Question

What is the oxidation number of manganese (Mn) in potassium permanganate (KMnO4)? +3 +5 +6 +7 after titration

Answer 1

You know K+ has a charge of +1, meaning MnO4 must collectively have a charge of -1 to give a neutral compound.

Since this isn't a peroxide or molecular oxygen, the oxidation state of each O is -2. You have 4, giving a net charge of -8. Since the total charge is only -1, Mn must have an oxidation state of +7 to cancel the extra negative charges.

Answer 2

If it makes it easier, I like to solve these using a mathematical approach, (although most people prefer not to involve any mathematics). I like to think of it as an equation and solving for, x, where x represents the oxidation state of your central atom. It's also crucial that you understand the charges for some of the elements and for more advance chemistry courses, of your polyatomic ions.

I would set it up like this:

x + 4(-2) = 0

Now, let me explain what everything is. As matt101 said, there is no charge on this compound and thus it has a TOTAL charge of ZERO, which is why I set it equal to zero.
The FOUR is because there are 4 oxygen atoms, and each of them carry a charge of -2! So in essence, it's

(central atom oxidation state) + (number of attached atoms)(charge of ONE attached atom = total charge

To help you better understand, for example, take VO2 +

x + 2(-2) = +1

I have two oxygens, oxygen carries a formal charge of (-2), and the total charge of the atom is + (also means +1).

Rearranging to solve for x:
x-2 = +1
x = +1+2
x = 3

therefore, vanadium (V) has oxidation state of 3. Best of luck. Cheers.

Answer 3

I do not understand the "after titration" part of the question.

After titration, \(MnO_4^-\) turns into \(Mn^{2+}\)

So Mn has two different o.n. in the two different compounds.