Chemistry S & P Block Elements SZABMU 2024
PMDC Verified Question 7 of 104
The correct stability order of \( \text{M}^{+4} \) cations is (Out of syllabus)
A
\( \text{Ge}^{+4} < \text{Pb}^{+4} < \text{Sn}^{+4} \)
B
\( \text{Ge}^{+4} > \text{Pb}^{+4} > \text{Sn}^{+4} \)
C
\( \text{Ge}^{+4} < \text{Sn}^{+4} < \text{Pb}^{+4} \)
D
\( \text{Ge}^{+4} > \text{Sn}^{+4} > \text{Pb}^{+4} \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option D: \( \text{Ge}^{+4} > \text{Sn}^{+4} > \text{Pb}^{+4} \)
Concept:

In the heavier p-block elements (like Group IVA), the "inert pair effect" causes the lower oxidation states (e.g., +2) to become more stable than the higher oxidation states (e.g., +4) as you move down the group.

Solution:

  • The elements in Group IVA, going down, are Germanium (Ge), Tin (Sn), and Lead (Pb).
  • As we move from Ge to Pb, the ns² valence electrons (the 4s, 5s, 6s electrons) become increasingly resistant to participating in chemical bonding. This is the inert pair effect, primarily caused by the poor shielding of intervening d and f orbitals.
  • Because the s-electrons refuse to ionize in heavy elements like Lead, the +4 oxidation state becomes extremely unstable (Pb+4 is a vicious oxidizing agent that wants to revert to Pb+2).
  • Conversely, Germanium (higher up) comfortably forms stable +4 compounds.
  • Therefore, the stability of the +4 state strictly decreases down the group: \( \text{Ge}^{+4} > \text{Sn}^{+4} > \text{Pb}^{+4} \).


Why other options are incorrect:

  • Options A, B, and C provide incorrect trends that contradict the inert pair effect. Pb+4 is the least stable, not the most or intermediate.

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