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(C) According to Faraday's law of electrolysis,the mass deposited $(W)$ is given by $W = \frac{M 	imes Q}{n 	imes F}$,where $M$ is the atomic mass,$

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$1, 3, 2$

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(C) According to Faraday's law of electrolysis,the mass deposited $(W)$ is given by $W = \frac{M 	imes Q}{n 	imes F}$,where $M$ is the atomic mass,$Q$ is the quantity of electricity,$n$ is the valency,and $F$ is Faraday's constant.Since $Q$ and $F$ are constant,$W \propto \frac{M}{n}$,which implies $n \propto \frac{M}{W}$.For element $X$: $n_X \propto \frac{7}{2.1} = 3.33$.For element $Y$: $n_Y \propto \frac{27}{2.7} = 10$.For element $Z$: $n_Z \propto \frac{48}{7.2} = 6.66$.The ratio of valencies $n_X : n_Y : n_Z = 3.33 : 10 : 6.66$.Dividing by $3.33$,we get the ratio $1 : 3 : 2$.

Column $I$ (Reduction process)	Column $II$ (Charge required)
$(a)$ $1$ mol of $MnO_4^-$ to $Mn^{2+}$	$(p)$ $193000$ $C$
$(b)$ $1$ mole of $Cr_2O_7^{2-}$ to $Cr^{3+}$	$(q)$ $289500$ $C$
$(c)$ $1$ mole of $Sn^{4+}$ to $Sn^{2+}$	$(r)$ $482500$ $C$
$(d)$ $1$ mole of $Al^{3+}$ to $Al$	$(s)$ $579000$ $C$

The same quantity of electricity was passed through solutions of salts of elements $X$,$Y$,and $Z$ with atomic masses $7$,$27$,and $48$ respectively. The masses of $X$,$Y$,and $Z$ deposited were $2.1 \ g$,$2.7 \ g$,and $7.2 \ g$ respectively. The valencies of $X$,$Y$,and $Z$ respectively are:

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