I am having problem with this question including topics of redox reactions and charges 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.
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 $$\ce{Fe^{x+} + 6 CN^-  -&gt; [Fe(CN)_6]^{4-}}$$ 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 $\ce{CN^-}$.