesium (Mg), Rubidium (Rb), and Sodium (Na).

Let's recall the behavior of alkali metals (Group 1) when they react with oxygen:
- **Lithium** is very small and has a high charge density. It forms a normal oxide: $4Li + O_2 \rightarrow 2Li_2O$.
- **Sodium** is slightly larger and forms a peroxide: $2Na + O_2 \rightarrow Na_2O_2$.
- The larger alkali metals—**Potassium (K), Rubidium (Rb), and Cesium (Cs)**—form superoxides: $M + O_2 \rightarrow MO_2$.

Magnesium, being an alkaline earth metal, forms a normal oxide ($MgO$). Looking at our list, Lithium, Sodium, and Magnesium are out. The only metal capable of forming a superoxide is **Rubidium**.

### The Science of Stability
You might be wondering, *why* do only the larger alkali metals form superoxides? It all comes down to a beautiful concept called **lattice energy** and size compatibility.

The superoxide ion ($O_2^-$) is a relatively large, bulky anion. In solid-state chemistry, a crystal lattice is most stable when the sizes of the cation and anion are comparable. A large anion is best stabilized by a large cation.

Rubidium ($Rb^+$) is a large cation, so it can comfortably pack together with the large superoxide ion, creating a highly stable crystal structure. If a small cation like Lithium tried to bond with a superoxide ion, the size mismatch would lead to an unstable lattice. The small Lithium ion prefers the small oxide ion ($O^{2-}$).

### Final Calculation and Conclusion
Let's write down the complete chemical equations to seal the deal. First, Rubidium burns in excess oxygen to form Rubidium superoxide:
$$Rb + O_2 \rightarrow RbO_2$$

Next, the Rubidium superoxide undergoes hydrolysis:
$$2RbO_2 + 2H_2O \rightarrow 2RbOH + H_2O_2 + O_2$$

The products perfectly match the description in the question. The mystery metal is undeniably Rubidium.

Animated Solution for Chemistry - s and p-Block Elements: A metal on combustion in excess air forms X. X upon hydrolysis with water yields $H_{2} O_{2}$ and $O_{2}$ along with another product. The metal is

Select Answer:

Visualized Solution

$Reaction Sequence$

$Identifying Compound X$

$Alkali Metals and Oxygen$

$Lattice Energy Effect$

$Complete Reactions$

$What if it was Sodium?$

The Sigma Insight: Alkali Metals

Analyzing the Setup Imagine you are in a chemistry lab

The Clue in the Bubbles The most critical piece of evidence lies in the products of the hydrolysis

The Lineup of Suspects Now that we know we are looking for a metal that forms a superoxide when burned in excess oxygen, let's examine our suspects

The Science of Stability

Final Calculation and Conclusion Let's write down the complete chemical equations to seal the deal

Similar Questions

On combustion of Li, Na and K in excess of air, the major oxides formed, respectively, are

The main oxides formed on combustion of Li, Na and K in excess of air respectively are

A s-block element (M) reacts with oxygen to form an oxide of the formula $MO_{2}$ . The oxide is pale yellow in colour and paramagnetic. The element (M) is

The metal that forms nitride by reacting directly with $N_{2}$ of air, is

The products obtained on heating $LiNO_{3}$ will be

Reaction of an inorganic sulphite X with dilute $H_{2} SO_{4}$ generates compound Y. Reaction of Y with NaOH gives X. Further, the reaction of X with Y and water affords compound Z. Y and Z respectively, are

Which of the following liberates $O_{2}$ upon hydrolysis?

A hydrated solid $X$ on heating initially gives a monohydrated compound $Y$ . $Y$ upon heating above $373 K$ leads to an anhydrous white powder $Z$ . $X$ and $Z$ , respectively, are

The incorrect statement is

Two elements $A$ and $B$ have similar chemical properties. They don't form solid hydrogencarbonates, but react with nitrogen to form nitrides. $A$ and $B$ , respectively, are

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Reaction Sequence

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Identifying Compound X

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Alkali Metals and Oxygen

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Lattice Energy Effect

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Complete Reactions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What if it was Sodium?