What is the oxidation number of Mn in MnO4 -1?

+2

Oxidation number in O is 2-, for 4 O we have 8-. The sum of all oxidation numbers is 1-.

Conclusion: Mn has oxidation number 7+

Manganate ions, or MnO4, have a charge of -1. Oxygen will usually have an oxidation number of -2.

So if we do the maths, (letting the charge of the Manganese ion be X),

X + 4(-2) = -1

X= +7

So the oxidation number of Mn in the MnO4 ion is +7.

Hope this helps!

• Let oxidation no of Mn be x . oxidation no of O is (–2).
• The algebraic sum of oxidation no. of all atoms in a molecule is charge on the molecule . Here the charge is (–2).
• Therefore,
• X + (–2×4) =–2
• X –8 = –2
• X= –2+8
• X= + 6
• Therefore ,oxidation number of Mn is +6

Summation of oxidation states in every compound is zero.

ie. In MnSO4

Mn + S + O4 = 0

Mn + 6 + (-2x4) = 0

Mn + 6 + - 8 = 0

Mn - 2 = 0

Mn = +2

Therefore the oxidation state of manganese in MnSO4 is +2

Take Mn = x

Overall ox. No = -1

So,

-1 = x - 4 (Oxygen )

-1 = x - 4 X2

-1 = x - 8

x. = 8–1= 7

So oxidation no. Of Mn in( MnO⁴)-1 is 7.

• Let oxidation number of Mn in MnO4- be x and oxidation number of O is –2
• The algebraic sum of oxidation no. of all atoms in a molecule is charge on the molecule .Here charge is (–1).
• Therefore, X +(–2× 4)=–1
• X–8=–1
• X=–1+8
• X= +7
• Therefore, oxidation number of Mn is +7

The ans is +6

Let the oxidation no of Mn be x.

As the above compound has no net charge therefore sum of all oxidation states must be 0.

(+1)*2+x+(-2)*4=0

Oxidation state of - K-(+1 ) 0- (-2)

2+x-8=0

x=+6

If the oxidation number of Mn is considered to be x,

then,

x+(-2*4)=-1

(-2 as oxidation number of oxygen is -2)

x-8=-1

Make x the subject, and voila!

x=+7

+2.

It’s indicated in brackets thanks to Stock’s nomenclature.