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Question #44330

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The Question

An aqueous solution of a Period 6, Group 13 metal nitrate, $\ce{M(NO3)3}$ , is treated with excess aqueous $\ce{KI}$ at $298\text{ K}$ , resulting in the precipitation of a dark, diamagnetic solid $P$ . Solid $P$ is isomorphous with $\ce{CsI3}$ . Separately, when the Period 6, Group 14 metal dioxide, $\ce{M'O2}$ , is treated with excess aqueous $\ce{HI}$ , it yields a yellow precipitate $Q$ and a dark brown solution $R$ . Solid $Q$ adopts a $\ce{CdI2}$ -type layered structure.

Which of the following correctly identifies the structures of $P$ and $Q$ , and the fundamental electronic phenomenon responsible for their formation?

Options

A.
$P = \ce{M^{3+}(I^{-})3}$ , $Q = \ce{M'^{4+}(I^{-})4}$ ; The $6s^2$ electrons are stabilized by the strong shielding of $4f$ and $5d$ orbitals, favoring the highest group oxidation states.
B.
$P = \ce{M^{+}(I3^{-})}$ , $Q = \ce{M'^{2+}(I^{-})2}$ ; The poor shielding of $4f$ and $5d$ orbitals increases $Z_{\text{eff}}$ on the $6s^2$ electrons, making $\ce{M^{3+}}$ and $\ce{M'^{4+}}$ strong oxidizing agents.
C.
$P = \ce{M^{+}(I^{-}) \cdot I2}$ , $Q = \ce{M'^{2+}(I3^{-})2}$ ; The lanthanide contraction decreases the ionic radii, causing steric crowding that prevents the coordination of simple iodide ions.
D.
$P = \ce{M^{3+}(I3^{-})3}$ , $Q = \ce{M'^{4+}(I3^{-})4}$ ; The high polarizability of $\ce{I^{-}}$ induces a partial reduction of the metal centers, stabilizing the polyiodide complexes via back-bonding.

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

Screening Effect in Inner Shells|Effective Nuclear Charge (Zeff)|Inert Pair Effect

Why This Question Is Tricky

Common Mistakes

The question tests the 'Inert Pair Effect' and its chemical consequences on the redox stability of heavier p-block elements.

Students often fall into the trap of assuming that Group 13 and 14 elements will form stable +3 and +4 iodides (like $\ce{AlI3}$ or $\ce{SnI4}$ ).

However, the hidden variable is the relativistic contraction and poor shielding of $4f/5d$ orbitals, which drastically increases $Z_{\text{eff}}$ on the $6s$ electrons.

This makes $\ce{Tl^{3+}}$ and $\ce{Pb^{4+}}$ powerful oxidizing agents that cannot coexist with reducing anions like $\ce{I-}$ .

The trap is expecting $\ce{TlI3}$ to be a simple $\ce{Tl^{3+}}$ salt, whereas it is actually a $\ce{Tl+}$ polyiodide.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Classification of Elements and Periodicity in Properties - Class 11 Part 1
Concept	Screening Effect in Inner Shells
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free