# Polyatomics -How Do They Work?

In class, we recently learned about Ionic nomenclature. Specifically, polyatomics and monoatomics. Even though our teacher did a decent job (I guess) of explaining it to us, I'm still very confused.

I have multiply questions:

Question: How would you know what type a compound is when given only the formula?

For example: Our teacher started with $\text{AuCl}$ and wrote down Gold (I) Chloride as the nomenclature. How did he get from $\text{AuCl}$ to Gold (I)?

There are two possibilities: Gold (I) and Gold (III).

Another example: Another example would be $\text{Pb}_3\left(\text C_6\text H_5\text{O}_7\right)$. Which apparently corresponds with Lead (II) Citrate. What does the II inside the parenthesis mean?

And below, I have provided some more examples. Please leave some for me! :P

Examples: Write the formulas for the following compounds:

• Copper (II) Chloride

• Vandium (III) Selenide

• Beryllium Oxide

• Potassium Permanganate

• Ammonium Nitrate

• Tin (II) Sulfite

• Chromium (IV) Cyanide

• Aluminum Arsenide

I'm lost at this point. Any help and a few suggestions would help a lot!

• Has your teacher yet talked about (preferred) oxidation states of the elements in compounds? He should, because this all makes no sense at all without knowing that oxygen is usually -II, halogens -I, alkali metals +I etc. – Karl Nov 3 '16 at 20:44
• @Karl That doesn't seem too familiar. Maybe he has taught us, but I wasn't listening, maybe he didn't... But either way, I don't quite remember that. – Frank Nov 3 '16 at 21:29
• Well, you will be screwed in chemistry without oxidation states, so check on it. – Karl Nov 3 '16 at 21:37

Our teacher started with $\ce{AuCl}$ and wrote down Gold (I) Chloride as the nomenclature. How did he get from $\ce{AuCl}$ to Gold (I)?

There are two possibilities: Gold (I) and Gold (III).

True, but there is only one possibility for chloride ;)

Karl's comment under your question gives the crucial hint:

1. Chloride is (always) $\ce{Cl-}$; it has a charge of (-1).
If your teacher would have written $\ce{AuCl3}$, you would know that there are 3 chloride ions with a total charge of $3\times(-1)=-3$. Consequently, you would know that you're having $\ce{Au(III)}$ here. The copper (II) cloride example belongs to the same category and if you like, you can try iron(II) bromide, iron(III) chloride and copper(I) iodide.
2. Karl also mentioned oxygen. You can safely assume one or more $\ce{O^{2-}}$ units in your metal oxides, depending on the oxidation state of the metal. Can you assign the oxidation state in $\ce{Fe2O3}$?