# How do we know which ions each element forms? And how do assign charges in complex ions?

Sep 25, 2015

Use the periodic table.

#### Explanation:

Periodic table is arranged such that most common oxidation state of the elements are represented as column numbers. All elements with same valence electrons in the outer shell are arranged column-wise. Remembering the position of the element or even the group of the element is sufficient. The left half of the periodic table has increasing positive valencies while the right half has decreasing negative valencies.

Sep 25, 2015

You can remember by their position on the Periodic Table, and by some simple arithmetic.

#### Explanation:

You have listed 1 element, $C {l}_{2}$, and 2 anions, $C {l}^{-}$, and $C l {O}_{3}^{-}$. In each case the charge (or oxidation state) of the original species (anion, neutral, or cation), MUST equal the sum of the individual oxidation states of the constituent atoms.

To continue, we consider elemental chlorine, $C l - C l$; if we break the bond the elements (this consists of 2 electrons), the electrons will go to the most electronegative element. Because this is an element-element bond, we assume that each atom gets $1$ electron. This would not be the case in the interhalogen compound, $C l - F$.

i.e. $C l - C l \rightarrow 2 C {l}^{\cdot}$.

Each chlorine atom now has 7 electrons, and is therefore electronically neutral. It has a zero oxidation state. If it were a chloride ion, $C {l}^{-}$, there are 8 electrons around the nucleus, and therefore this species bears a negative charge, and thus the oxidation state of the elemental ion is $- 1$.

As for compound ions, let's go to chlorates, salts of chloric acid, $H C l {O}_{3}$, and we proceed the same way. We have $C l {O}_{3}^{-}$, with 1 negative charge. The sum of the individual elemental oxidation states MUST be equal to $- 1$. Oxygen normally has an oxidation state of $- I I$, and it does here: $3 \times \left(- 2\right)$ $+$ $C {l}_{O N}$ $=$ $- 1$, which was the charge on the ion. It doesn't take rigorous algebra to calculate that the oxidation number of $C l$ in chlorate ion is equal to $V +$. That is $5 + \left(- 2 \times 3\right) = - 1$, which was the overall charge on the ion.

We would have got the same result and considered the parent chloric acid, $H - O C {l}_{3}$, and broken the bond between hydrogen and oxygen to give ${H}^{+}$ and $C l {O}_{3}^{-}$. So now what is the oxidation state of $C l$ in perchloric acid, $H C l {O}_{4}$? And what is the oxidation state of $C l$ in the interhalogen, $C l - F$?