Reviewing for exams and completely forgot how to do this and looking back at my notes doesn't help for this one... Can you please explain how to assign oxidation numbers for Pb(OH)4^2- I guess what's confusing me is the hydroxide... I haven't come across brackets until now. How would that work? I know that since there is a charge on the compound, it would all be set equal to -2. And I know that normally, H would get a +1 and O would get a -2, but then to cancel the H, O might need to be -1, but then I thought that'd be wrong because hydroxide has a -1 charge... And I don't even know if I'm supposed to do it separately like that? I mean, do I just treat OH as one thing and give it -1 as it's "oxidation number" (I feel like that'd be wrong too because -1 is its actual charge, not a hypothetical number..)

Anyway, would appreciate some help.

ok. You are confusing with the charge and the oxidation number. To me, both are the same, except that they are represented differently; the charge is written or shown in the chemical formula of the ion using a number followed by a sign (2- or 3- etc).

The oxidation number shows the degree at which the ion is oxidized and written as an integer shown in the charge of the ion i.e. a sign followed by a number. In the case of OH-, the charge is 1- of the whole ion so the oxidation number is -1. But then you can also have individual oxidation numbers; for oxygen and hydrogen. As you said, oxygen always have -2 oxidation number (charge of 2-) and H has +1 (charge of 1-). This is correct, because, combining O2- with H+ will give OH- right??? because -2+1 = -1 which is shown as a - sign in OH-.

This is similar in the case of complex chemical formulas. The idea is to add all the individual oxidation numbers and equate to the overall oxidation number of the ion. Pb(OH)4^2- means that there are 4 OH- ions (that is why OH is in bracket) attached to one Pb metal ion, giving an overall charge of the complex ion to be 2- (or an overall oxidation number of -2). oxygen has -2 O.N and H is +1. So Pb O.N can be calculated by adding all the individual oxidation number and equate it to the overall oxidation number;

O.N (Pb) + O.N (OH)x4 = -2 (not 2- as we are now dealing with oxidation numbers).

so, rearrange for O.N (Pb)

= -2 - (-1x4) = -2 + 4 = +2

Pb is therefore Pb2+.

OH^- is -1 so 4*-1 = -4.

So Pb must be 2+ so that 2-4 = -2 on the ion.

You want to do it individually? Then
O = -2 x 4 = -8
H = +1 x 4 = +4
So Pb must be 2+ so that
+2-8+4 = -2 on the ion.

Assigning oxidation numbers is a process to determine the hypothetical charge that an atom would have if all the bonds in a compound were ionic in nature. To determine the oxidation numbers for each element in Pb(OH)4^2-, let's break down the process step by step.

1. Start with the element that is most likely to have a fixed oxidation state. In this case, it's hydrogen (H) and oxygen (O). Hydrogen is generally assigned an oxidation number of +1, and oxygen is assigned an oxidation number of -2.

2. Since there are four hydrogen atoms in the compound, it would contribute a total positive charge of +4 (4H x +1 = +4).

3. The compound has an overall charge of -2, so now we need to distribute the remaining negative charge (-2) between the remaining atoms.

4. We can consider the hydroxide group (OH) as a single entity because it is a polyatomic ion. The overall charge of hydroxide is -1. Since there is one hydroxide group in the compound, it would contribute a total negative charge of -1.

5. The sum of the oxidation numbers in a compound must equal the overall charge. Since hydrogen contributes a positive charge (+4) and hydroxide contributes a negative charge (-1), the lead (Pb) atom must have an oxidation number that balances these charges.

6. To find the oxidation number of lead, let's use the equation:

(+4 from hydrogen) + (x from lead) + (-1 from hydroxide) = -2

Simplifying the equation, we get:

x + 3 = -2

Subtracting 3 from both sides, we find:

x = -5

Therefore, the oxidation number of lead (Pb) in Pb(OH)4^2- is -5.

To summarize, the oxidation numbers of the elements in Pb(OH)4^2- are as follows:
- Hydrogen (H) = +1
- Oxygen (O) = -2
- Lead (Pb) = -5

Remember, oxidation numbers are just hypothetical numbers assigned for calculation purposes. They're not the same as the actual charge of the ion or atom.