To characterize the state of elements in compounds, the concept of oxidation state was introduced.

DEFINITION

The number of electrons displaced from an atom of a given element or to an atom of a given element in a compound is called oxidation state.

A positive oxidation state indicates the number of electrons that are displaced from a given atom, and a negative oxidation state indicates the number of electrons that are displaced toward a given atom.

From this definition it follows that in compounds with non-polar bonds the oxidation state of elements is zero. Examples of such compounds are molecules consisting of identical atoms (N 2, H 2, Cl 2).

The oxidation state of metals in the elemental state is zero, since the distribution of electron density in them is uniform.

In simple ionic compounds, the oxidation state of their constituent elements is equal to electric charge, since during the formation of these compounds there is an almost complete transfer of electrons from one atom to another: Na +1 I -1, Mg +2 Cl -1 2, Al +3 F -1 3, Zr +4 Br -1 4.

When determining the oxidation state of elements in compounds with polar covalent bonds, their electronegativity values ​​are compared. Since when a chemical bond is formed, electrons are displaced to the atoms of more electronegative elements, the latter have a negative oxidation state in compounds.

Lowest oxidation state

For elements that exhibit different oxidation states in their compounds, there are concepts of highest (maximum positive) and lowest (minimum negative) oxidation states. Lowest oxidation state chemical element usually numerically equal to the difference between the number of the group in the Periodic Table of D.I. Mendeleev, in which the chemical element is located, and the number 8. For example, nitrogen is in the VA group, which means its lowest oxidation state is (-3): V-VIII = - 3; sulfur is in the VIA group, which means its lowest oxidation state is (-2): VI-VIII = -2, etc.

Examples of problem solving

EXAMPLE 1

DEFINITION

Oxidation state is a quantitative assessment of the state of an atom of a chemical element in a compound, based on its electronegativity.

She accepts both positive and negative values. To indicate the oxidation state of an element in a compound, you need to place an Arabic numeral with the corresponding sign (“+” or “-”) above its symbol.

It should be remembered that the oxidation state is a value that does not have physical meaning, since it does not reflect the real charge of the atom. However, this concept is very widely used in chemistry.

Table of oxidation states of chemical elements

The maximum positive and minimum negative oxidation state can be determined using the Periodic Table D.I. Mendeleev. They are equal to the number of the group in which the element is located and the difference between the value of the “highest” oxidation state and the number 8, respectively.

If we consider chemical compounds more specifically, in substances with non-polar bonds the oxidation state of elements is zero (N 2, H 2, Cl 2).

The oxidation state of metals in the elemental state is zero, since the distribution of electron density in them is uniform.

In simple ionic compounds, the oxidation state of the elements included in them is equal to the electric charge, since during the formation of these compounds there is an almost complete transition of electrons from one atom to another: Na +1 I -1, Mg +2 Cl -1 2, Al +3 F - 1 3 , Zr +4 Br -1 4 .

When determining the oxidation state of elements in compounds with polar covalent bonds, their electronegativity values ​​are compared. Since when a chemical bond is formed, electrons are displaced to the atoms of more electronegative elements, the latter have a negative oxidation state in compounds.

There are elements that are characterized by only one oxidation state value (fluorine, metals of groups IA and IIA, etc.). Fluorine, characterized by highest value electronegativity, in compounds it always has a constant negative oxidation state (-1).

Alkaline and alkaline earth elements, which are characterized by a relatively low electronegativity value, always have a positive oxidation state equal to (+1) and (+2), respectively.

However, there are also chemical elements that are characterized by several oxidation states (sulfur - (-2), 0, (+2), (+4), (+6), etc.).

