Arrange the following alkali metal carbonates in increasing order of their thermal stability:
$L i_{2} C O_{3}, R b_{2} C O_{3}, C s_{2} C O_{3}, N a_{2} C O_{3}, K_{2} C O_{3}$

L i_{2} C O_{3} < N a_{2} C O_{3} < K_{2} C O_{3} < R b_{2} C O_{3} < C s_{2} C O_{3}

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C s_{2} C O_{3} < N a_{2} C O_{3} < K_{2} C O_{3} < R b_{2} C O_{3} < L i_{2} C O_{3}

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L i_{2} C O_{3} < K_{2} C O_{3} < N a_{2} C O_{3} < R b_{2} C O_{3} < C s_{2} C O_{3}

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None of these

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Solution

The correct option is A $L i_{2} C O_{3} < N a_{2} C O_{3} < K_{2} C O_{3} < R b_{2} C O_{3} < C s_{2} C O_{3}$
Thermal stability is a stability of a molecule at high temperature. A molecule with more stability has more resistance to decomposition at high temperatures.
As we move from top to bottom in a group, the thermal stability increases.
Thermal stability decreases from Li to Cs.
So, the correct order of thermal stability will be-
$L i_{2} C O_{3} < N a_{2} C O_{3} < K_{2} C O_{3} < R b_{2} C O_{3} < C s_{2} C O_{3}$ .
Hence, option $(A)$ is correct.

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