Oxidation reaction involves loss of electrons, and reduction reaction involves gain of electrons. The reaction in which a species disproportinates into two oxidation states ( lower and higher) is called disproportionation reaction.
Which of the following statements is wrong?

Options

A. An acidified `K_(2)Cr_(2)O_(7)` paper on being exposed to `SO_(2)` turns green.
B. Mercuric chloride and stannous chloride cannot exist as such.
C. Iron turning on addition to `CuSO_(4)` solution decolourises the blue colour.
D. `[CuI_(4)]^(2-)` is formed but `[CuCl_(4)]^(2-)` is not.

Answer

D

Explanation

a. `Cr_(2)O_(7)^(2-)` (orange red) oxidises `SO_(2)` to `SO_(4)^(2-)` and is itself reduced to `Cr^(+3)` (green).
`underset(("Orange"))(Cr_(2)O_(7)^(2-)+3H^(o+)+3SO_(2))rarrunderset(("Green"))(2Cr^(3+)+3SO_(4)^(2-)+4H_(2)O`
b. Both reacts together
`overset(+2)(Hg)Cl_(2)+overset(+2)(SnCl_(2))rarroverset(+1)underset(("White ppt"))(HgCl_(2))darr+overset(+4)(SnCl_(4))`
`overset(+1)(Hg_(2))Cl_(2)+overset(+2)(SnCl_(2))rarrunderset((Grey))(2Hg)darr+overset(+4)(SnCl_(4))`
c. `overset(0)Fe+underset((Blue))overset(+22)(CuSO_(4))rarrFe^(2+)+overset(0)(Cu)`
d. `[CuCl_(4)]^(2-)` is formed but `[CuI_(4)]^(2-)` is not.
`I^(ө)`ion reduces `Cu^(2+)` to `overset(+1)(Cul)` and itself undergoes oxidation to form `I_(2)`. However, `CI^(ө)` does not reduce `Cu^(2+)`.