Explanation:
(d)Potassium ferrocynide \({K_4}[Fe\,{(CN)_6}]\) will ionize as \({K_4}[Fe\,{(CN)_6}]\)
\(\mathop {[Co\,{{(N{H_3})}_5}S{O_4}]\,Br}\limits_{0.02\,{\text{mole}}} + AgN{O_3} \to \)\(\mathop {[Co\,{{(N{H_3})}_5}.\,S{O_4}]\,N{O_3}}\limits_{0.02\,{\text{mole }}(y)} + AgBr\)
\(\mathop {[Co\,{{(N{H_3})}_5}B{r_2}]S{O_4}}\limits_{0.02\,{\text{mole}}} + BaC{l_2} \to \)\(\mathop {[Co\,{{(N{H_3})}_5}Br]\,C{l_2}}\limits_{0.02\,{\text{mole}}\,(z)} + BaS{O_4}\)
On using one lit. solution, we will get \( 0.01\) \(mole\) \(y \) and \(0.01\) \(mole\) \(z.\)
\(4{K^ + } + {[Fe\,\,{(CN)_6}]^{4 - }}\)
So, it will give five ions in solution.