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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: A) $\text{ P }>\text{ Q }>\text{ R}$B) $\text{ P }>\text{R }>\text{ Q }$C) $\text{ R }>\text{ P }>\text{ Q}$D) $\text{ Q }>\text{ P }>\text{ R}$

Last updated date: 16th Jun 2024
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Hint: The metal oxides have the general formula as a$\text{ M(O}{{\text{)}}_{\text{n}}}\text{ }$. The metal oxides can be reduced by the other metals $\text{ M }\!\!'\!\!\text{ }$. Not every metal can reduce the metal oxide. The metal which has a higher reactivity towards the oxides easily reduces the metal.

Metal oxides are compounds such that one or more oxygen is bonded to the atom. The general formula of the metal oxide is a $\text{ M}{{\text{O}}_{\text{n}}}\text{ }$ , where ‘n’ is the total number of oxygen bonded to the metal.
We know that the reduction is a chemical process in which the compound or the atom gains the electrons. The metal in the metal oxide is having a positive oxidation state. On reduction reaction, the metal in the metal oxide accepts the electron and reduces itself to form a free metal.
The metal oxides are reduced by the metal. The reaction of the reduction of metal oxide in presence of the other metal is as shown below,
$\text{ M(O}{{\text{)}}_{\text{n}}}\text{ + M }\!\!'\!\!\text{ }\to \text{ M }\!\!'\!\!\text{ (O}{{\text{)}}_{\text{n}}}\text{ + M }$

The reduction of metal oxide by the other metal depends on the reactivity of the metal which reduces the metal oxide.
The reactivity series is an arrangement of metals which refers to the arrangement of the metal according to the decreasing order of their reactivities.
We have given the three metals P, Q, and R. It is given that the oxide of R metal is reduced by the metal P.
If we want to reduce the metallic oxide, then we have to use the metal which has higher reactivity than it. Thus, if metal oxide R is reduced by the P, means that the metal P has higher reactivity than the R. The reaction between metal oxide of R and P is given as follows,
$\text{ R(O}{{\text{)}}_{\text{n}}}\text{ + P }\to \text{ P(O}{{\text{)}}_{\text{n}}}\text{ + R }$
Thus, $\text{ P }>\text{ R }$ equation (1)

Here, we are given that the metal oxide of Q is reduced by the metal R. This means that the reactivity of R is higher than the reactivity of metal Thus, metal R can easily displace the metal from the meat oxide and form a metal oxide$\text{ R(O}{{\text{)}}_{\text{n}}}\text{ }$. The reaction is as shown below,
$\text{ Q(O}{{\text{)}}_{\text{n}}}\text{ + R }\to \text{ R(O}{{\text{)}}_{\text{n}}}\text{ + Q }$
Thus, $\text{ R }>\text{ Q }$ equation (2)
If we compare the reaction (1) and reaction (2). We can say that the reactivity of P is higher than the R than the Q.
The order of reactivities towards oxide is,
$\text{ P }>\text{R }>\text{ Q }$
So, the correct answer is “Option B”.

Note: Note that, in reactivity series, the metals which are at the top of the reactivity series are powerful reducing agents. These can be easily oxidised. Therefore, the metals which have low ranking are displaced by the high ranking metals. There are other methods for the reduction of metal oxide, such as reduction of non-metal carbon. The hydrogen acts as a reference to determine the reactivities of metal.