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$\ce{[Cr(H2O)4Cl2]Cl}$, Tetraaquadichlorochromium(III) chloride

According to a book, the (III) indicates the oxidation state of chromium, but it doesn't explain why. Cr has 6 valence electrons. How does it get to +3?

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    $\begingroup$ I have cleaned up your formula using the chemistry plugin for mathjax (see the faq for more info). $\endgroup$ – Ben Norris Jun 9 '13 at 0:21
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I think the easiest way to see this is to count the formal charges on the ligands.

The four waters do not have net charge. The remaining ligands--the three Cl's--are likely $\ce{Cl-}$, and so their total charge contribution is $-3$.

The only way the complex can be neutral is if the chromium has a $+3$ charge: hence Chromium (III).

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    $\begingroup$ By the way, since chromium is a transition metal, it is able to form multiple ions, such as 2+, +6 or even -1. There are no rules about ionisation on transition metals, and the only way to discover their possible ionisations is going through every specific case of ions. $\endgroup$ – Pedro823 Jul 31 '14 at 18:20
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Eric is correct about the easiest method to determine the most likely oxidation state of Cr. As for the rest of your question, the oxidation state of Cr in this compound means that it has three electrons less than in its neutral state, i.e. three valence electrons instead of six. This +3 oxidation state is the most stable one for chromium, although many other states are also possible. Note that a wide range of oxidation states is characteristic of d-block elements such as chromium.

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Most elements have several oxidation states in the everyday life. The simplest one, hydrogen, is usually +1 but it is 0 in H2, or even -1 when bonded to an element less electronegative than itself such as a metal.

Remember that valence is different from oxidation state (see https://en.wikipedia.org/wiki/Valence_(chemistry) ).

Off-topic but always good to know: chromium VI (6+) is extremely toxic and cancerogenous, whereas chromium III (3+) is much less dangerous.

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  • $\begingroup$ This answer is all useful and true, but it does not address the OP's question of how to determine the oxidation state of chromium in the compound shown. $\endgroup$ – Ben Norris Jul 23 '16 at 19:11
  • $\begingroup$ Well his question was more about "how can chromium be 3+" than about "formulas must be wrong". But I understand your concern. In fact, the "oxidation number" is completely artificial and part of our (poor) understanding of chemistry. $\endgroup$ – SteffX Jul 23 '16 at 19:23

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