In naming Ionic Compounds, how do I find the Roman Numeral?

Question

I believe finding the Roman Numeral involves something like

Example: Co(NO3)3

_____(1) + -1 (3)

However, I do not know how to get the numbers in the parenthesis. I am naming a few ionic compounds that require a Roman Numeral, but I don't what to put in the parenthesis's.

____ ( # ) + ___ ( # )

I know the numbers outside of the parenthesis's are the ions.

Answer 1

1   I

2   II

3   III

4   IV

5   V

6   VI

7   VII

8   VIII

9   IX

10   X

Answer 2

In your example, we have cobalt (III) nitrate.

Nitrate has an oxidation number of -1.  Three of them is -3.  Since there is one Co, it must have an oxidation number of +3, since the sum of the oxidation numbers of a compound is zero.  

We use Roman numerals when naming compounds using the Stock system of chemical nomenclature.  (A fancy word for "naming".)  The stock system is used to name compounds of a metal cation (element with a positive oxidation number), not compounds of nonmetals.

ie.

MnCl2 = manganese (II) chloride, metal compound, use Stock system

SCl2 = sulfur dichloride - two nonmetals, use Greek prefixes

We use a Roman numeral when the metal has more than one oxidation number.  The Roman numeral corresponds to the oxidation number of the metal.  

FeCl3 = iron (III) chloride - iron has oxidation numbers of +2 and +3

AlCl3 = aluminum chloride - Al only has one oxidation number

Rules for Assigning Oxidation Numbers

which can be found at:

http://www.cartage.org.lb/en/themes/Scie...

    * The oxidation number of an atom is zero in a neutral substance that contains atoms of only one element. Thus, the atoms in O2, O3, P4, S8, and aluminum metal all have an oxidation number of 0.

    * The oxidation number of monatomic ions is equal to the charge on the ion. The oxidation number of sodium in the Na+ ion is +1, for example, and the oxidation number of chlorine in the Cl- ion is -1.

    * The oxidation number of hydrogen is +1 when it is combined with a nonmetal. Hydrogen is therefore in the +1 oxidation state in CH4, NH3, H2O, and HCl.

    * The oxidation number of hydrogen is -1 when it is combined with a metal. Hydrogen is therefore in the -1 oxidation state in LiH, NaH, CaH2, and LiAlH4.

    * The metals in Group IA form compounds (such as Li3N and Na2S) in which the metal atom is in the +1 oxidation state.

    * The elements in Group IIA form compounds (such as Mg3N2 and CaCO3) in which the metal atom is in the +2 oxidation state.

    * Oxygen usually has an oxidation number of -2. Exceptions include molecules and polyatomic ions that contain O-O bonds, such as O2, O3, H2O2, and the O22- ion.

    * The nonmetals in Group VIIA often form compounds (such as AlF3, HCl, and ZnBr2) in which the nonmetal is in the -1 oxidation state.

    * The sum of the oxidation numbers of the atoms in a molecule is equal to the charge on the molecule.

    * The most electronegative element in a compound has a negative oxidation number.

Answer 3

so you only need to use a roman numeral when dealing with transition metals like Co in your example. Simply add up the anion charges, in this case each nitrate ion has a -1 charge, so the total negative charge is -1 x 3, or -3. Since there is no indication that the molecule has a an overall charge (it's neutral, and total charge should add up to zero) then Co must have a +3 charge. So you would write Co(III) and just call it "cobalt 3."

hope this helped.