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Question #44680

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The Question

An unknown solid element $M$ reacts with oxygen at $298\text{ K}$ to form a stable oxide $X$ . Quantitative analysis reveals that $10.45\text{ g}$ of $M$ combines exactly with $1.20\text{ g}$ of oxygen to form $X$ . A theoretical higher oxide $Y$ of element $M$ is proposed, which obeys the Law of Multiple Proportions. The ratio of the mass of oxygen combining with $1\text{ g}$ of $M$ in oxide $Y$ to that in oxide $X$ is exactly $5:3$ .

Given that $M$ is a p-block element, which of the following statements correctly identifies the acid-base nature of oxide $X$ and the thermodynamic reality of oxide $Y$ at $298\text{ K}$ ?

Options

A.
$X$ is amphoteric; $Y$ is a stable covalent network solid because higher oxidation states strongly polarize the oxide ions according to Fajans' rules.
B.
$X$ is basic; $Y$ is highly unstable and acts as a powerful oxidizing agent due to the relativistic contraction of the valence $s$ -orbital and poor shielding by inner $f$ and $d$ electrons.
C.
$X$ is acidic; $Y$ is a stable ionic solid because the Law of Multiple Proportions dictates the existence of discrete $M^{5+}$ and $O^{2-}$ ions in a crystal lattice.
D.
$X$ is basic; $Y$ is unstable primarily because element $M$ lacks accessible $d$ -orbitals to expand its valency to $+5$ .

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Law of Multiple Proportions|Acid-Base Nature of Oxides|Screening Effect in Inner Shells

Why This Question Is Tricky

Common Mistakes

The question tests the intersection of stoichiometry (Law of Multiple Proportions) and advanced periodic trends (Inert Pair Effect).

Students often fall into the trap of assuming that if a higher oxide's empirical formula can be mathematically derived using the Law of Multiple Proportions, the compound must be a stable entity governed by standard rules like Fajans' rules (which would incorrectly suggest a stable covalent network for a +5 state).

The 'hidden variable' is the relativistic contraction of the 6s orbital in heavy Period 6 elements (Bismuth), which makes the +5 oxidation state thermodynamically unstable and highly oxidizing, overriding simple stoichiometric predictions.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Some Basic Concepts of Chemistry - Class 11 Part 1
Concept	Law of Multiple Proportions
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free