Determine the oxidation state of atoms. Valency of chemical elements

Chemistry preparation for cancer and DPA
Comprehensive edition

PART AND

GENERAL CHEMISTRY

CHEMICAL BONDING AND STRUCTURE OF SUBSTANCE

Oxidation state

The oxidation state is the conditional charge on an atom in a molecule or crystal that would arise on it when all the polar bonds created by it were ionic in nature.

Unlike valence, oxidation states can be positive, negative, or zero. In simple ionic compounds, the oxidation state coincides with the charges of the ions. For example, in sodium chloride NaCl (Na + Cl - ) Sodium has an oxidation state of +1, and Chlorine -1, in calcium oxide CaO (Ca +2 O -2). Calcium exhibits an oxidation state of +2, and Oxysene - -2. This rule applies to all basic oxides: the oxidation state of a metal element is equal to the charge of the metal ion (Sodium +1, Barium +2, Aluminum +3), and the oxidation state of Oxygen is -2. The oxidation state is indicated by Arabic numerals, which are placed above the symbol of the element, like valence, and the sign of the charge is indicated first, and then its numerical value:

If the modulus of the oxidation state is equal to one, then the number “1” can be omitted and only the sign can be written: Na + Cl - .

Oxidation number and valence are related concepts. In many compounds, the absolute value of the oxidation state of elements coincides with their valency. However, there are many cases where the valence differs from the oxidation state.

In simple substances - non-metals, there is a covalent non-polar bond; the shared electron pair is displaced to one of the atoms, therefore the oxidation state of elements in simple substances is always zero. But the atoms are connected to each other, that is, they exhibit a certain valence, as, for example, in oxygen the valence of Oxygen is II, and in nitrogen the valence of Nitrogen is III:

In the hydrogen peroxide molecule, the valency of Oxygen is also II, and that of Hydrogen is I:

Definition of possible degrees oxidation of elements

The oxidation states that elements can exhibit in various compounds can in most cases be determined by the structure of the outer electronic level or by the element’s place in the Periodic Table.

Atoms of metallic elements can only donate electrons, so they exhibit positive oxidation states in compounds. Its absolute value in many cases (except d -elements) is equal to the number of electrons in the outer level, that is, the group number in the Periodic Table. Atoms d -elements can also donate electrons from a higher level, namely from unfilled d -orbitals. Therefore for d -elements, determining all possible oxidation states is much more difficult than for s- and p-elements. It is safe to say that the majority d -elements exhibit an oxidation state of +2 due to electrons in the outer electron level, and the maximum oxidation state in most cases is equal to the group number.

Atoms of nonmetallic elements can exhibit both positive and negative oxidation states, depending on which atom of the element they form a bond with. If an element is more electronegative, then it exhibits a negative oxidation state, and if it is less electronegative, it exhibits a positive oxidation state.

The absolute value of the oxidation state of non-metallic elements can be determined by the structure of the outer electronic layer. An atom is capable of accepting so many electrons that eight electrons are located on its outer level: non-metallic elements of group VII accept one electron and exhibit an oxidation state of -1, group VI - two electrons and exhibit an oxidation state of -2, etc.

Non-metallic elements are capable of donating a different number of electrons: a maximum of as many as are located at the outer energy level. In other words, the maximum oxidation state of non-metallic elements is equal to the group number. Due to the circulation of electrons at the outer level of atoms, the number of unpaired electrons that an atom can give up in chemical reactions varies, so non-metallic elements are capable of exhibiting different intermediate values ​​of oxidation state.

Possible oxidation states s- and p-elements

PS Group
Highest oxidation state
Intermediate oxidation state
Lower oxidation state

Determination of oxidation states in compounds

Any electrically neutral molecule, therefore the sum of the oxidation states of the atoms of all elements must be equal to zero. Let us determine the degree of oxidation in sulfur(I) V) oxide SO 2 tauphosphorus (V) sulfide P 2 S 5.

