Chemistry S & P Block Elements NUMS 2024
PMDC Verified Question 21 of 104
When a metal carbonate is heated at 100°C, which of the following compound will readily decompose?
A
\( \text{BaCO}_3 \)
B
\( \text{BeCO}_3 \)
C
\( \text{MgCO}_3 \)
D
\( \text{SrCO}_3 \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option B: \( \text{BeCO}_3 \)
Concept:

The thermal stability of Group II-A (alkaline earth metal) carbonates increases as you move down the group due to decreasing polarizing power of the metal cation.

Solution:

  • Beryllium (Be) is located at the very top of Group 2.
  • The \( \text{Be}^{2+} \) cation has an exceptionally small ionic radius and a high +2 charge, giving it a massive charge density (polarizing power).
  • When bonded to a large polyatomic carbonate ion (\( \text{CO}_3^{2-} \)), the tiny \( \text{Be}^{2+} \) strongly distorts (polarizes) the carbonate's electron cloud.
  • This distortion heavily weakens the internal C-O bonds of the carbonate ion.
  • Because of this severe instability, Beryllium carbonate (\( \text{BeCO}_3 \)) decomposes incredibly readily into BeO and \( \text{CO}_2 \), even at relatively low temperatures like 100°C (in fact, it is difficult to keep stable even at room temperature in a dry atmosphere).


Why other options are incorrect:

  • As we move down the group (Mg, Sr, Ba), the metal cations become progressively larger. Their polarizing power drops, meaning they distort the carbonate ion much less. Thus, they require much higher temperatures to decompose.

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