To make it easier to remember how many and what oxidation states are characteristic of a particular chemical element, use tables of oxidation states of chemical elements, which look like this:

Serial number

Russian / English Name

Chemical symbol

Oxidation state

Hydrogen

Helium

Lithium

Beryllium

(-1), 0, (+1), (+2), (+3)

Carbon

(-4), (-3), (-2), (-1), 0, (+2), (+4)

Nitrogen / Nitrogen

(-3), (-2), (-1), 0, (+1), (+2), (+3), (+4), (+5)

Oxygen

(-2), (-1), 0, (+1), (+2)

Fluorine

Sodium/Sodium

Magnesium / Magnesium

Aluminum

Silicon

(-4), 0, (+2), (+4)

Phosphorus / Phosphorus

(-3), 0, (+3), (+5)

Sulfur/Sulfur

(-2), 0, (+4), (+6)

Chlorine

(-1), 0, (+1), (+3), (+5), (+7), rarely (+2) and (+4)

Argon / Argon

Potassium/Potassium

Calcium

Scandium / Scandium

Titanium

(+2), (+3), (+4)

Vanadium

(+2), (+3), (+4), (+5)

Chrome / Chromium

(+2), (+3), (+6)

Manganese / Manganese

(+2), (+3), (+4), (+6), (+7)

Iron

(+2), (+3), rare (+4) and (+6)

Cobalt

(+2), (+3), rarely (+4)

Nickel

(+2), rare (+1), (+3) and (+4)

Copper

+1, +2, rare (+3)

Gallium

(+3), rare (+2)

Germanium / Germanium

(-4), (+2), (+4)

Arsenic/Arsenic

(-3), (+3), (+5), rarely (+2)

Selenium

(-2), (+4), (+6), rarely (+2)

Bromine

(-1), (+1), (+5), rarely (+3), (+4)

Krypton / Krypton

Rubidium / Rubidium

Strontium / Strontium

Yttrium / Yttrium

Zirconium / Zirconium

(+4), rare (+2) and (+3)

Niobium / Niobium

(+3), (+5), rare (+2) and (+4)

Molybdenum

(+3), (+6), rare (+2), (+3) and (+5)

Technetium / Technetium

Ruthenium / Ruthenium

(+3), (+4), (+8), rare (+2), (+6) and (+7)

Rhodium

(+4), rare (+2), (+3) and (+6)

Palladium

(+2), (+4), rarely (+6)

Silver

(+1), rare (+2) and (+3)

Cadmium

(+2), rare (+1)

Indium

(+3), rare (+1) and (+2)

Tin/Tin

(+2), (+4)

Antimony / Antimony

(-3), (+3), (+5), rarely (+4)

Tellurium / Tellurium

(-2), (+4), (+6), rarely (+2)

(-1), (+1), (+5), (+7), rarely (+3), (+4)

Xenon / Xenon

Cesium

Barium / Barium

Lanthanum / Lanthanum

Cerium

(+3), (+4)

Praseodymium / Praseodymium

Neodymium / Neodymium

(+3), (+4)

Promethium / Promethium

Samarium / Samarium

(+3), rare (+2)

Europium

(+3), rare (+2)

Gadolinium / Gadolinium

Terbium / Terbium

(+3), (+4)

Dysprosium / Dysprosium

Holmium

Erbium

Thulium

(+3), rare (+2)

Ytterbium / Ytterbium

(+3), rare (+2)

Lutetium / Lutetium

Hafnium / Hafnium

Tantalum / Tantalum

(+5), rare (+3), (+4)

Tungsten/Tungsten

(+6), rare (+2), (+3), (+4) and (+5)

Rhenium / Rhenium

(+2), (+4), (+6), (+7), rare (-1), (+1), (+3), (+5)

Osmium / Osmium

(+3), (+4), (+6), (+8), rare (+2)

Iridium / Iridium

(+3), (+4), (+6), rarely (+1) and (+2)

Platinum

(+2), (+4), (+6), rare (+1) and (+3)

Gold

(+1), (+3), rarely (+2)

Mercury

(+1), (+2)

Thalium / Thallium

(+1), (+3), rarely (+2)

Lead/Lead

(+2), (+4)

Bismuth

(+3), rare (+3), (+2), (+4) and (+5)

Polonium

(+2), (+4), rarely (-2) and (+6)

Astatine

Radon / Radon

Francium

Radium / Radium

Actinium

Thorium

Proactinium / Protactinium

Uranium / Uranium

(+3), (+4), (+6), rare (+2) and (+5)

Examples of problem solving

EXAMPLE 1

Answer We will alternately determine the oxidation state of phosphorus in each of the proposed transformation schemes, and then select correct option answer.
  • The oxidation state of phosphorus in phosphine is (-3), and in orthophosphoric acid - (+5). Change in the oxidation state of phosphorus: +3 → +5, i.e. first answer option.
  • The oxidation state of a chemical element in a simple substance is zero. The oxidation degree of phosphorus in the oxide of composition P 2 O 5 is (+5). Change in the oxidation state of phosphorus: 0 → +5, i.e. third answer option.
  • The oxidation degree of phosphorus in the acid composition HPO 3 is (+5), and H 3 PO 2 is (+1). Change in the oxidation state of phosphorus: +5 → +1, i.e. fifth answer option.

