Oxidation number |

**used to specify some oxidation-reduction reactions. The definition of oxidation and reduction based on loss or gain of electrons is limited to the scope. The definition holds goods for ionic compounds.**

*Oxidation number*- Examples, the formation of water from hydrogen and oxygen, can not be covered by the electronic concept since water is not an ionic compound.

- 2H

_{2}+ O

_{2}→ 2H

_{2}O

- It may recall classically we could still say that hydrogen is oxidized to H

_{2}O. In the same seance burning of magnesium in oxygen is considered oxidation. Similarly, hydrogen and chlorine react to form a covalent molecule hydrogen chloride.

- 2H

_{2}+ Cl

_{2}→ 2HCl

- The Above reaction hydrogen is oxidized or chlorine is reduced but the resulting compound is covalent one, thus the reaction cannot be covered by the electronic concept. In order to cover such reactions also under oxidation-reduction, the concept of

**is developed and it is defined as,**

*oxidation number*###
**Oxidation number**

- The

**of an element in a compound is the formal charge (positive or negative) which would be assigned to the element if all the bonds in the compounds were ionic bonds.**

*oxidation number*- All the compounds are treated as though they were ionic merely because of the case of counting

**. The**

*oxidation numbers***, therefore**

*oxidation number*__,__is arbitrary.

- Electronegativity concept is utilized in adducing a formal charge to an atom. The less electronegative of the partners of a binary compound is arbitrarily assigned a positive

**and more electronegative one a negative**

*oxidation number**oxidation number*.

- Florine, being the most electronegative, has always a negative oxidation number. The alkali metals are highly electropositive, have low electronegativity, hence have been assigned positive oxidation numbers.

###
*Method of finding out the oxidation number*

*Method of finding out the oxidation number*

The following general rules are to be observed for the assignment of oxidation numbers.

- Atoms of diatomic molecules like H₂, Cl₂, O₂, etc or of metallic elements like Zn, Cu, Na, etc are assigned zero oxidation numbers since the same elements of similar electronegativity are involved in the bonding.

H ➖H

- The oxidation number of the above molecules are zero because two hydrogen atom of the same electronegativity is involved for bonding.

- Except for metal hydrides, the oxidation number of hydrogen is +1. In alkali metal hydrides, LiH, NaH, CsH, etc, the oxidation number of hydrogen is -1.

NaH → Naᐩ + H⁻

(Here oxidation number of H is -1)

(Here oxidation number of H is -1)

HCl → Hᐩ + Cl⁻

(Here oxidation number of H is +1)

(Here oxidation number of H is +1)

- The oxidation number of metal is positive.

CuO → Cuᐩ² + O⁻²

(Here Oxidation number of Cu is +2)

(Here Oxidation number of Cu is +2)

- Oxygen has normally an oxidation number -2. For examples, CuO → Cu⁺² + O⁻², here oxidation number of O is -2.

- In peroxide (H₂O₂, Na₂O₂), the oxidation number of oxygen is -1 since hydrogen has to be assigned +1.

- In H₂O, the oxidation number of oxygen is -2, but in H₂O₂, the oxidation number of H is +1 and the oxidation number of oxygen is -1.

- In Na₂O₂, the oxidation number of Na is +1 and the oxidation number of oxygen is -1.

- In BaO₂, the oxidation number of oxygen is -1 because the oxidation number of Ba is +2.

- In fluorine monoxide(F₂O) oxygen has an oxidation number +2 because fluorine is more electronegative than oxygen.

- In superoxides (KO₂), the oxidation number of oxygen is -1/2.

- The oxidation number of an ion is equal to its charge.

NaCl → Na⁺ + Cl⁻

- The charge and oxidation number of Na⁺ and Cl⁻ are +1 and -1 respectively.

MgBr₂ → Mg⁺² + 2Br⁻

- Here the charge and oxidation number of Mg⁺² and Br⁻ are +2 and -1 respectively.

- The algebraic sum of the oxidation numbers of all the atoms in a compound must be zero and the oxidation number of many atomic ions equal to its charge.

- In HCl oxidation number of hydrogen is +1 and the oxidation number of chlorine is -1. And the sum of these = (+1) + (-1) = 0.

- In MnO₄⁻ ion sum of the oxidation number of Mn and oxygen equal to -1.

###
*Oxidation number of an element in a compound*

####
*Oxidation number of Mn in KMnO₄*

- Let the oxidation number of Mn in KMnO₄ is x. Thus according to the above rule,

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

or, x = +7

Thus, the oxidation number of Mn in KMnO₄ is +7.

or, x = +7

Thus, the oxidation number of Mn in KMnO₄ is +7.

####
*Oxidation number of Mn in MnO₄⁻²*

- Let the oxidation number of Mn in MnO₄⁻² is x and the oxidation number of oxygen is -2(according to the above rule).

- Thus the sum of the oxidation number of MnO₄⁻² = Charge of the MnO₄⁻².

