The decomposition temperature of BaCO₃ (A), MgCO₃ (B) and CaCO₃ (C) follow the order

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B > C > A

C > B > A

C > A > B

A > C > B

answer is B.

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Detailed Solution

$\large \xrightarrow{{\;\;BeC{O_3}\; < \;MgC{O_3}\; < \;CaC{O_3}\; < \;SrC{O_3}\; < \;BaC{O_3}}}\;:\;Thermal\;stability\;order$
As size of metal increases stability of carbonate increases
Since a larger size cation will stabilise the large size anion $\large \left [ CO^{-2}_3 \right ]$
$\large Thermal\;stability\; \propto \;\frac{1}{{Decomposition}}$
Therefore, greater the stability greater will be decomposition temperature
Decomposition temperature order: $\large \;BeC{O_3}\; < \;\mathop {MgC{O_3}}\limits_B \; < \;\mathop {CaC{O_3}}\limits_c \; < \;SrC{O_3}\; < \;\mathop {BaC{O_3}}\limits_A$