Determine the oxidation state of each element in Cr(C2H3O2)3. Cr: C:  H:  O

To determine the oxidation state of each element in Cr(C2H3O2)3, we need to understand the rules of oxidation numbers.

1. The oxidation state of an atom in its elemental state is always zero.

2. The oxidation state of a monoatomic ion is equal to its charge.

3. Oxygen usually has an oxidation state of -2 in its compounds, except in peroxides where it is -1.

4. Hydrogen usually has an oxidation state of +1 in its compounds, except when it is bonded to metals in binary compounds (then it is -1).

5. The sum of the oxidation states of all atoms in a neutral molecule or formula unit is zero.

Now, let's apply these rules to Cr(C2H3O2)3.

1. The compound is neutral, so the sum of the oxidation states of all atoms must be zero.

2. Each acetate ion (C2H3O2-) has a charge of -1. Since there are three acetate ions, the total charge contributed by the acetate ions is -3.

3. Therefore, the oxidation state of Cr must be +3 to balance out the -3 from the acetate ions.

4. In the acetate ion (C2H3O2-), there are 2 carbon atoms, 3 hydrogen atoms, and 2 oxygen atoms.

5. Oxygen has an oxidation state of -2, so the total charge contributed by the oxygen atoms is -4 (2 atoms * -2 charge/atom).

6. Hydrogen has an oxidation state of +1, so the total charge contributed by the hydrogen atoms is +3 (3 atoms * +1 charge/atom).

7. The sum of the oxidation states in the acetate ion must be -1 (the charge of the ion). So, the total charge contributed by the carbon atoms is -1 - (-4 + 3) = -1 - (-1) = 0. Since there are 2 carbon atoms, each carbon atom has an oxidation state of 0/2 = 0.

So, the oxidation states of the elements in Cr(C2H3O2)3 are:

Cr: +3 C: 0 H: +1 O: -2