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.

  • 3
    $\begingroup$ 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. $\endgroup$ – Aniruddha Deb Jun 16 at 16:55
  • $\begingroup$ @AniruddhaDeb that isn't really very helpful as that's not what I was asking. But allright, thanks $\endgroup$ – Aditya_math Jun 16 at 19:32
  • $\begingroup$ @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 $\endgroup$ – Aditya_math Jun 18 at 6:40
  • $\begingroup$ Know how to draw cyanide (it’s not hard … only two atoms) and then the linked dupe will help you. $\endgroup$ – Jan Jun 26 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 $\ce{CN^-}$.

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  • $\begingroup$ Thanks, but I was asking if there was any way to find out that CN has a negative charge apart from just knowing it. Thanks for your help though $\endgroup$ – Aditya_math Jun 16 at 19:34
  • $\begingroup$ If you try to join one C atom and one N atom, you should see that there is a triple bond between C and N. This being done, a doublet is remaining on the N atom, and this N atom has now an octet around it, which is supposed to be good. But a free electron remains on the Carbon atom, and the outer valence shell of C has only 7 electrons. A good way to stabilize this C atom is to approach an atom like sodium which can easily loose and give an electron, making CN negative. This being done, C has now 8 electrons in its outer shell, and is stable : this is the CN- ion. $\endgroup$ – Maurice Jul 18 at 15:56
  • $\begingroup$ thanks so much.... I understand it now $\endgroup$ – Aditya_math Jul 20 at 13:41

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