Question:
Identify the carbonate which is thermally very stable.
Identify the carbonate which is thermally very stable.
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Thermal stability of carbonates increases with increase in cation size and decrease in polarizing power.
Updated On: Jun 20, 2026
- Li\(_2\)CO\(_3\)
- Na\(_2\)CO\(_3\)
- BeCO\(_3\)
- CaCO\(_3\)
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The Correct Option is B
Solution and Explanation
Step 1: Understand thermal stability of carbonates.
Thermal stability of metal carbonates depends on the cation size and polarizing power. According to Fajan’s rule, smaller and highly charged cations distort the carbonate ion more, making it less stable. Larger alkali metal cations stabilize carbonate ions better.
Step 2: Trend across periodic table.
Down Group 1, thermal stability increases: \[ Li_2CO_3 < Na_2CO_3 < K_2CO_3 \] This is because polarizing power decreases as cation size increases. Hence decomposition becomes more difficult.
Step 3: Analyze Li\(_2\)CO\(_3\).
Li\(^+\) is very small and highly polarizing. It strongly distorts CO\(_3^{2-}\), making Li\(_2\)CO\(_3\) less stable and easily decomposes on heating.
Step 4: Analyze Na\(_2\)CO\(_3\).
Na\(^+\) is larger than Li\(^+\), so it has lower polarizing power. It stabilizes carbonate ion better, making Na\(_2\)CO\(_3\) thermally more stable than Li\(_2\)CO\(_3\).
Step 5: Analyze BeCO\(_3\) and CaCO\(_3\).
Be\(^{2+}\) has very high charge density and strongly polarizes carbonate, making BeCO\(_3\) highly unstable. CaCO\(_3\) is more stable than BeCO\(_3\) but still less stable than alkali carbonates like Na\(_2\)CO\(_3\).
Step 6: Final comparison and conclusion.
Among the given options, Na\(_2\)CO\(_3\) shows maximum thermal stability because Na\(^+\) has optimal size and low polarizing power compared to others listed.
Final Answer: \[ \boxed{\text{Na}_2\text{CO}_3} \]
Thermal stability of metal carbonates depends on the cation size and polarizing power. According to Fajan’s rule, smaller and highly charged cations distort the carbonate ion more, making it less stable. Larger alkali metal cations stabilize carbonate ions better.
Step 2: Trend across periodic table.
Down Group 1, thermal stability increases: \[ Li_2CO_3 < Na_2CO_3 < K_2CO_3 \] This is because polarizing power decreases as cation size increases. Hence decomposition becomes more difficult.
Step 3: Analyze Li\(_2\)CO\(_3\).
Li\(^+\) is very small and highly polarizing. It strongly distorts CO\(_3^{2-}\), making Li\(_2\)CO\(_3\) less stable and easily decomposes on heating.
Step 4: Analyze Na\(_2\)CO\(_3\).
Na\(^+\) is larger than Li\(^+\), so it has lower polarizing power. It stabilizes carbonate ion better, making Na\(_2\)CO\(_3\) thermally more stable than Li\(_2\)CO\(_3\).
Step 5: Analyze BeCO\(_3\) and CaCO\(_3\).
Be\(^{2+}\) has very high charge density and strongly polarizes carbonate, making BeCO\(_3\) highly unstable. CaCO\(_3\) is more stable than BeCO\(_3\) but still less stable than alkali carbonates like Na\(_2\)CO\(_3\).
Step 6: Final comparison and conclusion.
Among the given options, Na\(_2\)CO\(_3\) shows maximum thermal stability because Na\(^+\) has optimal size and low polarizing power compared to others listed.
Final Answer: \[ \boxed{\text{Na}_2\text{CO}_3} \]
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