Sulfur(I V) oxide SO 2 formed by atoms of two elements. Of these, Oxygen has the largest electronegativity, so Oxygen atoms will have a negative oxidation state. For Oxygen it is equal to -2. In this case, Sulfur has a positive oxidation state. Sulfur can exhibit different oxidation states in different compounds, so in this case it must be calculated. In a molecule SO 2 two Oxygen atoms with an oxidation state of -2, so the total charge of the Oxygen atoms is -4. In order for the molecule to be electrically neutral, the Sulfur atom has to completely neutralize the charge of both Oxygen atoms, therefore the oxidation state of Sulfur is +4:

In the molecule there is phosphorus ( V) sulfide P 2 S 5 The more electronegative element is Sulfur, that is, it exhibits a negative oxidation state, and Phosphorus has a positive oxidation state. For Sulfur, the negative oxidation state is only 2. Together, the five atoms of Sulfur carry a negative charge of -10. Therefore two Phosphorus atoms have to neutralize this charge with a total charge of +10. Since there are two Phosphorus atoms in the molecule, each must have an oxidation state of +5:

It is more difficult to calculate the oxidation state in non-binary compounds - salts, bases and acids. But for this you should also use the principle of electrical neutrality, and also remember that in most compounds the oxidation state of Oxygen is -2, Hydrogen +1.

Let's look at this using potassium sulfate as an example. K2SO4. The oxidation state of Potassium in compounds can only be +1, and Oxygen -2:

Using the principle of electrical neutrality, we calculate the oxidation state of Sulfur:

2(+1) + 1 (x) + 4 (-2) = 0, whence x = +6.

When determining the oxidation states of elements in compounds, the following rules should be followed:

1. The oxidation state of an element in a simple substance is zero.

2. Fluorine is the most electronegative chemical element, therefore the oxidation state of Fluorine in all compounds is equal to -1.

3. Oxygen is the most electronegative element after Fluorine, therefore the oxidation state of Oxygen in all compounds except fluorides is negative: in most cases it is -2, and in peroxides - -1.

4. The oxidation state of Hydrogen in most compounds is +1, and in compounds with metal elements (hydrides) - -1.

5. The oxidation state of metals in compounds is always positive.

6. A more electronegative element always has a negative oxidation state.

7. The sum of the oxidation states of all atoms in a molecule is zero.

In studying ionic and covalent polar chemical bonds, you were introduced to complex substances consisting of two chemical elements. Such substances are called bi-paired (from the Latin bi - “two”) or two-element.

Let us recall the typical bpnar compounds that we cited as an example to consider the mechanisms of formation of ionic and covalent polar chemical bonds: NaHl - 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 became an ion with a charge of -1. Schematically, the process of converting atoms into ions can be depicted as follows:

In the HCl molecule, the bond is formed due to the pairing of unpaired outer 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:

During a chemical interaction, the shared electron pair is shifted towards the more electronegative chlorine atom:

Such conditional charges are called oxidation state. 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.

is the conditional charge of the atoms of a chemical element in a compound, calculated on the basis of the assumption that all compounds (both ionic and covalently polar) consist only of 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 received electrons from other atoms or to which common electron pairs are displaced, that is, atoms of more electronegative elements, have a negative oxidation state. Fluorine always has an oxidation state of -1 in all compounds. Oxygen, the second most electronegative element after fluorine, 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 shared electron pairs are drawn, that is, atoms of less electronegative elements. Metals always have a positive oxidation state. Metals of main subgroups:

Group I in all compounds the oxidation state is +1,
Group II is equal to +2. Group III - +3, for example:

In compounds, the total oxidation state is always zero. Knowing this and the oxidation state of one of the elements, you can always find the oxidation state of another element using the formula of a binary compound. For example, let's find the oxidation state of chlorine in the compound Cl2O2. Let's denote the oxidation state -2
oxygen: Cl2O2. Therefore, seven oxygen atoms will have a total negative charge (-2) 7 =14. Then the total charge of two chlorine atoms will be +14, and of one chlorine atom:
(+14):2 = +7.

