Consider $\ce{[Cr(H2O)6]^3+}$, the charge on the Chromium before the water molecules react is $3+$, clearly each water ligand contributes one lone pair, meaning the formal charge should be $3-6=-3$ (each bonding pair is identical to having an extra singular electron in the shell) , but the charge is actually $+3$. What am I doing wrong?

  • $\begingroup$ Following your way of looking at it if each water gives one electron to the chromium what is the resulting charge on each water? And so what does that make the total charge on the complex? $\endgroup$
    – Ian Bush
    Commented Feb 13 at 19:01
  • $\begingroup$ Charge is +1 which gives net charge of +3, so it works. What is the standard way of working out charge on a complex ion? $\endgroup$
    – nav op
    Commented Feb 13 at 19:52
  • $\begingroup$ Well, you just pointed out that formal charge isn't terribly useful. In complexes ox. state or "valency" is written for central atoms. Overall charge is fine, but actual partial charges are non-trivial to find. $\endgroup$
    – Mithoron
    Commented Feb 13 at 23:08

1 Answer 1


You are doing nothing wrong, but you are confusing the formal charge with the oxidation state. The formal charge is the charge that an atom would have if all the bonding electrons were equally shared between the atoms. The oxidation state is the hypothetical charge that an atom would have if all the bonds were ionic, meaning that the more electronegative atom takes all the bonding electrons.

In the [Cr(H2O)6]3+ complex, the formal charge of Cr is -3, as you correctly calculated. This means that Cr has three more electrons than its valence shell can accommodate. However, the oxidation state of Cr is +3, because we assume that all the bonds are ionic and that the oxygen atoms of the water molecules take all the bonding electrons. This means that Cr has lost three electrons from its valence shell.

The formal charge and the oxidation state are not the same thing, and they do not have to be equal. The formal charge is a way of keeping track of the electrons in a covalent bond, while the oxidation state is a way of assigning charges based on electronegativity differences. The formal charge is useful for predicting the reactivity and stability of a molecule, while the oxidation state is useful for balancing redox reactions and describing the electron transfer processes.

I hope this explanation was clear and helpful. If you want to learn more about the formal charge and the oxidation state, you can check out the web links that I have provided. Thank you for your attention. 😊

Reference: (1) Chemistry of Chromium - Chemistry LibreTexts. https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_%28Inorganic_Chemistry%29/Descriptive_Chemistry/Elements_Organized_by_Block/3_d-Block_Elements/Group_06%3A_Transition_Metals/Chemistry_of_Chromium/Chemistry_of_Chromium. (2) Introduction to Crystal Field Theory - Chemistry LibreTexts. https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_%28Inorganic_Chemistry%29/Crystal_Field_Theory/Introduction_to_Crystal_Field_Theory. (3) What is the oxidation state of chromium in Cr(H2O)6 (3+)? - Socratic. https://socratic.org/questions/what-is-the-oxidation-state-of-chromium-in-cr-h2o-6-3.


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