Characteristics of cations with KCN and excess KCN

Question

How many cations from $\ce{Ag+,Pb^2+,Cu^2+,Cd^2+,Fe^2+,Fe^3+}$ form precipitate with $\ce{KCN}$ but also get dissolved in excess $\ce{KCN}$?

I have just been introduced to Qualitative Analysis of a few cations and came across this problem. All of this is rather memory based applications for me because I'm preparing for a test, so I knew a few of the reactions but I'm not sure about them:

(I have a feeling that $\ce{Pb^2+}$ would not satisfy the conditions given in the problem)

$$\ce{Ag+ ->[KCN] \underset{White ppt}{AgCN} ->[excess KCN] K[Ag(CN)2]}$$

$$\ce{Cu^2+->[KCN] \underset{Yellow Solution}{Cu(CN)2} ->[KCN] \underset{White ppt}{CuCN} ->[excess KCN] K3[Cu(CN)4]}$$

$$\ce{Cd^2+->[KCN] \underset{White ppt}{Cd(CN)2} ->[excess KCN] K2[Cd(CN)4]}$$

$$\ce{Fe^2+->[KCN] \underset{Green ppt}{Fe(CN)2} ->[excess KCN]K4[Fe(CN)6]}$$

I'm not quite sure about $\ce{Pb^2+}$ or $\ce{Fe^3+}$ , I would appreciate if someone could help me out with the equations for these two and if the equations written above are right or not?

Answer 1

_{Converting my comment into an answer.}

---

You are correct with your analysis. Lead(II) is the only cation to not form a complex with excess cyanide. It forms the white $\ce{Pb(CN)2}$ and is insoluble in excess cyanide which helps is distinguishing with mercury(I) and silver(I) which forms the complex.

Copper(II) and cadmium(II) also forms the cyanide complex (with the said path you wrote) and the good thing is that this test can be used to separate them. $\ce{[Cu(CN)4]^3-}$ is so stable that it hardly has any free copper(I) ions and hydrogen sulfide won't precipitate out copper(I) sulfide while $\ce{[Cd(CN)4]^3-}$ is not so stable and it will precipitate out yellow cadmium sulfide on adding hydrogen sulfide.

$$\ce{[Cd(CN)4]^3- + H2S -> \underset{yellow ppt.}{CdS \downarrow} + 2H+ + 4CN-}$$

Both Fe(II) and Fe(III) will form hexacyanoferrate salts on excess cyanide through intermediate iron(II)/iron(III) cyanide.

$$\ce{Fe^2+ + 2CN- -> \underset{yellow-brown ppt.}{Fe(CN)2 \downarrow} ->[4CN-] \underset{pale-yellow soln.}{[Fe(CN)6]^4-}}$$

$$\ce{Fe^3+ + 3CN- -> \underset{reddish-brown ppt.}{Fe(CN)3 \downarrow} ->[3CN-] \underset{yellow soln.}{[Fe(CN)6]^3-}}$$

Free $\ce{Fe^3+}$ might react with ferricyanide to form brown iron(III) ferricyanide (un)commonly known as "Prussian brown" but it is hard for free $\ce{Fe^2+}$ to react with ferrocyanide to form iron(II) ferrocyanide as it quickly oxidize in air to form Prussian blue.