EXAMPLE 2

Exercise The oxidation state (-3) of carbon in the compound is: a) CH 3 Cl; b) C 2 H 2; c) HCOH; d) C 2 H 6.
Solution In order to give the correct answer to the question posed, we will alternately determine the degree of carbon oxidation in each of the proposed compounds.

a) the oxidation state of hydrogen is (+1), and that of chlorine is (-1). Let us take the oxidation state of carbon as “x”:

x + 3×1 + (-1) =0;

The answer is incorrect.

b) the oxidation state of hydrogen is (+1). Let us take the oxidation state of carbon as “y”:

2×y + 2×1 = 0;

The answer is incorrect.

c) the oxidation state of hydrogen is (+1), and that of oxygen is (-2). Let us take the oxidation state of carbon as “z”:

1 + z + (-2) +1 = 0:

The answer is incorrect.

d) the oxidation state of hydrogen is (+1). Let us take the oxidation state of carbon as “a”:

2×a + 6×1 = 0;

Correct answer.
Answer Option (d)

When studying ionic and covalent polar chemical bonds, you became familiar with complex substances, consisting of two chemical elements. Such substances are called binary (from the Latin bi - two) or two-element.

Let us recall the typical binary compounds that we cited as an example to consider the mechanisms of formation of ionic and covalent polar chemical bonds: NaCl - sodium chloride and HCl - hydrogen chloride.

In the first case, the bond is ionic: the sodium atom transferred its outer electron to the chlorine atom and turned into an ion with a charge of +1, and the chlorine atom accepted an electron and turned into an ion with a charge of -1. Schematically, the process of converting atoms into ions can be depicted as follows:

In the hydrogen chloride molecule HC1, a chemical bond is formed due to the pairing of unpaired external electrons and the formation of a common electron pair of hydrogen and chlorine atoms:

It is more correct to imagine the formation of a covalent bond in a hydrogen chloride molecule as the overlap of the one-electron s-cloud of the hydrogen atom with the one-electron p-cloud of the chlorine atom:

At chemical interaction the common electron pair is shifted towards the more electronegative chlorine atom: , i.e., the electron will not completely transfer from the hydrogen atom to the chlorine atom, but partially, thereby determining the partial charge of atoms 5 (see § 12): . If we imagine that in the molecule of hydrogen chloride HCl, as well as in sodium chloride NaCl, the electron has completely transferred from the hydrogen atom to the chlorine atom, then they would receive charges +1 and -1: . Such conventional charges are called oxidation states. When defining this concept, it is conventionally assumed that in covalent polar compounds the bonding electrons are completely transferred to a more electronegative atom, and therefore the compounds consist only of positively and negatively charged ions.

The oxidation number can have negative, positive or zero values, which are usually placed above the element symbol at the top, for example:

Those atoms that have accepted electrons from other atoms or to which common electron pairs are displaced, i.e., atoms of more electronegative elements, have a negative oxidation state. Fluorine always has an oxidation state of -1 in all compounds. Oxygen, the second element after fluorine in terms of electronegativity, almost always has an oxidation state of -2, except for compounds with fluorine, for example:

A positive oxidation state is assigned to those atoms that donate their electrons to other atoms or from which common electron pairs are drawn, i.e., atoms of less electronegative elements. Metals in compounds always have a positive oxidation state. For metals of the main subgroups: group I (group IA) in all compounds the oxidation state is +1, group II (group IIA) is +2, group III (group IIIA) is +3, for example:

but in compounds with metals, hydrogen has an oxidation state of -1:

Atoms in molecules have a zero oxidation state simple substances and atoms in a free state, for example:

Close to the concept of “oxidation state” is the concept of “valence,” which you became familiar with when considering covalent chemical bond. However, this is not the same thing.

