s-Block Elements

The correct order of decreasing stability of carbonates is:

$ \mathrm{BaCO}_3 > \mathrm{SrCO}_3 > \mathrm{CaCO}_3 > \mathrm{MgCO}_3$

Therefore, the answer is Option B.

Let's understand why this is the case. The stability of carbonates is related to the size of the metal cation. As we move down the group in the periodic table, the size of the metal cation increases. A larger cation has a weaker attraction to the carbonate anion. This weaker attraction makes the carbonate less stable.

Here's a breakdown:

• Ba²⁺ is the largest cation and therefore has the weakest attraction to the carbonate anion, making BaCO₃ the most stable.


• Sr²⁺ is smaller than Ba²⁺, leading to a stronger attraction and less stability compared to BaCO₃.


• Ca²⁺ is smaller than Sr²⁺, making CaCO₃ less stable than SrCO₃.


• Mg²⁺ is the smallest cation, resulting in the strongest attraction and the least stable carbonate, MgCO₃.

The thermal stability of carbonates increases down the group. So, the order of thermal stability of carbonates of alkaline earth metals is

BaCO₃ > CaCO₃ > MgCO₃ > BeCO₃.

Therefore, BaCO₃ is thermodynamically most stable and BeCO₃ is least stable.

Sodium iodide reacts with ammonia to produce tetraamminesodium iodide. The reaction is conducted in liquid ammonia. The reaction is as follows

$\mathop {NaI}\limits_{Sodium\,iodide} + \mathop {4N{H_3}}\limits_{Ammonia} \buildrel {} \over \longrightarrow \mathop {[Na{{(N{H_3})}_4}]I}\limits_{Tetraam{\mathop{\rm mine}\nolimits} - sodium\,iodide} $