Oxidation number of elements

Oxidation number and oxidation state of elements

The oxidation number of an element defined as the formal positive or negative charge which would be assigned to the element, if all the bonds in the compounds were ionic bonds.

The formation of water by the hydrogen bonding with oxygen, can not be covered by the electronic concept because water is a covalent compound. But classically we can still say hydrogen oxidized to form water.

2H2 + O2 → 2H2O

Similarly, hydrogen and chlorine react to form a covalent molecule HCl.

2H2 + Cl2 → 2HCl

The formation of HCl, hydrogen oxidized or chlorine reduced and easily explained by the oxidation number concept.

How to find the oxidation number of elements?

Oxidation number and state of elements
Elements on the periodic table

The following general rules are to be observed for the assignment of 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. Since the same elements with similar electronegativity involved in the chemical bonding.
  • Except for metal hydrides, in HF, HCl, HBr, HI, etc, the oxidation number of hydrogen +1. In alkali metal hydrides, lithium hydride, sodium hydride, cesium hydride, etc, the oxidation state of hydrogen -1.
  • Metal always shows a positive oxidation state.
  • Oxygen has normally an oxidation number -2. 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.

Alkali and alkaline earth hydrides

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

Thus this electronic configuration justifies the monovalent cation of hydrogen. So hydrogen can easily losses this electron to show the +1 oxidation state.

H → H+ + e

But the alkali and alkaline earth metals are highly electropositive with very low ionization energy. Thus like the halogens in alkali halides, hydrogen presents as the negative hydride anion in alkali and alkaline earth metals. Electrolysis of LiH, CsH, and CaH2, etc liberated hydrogen at the anode.

CaH2 → Ca+2 + 2H

Compound Oxidation Numbers
HCl Cl =-1 H = +1
NaH Na = +1 H = -1
CsH2 Cs = +1 H = -1
CaH2 Ca = +2 H = -1

The negation oxidation state also explains by its electron configuration.  Hydrogen is just one electron short of the next noble gas helium. Thus it takes one electron to form H ion.

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

The oxidation state of superoxides and peroxides

The alkalies readily react with oxygen to form the number of binary compounds: monoxides(M2O), peroxides(M2O2) and superoxides(MO2).

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.

Thus Lithium only forms monoxide, sodium up to peroxides and another alkali up to superoxides.

In monoxides, the oxygen is -2 state, but for superoxides and peroxides, the oxidation number of oxygen is different.

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

Compound Oxidation numbers
H2O H=+1 O=-2
O2F2 O=+1 F = -1
KO2 K=+1 O=-½
OF2 O=+½ F=-1

How to calculate the oxidation number of an element?

Manganese 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

Let the oxidation of manganese KMnO4-2 ion = x and oxygen  = -2.

∴ x +4 (-2) = -2
or, x = +6

Chromium in dichromate ion

Let the oxidation number of chromium in Cr2O7-2 ion = x.

∴ 2x + 7(-2) = -2
or, x = +6

Both nitrogen in NH4NO3

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

∴ x+3(+1) = +1
or, x=0

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

Oxidation number 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 oxidation number of one chlorine = +1 and other =-1.

Phosphorus in hypophosphorous acid

Let the oxidation number 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 oxidation numbers 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

The 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.

The oxidation state of 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 number and secondary valency coordination number.

In [Cr(NH3)6]Cl3, let the oxidation numbers 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

The oxidation state of iron in Fe(CO)5

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

The 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)

The oxidation number of 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?


Oxidation number of chromium
Structure of chromium

Due to the peroxy linkage oxidation number of chromium in CrO5 = +6.