Question

Match List-I with change in oxidation state given in List-II.

Answer 1

In $B{i}_2{S}_3$:
change in oxidation state of $S$ $=0-(-2)=2$ and
change in oxidation state of $Bi=(+5)-(+3)=2$ so
Eq. wt of $B{i}_2{S}_3=\frac{M}{6+2\times 2=10}$
In $A{l}_2(C{r}_2{O}_7{)}_3$:
change in oxidation state of $Cr$ $=6\times +6-\left(3\right)=18$
Eq. wt of $A{l}_2(C{r}_2{O}_7{)}_3=\frac{M}{6\times 3=18}$
In $Fe{S}_2$:
change in oxidation state of $S$ $=4-(-1)=5$ and
change in oxidation state of $Fe=(+3)-(+2)=1$.
So, eq. $Fe{S}_2=\frac{M}{1+2\ast 5=11}$
In $Mn(N{O}_3{)}_2$:
change in oxidation state of S $=6-(+2)=4$ and
change in oxidation state of $N=(+2)-(+5)=-3$ so
Eq. wt of $Mn(N{O}_3{)}_2=\frac{M}{|4+2\ast -3)|=2}$