# I am having problem with this question including topics of redox reactions and charges [duplicate]

What is the oxidation number of $$\ce{Fe}$$ in $$\ce{K4[Fe(CN)6]}$$?

This seems to be a pretty easy problem. Just input the oxidation numbers of everything and equate for 'x'. But the problem is that I don't know the charge on the Cyanide ion right there. Now, I knew from doing many questions that $$\ce{Fe}$$ usually has +2 or +3 so i guessed +2. But I don't want to be guessing on the actual day.

I essentially need to know the best way to find charges on polyatomic ions because there's just loads of them and I don't think I can remember over 300 oxidation states. I know the charge can be calculated by adding up Oxidation states but most elements have variable Oxidation states too so I am really confused.

Even if you know a certain website that may be useful, please link it. Any help is appreciated.

• Maybe read your textbook? Cyanide ligand is clearly mentioned to be $\ce{CN-}$ in your text (which I'm guessing is NCERT). You have to remember the basic oxidation states of the most common ligands. – Aniruddha Deb Jun 16 '20 at 16:55
• @AniruddhaDeb that isn't really very helpful as that's not what I was asking. But allright, thanks – Aditya_math Jun 16 '20 at 19:32
• @AniruddhaDeb Hello, do you know which class's NCERT contains those polyatomic ion charges. I'll buy that book so I can read that too. Thanks – Aditya_math Jun 18 '20 at 6:40
• Know how to draw cyanide (it’s not hard … only two atoms) and then the linked dupe will help you. – Jan Jun 26 '20 at 17:31

As the formula of the substance is known, you should first state the formula of the ions produced when the substance is dissolved into water. Here, $$\ce{K_4Fe(CN)_6}$$ gets dissolved in water and produced $$4$$ ions $$\ce{K^+}$$ so that the $$4$$ corresponding negative charges must be fixed on the remaining anion, which has the formula $$\ce{[Fe(CN)_6]^{4-}}$$ with $$4$$ negative charges. Now you have to think how to synthesize this complex anion from $$\ce{1 Fe^x+}$$ ion and $$\ce{6 CN^-}$$ ions. The equation is $$\ce{Fe^{x+} + 6 CN^- -> [Fe(CN)_6]^{4-}}$$ Counting the charges gives x positive charges from $$\ce{Fe^{x+}}$$, 6 negative charges form the six cyanide ions, and this makes a total of $$\ce{4-}$$. So the initial $$\ce{Fe}$$ ion must have been charged $$+2$$
The nature of the ion $$\ce{CN^-}$$ remains to be explained : Well ! The carbon atom has 4 independent valence electrons. Nitrogen has 5 valence electrons, but 2 of them are taken in a doublet. So there is only 3 available electrons in N atoms for bonding. When 1 C is attached to 1 N, this makes 3 covalent bonds between C and N. As a consequence one electron is still available on the carbon atom. This is not a stable compound. The carbon atom needs one more electron to get 8 electrons in its outer layer. This electron can be offered by one H atom, which would make a covalent molecule HCN, with one covalence between H and C, and three covalences between C and N. But this last electron can also be given by a sodium Na or a potassium atom K. This would produce the compound NaCN or KCN, where Na (or K) and the carbon atom are bound in a ionic bond between $$\ce{Na^+}$$ (or $$\ce{K^+}$$) and the ion $$\ce{CN^-}$$. This is the origin of the ion cyanide $$\ce{CN^-}$$.