I have to write the complete balanced molecular, ionic, and net ionic equations for the reaction of nickel II chloride + sodium phosphate. What product, if any, is a precipitiate? I started by getting NiCl2 + Na2(PO4) = (PO4)Ni + NaCl. Balancing it I got NiCl2 + Na2(PO4)Ni + 2NaCl. How do you go about figuring out the ionic and net ionic equations? Do I break them up into oxidation numbers? Ni has a +2 oxidarion number or am I totally going in the wrong direction? Thanks so much for your time

Sodium phosphate is Na3PO4.

Write the balanced equation using that formula. That gives you the balanced MOLECULAR equation.
Now separate the molecules into ions using the following guidelines.
Gases are written as the molecule.
Precipitates (Ni3(PO4)2 is a ppt) are written as the molcule.
Weakly ionized substances (H2O is an example) are written as molecules.
Everything else is written as ions.

Next, look at ions on the left and right. Cancel ions common to both sides.
The result is the net ionic equation.

i worked it out and got

NiCl2+ Na3(PO4)= Ni(PO4)+ 3NaCl2 for the balanced moloecualr equation.
Ni +2
Cl2 -1
Na3 +1
PO4 -3

Ni+2 = Cl2-1 + Na3+1 + (PO4)-3
Ni+2 + (PO4)-3 = Ni=2 (PO4)-3 for the net ionic equation

would the precipate be nickel II phosphate or Nickel III phosphate. being if i did thid correctly

I gave you the formula for nickel phosphate AND told you it was a precipitate in my first response but you must have copied it down wrong. Here is how you do it. I'm adding (aq) to show the ones in solution and (s) to show the solid precipitate.

3NiCl2(aq) + 2Na3PO4(aq) ==> Ni3(PO4)2(s) + 6NaCl(aq)

The above is the balanced molecular equation. Now we separate it into ions with the rules I gave in my first response. To save time I will omit the (aq) after the ions.
3Ni^+2 + 6Cl^- + 6Na^+ + 2PO4^-3 ==>Ni3(PO4)2(s) + 6Na^+ + 6Cl^-

The above is the balanced ionic equation. The only thing I've done is to separate those aqueous solution into ion if they dissolve but show the ppt as a solid (s). Now to make the NET ionic equation, we cancel those ions common to both sides. Those are 6Cl^- and 6 Na^+ so the net ionic equation is
3Ni^+2(aq) + 2PO4^-3(aq) ==> Ni3(PO4)2(s)

To write the complete balanced molecular equation, you first write the formulas for the reactants and products, then balance the equation by adjusting the coefficients to make sure the number and type of atoms on both sides are equal. In your equation, you started with NiCl2 + Na2(PO4)Ni = (PO4)Ni + NaCl.

However, it seems like you made a mistake in writing the formulas. Nickel (II) chloride is actually NiCl2, and sodium phosphate is Na3PO4. So, the correct molecular equation for the reaction is:

NiCl2 + Na3PO4 -> Ni3(PO4)2 + 2NaCl

Next, to write the ionic equation, you need to break apart the soluble compounds into their respective ions. In this case, NiCl2 dissociates into Ni2+ and 2Cl- ions, and Na3PO4 dissociates into 3Na+ and PO4^3- ions. The equation becomes:

2Ni2+ + 3PO4^3- + 6Na+ + 6Cl- -> Ni3(PO4)2 + 6Na+ + 6Cl-

Finally, you can cancel out the spectator ions (the ions that are present on both sides of the equation) to write the net ionic equation. In this case, the Na+ and Cl- ions are present on both sides and cancel out, leaving:

2Ni2+ + 3PO4^3- -> Ni3(PO4)2

So, the net ionic equation is: 2Ni2+ + 3PO4^3- -> Ni3(PO4)2

The precipitate is the solid that forms during a chemical reaction. In this case, the net ionic equation shows the formation of Ni3(PO4)2, which is insoluble (does not dissolve) in water. Therefore, Ni3(PO4)2 is the precipitate in this reaction.