Oxidation number of periodic elements

The oxidation number of periodic elements defined as the formal positive or negative charge when all the bonds in the compounds were ionic. Therefore some general rules contracted to find the oxidation number of periodic elements.

To describe some chemical reaction we use oxidation number rules because some of the reactions can not be explained by the electronic or classical concepts.

The formation of water by the hydrogen bonding with oxygen and formation of HCl from H2 and Cl2 can not be explained by the electronic or classical concepts.

Therefore to explain these types of reactions we use O.N rules. These rules also easily predict the oxidizing agent and reducing agent in a chemical reaction.

How to find the oxidation number of elements?

Oxidation number and state of elements

The following general rules are observed to find the oxidation number of elements

  • The atom of the diatomic molecules like hydrogen, chlorine, oxygen, etc and metallic element like zinc, copper, sodium, etc is assigned zero oxidation number.
    Because these same elements forming a chemical bond with electronegativity difference zero.
  • Except for metal hydrides the oxidation number of hydrogen +1. Hence alkali metal hydrides like lithium hydride, sodium hydride, cesium hydride, etc, the oxidation state of hydrogen -1.
  • Metal always possess a positive oxidation state.
  • Oxygen has normally -2 oxidation number. But in peroxide and superoxides the oxidation number of oxygen -1 and -1/2 respectively.
  • The oxidation number of an ion equal to its charge.
  • The algebraic sum of the oxidation number of all the atoms in a compound must be zero and the many atomic ions equal to its charge.

Oxidation numbers of hydrogen

Like alkalines hydrogen has a single electron in outer valence shall and like halogen it just one electron short of next noble gas helium. The electron configuration of hydrogen

Hydrogen: 1S1

Therefore hydrogen can easily losses this electron to show the +1 oxidation number and gaining one electron to show -1 oxidation number.

H → H+ + e
H + e → H

Oxidation numbers of alkali and alkaline metals

Alkali and alkaline earth metals are highly electropositive with very low ionization energy. Therefore like the halogens in alkali halides, halogen shows negative oxidation numbers but alkali and alkaline earth metals show the positive oxidation number. Thus electrolysis of LiH, CsH, and CaH2, etc liberated hydrogen at the anode.

Compound O.N.
HCl Cl =-1 H = +1
NaH Na = +1 H = -1
CsH2 Cs = +1 H = -1
CaH2 Ca = +2 H = -1

Al alkali metal hydrides are ionic crystalline solid. They are formed by the direct combination of hydrogen and the alkalies at 3500 C 6000 C.

Alkali + H2 → Alkali Hydrides

Oxidation number of superoxides and peroxides

The alkalies readily react with oxygen to form the number of binary compounds: monoxides(M2O), peroxides(M2O2) and superoxides(MO2) shows a different oxidation number of O-atom.

Alkali and alkaline earth elements are the only known example for the formation of superoxides. But with the increasing size of the alkali ion, the stability of superoxides also increases. Therefore Lithium only forms monoxide, sodium up to peroxides and another alkali up to superoxides.

Fluorine is more electronegative than oxygen. Thus it also forms monoxides and peroxides.

Compound O.N
H2O H=+1 O=-2
O2F2 O=+1 F = -1
KO2 K=+1 O=-½
OF2 O=+½ F=-1

Oxidation numbers of atoms in compounds

Oxidation number of Mn in potassium KMnO4

Let the oxidation number of manganese in KMnO4 = x. Thus according to the above rule

(+1) + x + 4(-2) = 0
or, x = +7

Oxidation number of chromium in dichromate ion

Let the O.N. of chromium in Cr2O7-2 ion = x.

∴ 2x + 7(-2) = -2
or, x = +6

Oxidation numbers of both nitrogen in NH4NO3

NH4NO3 present as a cation NH4+ and NO3 ion and let  the O.N of nitrogen in NH4+ = x and NO3 = y.

∴ x+3(+1) = +1
or, x=0

y+3(-2) =-1
or, y= +5

Oxidation numbers of Cl in Ca(OCl)Cl

In Ca(OCl)Cl compound one chlorine combine with oxygen to form OCl ion and one chlorne in Cl ion. Thus the O.N. of one chlorine = +1 and other =-1.

Phosphorus in hypophosphorous acid

Let the O.N. of phosphorus in H4P2O7 = x.

∴ 4(+1) + 2x + 7(-2) = 0
or, x = +5

What is the oxidation number of barium in Ba(H2PO2)2?

The O.N. of hydrogen and oxygen in Ba(H2PO2)2 are +1 and -2 respectively and let Ba = x.

∴ (+2) + 2{2(+1)+x+2(-2)} = 0
or, x = +1

Oxidation state of sulfur in sulphuric acid

Let the oxidation state of sulfur in H2SO4 = x and the oxidation state of hydrogen +1 and oxygen -2.

∴ 2(+1) + x + 4(-2) = 0
or, x = +6

Thus the oxidation stater of sulfur in H2SO4 = +6.

Metals in coordination compounds

Metal ions ion in coordination compound possesses two kinds of valencies primary and secondary valency.

  1. Primary valency satisfied by the requisite number of anions or negative groups being present inside and outside the coordination sphere.
  2. But secondary valency indicated the capacity of metal ion or groups around itself in the first sphere of coordination zone.

Now primary valency equated with the oxidation state or numbers and secondary valency coordination numbers.

In [Cr(NH3)6]Cl3, let the O.N. of Cr = x and NH3 and Cl = 0 and -1 respectively.

∴ x + 0 +3(-1) = 0
or, x = +3

Calculate the oxidation state of iron in [Fe(H2O)5(NO)+]SO4.

Let the oxidation state of iron in [Fe(H2O)5NO+]SO4 = x and water, NO+ and sulfate ion = 0, +1and -2 respectively.

∴ x+5(0)+(+1)-2=0
or, x = +1

Oxidation state of iron in Fe(CO)5

Both the carbonyl and metal has zero oxidation state in  Fe(CO)5

Oxidation state of carbon in acetone

Let the oxidation state of carbon in acetone = x and hydrogen and oxygen +1 and -2 respectively.

∴ 3x + 6(+1) + (-2) = 0
or, x = -(4/3)

Atoms in organic compounds

Sugar, glucose, formaldehyde, etc are the examples of organic compounds where the oxidation number of carbon on this compound is zero.

For glucose (C12H22O11) = x.

∴ 6x + 12(+1) + 6(-2) = 0
or, x = 0

What is the oxidation number of chromium in CrO5?


Due to the peroxy linkage O.N. of chromium in CrO5 = +6.