The correct sequence of thermal stability of the following carbonates is
(a) BaCO3

Q: Why does thermal stability increase down the group for carbonates, but decrease for hydrides?

It's all about size matching! Carbonate is a large polyatomic anion, so it forms a more stable lattice with large cations (like Barium). Hydride is a tiny anion, so it forms a more stable lattice with small cations (like Magnesium).

Q: What exactly happens during the thermal decomposition of these carbonates?

When heated sufficiently, the metal carbonate breaks down into a solid metal oxide and releases carbon dioxide gas. For example: $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$.

Q: Does this same trend apply to Group 1 (alkali metal) carbonates?

Yes! Group 1 carbonates also become more thermally stable as you move down the group. However, because Group 1 cations have only a $+1$ charge, they have lower polarizing power than Group 2 cations, making Group 1 carbonates generally much more stable to heat overall.

Q: How exactly does Fajan's rule explain the weakening of the bond?

A small cation with high polarizing power pulls the electron density of the oxygen atom in the carbonate ion towards itself. This shifts electron density away from the C-O bond, weakening it and making it easier to break when thermal energy is applied.

Q: Is lattice energy or Fajan's rule the better explanation here?

Both are valid and lead to the exact same conclusion! Fajan's rule focuses on the covalent character and bond weakening, while lattice energy focuses on the thermodynamic stability of packing large ions together. You can use whichever concept you find more intuitive.

Question

The correct sequence of thermal stability of the following carbonates is

Options:

(A) $	ext{BaCO}_3 < 	ext{CaCO}_3 < 	ext{SrCO}_3 < 	ext{MgCO}_3$

(B) $	ext{MgCO}_3 < 	ext{CaCO}_3 < 	ext{SrCO}_3 < 	ext{BaCO}_3$

(C) $	ext{MgCO}_3 < 	ext{SrCO}_3 < 	ext{CaCO}_3 < 	ext{BaCO}_3$

(D) $	ext{BaCO}_3 < 	ext{SrCO}_3 < 	ext{CaCO}_3 < 	ext{MgCO}_3$

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Heat and Carbonates

Imagine you are holding a piece of chalk (calcium carbonate) and a piece of magnesite (magnesium carbonate). If you were to throw both into a blazing furnace, which one would break down first? This is the essence of thermal stability. In this problem, we are asked to rank the thermal stability of Group 2 (alkaline earth metal) carbonates: $	ext{MgCO}_3$, $	ext{CaCO}_3$, $	ext{SrCO}_3$, and $	ext{BaCO}_3$.

When a carbonate decomposes, it breaks apart into a metal oxide and carbon dioxide gas:

$$ 	ext{MCO}_3 \xrightarrow{\Delta} 	ext{MO} + 	ext{CO}_2 $$

The question is: how tightly does the metal hold onto the carbonate ion before the heat rips them apart?

### The Bully Cation: Fajan's Rule in Action

To understand this, we need to look at the players involved. The carbonate ion, $	ext{CO}_3^{2-}$, is a large, fluffy polyatomic anion. Its electron cloud is spread out over three oxygen atoms.

Now, let's look at the metal cations. As we move down Group 2 from Magnesium to Barium, the number of electron shells increases. This means the ionic radius grows:

$$ 	ext{Mg}^{2+} < 	ext{Ca}^{2+} < 	ext{Sr}^{2+} < 	ext{Ba}^{2+} $$

Here is where **Fajan's Rule** comes into play. Fajan's rule tells us that a small, highly charged cation is like a tiny bully. It has a very high charge density (polarizing power). When the tiny $	ext{Mg}^{2+}$ ion gets close to the large, fluffy $	ext{CO}_3^{2-}$ ion, it aggressively pulls the carbonate's electron cloud towards itself.

This severe distortion weakens the internal carbon-oxygen bonds within the carbonate ion. Because the bonds are already strained by the magnesium bully, it takes very little thermal energy (heat) to snap the bond and release $	ext{CO}_2$. Therefore, $	ext{MgCO}_3$ is highly unstable to heat.

### The Gentle Giant: Barium's Approach

On the other end of the spectrum, we have the Barium ion, $	ext{Ba}^{2+}$. Barium is a massive cation. Because its $+2$ charge is spread over a much larger volume, its polarizing power is very low.

When $	ext{Ba}^{2+}$ sits next to the $	ext{CO}_3^{2-}$ ion, it doesn't distort the electron cloud much at all. The carbon-oxygen bonds remain strong and intact. As a result, you have to pump a massive amount of heat into $	ext{BaCO}_3$ to force it to decompose.

### The Perfect Match: Lattice Energy

We can also view this through the lens of **Lattice Energy**. A fundamental rule of solid-state chemistry is that *large cations stabilize large anions*, and *small cations stabilize small anions*.

Since the carbonate ion is a large anion, it forms a much more stable, tightly packed crystal lattice with a large cation like $	ext{Ba}^{2+}$ than it does with a small cation like $	ext{Mg}^{2+}$. This high lattice stability directly translates to high thermal stability.

### Final Conclusion

Putting it all together, as we move down the group, the cation size increases, polarizing power decreases, and the stability of the lattice with the large carbonate ion increases.

Thus, the thermal stability strictly increases down the group:

$$ 	ext{MgCO}_3 < 	ext{CaCO}_3 < 	ext{SrCO}_3 < 	ext{BaCO}_3 $$

This perfectly matches option (b).

Animated Solution for Chemistry - s and p-Block Elements: The correct sequence of thermal stability of the following carbonates is

Select Answer:

Visualized Solution

\text{Thermal Stability of Carbonates}

\text{Size of Cations}

\text{Polarizing Power}

\text{Distortion of } \text{CO}_3^{2-}

\text{Thermal Decomposition}

\text{Correct Sequence}

\text{Lattice Energy Perspective}

The Sigma Insight: Alkaline Metals

The Setup

The Bully Cation

The Gentle Giant

The Perfect Match

Final Conclusion

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