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). Let's write down the signs of aluminum and carbon next to AlC, with the sign of aluminum first, since it is a metal. Using the periodic table of elements, we determine the number of outer electrons: Al has 3 electrons, C has 4. The aluminum atom will give up its 3 outer electrons to carbon and receive an oxidation state of +3, equal to the charge of the ion. The carbon atom, on the contrary, will take the 4 electrons missing to the “cherished eight” and receive an oxidation state of -4.

Let's write these values ​​into the formula: AlC, and find the least common multiple for them, it is equal to 12. Then we calculate the indices:

Knowing the oxidation states of elements is also necessary in order to be able to correctly name a chemical compound.

Names of binary compounds consist of two words - the names of the chemical elements that form them. The first word denotes the electronegative part of the compound - nonmetal; its Latin name with the suffix -id is always in the nominative case. The second word denotes the electropositive part - a metal or less electronegative element; its name is always in the genitive case. If an electropositive element exhibits different degrees of oxidation, then this is reflected in the name, indicating the degree of oxidation with a Roman numeral, which is placed at the end.

In order for chemists from different countries to understand each other, it was necessary to create a unified terminology and nomenclature of substances. The principles of chemical nomenclature were first developed by French chemists A. Lavoisier, A. Fourqutois, L. Guiton and C. Berthollet in 1785. Currently, the International Union of Pure and Applied Chemistry (IUPAC) coordinates the activities of scientists from several countries and issues recommendations on the nomenclature of substances and terminology used in chemistry.

Topics of the Unified State Examination codifier: Electronegativity. Oxidation state and valency of chemical elements.

When atoms interact and form, electrons between them are in most cases unevenly distributed, since the properties of the atoms differ. More electronegative the atom attracts electron density more strongly to itself. An atom that has attracted electron density to itself acquires a partial negative charge δ — , its “partner” is a partial positive charge δ+ . If the difference in electronegativity of the atoms forming a bond does not exceed 1.7, we call the bond covalent polar . If the difference in electronegativities forming a chemical bond exceeds 1.7, then we call such a bond ionic .

Oxidation state is the auxiliary conditional charge of an element atom in a compound, calculated from the assumption that all compounds consist of ions (all polar bonds are ionic).

What does "conditional charge" mean? We simply agree that we will simplify things a little: we will consider any polar bonds to be completely ionic, and we will assume that the electron is completely leaving or coming from one atom to another, even if in fact this is not the case. And a conditionally electron leaves from a less electronegative atom to a more electronegative one.

For example, in the H-Cl bond we believe that hydrogen conditionally “gave up” an electron, and its charge became +1, and chlorine “accepted” an electron, and its charge became -1. In fact, there are no such total charges on these atoms.

Surely, you have a question - why invent something that doesn’t exist? This is not an insidious plan of chemists, everything is simple: this model is very convenient. Ideas about the oxidation state of elements are useful when compiling classifications chemical substances, description of their properties, compilation of formulas of compounds and nomenclature. Oxidation states are especially often used when working with redox reactions.

There are oxidation states higher, inferior And intermediate.

Higher the oxidation state is equal to the group number with a plus sign.

Lowest is defined as the group number minus 8.

AND intermediate An oxidation number is almost any whole number ranging from the lowest oxidation state to the highest.

For example, nitrogen is characterized by: the highest oxidation state is +5, the lowest 5 - 8 = -3, and intermediate oxidation states from -3 to +5. For example, in hydrazine N 2 H 4 the oxidation state of nitrogen is intermediate, -2.

Most often, the oxidation state of atoms in complex substances is indicated first by a sign, then by a number, for example +1, +2, -2 etc. When talking about the charge of an ion (assuming that the ion actually exists in a compound), then first indicate the number, then the sign. For example: Ca 2+ , CO 3 2- .

