Question

Name the carbonate which is not decomposed by heating?
(A) Sodium carbonate
(B) Potassium carbonate
(C) Magnesium carbonate
(D) None of these

Answer 1

We know that metal carbonates are the type of bases which are formed when a metal reacts with a carbonate. The stability of the metal carbonates is the same as the reactivity series of metals. So, we can say that the stability of metal carbonates increases as we move upwards in the reactivity series of metals.

Complete answer: As we know that carbonates of alkaline earth metal i.e. group $(II)$ on heating easily decomposes into carbon dioxide and metal oxide so, as here in magnesium carbonate as Magnesium is present which belongs to group $(II)$ will also decompose when heat will be passed into it.
Hence, we write the reaction as: -
$MgC{O_3} + \Delta \to MgO + C{O_2}$
Thus, it will form its respective metal oxide i.e. magnesium oxide and carbon dioxide will be evolved.
Also $MgC{O_3}$ is a salt of weak acid and a weak base so it can easily be decomposed on heating.
On the other hand we see that as Sodium and Potassium belongs to alkali metals or group $(I)$ metals so the carbonates formed with the alkali group metals will not decompose on heating as they are thermally stable except $L{i_2}C{O_3}$ which is thermally unstable.
Sodium and potassium both are very reactive metals so the carbonates of both the metals are very stable and require a high temperature to decompose into their respective metal oxides and carbon dioxide. In the process of formation of $N{a_2}C{O_3}$ and ${K_2}C{O_3}$ , strong bases i.e. $NaOH$ and $KOH$ are used which have the capability to hold carbon dioxide.

So here we are having two correct options: - option A and option B.

Note: $L{i_2}C{O_3}$ does not decomposes as we know that the size of the alkali metals increases as we move down the group so $Li$ is the smallest in group $(I)A$ . Thus due to the small size and high charge density of $Li +$ ion , it has a strong polarizing power so it will not be able to stabilize the polarization of $C{O_3}^{2 - }$ ion and will therefore make $L{i_2}C{O_3}$ unstable.