The concept of “valency” is applicable to substances that have molecular structure. The vast majority organic matter, which you will become familiar with in grade 10, has exactly this structure. In the basic school course, you study inorganic chemistry, the subject of which is substances of both molecular and non-molecular, for example ionic, structure. Therefore, it is preferable to use the concept of “oxidation state”.

What is the difference between valency and oxidation state?

Often valency and oxidation number coincide numerically, but valency does not have a charge sign, but oxidation number does. For example, monovalent hydrogen has the following oxidation states in various substances Oh:

It would seem that monovalent fluorine, the most electronegative element, should have complete coincidence of oxidation state and valency values. After all, his atom is capable of forming only one single covalent bond, since it lacks one electron to complete the outer electron layer. However, there is a difference here too:

The valency and oxidation state differ even more if they do not coincide numerically. For example:

In compounds, the total oxidation state is always zero. Knowing this and the oxidation state of one of the elements, you can find the oxidation state of another element using the formula, for example, a binary compound. So, let's find the oxidation state of chlorine in the compound C1 2 O 7.

Let us denote the oxidation state of oxygen: . Therefore, seven oxygen atoms will have a total negative charge of (-2) × 7 = -14. Then the total charge of two chlorine atoms will be equal to +14, and of one chlorine atom: (+14) : 2 = +7. Therefore, the oxidation state of chlorine is .

Similarly, knowing the oxidation states of elements, you can create a formula for a compound, for example, aluminum carbide (a compound of aluminum and carbon).

It is easy to see that you worked similarly with the concept of “valence” when you derived the formula of a covalent compound or determined the valence of an element from the formula of its compound.

The names of binary compounds are formed from two words - the names of the chemical elements included in their composition. The first word denotes the electronegative part of the compound - a nonmetal; its Latin name with the suffix -id always appears in nominative case. The second word denotes the electropositive part - a metal or less electronegative element; its name always appears in genitive case:

For example: NaCl - sodium chloride, MgS - magnesium sulfide, KH - potassium hydride, CaO - calcium oxide. If an electropositive element exhibits different oxidation states, then this is reflected in the name, indicating the oxidation degree with a Roman numeral, which is placed at the end of the name, for example: - iron (II) oxide (read “iron oxide two”), - iron (III) oxide (read “iron oxide three”).

If a compound consists of two non-metal elements, then the suffix -id is added to the name of the more electronegative of them, and the second component is placed after this in the genitive case. For example: - oxygen fluoride (II), - sulfur oxide (IV) and - sulfur oxide (VI).

In some cases, the number of atoms of elements is indicated using the names of numerals on Greek- mono, di, tri, tetra, penta, hexa, etc. For example: - carbon monoxide, or carbon monoxide (II), - carbon dioxide, or carbon monoxide (IV), - lead tetrachloride, or lead chloride (IV) ).

To chemists different countries understood each other, it was necessary to create a unified terminology and nomenclature of substances. Principles chemical nomenclature were first developed by French chemists A. Lavoisier, A. Fourcroix, L. Guiton de Mervo and C. Berthollet in 1785. Currently, the International Union of Pure and Applied Chemistry (IUPAC) coordinates the activities of scientists from different countries and issues recommendations on the nomenclature of substances and terminology used in chemistry.

Key words and phrases

  1. Binary, or two-element, compounds.
  2. Oxidation state.
  3. Chemical nomenclature.
  4. Determination of oxidation states of elements using the formula.
  5. Drawing up formulas of binary compounds according to the oxidation states of elements.

Working with a computer

  1. Refer to the electronic application. Study the lesson material and complete the assigned tasks.
  2. Find email addresses on the Internet that can serve as additional sources that reveal the content of keywords and phrases in the paragraph. Offer your help to the teacher in preparing a new lesson - send a message by keywords and phrases in the next paragraph.