To find oxidation states, use the following rules :

1. Oxidation state of atoms in simple substances equal to zero;
2. IN neutral molecules the algebraic sum of oxidation states is zero, for ions this sum is equal to the charge of the ion;
3. Oxidation state alkali metals (elements of group I of the main subgroup) in compounds is +1, oxidation state alkaline earth metals (elements of group II of the main subgroup) in compounds is +2; oxidation state aluminum in connections it is equal to +3;
4. Oxidation state hydrogen in compounds with metals (- NaH, CaH 2, etc.) is equal to -1 ; in compounds with non-metals () +1 ;
5. Oxidation state oxygen equal to -2 . Exception make up peroxides– compounds containing the –O-O- group, where the oxidation state of oxygen is equal to -1 , and some other compounds ( superoxides, ozonides, oxygen fluorides OF 2 and etc.);
6. Oxidation state fluoride in all complex substances is equal -1 .

Listed above are situations when we consider the oxidation state constant . All other chemical elements have an oxidation state — variable, and depends on the order and type of atoms in the compound.

Examples:

Exercise: determine the oxidation states of the elements in the potassium dichromate molecule: K 2 Cr 2 O 7 .

Solution: The oxidation state of potassium is +1, the oxidation state of chromium is denoted as X, the oxidation state of oxygen is -2. The sum of all oxidation states of all atoms in a molecule is equal to 0. We get the equation: +1*2+2*x-2*7=0. Solving it, we get the oxidation state of chromium +6.

In binary compounds, the more electronegative element has a negative oxidation state, and the less electronegative element has a positive oxidation state.

note that The concept of oxidation state is very arbitrary! The oxidation number does not indicate the real charge of an atom and has no real physical meaning. This is a simplified model that works effectively when we need, for example, to equalize the coefficients in the equation of a chemical reaction, or to algorithmize the classification of substances.

Oxidation number is not valence! The oxidation state and valency do not coincide in many cases. For example, the valence of hydrogen in the simple substance H2 is equal to I, and the oxidation state, according to rule 1, is equal to 0.

These are the basic rules that will help you determine the oxidation state of atoms in compounds in most cases.

In some situations, you may have difficulty determining the oxidation state of an atom. Let's look at some of these situations and look at how to resolve them:

1. In double (salt-like) oxides, the degree of an atom is usually two oxidation states. For example, in iron scale Fe 3 O 4, iron has two oxidation states: +2 and +3. Which one should I indicate? Both. To simplify, we can imagine this compound as a salt: Fe(FeO 2) 2. In this case, the acidic residue forms an atom with an oxidation state of +3. Or the double oxide can be represented as follows: FeO*Fe 2 O 3.
2. In peroxo compounds, the oxidation state of oxygen atoms connected by covalent nonpolar bonds, as a rule, changes. For example, in hydrogen peroxide H 2 O 2 and alkali metal peroxides, the oxidation state of oxygen is -1, because one of the bonds is covalent nonpolar (H-O-O-H). Another example is peroxomonosulfuric acid (Caro acid) H 2 SO 5 (see figure) contains two oxygen atoms with an oxidation state of -1, the remaining atoms with an oxidation state of -2, so the following entry will be more understandable: H 2 SO 3 (O2). Chromium peroxo compounds are also known - for example, chromium (VI) peroxide CrO(O 2) 2 or CrO 5, and many others.
3. Another example of compounds with ambiguous oxidation states is superoxides (NaO 2) and salt-like ozonides KO 3. In this case, it is more appropriate to talk about the molecular ion O 2 with a charge of -1 and O 3 with a charge of -1. The structure of such particles is described by some models that are taught in the Russian curriculum in the first years of chemical universities: MO LCAO, the method of superimposing valence schemes, etc.
4. In organic compounds, the concept of oxidation state is not very convenient to use, because There are a large number of covalent nonpolar bonds between carbon atoms. However, if you draw the structural formula of a molecule, the oxidation state of each atom can also be determined by the type and number of atoms to which that atom is directly bonded. For example, the oxidation state of primary carbon atoms in hydrocarbons is -3, for secondary atoms -2, for tertiary atoms -1, and for quaternary atoms - 0.

Let's practice determining the oxidation state of atoms in organic compounds. To do this, it is necessary to draw the complete structural formula of the atom and highlight the carbon atom with its closest environment - the atoms with which it is directly connected.

