# What is the formula for copper (ll) phosphate?

Oct 16, 2016

"Cu"_3("PO"_4)_2

#### Explanation:

The first thing to note here is that the name of the compound provides you with the charge of the cation.

Copper, $\text{Cu}$, a transition metal, can have multiple oxidation states, which is why compounds that contain copper are written using Roman numerals.

These Roman numerals describe the oxidation state of the transition metal in a given compound. In this case, the Roman numeral (II) means that copper has a $+ 2$ oxidation state, i.e. it has an overall $2 +$ positive charge.

You can thus say that the cation will be

${\text{Cu}}^{2 +} \to$ the copper(II) cation

Now for the anion. Phosphate is the name given to a polyatomic ion that contains

• one atom of phosphorus, $1 \times \text{P}$
• four atoms of oxygen, $4 \times \text{O}$

As you can see, the phosphate anion carries a $3 -$ negative charge. This means that your anion will be

${\text{PO}}_{4}^{3 -} \to$ the phosphate anion

Now, ionic compounds must be electrically neutral, i.e. the overall positive charge coming from the cations must be balanced by the overall negative charge coming from the anions.

In this case, you have

$\text{Ca"^(2+)" }$ and ${\text{ ""PO}}_{4}^{3 -}$

You will need $3$ copper(II) cations to get an overall positive charge of $6 +$ and $2$ phosphate anions to get an overall negative charge of $6 -$, and so the chemical formula for this compound will be

$\textcolor{b l u e}{3} \times {\left[{\text{Cu"^(color(red)(2+))] + color(red)(2) xx ["PO"_ 4^(color(blue)(3-))] -> "Cu"_ 3("PO}}_{4}\right)}_{2}$

Therefore, you can say that you have

"Cu"_3("PO"_4)_2 -> copper(II) phosphate