Questions and tasks

  1. Write down the formulas of nitrogen oxides (II), (V), (I), (III), (IV).
  2. Give the names of binary compounds whose formulas are: a) C1 2 0 7, C1 2 O, C1O 2; b) FeCl 2, FeCl 3; c) MnS, MnO 2, MnF 4, MnO, MnCl 4; r) Cu 2 O, Mg 2 Si, SiCl 4, Na 3 N, FeS.
  3. Find in reference books and dictionaries all possible names of substances with formulas: a) CO 2 and CO; b) SO 2 and SO 3. Explain their etymology. Give two names of these substances according to international nomenclature in accordance with the rules set out in the paragraph.
  4. What other name can be given to ammonia H 3 N?
  5. Find the volume that they have at n. u. 17 g of hydrogen sulfide.
  6. How many molecules are there in this volume?
  7. Calculate the mass of 33.6 m3 of methane CH 2 at air. u. and determine the number of its molecules contained in this volume.
  8. Determine the oxidation state of carbon and write down structural formulas the following substances, knowing that carbon is in organic compounds always tetravalent: methane CH 4, carbon tetrachloride CC1 4, ethane C 2 H 4, acetylene C 2 H 2.

The oxidation state is the conditional charge of the atoms of a chemical element in a compound, calculated from the assumption that all bonds have ion type. Oxidation states can have a positive, negative or zero value, therefore the algebraic sum of the oxidation states of elements in a molecule, taking into account the number of their atoms, is equal to 0, and in an ion - the charge of the ion.

This list of oxidation states shows all known oxidation states of the chemical elements of the periodic table. The list is based on Greenwood's table with all additions. The lines that are highlighted in color contain inert gases whose oxidation state is zero.

1 −1 H +1
2 He
3 Li +1
4 -3 Be +1 +2
5 −1 B +1 +2 +3
6 −4 −3 −2 −1 C +1 +2 +3 +4
7 −3 −2 −1 N +1 +2 +3 +4 +5
8 −2 −1 O +1 +2
9 −1 F +1
10 Ne
11 −1 Na +1
12 Mg +1 +2
13 Al +3
14 −4 −3 −2 −1 Si +1 +2 +3 +4
15 −3 −2 −1 P +1 +2 +3 +4 +5
16 −2 −1 S +1 +2 +3 +4 +5 +6
17 −1 Cl +1 +2 +3 +4 +5 +6 +7
18 Ar
19 K +1
20 Ca +2
21 Sc +1 +2 +3
22 −1 Ti +2 +3 +4
23 −1 V +1 +2 +3 +4 +5
24 −2 −1 Cr +1 +2 +3 +4 +5 +6
25 −3 −2 −1 Mn +1 +2 +3 +4 +5 +6 +7
26 −2 −1 Fe +1 +2 +3 +4 +5 +6
27 −1 Co +1 +2 +3 +4 +5
28 −1 Ni +1 +2 +3 +4
29 Cu +1 +2 +3 +4
30 Zn +2
31 Ga +1 +2 +3
32 −4 Ge +1 +2 +3 +4
33 −3 As +2 +3 +5
34 −2 Se +2 +4 +6
35 −1 Br +1 +3 +4 +5 +7
36 Kr +2
37 Rb +1
38 Sr +2
39 Y +1 +2 +3
40 Zr +1 +2 +3 +4
41 −1 Nb +2 +3 +4 +5
42 −2 −1 Mo +1 +2 +3 +4 +5 +6
43 −3 −1 Tc +1 +2 +3 +4 +5 +6 +7
44 −2 Ru +1 +2 +3 +4 +5 +6 +7 +8
45 −1 Rh +1 +2 +3 +4 +5 +6
46 Pd +2 +4
47 Ag +1 +2 +3
48 Cd +2
49 In +1 +2 +3
50 −4 Sn +2 +4
51 −3 Sb +3 +5
52 −2 Te +2 +4 +5 +6
53 −1 I +1 +3 +5 +7
54 Xe +2 +4 +6 +8
55 Cs +1
56 Ba +2
57 La +2 +3
58 Ce +2 +3 +4
59 Pr +2 +3 +4
60 Nd +2 +3
61 Pm +3
62 Sm +2 +3
63 Eu +2 +3
64 Gd +1 +2 +3
65 Tb +1 +3 +4
66 Dy +2 +3
67 Ho +3
68 Er +3
69 Tm +2 +3
70 Yb +2 +3
71 Lu +3
72 Hf +2 +3 +4
73 −1 Ta +2 +3 +4 +5
74 −2 −1 W +1 +2 +3 +4 +5 +6
75 −3 −1 Re +1 +2 +3 +4 +5 +6 +7
76 −2 −1 Os +1 +2 +3 +4 +5 +6 +7 +8
77 −3 −1 Ir +1 +2 +3 +4 +5 +6
78 Pt +2 +4 +5 +6
79 −1 Au +1 +2 +3 +5
80 Hg +1 +2 +4
81 Tl +1 +3
82 −4 Pb +2 +4
83 −3 Bi +3 +5
84 −2 Po +2 +4 +6
85 −1 At +1 +3 +5
86 Rn +2 +4 +6
87 Fr +1
88 Ra +2
89 Ac +3
90 Th +2 +3 +4
91 Pa +3 +4 +5
92 U +3 +4 +5 +6
93 Np +3 +4 +5 +6 +7
94 Pu +3 +4 +5 +6 +7
95 Am +2 +3 +4 +5 +6
96 Cm +3 +4
97 Bk +3 +4
98 Cf +2 +3 +4
99 Es +2 +3
100 Fm +2 +3
101 MD +2 +3
102 No +2 +3
103 Lr +3
104 Rf +4
105 Db +5
106 Sg +6
107 Bh +7
108 Hs +8