• To simplify calculations, you can use the solubility table - it shows the charges of the most common ions. In most Russian chemistry exams (USE, GIA, DVI), the use of a solubility table is permitted. This is a ready-made cheat sheet, which in many cases can significantly save time.
• When calculating the oxidation state of elements in complex substances, we first indicate the oxidation states of elements that we know for sure (elements with a constant oxidation state), and denote the oxidation state of elements with a variable oxidation state as x. The sum of all charges of all particles is zero in a molecule or equal to the charge of an ion in an ion. From this data it is easy to create and solve an equation.

Many school textbooks and manuals teach how to create formulas based on valencies, even for compounds with ionic bonds. To simplify the procedure for drawing up formulas, this, in our opinion, is acceptable. But you need to understand that this is not entirely correct due to the above reasons.

A more universal concept is the concept of oxidation state. Using the values ​​of the oxidation states of atoms, as well as the valency values, you can compose chemical formulas and write down formula units.

Oxidation state- this is the conditional charge of an atom in a particle (molecule, ion, radical), calculated in the approximation that all bonds in the particle are ionic.

Before determining oxidation states, it is necessary to compare the electronegativity of the bonded atoms. An atom with a higher electronegativity value has a negative oxidation state, and an atom with a lower electronegativity has a positive oxidation state.

In order to objectively compare the electronegativity values ​​of atoms when calculating oxidation states, in 2013 IUPAC recommended using the Allen scale.

* So, for example, according to the Allen scale, the electronegativity of nitrogen is 3.066, and chlorine is 2.869.

Let us illustrate the above definition with examples. Let's compose the structural formula of a water molecule.

Covalent polar O-H bonds are indicated in blue.

Let's imagine that both bonds are not covalent, but ionic. If they were ionic, then one electron would transfer from each hydrogen atom to the more electronegative oxygen atom. Let's denote these transitions with blue arrows.

*In thatexample, the arrow serves to visually illustrate the complete transfer of electrons, and not to illustrate the inductive effect.

It is easy to notice that the number of arrows shows the number of electrons transferred, and their direction indicates the direction of electron transfer.

There are two arrows directed at the oxygen atom, which means that two electrons are transferred to the oxygen atom: 0 + (-2) = -2. A charge of -2 is formed on the oxygen atom. This is the oxidation state of oxygen in a water molecule.

Each hydrogen atom loses one electron: 0 - (-1) = +1. This means that hydrogen atoms have an oxidation state of +1.

The sum of oxidation states always equals the total charge of the particle.

For example, the sum of oxidation states in a water molecule is equal to: +1(2) + (-2) = 0. The molecule is an electrically neutral particle.

If we calculate the oxidation states in an ion, then the sum of the oxidation states is, accordingly, equal to its charge.

The oxidation state value is usually indicated in the upper right corner of the element symbol. Moreover, the sign is written in front of the number. If the sign comes after the number, then this is the charge of the ion.

For example, S -2 is a sulfur atom in the oxidation state -2, S 2- is a sulfur anion with a charge of -2.

S +6 O -2 4 2- - values ​​of the oxidation states of atoms in the sulfate anion (the charge of the ion is highlighted in green).

Now consider the case when the compound has mixed bonds: Na 2 SO 4. The bond between the sulfate anion and sodium cations is ionic, the bonds between the sulfur atom and the oxygen atoms in the sulfate ion are covalent polar. Let's write down the graphic formula of sodium sulfate, and use arrows to indicate the direction of electron transition.

*Structural formula displays the order of covalent bonds in a particle (molecule, ion, radical). Structural formulas are used only for particles with covalent bonds. For particles with ionic bonds, the concept of a structural formula has no meaning. If the particle contains ionic bonds, then a graphical formula is used.

We see that six electrons leave the central sulfur atom, which means the oxidation state of sulfur is 0 - (-6) = +6.

The terminal oxygen atoms each take two electrons, which means their oxidation states are 0 + (-2) = -2

The bridging oxygen atoms each accept two electrons and have an oxidation state of -2.

