The valency factor of `I_(2)` when, (i) it is formed by the reaction of potassium iodide and potassium iodate in acid medium and (ii) when it reacts with hypo, are respectively:

Options

A. (a)`2, 2`
B. (b)`5/3, 2`
C. (c )`3/5, 2`
D. (d)`5, 2`

Answer

B

Explanation

`IO_(3)^(-)+I^(-)+H^(+) rarr I_(2)+H_(2)O`
`I_(2)` can be said to have undergone disproportionation.
:. V.f. of `I_(2)` for oxidation, `n_(1)=2(5-0)=10`
and v.f. of `I_(2)` for reduction, `n_(2)=2(0-(-1))=2`
`:. V.f. of I_(2) (overall) (n_(1)n_(2))/(n_(1)+N_(2))=(10xx2)/(10+2)=5/3`
(ii) `I_(2)+2Na_(2)S_(2)O_(3) rarr Na_(a)S_(4)O_(6)+2Nal`
`I_(2)+2e^(-) rarr 2I^(-) implies v.f.=2`