Question

What is the increasing order of equivalent weights of the following oxidising agents?
(i) $KMn{O}_4⟶M{n}^{2+}$
(ii) $K_{2}C{r}_2{O}_7⟶C{r}^{3+}$
(iii) $KCl{O}_3⟶C{l}^{-}$
[Atomic wt.:
$K=39,Cr=52,Mn=55,Cl=35.5,O=16$]

• A. $KCl{O}_3<KMn{O}_4<K_{2}C{r}_2{O}_7$
• B. $K_{2}C{r}_2{O}_7<KMn{O}_4<KCl{O}_3$
• C. $KMn{O}_4<K_{2}C{r}_2{O}_7<KCl{O}_3$
• D. $KCl{O}_3<K_{2}C{r}_2{O}_7<KMn{O}_4$

Answer 1

The correct option is A $KCl{O}_3<KMn{O}_4<K_{2}C{r}_2{O}_7$

$\left(i\right)K\stackrel{7+}{M}n{O}_4⟶M{n}^{2+}$ $\Rightarrow E_1=\frac{39+55+64}{5}=31.6$
$\left(ii\right)K_2\stackrel{6+}{C{r}_2}{O}_7⟶C{r}^{3+}$ $\Rightarrow E_2=\frac{39\times 2+52\times 2+16\times 7}{6}=49$
$\left(iii\right)K\stackrel{5+}{C}l{O}_3⟶C{l}^{-}$
$\Rightarrow E_3=\frac{39+35.5+48}{6}=20.4$