It is also possible to determine the degree of oxidation using a structural-graphical formula, where covalent bonds are indicated by dashes, and the charge of ions is indicated.

In this formula, the bridging oxygen atoms already have single negative charges and an additional electron comes to them from the sulfur atom -1 + (-1) = -2, which means their oxidation states are equal to -2.

The degree of oxidation of sodium ions is equal to their charge, i.e. +1.

Let us determine the oxidation states of elements in potassium superoxide (superoxide). To do this, let’s create a graphical formula for potassium superoxide and show the redistribution of electrons with an arrow. The O-O bond is a covalent non-polar bond, so it does not indicate the redistribution of electrons.

* Superoxide anion is a radical ion. The formal charge of one oxygen atom is -1, and the other, with an unpaired electron, is 0.

We see that the oxidation state of potassium is +1. The oxidation state of the oxygen atom written opposite potassium in the formula is -1. The oxidation state of the second oxygen atom is 0.

In the same way, you can determine the degree of oxidation using the structural-graphic formula.

The circles indicate the formal charges of the potassium ion and one of the oxygen atoms. In this case, the values ​​of formal charges coincide with the values ​​of oxidation states.

Since both oxygen atoms in the superoxide anion have different oxidation states, we can calculate arithmetic mean oxidation state oxygen.

It will be equal to / 2 = - 1/2 = -0.5.

Values ​​for arithmetic mean oxidation states are usually indicated in gross formulas or formula units to show that the sum of the oxidation states is equal to the total charge of the system.

For the case with superoxide: +1 + 2(-0.5) = 0

It is easy to determine oxidation states using electron-dot formulas, in which lone electron pairs and electrons of covalent bonds are indicated by dots.

Oxygen is an element of group VIA, therefore its atom has 6 valence electrons. Let's imagine that the bonds in a water molecule are ionic, in this case the oxygen atom would receive an octet of electrons.

The oxidation state of oxygen is correspondingly equal to: 6 - 8 = -2.

A hydrogen atoms: 1 - 0 = +1

The ability to determine oxidation states using graphic formulas is invaluable for understanding the essence of this concept; this skill will also be required in a course in organic chemistry. If we are dealing with inorganic substances, then it is necessary to be able to determine oxidation states using molecular formulas and formula units.

To do this, first of all you need to understand that oxidation states can be constant and variable. Elements exhibiting constant oxidation states must be remembered.

Any chemical element is characterized by higher and lower oxidation states.

Lowest oxidation state- this is the charge that an atom acquires as a result of receiving the maximum number of electrons on the outer electron layer.

In view of this, the lowest oxidation state has a negative value, with the exception of metals, whose atoms never accept electrons due to low electronegativity values. Metals have a lowest oxidation state of 0.

Most nonmetals of the main subgroups try to fill their outer electron layer with up to eight electrons, after which the atom acquires a stable configuration ( octet rule). Therefore, in order to determine the lowest oxidation state, it is necessary to understand how many valence electrons an atom lacks to reach the octet.

For example, nitrogen is a group VA element, which means that the nitrogen atom has five valence electrons. The nitrogen atom is three electrons short of the octet. This means the lowest oxidation state of nitrogen is: 0 + (-3) = -3

Valence is a complex concept. This term underwent a significant transformation simultaneously with the development of the theory of chemical bonding. Initially, valency was the ability of an atom to attach or replace a certain number of other atoms or atomic groups to form a chemical bond.

A quantitative measure of the valence of an element’s atom was the number of hydrogen or oxygen atoms (these elements were considered mono- and divalent, respectively) that the element attaches to form a hydride of the formula EH x or an oxide of the formula E n O m.

Thus, the valence of the nitrogen atom in the ammonia molecule NH 3 is equal to three, and the sulfur atom in the H 2 S molecule is equal to two, since the valence of the hydrogen atom is equal to one.

In the compounds Na 2 O, BaO, Al 2 O 3, SiO 2, the valencies of sodium, barium and silicon are 1, 2, 3 and 4, respectively.

