The oxide of 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 metal $P$ , $Q$ and $R$ with oxygen is:

P > Q > R

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P > R > Q

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R > P > Q

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Q > P > R

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Solution

The correct option is C $P > R > Q$
Metal $P$ can reduce metal $R$ . Hence, the reactivity of metal $P$ towards oxygen is higher than the reactivity of metal $R$ towards oxygen.
Similarly, metal $R$ can reduce metal $Q$ . Hence, the reactivity of metal $R$ towards oxygen is higher than the reactivity of metal $Q$ towards oxygen.
Thus, the decreasing order of the reactivity of metal, $P, Q$ and $R$ with oxygen is $P > R > Q$ .

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The oxide of 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 metal P, Q and R with oxygen is:

The oxide of 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 metal $P$ , $Q$ and $R$ with oxygen is: