A polyvalent metal weighing 0.1 g and having atomic weight of 51g reacted with dilute `H_2 SO_4`, to give 43.9 mL of hydrogen at STP The solution containing the metal in this lower oxidation state, was found to require 58.8 mL of 0.1 N `KMnO_4` solution for complete oxidation. What are oxidation state of metal?

Options

A. 3 and 5
B. 3 and 3
C. 5 and 6
D. 2 and 5

Answer

D

Explanation

The reaction involved is Metal `(M)+H_2 SO_4 to H_2`
In the reaction `M-xe^(-) to M^(x+) ,x=2 `or oxidation state is +2. Now `M^(2+)` will acquire higher oxidation state when oxidized by `KMnO_4`
` M^(2+) - ne^(-) to M^(2 +n)`
Now, milliequiv. of `M^(2+)` = milliequiv. of `KMnO_4 ,(0.1)/(50//n) xx 1000 = 0.1 xx 60implies n =2`
So oxidation state of M is `M^(2+n) -= M^(+5 )s`. Hence, the possible valencies of the metal arc +2 and +5.