The concept of valency was introduced into chemistry before the structure of the atom became known, namely in 1853 by the English chemist Frankland. It has now been established that the valence of an element is closely related to the number of outer electrons of the atoms, since the electrons of the inner shells of the atoms do not participate in the formation of chemical bonds.

In the electronic theory of covalent bonds it is believed that valence of an atom is determined by the number of its unpaired electrons in the ground or excited state, participating in the formation of common electron pairs with electrons of other atoms.

For some elements, valence is a constant value. Thus, sodium or potassium in all compounds is monovalent, calcium, magnesium and zinc are divalent, aluminum is trivalent, etc. But most chemical elements exhibit variable valency, which depends on the nature of the partner element and the conditions of the process. Thus, iron can form two compounds with chlorine - FeCl 2 and FeCl 3, in which the valence of iron is 2 and 3, respectively.

Oxidation state- a concept that characterizes the state of an element in a chemical compound and its behavior in redox reactions; numerically, the oxidation state is equal to the formal charge that can be assigned to an element, based on the assumption that all the electrons in each of its bonds have transferred to a more electronegative atom.

Electronegativity- a measure of the ability of an atom to acquire a negative charge when forming a chemical bond or the ability of an atom in a molecule to attract valence electrons involved in the formation of a chemical bond. Electronegativity is not an absolute value and is calculated by various methods. Therefore, the electronegativity values ​​given in different textbooks and reference books may differ.

Table 2 shows the electronegativity of some chemical elements on the Sanderson scale, and Table 3 shows the electronegativity of elements on the Pauling scale.

The electronegativity value is given below the symbol of the corresponding element. The higher the numerical value of an atom's electronegativity, the more electronegative the element is. The most electronegative is the fluorine atom, the least electronegative is the rubidium atom. In a molecule formed by atoms of two different chemical elements, the formal negative charge will be on the atom whose numerical value of electronegativity is higher. Thus, in a molecule of sulfur dioxide SO2, the electronegativity of the sulfur atom is 2.5, and the electronegativity of the oxygen atom is greater - 3.5. Therefore, the negative charge will be on the oxygen atom, and the positive charge will be on the sulfur atom.

In the ammonia molecule NH 3, the electronegativity value of the nitrogen atom is 3.0, and that of the hydrogen atom is 2.1. Therefore, the nitrogen atom will have a negative charge, and the hydrogen atom will have a positive charge.

You should clearly know the general trends in electronegativity changes. Since an atom of any chemical element tends to acquire a stable configuration of the outer electronic layer - an octet shell of an inert gas, the electronegativity of elements in a period increases, and in a group the electronegativity generally decreases with increasing atomic number of the element. Therefore, for example, sulfur is more electronegative compared to phosphorus and silicon, and carbon is more electronegative compared to silicon.

When drawing up formulas for compounds consisting of two non-metals, the more electronegative of them is always placed to the right: PCl 3, NO 2. There are some historical exceptions to this rule, for example NH 3, PH 3, etc.

The oxidation number is usually indicated by an Arabic numeral (with a sign in front of the number) located above the element symbol, for example:

To determine the degree of oxidation of atoms in chemical compounds, the following rules are followed:

1. The oxidation state of elements in simple substances is zero.
2. The algebraic sum of the oxidation states of atoms in a molecule is zero.
3. Oxygen in compounds exhibits mainly an oxidation state of –2 (in oxygen fluoride OF 2 + 2, in metal peroxides such as M 2 O 2 –1).
4. Hydrogen in compounds exhibits an oxidation state of + 1, with the exception of hydrides of active metals, for example, alkali or alkaline earth ones, in which the oxidation state of hydrogen is – 1.
5. For monoatomic ions, the oxidation state is equal to the charge of the ion, for example: K + - +1, Ba 2+ - +2, Br – - –1, S 2– - –2, etc.
6. In compounds with a covalent polar bond, the oxidation state of the more electronegative atom has a minus sign, and the less electronegative atom has a plus sign.
7. In organic compounds, the oxidation state of hydrogen is +1.

