Question

What is the oxidation number of all elements in
MnSO4

Answer 1

I already know the rules in finding the oxidation number I just don't know how to get it in this one.

Answer 2

Sulfate has a charge of -2

Answer 3

Thus we can conclude that Magnesium would have a charge of +2

Answer 4

I would assume

Answer 5

It is from KMnO4 + KCl + H2SO4 = K2SO4 + MnSO4+Cl2+H2O
I already determined the other oxidation number from this except for MnSO4

Answer 6

How about for oxygen? isn't it it should have an oxidation number of -2 ?

Answer 7

MnSO4 +H2O ----> Mn^(2+) + SO4^(2-)

Answer 8

Oxygen,
-2, -2, -2, -2
Sulpher is a less electronegative element so it will give up its electrons more readily
S has +6

6 -2 -2 -2 - 2 = -2 charge total

Answer 9

I could be wrong, as I'm going off what I have read on Wikipedia and my knowledge of chemistry

Answer 10

I'm basing this on the fact in

Oxygen Diflouride OF2

Oxygen has a charge of +2 instead of -2 because fluoride is more electronegative than oxygen

Answer 11

and since PO4 is a polyatomic anion, I would expect it to maintain its integrity while bonded to manganese, although this is speculation on my part I hope it is helpful.

Answer 12

If you have any questions feel free to ask

Answer 13

So the oxidation numbers will be Mn(+2)S(+6)O4(-2)
I based in your statement "Sulfate has a charge of -2 Thus we can conclude that Magnesium would have a charge of +2"

Answer 14

Is it correct?

Answer 15

O4 should probably have a charge of -8 if S has a charge of +6

Answer 16

but your conclusion seems correct other wise, Remember Oxygen can attract and hold electrons better than sulfer, so sulfer basically becomes an electron source for the oxygen so they can be more like Neon, making Sulfer more like Neon as well instead of what it usually wants to be like Argon

Answer 17

I'm pretty sure the -2 charge is just distributed through out all the oxygen atoms until manganese comes along and donates 2 of its electrons

Answer 18

wait you are probably correct if it is asking for the charge of each individual atom separately :)

Answer 19

I am getting all the oxidation numbers separately for balancing.
Thanks :)

Answer 20

To be honest with you I have never had to answer a question like this I dont think

Answer 21

But I have taken a lot of chemistry courses in my life so I dont know that has to count for something

Answer 22

Did you mean you haven't balanced a reaction using the oxidation state change method ?

Answer 23

anyway, thanks for the help :)

Answer 24

yeah

Answer 25

I had some other method when I did these problems that I cannot remember, but yeah no problem thanks for the question chemistry problems subside the insanity for me :)