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Oxidation Number

Oxidation number of periodic table elements

Oxidation number or state of periodic table elements in a chemical compound or molecule is the formal charges (positive or negative) which assigned to the element if all the bonds in the compounds are ionic. In chemistry, the oxidation number or state is defined as the total number of electrons loses or gains by atoms or ions for the formation of the chemical bond. The less or more electronegative partner of a binary compound arbitrarily assigned positive or negative oxidation numbers or states of the periodic table elements. For example, halogens like fluorine (F), chlorine (Cl), bromine (Br) are highly electronegative but alkali or alkaline metals like sodium (Na), aluminum (Al), potassium (K), calcium (Ca) have highly electropositive. Therefore, halogen atoms commonly show negative and alkali or alkaline metals commonly show positive oxidation state or number.

Periodic table with oxidation number

Calculate oxidation number or state of periodic table chemical elements

Some general rules are used for the calculation of the oxidation numbers of s, p, d, and f-block elements in the periodic table. The s-block elements commonly show +1 and +2 oxidation numbers but p-block elements commonly show +3, +4, -3, -2 and -1 oxidation numbers. One of the most important properties that distinguish transition metals or d-block elements from non-transition elements is variable oxidation numbers or states.

How to find the oxidation number of elements?

To determine or balancing common redox reactions, we used the oxidation number rule because some of the reactions can not be explained by electronic or classical concepts. For example, a water molecule is formed by bonding hydrogen with oxygen, and hydrochloric acid is formed by bonding hydrogen with chlorine. The formation of water (H2O) and hydrochloric acid (HCl) molecules can not be explained from the classical definition but explained easily by oxidation number rules. The general rules which are used in this learning chemistry topic to find the oxidation number of the periodic table elements are given below,

  • Rule 1: The atoms of the diatomic molecules like chlorine (Cl2), oxygen (O2), hydrogen (H2), nitrogen (N2), etc, or of metallic elements like aluminum (Al), iron (Fe), zinc (Zn), copper (Cu), sodium (Na), calcium (Ca), etc are assigned zero oxidation number. Since the same elements of similar electronegativity are involved in the bonding of diatomic molecules.
  • Rule 2: The common oxidation number of hydrogen = +1. In alkali metal hydrides like lithium hydride, sodium hydride, cesium hydride, the oxidation state of hydrogen atom = -1. All the metal in a compound generally possesses a positive oxidation state.
  • Rule 3: The normal oxidation number of oxygen in a compound = -2 but in peroxides like hydrogen peroxide (H2O2) and superoxide, oxygen assign -1 and -1/2 state.
  • Rule 4: The oxidation numbers of the ions in polar molecules calculate by their charge. The algebraic sum of the oxidation numbers of all the atoms in a compound must be zero but in many atomic ions equal to its charge.

Oxidation numbers with examples

Examples of oxidation numbers in a compounds
Compound Oxidation number of elements
H2O2 H = +1 O = -1
CaH2 Ca = +2 H = -1
CHCl3 C = +2 H = +1 Cl = -1
Ba(MnO4)2 Ba = +2 Mn = +7 O = -2
K2MnO4 K = +1 Mn = +6 O = -2
H4P2O7 H = +1 P = +5 O = -2
CH2Cl2 C = 0 H = +1 Cl = -1

Oxidation number of hydrogen

The electron configuration of hydrogen, 1s1. Therefore, hydrogen has a single electron particle in outer quantum shall like alkaline earth metals and just one electron short of the next noble gas helium. Therefore, hydrogen can easily lose one electron to show the oxidation number +1. It can also gain one electron from alkali or alkaline earth metals to show the -1 state.

Alkali and alkaline metals

Alkali and alkaline earth metals are highly electropositive with very low ionization energy. Therefore, alkali and alkaline earth metals always represented positive oxidation numbers. For example, in alkali halides, halogen determines negative oxidation states but alkali and alkaline earth metals show the positive states.

The electrolysis of alkaline hydrides like lithium hydride (LiH), cesium hydride (CsH), and calcium hydride (CaH2) can liberated hydrogen gas at the anode. In sodium hydride (NaH), lithium hydride (LiH), cesium hydride (CsH), and calcium hydride (CaH2), hydrogen assign exceptional oxidation number = -1, since the common state of hydrogen = +1.

