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+.
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.
2 answers
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.
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.
3 answers