∴ x +4 (-2) = -2

or, x = +6

Thus, the oxidation number of Mn in MnO₄⁻² is +6.

or, x = +6

Thus, the oxidation number of Mn in MnO₄⁻² is +6.

####
*Oxidation number of Cr in Cr₂O₇⁻²*

Let the oxidation number of Cr in Cr₂O₇⁻² is x

∴ 2x + 7(-2) = -2

or, x = +6

Thus, the oxidation number of Cr in Cr₂O₇⁻² is +6.

∴ 2x + 7(-2) = -2

or, x = +6

Thus, the oxidation number of Cr in Cr₂O₇⁻² is +6.

####
*Oxidation number of S in H₂SO₄*

- Let the oxidation number of S in H₂SO₄ is x. According to the rule oxidation number of hydrogen is +1 and oxygen is -2.

∴ 2(+1) + x + 4(-2) = 0

or, x = +6

Thus, the oxidation number of S in H₂SO₄ is +6.

or, x = +6

Thus, the oxidation number of S in H₂SO₄ is +6.

####
*Oxidation number of C in CH₃COCH₃*

- Let the oxidation number of C in CH₃COCH₃ is x. And the oxidation number of hydrogen and oxygen is +1 and -2 respectively.

∴ 3x + 6(+1) + (-2) = 0

or, x = -(4/3)

Thus, the oxidation number of C in CH₃COCH₃ is 4/3.

or, x = -(4/3)

Thus, the oxidation number of C in CH₃COCH₃ is 4/3.

####
*Oxidation number of P in H₄P₂O₇*

Let the oxidation number of P in H₄P₂O₇ is x.

∴ 4(+1) + 2x + 7(-2) = 0

or, x = +5

Thus, the oxidation number of P in H₄P₂O₇ is +5.

∴ 4(+1) + 2x + 7(-2) = 0

or, x = +5

Thus, the oxidation number of P in H₄P₂O₇ is +5.

####
*Oxidation number of Fe in Fe(CO)₅*

Oxidation number CO is zero.

Thus the oxidation number of Fe also zero.

Thus the oxidation number of Fe also zero.

####
*Oxidation number of Cr in [Cr(NH₃)₆]Cl₃*

- Let the oxidation number of Cr in [Cr(NH₃)₆]Cl₃ is x. NH₃ is neutral thus the oxidation number is zero and the oxidation number of Chlorine is -1.

∴ x + 0 +3(-1) = 0

or, x = +3

Thus, the oxidation number of Cr in [Cr(NH₃)₆]Cl₃ is +3.

or, x = +3

Thus, the oxidation number of Cr in [Cr(NH₃)₆]Cl₃ is +3.

####
*Oxidation number of an element in a compound is zero*

*Oxidation number of an element in a compound is zero*

- Some organic compound where the oxidation number of carbon on this compound is zero.

Let, the oxidation number of carbon in Glucose (C12H22O11) is x.

∴ 6x + 12(+1) + 6(-2) = 0

or, x = 0

∴ 6x + 12(+1) + 6(-2) = 0

or, x = 0

Compound | Formula | Oxidation Number |

Sugar | C_{12}H_{22}O_{11} | 0 |

Glucose | C_{6}H_{12}O_{6} | 0 |

Formaldehyde | HCHO | 0 |

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*Questions answers on oxidation number*

*Question*

- Oxidation number of P in Ba(H

_{2}PO

_{2})

_{2}is - (a)+3, (b)+2, (c) +1, (d) -1.

*Answer*

- The oxidation number of Ba is +2, the oxidation number of hydrogen is +1 and the oxidation number of oxygen is -2.

Let the oxidation number of P is x.

∴ (+ 2) + 2{2(+1) + x +2(-2)} = 0

or, 2x - 2 = 0

or, x = +1

∴ (+ 2) + 2{2(+1) + x +2(-2)} = 0

or, 2x - 2 = 0

or, x = +1

*Question*

- Calculate the oxidation number of Iron in [Fe(H

_{2}O)

_{5}(NO)

^{+}]SO

_{4}.

*Answer*

- H₂O is neutral thus the oxidation number is zero, the oxidation number of (NO)ᐩ is +1 and the oxidation number of SO₄ is -2.

Let the oxidation number of Fe in [Fe(H₂O)₅(NO)ᐩ]SO₄ is x.

or, x - 1 = 0

or, x = +1

Thus, the oxidation number of Fe in [Fe(H₂O)₅(NO)ᐩ]SO₄ is +1.

or, x - 1 = 0

or, x = +1

Thus, the oxidation number of Fe in [Fe(H₂O)₅(NO)ᐩ]SO₄ is +1.

*Question*

- What is the Oxidation state of chromium in Cr₂O₅?

*Answer*

- Due to the peroxy linkage oxidation state of Cr in Cr₂O₅ is +6.