The oxide of a metal (R) can be reduced by th | vedclass

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Question

(B) metal with higher reactivity can reduce the oxide of a metal with lower reactivity.Given that metal $(P)$ can reduce the oxide of $(R)$,it implies

Vedclass · Accepted Answer

$P > R > Q$

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(B) metal with higher reactivity can reduce the oxide of a metal with lower reactivity.Given that metal $(P)$ can reduce the oxide of $(R)$,it implies that $(P)$ is more reactive than $(R)$,i.e.,$P > R$.Given that metal $(R)$ can reduce the oxide of $(Q)$,it implies that $(R)$ is more reactive than $(Q)$,i.e.,$R > Q$.Combining these two relationships,we get the order of reactivity as $P > R > Q$.

The oxide of a metal $(R)$ can be reduced by the metal $(P)$,and metal $(R)$ can reduce the oxide of metal $(Q)$. Then the decreasing order of the reactivity of metals $(P)$,$(Q)$,and $(R)$ with oxygen is:

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