The highest oxidation state of an element corresponds to the group number periodic table, where this element is located (exceptions are: Au+3 (group I), Cu+2 (II), from group VIII the oxidation state +8 can only be found in osmium Os and ruthenium Ru.

Oxidation states of metals in compounds

The oxidation states of metals in compounds are always positive, but if we talk about non-metals, then their oxidation state depends on which atom the element is connected to:

  • if with a non-metal atom, then the oxidation state can be either positive or negative. It depends on the electronegativity of the element's atoms;
  • if with a metal atom, then the oxidation state is negative.

Negative oxidation state of nonmetals

The highest negative oxidation state of non-metals can be determined by subtracting from 8 the number of the group in which the chemical element is located, i.e. the highest positive oxidation state is equal to the number of electrons in the outer layer, which corresponds to the group number.

Please note that the oxidation states of simple substances are 0, regardless of whether it is a metal or a non-metal.

Sources:

  • Greenwood, Norman N.; Earnshaw, A. Chemistry of the Elements - 2nd ed. - Oxford: Butterworth-Heinemann, 1997
  • Green Stable Magnesium(I) Compounds with Mg-Mg Bonds / Jones C.; Stasch A.. - Science Magazine, 2007. - December (issue 318 (No. 5857)
  • Science magazine, 1970. - Vol. 3929. - No. 168. - P. 362.
  • Journal of the Chemical Society, Chemical Communications, 1975. - pp. 760b-761.
  • Irving Langmuir The arrangement of electrons in atoms and molecules. - J.Am Magazine Chem. Soc., 1919. - Issue. 41.

In chemical processes main role atoms and molecules play, the properties of which determine the outcome chemical reactions. One of the important characteristics of an atom is the oxidation number, which simplifies the method of accounting for electron transfer in a particle. How to determine the oxidation state or formal charge of a particle and what rules do you need to know for this?

Definition

Any chemical reaction is caused by the interaction of atoms of different substances. The reaction process and its result depend on the characteristics of the smallest particles.

The term oxidation (oxidation) in chemistry means a reaction during which a group of atoms or one of them loses electrons or gains; in the case of acquisition, the reaction is called “reduction”.

The oxidation state is a quantity that is measured quantitatively and characterizes the redistributed electrons during a reaction. Those. During the process of oxidation, electrons in an atom decrease or increase, redistributing between other interacting particles, and the level of oxidation shows exactly how they are reorganized. This concept is closely related to the electronegativity of particles - their ability to attract and repel free ions.

Determining the level of oxidation depends on the characteristics and properties of a particular substance, so the calculation procedure cannot be unambiguously called easy or complex, but its results help conditionally record the processes of redox reactions. It should be understood that the resulting calculation result is the result of taking into account the transfer of electrons and has no physical meaning, and is not the true charge of the nucleus.

Important to know! Inorganic chemistry often uses the term valence instead of the oxidation state of elements, this is not a mistake, but it should be borne in mind that the second concept is more universal.

