Question

Reddish-brown (chocolate) ppt. is formed with:
(A) $C{u^{2 + }}$ and $Fe\left( {CN} \right)_6^{4 - }$
(B) $B{a^{2 + }}$ and $SO_4^{2 - }$
(C) $P{b^{2 + }}$ and ${I^ - }$
(D) None of these

Answer 1

In the given problem we need to find the reactants which on reacting form the reddish precipitate –brown (chocolate) in color. We will observe the product of the reaction to find the precipitate and its color. If we know the color of the complexes then it gives us an edge.

Complete step-by-step answer: First, we will understand the basic term precipitate. The term ‘Precipitate’ is the substance that is left and deposited in the solid form from the solution. The process of conversion of the solution into the solid form is commonly termed as precipitation. Now we will consider all the options one by one and we will consider all the reactions and their products formed. The product may be a colored precipitate or any colorless complex. So, let’s consider the reactions one by one.
(A)First, we will consider the reaction between $C{u^{2 + }}$ and $Fe\left( {CN} \right)_6^{4 - }$. The reaction can be written with the reactants $CuS{O_4}$ and ${K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$ to maintain the oxidation state of copper and iron. The product formed with the reaction is $C{u_2}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$ and ${K_2}S{O_4}$. The product $C{u_2}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$ is the reddish-brown chocolate color insoluble compound.
$2CuS{O_4} + {K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right] \to C{u_2}\left[ {Fe{{\left( {CN} \right)}_6}} \right] + 2{K_2}S{O_4}$
(B)Now the reaction is between barium chloride and sodium sulfate which leads to the formation of a white precipitate of barium sulfate with sodium chloride. The reaction can be written as $N{a_2}S{O_4}(aq) + BaC{l_2}(aq) \to BaS{O_4}\left( s \right) + 2NaCl(aq)$
(C) The reaction $P{b^{2 + }}$ and ${I^ - }$ results in the formation of a bright yellow precipitate of lead iodide.
Hence, we can conclude that Reddish-brown (chocolate) precipitate is formed with $C{u^{2 + }}$ and $Fe\left( {CN} \right)_6^{4 - }$.

Therefore, the correct option is (A) ,$C{u^{2 + }}$ and $Fe\left( {CN} \right)_6^{4 - }$.

Note: The compound $C{u_2}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$ formed is a reddish cubic crystal or powder. Some of the properties of reddish-brown precipitate are ferromagnetic and non-metal. Oxydimercuric ammonium iodide is also a reddish-brown precipitate.