# Ionic Compounds and finding Anions

K2HPO4 is an ionic compound with an -2 Anion and a +1 Cation. The thing I'm finding hard to understand is why the four Oxygen atoms don't add to the Anion charge.

E.g. H+1 P-3 O-2 O-2 O-2 O-2 = charge of -8.

I'm told I'm making this harder for myself but it's really frustrating not knowing why the 4 Oxygen atoms don't affect the charge.

Edit: I think I might have confused myself or confused my tutor because I'm mucking up the charges etc. Sorry if it's hard to understand

• Why, they do add. But P is neither minus nor 3. – Ivan Neretin Apr 2 '16 at 8:49

The Hydrogen Phosphate anion is made up of Covalent bonds. The overall charge of $-2$ that it has can be seen by its tetrahedral structure: The four Oxygen atoms do not each contribute a $-2$ charge to the molecule, but rather two of them contribute a $-1$ charge each.
I think you are confusing Oxidation state, or 'charge' and valence state. Oxidation state is the formal number of electrons added/removed from an atom when it forms a compound. Here, we have a covalent anion so the electrons are only formally removed, but it is still a useful formalism and can be used here to calculate the $-2$ charge :
Oxygen's valence state, the number of valence bonds used in bonding, is $-2$ and for Hydrogen it is $+1$ as you correctly said. Here, these are the same as their Oxidation States. But for Phosphorus, we must assign a $+5$ Oxidation state to it in order to obtain the overall $-2$ charge. Note that Phosphorus is not actually losing 5 electrons to form a $+5$ cation, this is a formalism, as aforementioned.
You CAN do this for $K^+$ and $HPO_4$$^-$$^2$ because these form an ionic compound, so the charges simply add: Two $K^+$ and one $HPO_4$$^-$$^2$ will combine to produce the overall $0$ charge ionic molecule (As shown in the diagram, but for $Na^+$ rather than $K^+$.