Superoxide and peroxide

Alkali and alkaline earth metals react with oxygen to form a list of binary compounds like monoxides (M2O), peroxides (M2O2), and superoxide (MO2). These elements are the only known example of the formation of superoxide.

The oxidation number of oxygen in alkali (lithium, sodium) and alkaline earth metals (magnesium, calcium) peroxide and superoxide are -1 and -½ respectively. Fluorine is more electronegative than oxygen. Therefore, fluorine in the periodic table chart forms monoxide and peroxide compounds with alkali and alkaline earth metals define the oxidation number = -1.

Oxidation number calculation

Common and most stable oxidation number or state of transition metals among the periodic table chemical elements

Oxidation number of Mn in KMnO4

Let the finding oxidation number of manganese (Mn) in potassium permanganate (KMnO4) = x. Therefore, according to the above rule, (+1) + x + 4(-2) = 0; or, x = +7.

Chromium in dichromate ion

Let the oxidation number of chromium in dichromate ion (Cr2O7-2) = x. Therefore, 2x + 7(-2) = -2; or, x = +6.

Both nitrogen in ammonium nitrate

Ammonium nitrate (NH4NO3) present as a cation NH4+ and NO3 ion, let the oxidation number of nitrogen in NH4+ = x and NO3 = y. Therefore, x + 3(+1) = +1; or, x = 0 and y + 3(-2) = -1;
or, y= +5.

Both chlorine atoms in bleaching powder

In bleaching powder, chemical formula Ca(OCl)Cl, one chlorine combine with oxygen to form OCl ion and another chlorine atom form Cl ion with the oxidation number of chlorine = +1 and -1 respectively.

Phosphorus in Pyrophosphoric acid

Let the oxidation state of phosphorus in pyrophosphoric acid ( H4P2O7) = x. Therefore, 4(+1) + 2x + 7(-2) = 0, or x = +5

Sulfur in sulfuric acid

Let the oxidation state or number of sulfur in sulfuric acid (H2SO4) = x. Therefore, 2(+1) + x + 4(-2) = 0; or x = +6.

Metals in coordination compounds

Metal ions ion in a coordination compound possesses two kinds of valency like primary and secondary valency. According to the Werner theory, primary valency is equated with the oxidation state and secondary valency coordination numbers of the coordination complex.

For example, in [Cr(NH3)6]Cl3 complex, the coordination number of chromium = 6 and oxidation number or state of chromium = +3, ammonia (NH3) molecule = 0, and chlorine ion (Cl) = -1. In the iron pentacarbonyl or Fe(CO)5 complex, the oxidation state of carbonyl (CO) and iron have zero.

Oxidation number of carbon in organic Compounds

Oxidation numbers of the list of hydrocarbon or carbon compounds like methane (CH4), methyl chloride (CH3Cl), dichloromethane (CH2Cl2), chloroform (CHCl3), and carbon tetrachloride (CCl4) are -4, -2, 0, +2, +4 respectively. But Sugar, glucose, formaldehyde, etc are examples in organic chemistry where the oxidation number or state of chemical element carbon on these compounds is always zero.

Carbon in acetone compounds

Let the oxidation numbers of carbon in acetone compounds = x and hydrogen and oxygen +1 and -2 respectively. According to the above rule, 3x + 6(+1) + (-2) = 0, or x = -(4/3).

Oxidation number problems

Problem: How to determine the oxidation state of phosphorus in Ba(H2PO2)2?

Solution: According to the rules, the oxidation state hydrogen and oxygen in Ba(H2PO2)2 are +1 and -2 respectively and phosphorus = x. Therefore, (+2) + 2{2(+1) + x + 2(-2)} = 0; or, x = +1.

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

Solution: Let the oxidation number of iron in [Fe(H2O)5NO+]SO4 = x and water, NO+ and sulfate ion = 0, +1, and -2 respectively. Therefore, x + 5(0) + (+1) – 2 = 0; or, x = +1.

Problem: How to finding the oxidation state of chromium in CrO5 in chemistry?

Solution: Due to the peroxy linkage oxidation number or state of chromium in CrO5 = +6.