Which one of the following will react most vigorously with water?

All four elements belong to Group 1 (alkali-metal family). Their reaction with water is a redox process:

$$2\,M\;(s) \;+\;2\,H_2O\;(l) \;\rightarrow\;2\,M^+\,(aq)\;+\;2\,OH^-\,(aq)\;+\;H_2\;(g)$$
where $$M$$ = Li, Na, K, Rb, Cs.

For any alkali metal, the overall enthalpy change depends mainly on three energetic factors:

(i) Ionisation enthalpy (IE)
(ii) Hydration enthalpy (ΔHhyd) of the ion $$M^+$$
(iii) Lattice (or cohesive) enthalpy of the metal (much smaller in magnitude)

Down the group:

• The atomic radius increases, so the outermost electron is held less tightly.
• Hence, first ionisation enthalpy falls sharply: $$IE_{Li} \gt IE_{Na} \gt IE_{K} \gt IE_{Rb}$$.
• Hydration enthalpy also decreases because the larger $$M^+$$ ions are less strongly solvated: $$|ΔH_{hyd}|_{Li^+} \gt |ΔH_{hyd}|_{Na^+} \gt |ΔH_{hyd}|_{K^+} \gt |ΔH_{hyd}|_{Rb^+}$$.
• The fall in ionisation enthalpy is numerically larger than the fall in hydration enthalpy, so the net energy released (and thus reactivity) increases down the group.

Therefore the order of vigour with which these metals react with water is:

$$Li \lt Na \lt K \lt Rb \lt Cs$$

Among the given choices, rubidium ($$Rb$$) lies farthest down the group, so it has the lowest ionisation enthalpy and therefore reacts most violently, often igniting the liberated $$H_2$$ immediately.

Option C which is: Rb