What are the oxidation numbers for In and Pb in the following compounds: InCl2 and Pb2O3?
(A) In = 2, Pb = 3
(B) In = 1, Pb = 3
(C) In = 2, Pb = 4
(D) In = 2, Pb = 2
(E) None of the above;
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32 comments:
A
i got A too
Think once more.
is it c then ?
Oxidation numbers for Pb in Pb2O3 is 2 and 4(PbO*PbO2)and for In is2 inInCl2.
None of the answers are right.Please post your answers.
d ?
-_-... what's the answer then?
B
Try to balance the equation. Cl is 1- and 0 is 2-
The correct answer is E.
For the metals in the lower part of groups 3-6 A, their is a tendency to form compounds with the valence x and x -2 where is x is the group number;
EX: Pb group 4 valences 2 and 4
In group 3 valences 3 and 1
Therefore when they aggregate together we have:
Pb2O3 a mixture of PbO and PbO2
InCl2 a mixture of InCl3 and InCl which can be written as In2Cl4- or InCl2[this is very confusing but if you checked the position of In in the periodical system you would have seen that it can not have valence 2]
So the oxidation number are Pb 2 and 4 and In 1 and 3
Is this a problem that could appear at SAT chemistry? Isn't it too hard for this type of exam?
yea..his question of the day's are supposed to be harder than the average SAT chemistry question
this problem would never appear on a chemistry SAT test and if it would the correct answer would be A. According to the rules regarding oxidation numbers which I have learned in my own chemistry class, as well as barrons, sparknotes, and PR, you are to suppose that in a compound oxygen has an oxidation number of 2- and the halogens have an oxidation number of 1-. Therefore, with some simple math, the answer would be A. I'm sure the blog owner is correct, but going that indepth will end up hurting you on the SAT because the test-makers oversimplify everything . Plus I don't think none of the above is ever an answer on the SAT subject test.
it IS august 15th, 2007. hope to see your site back up soon. i really like the sat questions of the day. thanks a mucho :]
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The blogowner is right. Indium and Thallium only forms compounds with O.S. number of +1 and +3
So, for InCl2, the actual formula would be In2Cl4, or In+1 and [InCl4]-, with the second In having +3 OS state.
Lead only forms compounds with +2 and +4 Os state (+2 is prefered).
The question is, this is supposed to be something you learn in an advanced inorganic chemistry, not in highschool..I think.
So, for SAT, I wonder if we can just answer A, or E...
the answer is E..because In +3 and Pb is +2...
its A .
Cl has a -1 charge. therefore In must have a +2 charge to balance the compound.
Pb2O3--> O has a -2 oxidation number and thus comes out to -6. to balance this Pb must be +3
and this is an extremely easy type of problem compared to the ones ive been doing in the barrons chem sat book..
the thing is ^^^
youre right if you follow oxudation rules.. however if the compounds are ions you have to take into account the ionic charge of each compound and thus you wouldnt balance the oxidation numbers to ). you would balance them to equal their ionic charge which wasnt given.
It is A
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A is the choice....
The answer: A. you have In and Cl2 and so Cl has an oxidation number of -1x2 (for the 2 molecules bount to In) so, in order to make the charges even out, In must have an oxidation number of +2. therefore +2, and -2 = 0. And for Pb2O3. Oxygen has an oxidation number of (-2) (a fact you would have to know). So, we take that number and multiply it by the number of molecules (3). getting -6. And so in order to balance that with the pb2, which means two molecules, we have 2(x) (x is the unknown oxidation number) -6 = 0, so x = +3. There you are!
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