Question

The following compounds have been arranged in order of their increasing thermal stabilities. Identify the correct order :
$K_{2}C{O}_3\left(I\right)MgC{O}_3\left(II\right)$
$CaC{O}_3\left(III\right)BeC{O}_3\left(lY\right)$

• A. I < II < III < IV
• B. IV < II < III < I
• C. IV < II < I< III
• D. II < IV < III < I

Answer 1

The correct option is B IV < II < III < I

If metal carbonate is heated, the carbon dioxide breaks free to leave the metal oxide.

How much we need to heat the carbonate before that happens depends on how polarised the ion was. If it is highly polarised, we need less heat than if it is only slightly polarised.

The smaller the positive ion is, the higher the charge density, and the greater effect it will have on the carbonate ion. As the positive ions get bigger as you go down the Group, they have less effect on the carbonate ions near them. To compensate for that, we have to heat the compound more in order to persuade the carbon dioxide to break free and leave the metal oxide.

In other words, as we go down the Group, the carbonates become more thermally stable.

Hence, option B is correct