How do you determine the oxidation numbers in these compounds?

**Answer 1**

To solve for the oxidation numbers you need to know the charge of the individual elements in their ionic form. For example:

Co in LiCoO2 – Li is in the first column (group) on the periodic table, so it has a +1 charge. O is in the second column to the left of the noble gases so it has a -2 charge. You have two O in O2, so a total negative charge of -4. Plus you have one +1 charge in Li, so a total positive charge of +1. Combine your charges, +1 + -4 and you get -3. The compound LiCoO2 has a 0 charge, as do all the other compounds on your list or the charge would be indicated. This means that Co must have a +3 charge to balance out the net -3 charge left over.

Now try the same thing with #2. Na has a +1 charge. H is tricky in this case because it normally has a +1 charge, however, when you see it written after a transition metal like that it commonly has a -1 charge. So, you have 4 H’s, thus four -1 charges. Now add your positive and negatives, +1 + -4 = -3. Therefore, Al must have a +3 charge to give the compound a net charge of 0.

Here are the other answers, try the work yourself to confirm.

3. +4 (this is another tricky one, the first 3 H’s have a negative charge but the last H is positive)

4. -3

5. +3

6. +6