For the given aqueous reactions, which of the statement(s) is(are) true? Excess KI + {K}₃[{Fe}({CN})₆] {{dilute }{H}₂{SO}₄ Brownish - yellow solution {{ZnSO}₄ White precipitate + {Brownish-yellow filtrate} {{Na}₂{S}₂{O}₃ Colourless solution [JEE Adv 2012 P2]

The first reaction is a redox reaction.

White precipitate is {Zn}₃[{Fe}({CN})₆]₂.

Addition of filtrate to starch solution gives blue colour.

White precipitate is soluble in NaOH solution.

Test Index

d and f block elements

Section: Chemistry

Question : 22 of 24

Marks: +1, -0

\mathrm{K}_3[\mathrm{Fe}(\mathrm{CN})_6]

\xrightarrow{\text{dilute }\mathrm{H}_2\mathrm{SO}_4}

\xrightarrow{\mathrm{ZnSO}_4}

\xrightarrow{\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3}

Solution:

The first reaction is

\mathrm{K}_3[\mathrm{Fe}(\mathrm{CN})_6] + \underset{\text{excess}}{\mathrm{KI}}

→

\mathrm{K}_4[\mathrm{Fe}(\mathrm{CN})_6] + \underset{\text{brownish yellow solution}}{\mathrm{KI}_3}

This is a redox reaction as both oxidation and reduction are taking place.

\underset{\text{soluble}}{\mathrm{K}_4[\mathrm{Fe}(\mathrm{CN})_6]} + \mathrm{ZnSO}_4

→

\underset{\text{White ppt.}}{\mathrm{K}_2\mathrm{Zn}_3[\mathrm{Fe}(\mathrm{CN})_6]_3} + \mathrm{NaOH}

→

\underset{\text{soluble}}{\mathrm{Na}_2[\mathrm{Zn}(\mathrm{OH})_4]^{2-}}

2\mathrm{I}_3^{-} + 2\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3

→

\underset{\text{clear solution}}{\mathrm{Na}_2\mathrm{S}_4\mathrm{O}_6} + 2\mathrm{NaI} + \underset{\text{(blue colour)}}{2\mathrm{I}_2}

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