Question

The correct order of increasing thermal stability of $K_2CO_3, MgCO_3, CaCO_3 $ and $ BeCO_3$ is

• A. $BeCO_3 < MgCO_3 < K_2CO_3 < CaCO_3$
• B. $BeCO_3 < MgCO_3 < CaCO_3 < K_2CO_3$
• C. $MgCO_3 < BeCO_3 < CaCO_3 < K_2CO_3 $
• D. $ K_2CO_3 < MgCO_3 < CaCO_3 < BeCO_3 $

Answer 1

The Correct Option is B

The question requires us to determine the correct order of increasing thermal stability for the given carbonates: K_2CO_3, MgCO_3, CaCO_3, and BeCO_3.

Thermal stability of carbonates increases with the increasing size and decreasing charge density of the metal cation. Carbonates decompose upon heating to form the oxide and carbon dioxide. The stability can be explained by the strength of the M-C bond, where 'M' is the metal in the carbonate:

1. Metal ions with a higher charge density (small ionic radius and higher charge) destabilize the carbonate ion to a greater extent due to strong covalent character in the M-O bond. This makes the carbonate less stable.
2. As we move down the group in the periodic table, the size of the metal cation increases, reducing the charge density and thus increasing the stability of the carbonate.

Now, let's analyze the given compounds:

• BeCO_3: Beryllium has a small ionic radius and a high charge density, making beryllium carbonate less stable. It decomposes to form beryllium oxide and carbon dioxide even at room temperature.
• MgCO_3: Magnesium has a larger ionic radius than beryllium, which results in greater thermal stability compared to beryllium carbonate.
• CaCO_3: Calcium is larger than magnesium, with even lower charge density, making calcium carbonate more thermally stable.
• K_2CO_3: Potassium is an alkali metal with a large ionic radius, resulting in the least charge density. As a result, potassium carbonate is the most thermally stable of the given options.

Based on the analysis, the correct order of increasing thermal stability is:

BeCO_3 < MgCO_3 < CaCO_3 < K_2CO_3

Thus, the correct answer is:

$BeCO_3 < MgCO_3 < CaCO_3 < K_2CO_3$