[travibe]

UO₂²⁺ - Oxidation for U
As₂O₃ - Oxidation for As
NaBiO₄ - Oxidation for Bi
HAsO₂ - Oxidation for As
Na₂S₂O₃ - Oxidation for S
XeOF₄ - Oxidation for Xe
K₄Fe(CN)₆ - Oxidation for Fe
Mg₂P₂O₇ - Oxidation for P
Na₂C₂O₄ - Oxidation for C
Ca(NO₃)₂ - Oxidation for N

Ok these are the answers i have gotten by advice from some friends and from reading the book to the best of my ability. (from top to bottom)
-2
+3
+7
+3
+2
+3
-2
+6
+3
-2
Now, im calculating all of these as though they all make a net charge of 0 unless otherwise stated.  Im getting conflicting answers with some of these (such as the last one, and the one including iron) and i think it has to do with my order of operations that im using to solve this.  Any suggestsions or pointers would be greatly appreciated!

(And thank you to Borek for explaining the use of subscripts and superscripts! Lifesaver!!)

Travis
Travibe@cox.net

[Mitch]

The first one should be +6

[Borek]

UO₂²⁺ - Oxidation for U

See Mitch answer.

As₂O₃ - Oxidation for As

OK

NaBiO₄ - Oxidation for Bi

OK

HAsO₂ - Oxidation for As

OK

Na₂S₂O₃ - Oxidation for S

This one is tricky. There are two different types of S here. One is a central atom, second replaces one of oxygens (and can be assigned the same charge oxygen is usually assigned).

XeOF₄ - Oxidation for Xe

Check your math/show your calculation.

K₄Fe(CN)₆ - Oxidation for Fe

Treat CN^- as entity and don't look for individual charges of C and N.

Mg₂P₂O₇ - Oxidation for P

Check your math/show your calculation.

Na₂C₂O₄ - Oxidation for C

OK

Ca(NO₃)₂ - Oxidation for N

Check your math/show your calculation. Try to calculate for HNO₃ - central atom in acid group doesn't change its charge in salts.

[xiankai]

since oxidation number assigns an imaginary "charge" to each atom, it is not neccesary for S in Na₂S₂O₃ to have several charges. just the average will do. i think +2 would be ok.

as for the CN^- thing, one main rule is that the atoms outlined in brackets have their own special charge, and this is where covalent bonding is displayed. so for common anions like CN^-, NO^-₃, SO₄^2-, u'll need to memorise their overall charge and not the charge of the individual atoms.

[Borek]

since oxidation number assigns an imaginary "charge" to each atom, it is not neccesary for S in Na₂S₂O₃ to have several charges. just the average will do. i think +2 would be ok.

Pretty interesting approach .

What charges are you going to assign to Fe in prussian blue,  Fe₄[Fe(CN)₆ ]₃?

[travibe]

wow thanks for the help.  For the ones that you asked to ssee my math lemme shwo you what i did.

6) (XeOF₄) i assigned it as a equation equalling 0 so for oxygen i used -2 for the charge and for Fluroine i used -1, equation was X-2+(4*-1)=0 so that would be X-6=0 wich would make x=6.
8 ) (Mg₂P₂O₇) assigned +2 charge for Magnesium, -2 charge for Oxygen, and solving for P so i set it up as (2*2)+(7*-2)+2x=0 --> 2X-10=0 -->2x=10 ---> X=5
10)(Ca(NO₃)₂) 2+2x-6 --> 2x-4=0 ---->2x=4 --->x=2

I dont know about the one that you said treat CN as a single entity..what does that mean for the charge that i should use on it then?  -1? Is there a general rule to follow for that compound?

[travibe]

alright i reworked some of those that i showed my work for and i got some different more appealing answers.

6)XeOF₄ i got XE as a charge of +6
7) K₄Fe(CN)₆ i got Fe as a charge of +2
8 ) Mg₂P₂O₇ i got P as a charge of +5
10) Ca(NO₃)₂ i got N as a charge of -2

i can show my math if needed...i cant eve ntell what i was doing wrong here, but its making more sense on the math now.

[Borek]

10)(Ca(NO₃)₂) 2+2x-6

-6?

I dont know about the one that you said treat CN as a single entity..what does that mean for the charge that i should use on it then?  -1?

Yes.

For salts or complexes - in general - when you see entity like SO₄^2- don't treat S and O seprately (unless you are asked about the charges that can be assigned to them) - treat whole group as -2.

[xiankai]

Pretty interesting approach .

What charges are you going to assign to Fe in prussian blue,  Fe₄[Fe(CN)₆ ]₃?

u got me there

i'll concede defeat