The concepts and rules for calculating the movement of electrons are the basis for classification chemicals(nomenclature), descriptions of their properties and drawing up communication formulas. But most often this concept is used to describe and work with redox reactions.

Rules for determining the degree of oxidation

How to find out the oxidation state? When working with redox reactions, it is important to know that the formal charge of a particle will always be equal to the value electron expressed as a numerical value. This feature is due to the assumption that the electron pairs forming a bond are always completely shifted towards more negative particles. It should be understood that we are talking about ionic bonds, and in the case of a reaction at electrons will be divided equally between identical particles.

The oxidation number can have both positive and negative values. The thing is that during the reaction the atom must become neutral, and for this it is necessary either to add a certain number of electrons to the ion, if it is positive, or to take them away if it is negative. To indicate this concept When writing formulas, an Arabic numeral with the corresponding sign is usually written above the element designation. For example, or etc.

You should know that the formal charge of metals will always be positive, and in most cases, you can use the periodic table to determine it. There are a number of features that must be taken into account in order to determine the indicators correctly.

Oxidation degree:

Having remembered these features, it will be quite simple to determine the oxidation number of elements, regardless of the complexity and number of atomic levels.

Useful video: determining the oxidation state

Mendeleev's periodic table contains almost all the necessary information for working with chemical elements. For example, schoolchildren use only it to describe chemical reactions. So, in order to determine the maximum positive and negative values ​​of the oxidation number, you need to check the designation of the chemical element in the table:

  1. The maximum positive is the number of the group in which the element is located.
  2. The maximum negative oxidation state is the difference between the maximum positive boundary and the number 8.

Thus, it is enough to simply find out the extreme boundaries of the formal charge of a particular element. This action can be performed using calculations based on the periodic table.

Important to know! One element can simultaneously have several different oxidation rates.

There are two main methods for determining the level of oxidation, examples of which are presented below. The first of them is a method that requires knowledge and ability to apply the laws of chemistry. How to arrange oxidation states using this method?

Rule for determining oxidation states

To do this you need:

  1. Determine whether a given substance is elemental and whether it is outside the bond. If so, then its oxidation number will be 0, regardless of the composition of the substance (individual atoms or multi-level atomic compounds).
  2. Determine whether the substance in question consists of ions. If so, then the degree of oxidation will be equal to their charge.
  3. If the substance in question is metal, then look at the indicators of other substances in the formula and calculate the metal readings using arithmetic operations.
  4. If the entire compound has one charge (essentially it is the sum of all particles of the elements represented), then it is enough to determine the indicators of simple substances, then subtract them from the total and get the metal data.
  5. If the relationship is neutral, then the total sum must be zero.

As an example, consider combining with an aluminum ion whose net charge is zero. The rules of chemistry confirm the fact that the Cl ion has an oxidation number of -1, and in this case there are three of them in the compound. This means that the Al ion must be +3 for the entire compound to be neutral.

This method is very good, since the correctness of the solution can always be checked by adding all the oxidation levels together.

The second method can be used without knowledge of chemical laws:

  1. Find data on particles for which there are no strict rules and the exact number of their electrons is unknown (this can be done by exclusion).
  2. Find out the indicators of all other particles and then find the desired particle from the total by subtraction.

Let's consider the second method using the example of the substance Na2SO4, in which the sulfur atom S is not determined, it is only known that it is different from zero.

To find what all oxidation states are equal to:

  1. Find known elements, keeping in mind traditional rules and exceptions.
  2. Na ion = +1, and each oxygen = -2.
  3. Multiply the number of particles of each substance by their electrons to obtain the oxidation states of all atoms except one.
  4. Na2SO4 contains 2 sodium and 4 oxygen; when multiplied, it turns out: 2 X +1 = 2 is the oxidation number of all sodium particles and 4 X -2 = -8 - oxygen.
  5. Add the results obtained 2+(-8) =-6 - this is the total charge of the compound without the sulfur particle.
  6. Represent the chemical notation as an equation: sum of known data + unknown number = total charge.
  7. Na2SO4 is represented as follows: -6 + S = 0, S = 0 + 6, S = 6.

Thus, to use the second method, it is enough to know the simple laws of arithmetic.