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Consider conversion of ${MnCl}_{2}$ into
(I) ${MnCl}_{3}$ (II) ${MnO}_{2}$ (III) ${MnO}_{4}^{2 -}$ (IV) ${MnO}_{4}^{-}$
Increasing order of equivalent weights of ${MnCl}_{2}$ in these conversions is

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I < II < III < IV

IV = III < II = I

IV < III < II < I

equivalent weight is independent of change in valency during a process

answer is C.

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Detailed Solution

(3)

Process	Change in Ox.number	Equivalent of ${MnCl}_{2}$
MnCl₂ $\to$ MnCl₃	1	M/1
MnCl₂ $\to$ MnO₂	2	M/2
MnCl₂ $\to$ ${MnO}_{4}^{- 2}$	4	M/4
MnCl₂ $\to$ ${MnO}_{4}^{-}$	4	M/5

Thus, M5 <M/4 <M/2 < M/1

IV < III < II < I

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