# When do you need to use roman numeral sin chemical compound?

Nov 22, 2017

When you assign oxidation states, and oxidation numbers to elements in compounds, salts, and molecules...

#### Explanation:

By definition, oxidation number is the charge left on the central atom when all the bonding pairs of electrons are conceptually broken, with the charge assigned to the most electronegative atom.....

When we do this for $H - C \equiv N$ we get $\stackrel{+ I}{H}$ and $\stackrel{+ I I}{C}$, and $\stackrel{- I I I}{N}$. Here nitrogen is tervalent; i.e. it formally has 8 valence electrons, and hence a formal charge of $- 3$.....As always, the charge on the molecule is EQUIVALENT to the SUM of the oxidation numbers, i.e. here $\text{ZERO}$.

And these follow.... the following rules...

$1.$ $\text{The oxidation number of a free element is always 0.}$

$2.$ $\text{The oxidation number of a mono-atomic ion is equal}$ $\text{to the charge of the ion.}$

$3.$ $\text{For a given bond, X-Y, the bond is split to give } {X}^{+}$ $\text{and}$ ${Y}^{-}$, $\text{where Y is more electronegative than X.}$

$4.$ $\text{The oxidation number of H is +1, but it is -1 in when}$ $\text{combined with less electronegative elements.}$

$5.$ $\text{The oxidation number of O in its}$ compounds $\text{is usually -2, but it is -1 in peroxides.}$

$6.$ $\text{The oxidation number of a Group 1 element}$ $\text{in a compound is +1.}$

$7.$ $\text{The oxidation number of a Group 2 element in}$ $\text{a compound is +2.}$

$8.$ $\text{The oxidation number of a Group 17 element in a binary compound is -1.}$

$9.$ $\text{The sum of the oxidation numbers of all of the atoms}$ $\text{in a neutral compound is 0.}$

$10.$ $\text{The sum of the oxidation numbers in a polyatomic ion}$ $\text{is equal to the charge of the ion.}$

And remember that oxidation states and numbers are fictitious entities....they do help in solving redox equations.....