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Which of the following is NOT correct?

(A) In an aqueous solution free Copper (I) disproportionates to give Copper (O) and Copper (II).

(B) Copper (II) reduces I$_2$ to I$^{-}$

(C) Strong heating of CuSO$_4$ gives CuO and SO$_3$

(a) (1) Only
(b) (2) Only
(c) (3) Only
(d) None of the above

1 Answer

Copper (II) Oxidizes I$^{-}$ to I${_2}$. The reaction is 2Cu$^{2+}$ + 4I$^{-} \rightarrow$ 2CuI + I$_2$
Hence B is correct
answered Mar 16, 2014 by balaji.thirumalai
edited Nov 10, 2017 by sharmaaparna1