I am doing very well with oxidation numbers but I have a few problems. I know all the rules but when I get an element where I can't use any of the constant oxidation numbers what do I do? For example: $\ce{HCNS,Ag2S,Bi2S3,KSCN}$ etc. ?


1 Answer 1


You shouldn’t really need to use the shortcut rules (fluorine having $\mathrm{-I}$, hydrogen having $\mathrm{+I}$ and oxygen having $\mathrm{+II}$ ‘except in peroxides’) with the exception of really simple compounds where it all falls into place immediately. Instead:

  1. If the compound is molecular or contains a molecular ion, draw the Lewis structure.

  2. Assign each bonding electron pair to the more electronegative bond partner.

  3. Count the electrons each atom ends up with, substract from the number of electrons it should have according to its position in the periodic table, invert that number’s sign.

  4. If at step one you are only left with ions, just take the ion’s charge as oxidation state.

  5. You did it right if the overall sum of oxidation states matches the molecular ion’s charge or zero for the overall structure.

(Note: I am simplifying here by assuming that compounds are either molecular or ionic. That, of course, is most certainly not the case but the very crude generalisation gets us very far in this case.)

To work through one of your examples:
$\ce{KSCN}$ is an ionic compound consisting of potassium cations and thiocyanate anions. According to step 4, we can assign potassium the oxidation state of $\mathrm{+I}$.

  1. The structure of thiocyanate is $$\ce{S=C=\overset{-}{N} <-> \overset{-}{S}-C#N}$$ In both cases, the terminal atoms have enough lone pairs to complete their octet.

  2. The electronegativites are $\ce{N}>\ce{S}>\ce{C}$. Thus, we assign the bonds so that:

    • sulphur and nitrogen each have eight valence electrons
    • carbon has zero

    This is independent of the resonance structure we used.

  3. Calculating:

    • Sulphur:
      eight electrons, should have six.
      $\displaystyle-(8-6) = \mathrm{-II}$

    • Carbon:
      zero electrons, should have four.
      $\displaystyle-(0-4) = \mathrm{+IV}$

    • Nitrogen:
      eight electrons, should have five.
      $\displaystyle-(8-5) = \mathrm{-III}$

  4. Has been done before.

  5. To check:

    $\displaystyle \mathrm{+I + -II + +IV + -III = \pm0}$ (for $\ce{KSCN}$)

    $\displaystyle \mathrm{-II + +IV - III = -I}$ (for $\ce{SCN-}$)


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