Question

# Equivalent mass of $KMn{{O}_{4}}$ in acidic, basic and neutral are in the ratio of:(a)- 3:5:15(b)- 5:3:1(c)- 5:1:3(d)- 3:15:5

Hint: $KMn{{O}_{4}}$ is a strong oxidizing agent. The Mn in $KMn{{O}_{4}}$ exists in +7 state. In acidic medium, it gets reduced to $M{{n}^{2+}}$ while in basic it gets reduced to $MnO_{4}^{2-}$. In a neutral medium it gets reduced to $Mn{{O}_{2}}$.

All elements combine with each other according to the laws of chemical combination, and the number of parts by which an element combines with 1 part by mass of hydrogen, or 8 parts by mass of oxygen, or 35.5 parts by mass of chlorine, or one gram equivalent of any other element, is the value of the equivalent mass of the element.
Equivalent mass is the molecular mass divided by the number of electrons gained.
The molecular mass of $KMn{{O}_{4}}$ is 158.04g.
In acidic medium
$MnO_{4}^{-}+8{{H}^{+}}+5{{e}^{-}}\to M{{n}^{2+}}+4{{H}_{2}}O$
\begin{align} & Equivalent\text{ }mass=\dfrac{Molecular\,Mass}{Electrons\,gained} \\ & \,\,\,\,\,=\dfrac{158.04}{5} \\ \end{align}
In basic medium
$MnO{{4}^{-}}+{{e}^{-}}\to MnO_{4}^{2-}$
\begin{align} & Equivalent\text{ }mass=\dfrac{Molecular\,Mass}{Electrons\,gained} \\ & \,\,\,\,\,=\dfrac{158.04}{1} \\ \end{align}
In neutral medium
$MnO_{4}^{-}+4{{H}^{+}}+3{{e}^{-}}\to Mn{{O}_{2}}+2{{H}_{2}}O$
\begin{align} & Equivalent\text{ }mass=\dfrac{Molecular\,Mass}{Electrons\,gained} \\ & \,\,\,\,\,=\dfrac{158.04}{3} \\ \end{align}
Ratio of equivalent mass$=\dfrac{158.04}{5}=\dfrac{158.04}{1}=\dfrac{158.04}{3}$
Ratio of equivalent mass = 3: 15: 5

So, the correct option is (d).

Note:
The formula for equivalent mass differs from literature to literature and can be modified according to our application. Few of the other formulas are:
Hydrogen displacement method
$Equivalent~Mass=\dfrac{Mass~of~the~Metal\times 11200}{Volume~of~Hydrogen~Liberated~\left( STP \right)}$
Oxide Method
$Equivalent~Mass=\dfrac{Mass~of~the~Metal\times 8}{Mass~of~Oxygen~in~the~Oxide}$
Chloride Method
$Equivalent~Mass=\dfrac{Mass~of~the~Metal\times 35.5}{Mass~of~Chlorine~in~the~Chloride}$