Question

$25\text{ml}$ of a $0.1\text{M}$ solution of a stable cation of transition metal $Z$ reacts exactly with $25\text{ml}$ of $0.04\text{M}$ acidified $KMn{O}_4$ solution. Which of the following is most likely to represent the change in oxidation state of $Z$ correctly ?

• A. $Z\to Z^{2+}$
• B. $Z^{2+}\to Z^{3+}$
• C. $Z^{3+}\to Z^{4+}$
• D. $Z^{2+}\to Z^{4+}$

Answer 1

The correct option is D $Z^{2+}\to Z^{4+}$

In acidic medium manganese change its oxidation state $=+7$
$\stackrel{+7}{M}n{O}_4^{-}\to \stackrel{2+}{Mn}$
n-factor$=7-2=5$

Let $Z$ undergoes change in oxidation number from $n_1\text{to}{n}_2(n_2>n_1)$ as a result of reaction with $KMn{O}_4$.

$\therefore$ Milli equivalent of $Z=$ Milli equivalent of $KMn{O}_4$

$M_1\times V_1\times n_{factor}$ of $Z$ $=M_2\times V_2\times n_{factor}$ of $KMn{O}_4$

On substituting the given values in formula.

$0.1\times 25\times (n_2-n_1)=0.04\times 25\times 5$

$\Rightarrow (n_2-n_1)=\frac{0.04\times 25\times 5}{0.1\times 25}=2$
$\therefore n_2-n_1=2$
Hence, the oxidation number of $Z$ increased by $2$.
As there is a cation present in the solution, then the reaction is
$Z{n}^{2+}\to Z{n}^{4+}$

Correct option is (D).