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The oxidation number for 'O2' is zero(0).

However, when calculating oxidation numbers in compounds containing oxygen, then the oxygen is used as a standard at '-2'.

So as an example using potassium permanganate (KMnO4), what is the oxidation number of manganese (Mn).

In solution this dissolves producing the permanganate ion (MnO4^(-).

Note the charge on the ion is '-1'.

Since there are 4 oxygens present , then the oxidation of the oxygen moiety is 4 X -2 = -8 .

So creating a little sum

Mn + -8(oxygen moiety) = -1 (overall charge on the ion).

Mn + -8 = -1

Add '8' to both sides

Mn = (+)7 is the oxidation number of manganese.

So the formula could be written as KMn(VII)O4 . (Note the Roman numerals for oxidation number).

BrO3 is an anion of minus one (-1) charge.

BrO3^-1

Using oxygen as the standard at '-2'

Then the oxygen moiety is 3 X -2 = -6

So doing a 'little' sum

-6 + Br = -1

Add '6' to both sides

Br = 5 (The oxidation number of Br).

NB Using '-2' for oxygen is a good standard for most molecules/ions. However, oxygen's oxidation number does vary , but not in very many molecules/ions.

C2H4O oxidation number is -1

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

2x+4-2=0

2x+2=0

2x=-2

x=-2/2

x=-1

K2Cr2O4

Using oxygen at '-2' as the standard.

Then the oxygen moiety is 4 x -2 = -8

Since K is only +1, and there are 2 x K , then this moiety is '+2'

The overall charge on the molecule is zero.

So we make a sum

- 2 + 2Cr - 8 = 0

2Cr = 8 - 2 = 6

Cr = +3 ( THe oxidation No. of Cr'.

To calculate oxidation numbers. Use oxygen at '-2' as the standard. It does vary in specific cases , but '-2' is the best starting point.

In H2SO4

There are 4 x oxygen , hence 4 x -2 = -8

So the oxygen moiety is '-8'. So the hydrogen and Sulphur MUST come to '+8'.

Since hydrogen is 'H^+' in Sulphuric Acid, its oxidation no. is '+1'

So there are 2 x hydrogen , hence 2 x +1 = +2

Since sulphuric acid is a neutrally charged compound, then

+2 + S - 8 = 0

Hence S = 8 - 2 = +6 THe oxidation no. of sulphur. in sulphuric acid.

NB The oxidation no. of sulphur can vary is different compounds.

NHO3 ???? I think you mean HNO3 ( Nitric Acid).

NB With all mineral acids the 'H' is the first letter.

To find the oxxidation number., use oxygen at '-2'

Since there are 3 oxygens, then the oxidation state of the oxygen component is 3 x -2 = -6 .

Since also hydrogen is always '+1' then we can constract a sum

1 + N -6 = 0 (NB We equate to zero because it is a neutrally charged molecule.

N - 5 = 0

N = 5 the oxidation state of nitrgoen .

Use oxygen at '-2' as the standard.

However, oxygen can vary, but initially use '-2'

So why is potassium permangate (VII) have the 'VII/7' in the name . It is the oxidation number of the manganese.

How do we gert there.

The formula is KMnO4

So there are four oxygens each at '-2' then overall the oxygen component is 4 x -2 = -8

Since the K is a Group '1' metal its oxidation is K^+ = +1

So creating a little sum,

+1 + Mn - 8 = 0 (Zero because it is a neutral molecule)

Hence

Mn - 7 = 0

Hence Mn = 7 ( 'VII' latin numerals are used for oxidation numbers).

NB Oxidation number refers to the numbers of electrons, of a given atom, involved in the bonding process.

NNB Sulphur has several oxidation numbers. They are: -

-2 in H2S

0 in elemental sulphur

+2 in SCl2

+4 in SO2

+6 in SO3 / H2SO4

K2SiO3 ( potassium silicate)

Use oxygen at '-2' as the starting point.

Since there are 3 x oxygen , then the oxygen component is 3 x -2 = -6

So the K & Si must sum to '+6'

Since K oxidises to '+1' as in 'K^+', then the 'K' component is 2 x 1 = 2

Hence we have 2 + Si + - 6 = 0 (neutral molecule)

Then by a simple sum Si is +4.

So the oxudation states of each element are

K = +1

Si = + 4

O = -2

This is the thiosulfate ion. Sulfur shows +6 oxidation number.

The oxidation No. is '3'.

To calculate oxidation numbers, remember the 'yardstick' of oxygen at '-2' . There are a few exceptions to this, but work from '-2'.

S2O3

There being 3 x oxygen present, the the oxygen moiety is '3 x -2 = -6

So the sulphur moiety must be '+6', in order to maintain charge neutrality.

Hence 2 x S = 2 X x = 6

x = 3

So sulphur oxidation is '+3'.

The oxidation number is a measure of the charge an atom would have if all its bonds were 100% ionic. It can be positive, negative or zero.

In the hydronium ion (H3O+), the oxidation number of carbon is +3.

The oxidation number of Na in NaCl is +1, while the oxidation number of Cl in NaCl is -1.

No, actually sodium chloride can be considered the "ash" of burning metallic sodium in a chlorine gas atmosphere. The ash of combustion will not undergo further combustion.

If the question is Cu2, then it is equivalent to Cu and the oxidation number for any element is zero.

If the question is Cu2+, then the oxidation number is +2.