Question

Which of the following orders are correct against the stated property?
I) NaO$_2 < KO$_2 < RbO$_2 < CsO$_2$ — stability
II) Mg(OH)$_2 < Ca(OH)$_2 < Sr(OH)$_2$ — basic strength
III) MgCO$_3 < CaCO$_3 < SrCO$_3$ — thermal stability

• A. I & III only
• B. II & III only
• C. I & II only
• D. I, II & III

Hint:

For group trends in alkali and alkaline earth metals, remember: thermal stability and basicity increase down the group, and larger cations stabilize larger anions better.

Answer 1

The Correct Option is D

I) Stability of superoxides increases down the group due to increase in cation size which stabilizes the larger anion O$_2^-$ better. Hence, NaO$_2 < KO$_2 < RbO$_2 < CsO$_2$ is correct.
II) Basic strength of group 2 hydroxides increases down the group due to increase in ionic character and solubility: Mg(OH)$_2 < Ca(OH)$_2 < Sr(OH)$_2$.
III) Thermal stability of carbonates also increases down the group due to increased lattice energy stability: MgCO$_3 < CaCO$_3 < SrCO$_3$.