A) \[Li\]
B) \[K\]
C) \[Na\]
D) \[Rb\]
Correct Answer: D
Solution :
\[2Rb+2{{H}_{2}}O\,\to \,2RbOH+{{H}_{2}}\] \[\xrightarrow{Li<Na<K<Rb<Cs}\] As we go down the group reactivity with \[{{H}_{2}}O\] increases.
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