The incorrect statement out of the following | vedclass

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(B) $1$. $Cu^{2+}$ reacts with $K_4[Fe(CN)_6]$ to form $Cu_2[Fe(CN)_6]$,which is a chocolate-brown precipitate.$2$. In the presence of $HCl$,the conce

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$Cu^{2+}$ and $Ni^{2+}$ ions give black precipitate with $H_2S$ in presence of $HCl$ solution.

Vedclass · Answer

(B) $1$. $Cu^{2+}$ reacts with $K_4[Fe(CN)_6]$ to form $Cu_2[Fe(CN)_6]$,which is a chocolate-brown precipitate.$2$. In the presence of $HCl$,the concentration of $S^{2-}$ ions is suppressed due to the common ion effect. $CuS$ $(K_{sp} \approx 10^{-36})$ precipitates,but $NiS$ $(K_{sp} \approx 10^{-21})$ does not precipitate because its ionic product remains lower than its $K_{sp}$. Thus,the statement that both give black precipitate is incorrect.$3$. $Fe^{3+}$ reacts with $SCN^-$ to form $[Fe(SCN)]^{2+}$,which is blood-red.$4$. $Cu^{2+}$ in a reducing flame forms metallic copper,which gives a red-coloured bead.

The incorrect statement out of the following is:

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