# Calculating the charges of anions and cations

I am new to chemistry. Could someone explain me how to calculate the charges on anions and cations. I've heard about some crossing method (the left subscript becomes the right superscript and the right subscript becomes the left super script.)

Example: $\ce{Rb2S -> 2Rb+ + S^2-}$

But what about: $\ce{K2MnO4}$ and $\ce{Zn(ClO)2}$

This trick doesn't seem to apply here, so how can I find out the charges?

• @Close flagger, I don't think this is a homework question. It looks more like questioning a [stupid] rule resulting from oversimplification to me. – M.A.R. Nov 1 '15 at 16:07

Some basic knowledge of the atoms in the periodic table is necessary to extract the ionic charges. In the case of $\ce{K2MnO4}$, for example, we know that the potassium ion is positive, and that the manganate ion is negative. We see that the compound is neutral: has zero net charge. So how can we place charges to get the molecule to be neutral, given the subscripts?

Potassium is a Group 1 element, meaning it only has 1 valence electron, placed in the 4s orbital. Potassium is eager to lose this electron to fulfil the octet rule. Therefore potassium ions have one less electron than protons, and the charge is +1. Since we have two $\ce{K^+}$ ions, the manganate ion must have a charge of -2 to make sure the molecule has zero net charge.

$$\ce{K2MnO4 <=>[\ce{H2O}] 2K+ + MnO4^2-}$$

Zinc is one of two transition metals that only have one (natural) oxidation state: +2. The electron configuration is [Ar]3d$^{10}$4s$^2$. When zinc becomes a cation, it loses the two 4s electrons. Knowing this, we see that the hypochlorite must have a charge of -1: the two hypochlorite -1 charges counter the 2+ charge of zinc.

$$\ce{Zn(ClO)2 <=>[\ce{H2O}] Zn^2+ + 2ClO^-}$$

I recommend not relying on simple rules, such as the one you mentioned. Rather, try to understand why the rule came about, and understand when the rule fails, and why. It really helps to be able to "read" the periodic table, as a lot of basic information is incorporated into it.

• Potassium problem: Is there a way to know that Mn is 2- except by deduction like you did? – privetDruzia Nov 1 '15 at 13:43
• @privetDruzia Each potassium ion has to have a 1+ charge, so $x$ in $\ce{[MnO4]^{$x$-}}$ has to be 2. – orthocresol Nov 1 '15 at 13:59
• Thx. This semi-deductive method seems to work indeed. But not for more "complex" molecules like: KnaCO3 or Ba(BrO2)NO2, ... Or am I wrong? – privetDruzia Nov 1 '15 at 14:05
• This may not be the best reply, but you should just memorize the charge of common ions. There are ways of estimating the charge in molecular ions, such as "Formal charge". The formal charge assumes, however, that all bond electrons are shared equally, regardless of electronegativity differences, so such an approach may not always be appropriate. In quantum mechanics, the electrons don't necessarily belong to a single atom, but rather the whole molecule. Thus, fractional charges are more common than integer charges. See this pdf: chem.ucla.edu/harding/tutorials/formalcharge.pdf – Yoda Nov 1 '15 at 15:37

Elements like K, Li, Cs and Rb can only have +1 charge on them (Due to very high second ionization energy). Also, they can't have negative charge due to their low electron gain enthalpy (they are metals and are electropositive in nature). Also, Oxygen can only have -2 (in oxides, eg: H2O), -1 (in peroxides, eg: H2O2) and -0.5 (in superoxides, eg: KO2). Also, H can only have +1 or -1(if H is more electronegative of the two elements, as in the case of hydrides, eg: LiH) and Fluorine can have only -1 (since it is the most electronegative element). Similarly, Ca, Mg.. have +2 charge. You need to remember these basic rules and a few others (refer to any inorganic chemistry text book and you can find them), to calculate the charges.

Take the case of K2MnO4 : K has +1. O has -2 (in is an oxide). Now, let the charge on Mn be x. So, x + 2*(+1) + 4*(-2) = 0 (Since the net charge on the species is zero). Hence, x = +6. I hope this answer helped.

• Note that alkali metals have been reported to have a negative oxidation state though. – M.A.R. Nov 1 '15 at 12:24
• Also, oxygen has an oxidation state of +2 in oxygen difluoride. – M.A.R. Nov 1 '15 at 12:28
• I clearly mentioned these are the few basic rules and the person who asked this question has to refer a book to know all the exceptions and other such rules. – ShankRam Nov 2 '15 at 13:51