Let us illustrate the above rules with several examples.

Example 1. Determine the degree of oxidation of elements in the oxides of potassium K 2 O, selenium SeO 3 and iron Fe 3 O 4.

Potassium oxide K 2 O. The algebraic sum of the oxidation states of atoms in a molecule is zero. The oxidation state of oxygen in oxides is –2. Let us denote the oxidation state of potassium in its oxide as n, then 2n + (–2) = 0 or 2n = 2, hence n = +1, i.e., the oxidation state of potassium is +1.

Selenium oxide SeO 3. The SeO 3 molecule is electrically neutral. The total negative charge of the three oxygen atoms is –2 × 3 = –6. Therefore, to reduce this negative charge to zero, the oxidation state of selenium must be +6.

Fe3O4 molecule electrically neutral. The total negative charge of the four oxygen atoms is –2 × 4 = –8. To equalize this negative charge, the total positive charge on the three iron atoms must be +8. Therefore, one iron atom must have a charge of 8/3 = +8/3.

It should be emphasized that the oxidation state of an element in a compound can be a fractional number. Such fractional oxidation states are not meaningful when explaining bonding in a chemical compound, but can be used to construct equations for redox reactions.

Example 2. Determine the degree of oxidation of elements in the compounds NaClO 3, K 2 Cr 2 O 7.

The NaClO 3 molecule is electrically neutral. The oxidation state of sodium is +1, the oxidation state of oxygen is –2. Let us denote the oxidation state of chlorine as n, then +1 + n + 3 × (–2) = 0, or +1 + n – 6 = 0, or n – 5 = 0, hence n = +5. Thus, the oxidation state of chlorine is +5.

The K 2 Cr 2 O 7 molecule is electrically neutral. The oxidation state of potassium is +1, the oxidation state of oxygen is –2. Let us denote the oxidation state of chromium as n, then 2 × 1 + 2n + 7 × (–2) = 0, or +2 + 2n – 14 = 0, or 2n – 12 = 0, 2n = 12, hence n = +6. Thus, the oxidation state of chromium is +6.

Example 3. Let us determine the degree of oxidation of sulfur in the sulfate ion SO 4 2–. The SO 4 2– ion has a charge of –2. The oxidation state of oxygen is –2. Let us denote the oxidation state of sulfur as n, then n + 4 × (–2) = –2, or n – 8 = –2, or n = –2 – (–8), hence n = +6. Thus, the oxidation state of sulfur is +6.

It should be remembered that the oxidation state is sometimes not equal to the valency of a given element.

For example, the oxidation states of the nitrogen atom in the ammonia molecule NH 3 or in the hydrazine molecule N 2 H 4 are –3 and –2, respectively, while the valency of nitrogen in these compounds is three.

The maximum positive oxidation state for elements of the main subgroups, as a rule, is equal to the group number (exceptions: oxygen, fluorine and some other elements).

The maximum negative oxidation state is 8 - the group number.

Training tasks

1. In which compound the oxidation state of phosphorus is +5?

1) HPO 3
2) H3PO3
3) Li 3 P
4) AlP

2. In which compound does the oxidation state of phosphorus equal to –3?

1) HPO 3
2) H3PO3
3) Li 3 PO 4
4) AlP

3. In which compound is the oxidation state of nitrogen equal to +4?

1) HNO2
2) N 2 O 4
3) N 2 O
4) HNO3

4. In which compound is the oxidation state of nitrogen equal to –2?

1) NH 3
2) N 2 H 4
3) N 2 O 5
4) HNO2

5. In which compound the oxidation state of sulfur is +2?

1) Na 2 SO 3
2)SO2
3) SCl 2
4) H2SO4

6. In which compound the oxidation state of sulfur is +6?

1) Na 2 SO 3
2) SO 3
3) SCl 2
4) H 2 SO 3

7. In substances whose formulas are CrBr 2, K 2 Cr 2 O 7, Na 2 CrO 4, the oxidation state of chromium is respectively equal to

1) +2, +3, +6
2) +3, +6, +6
3) +2, +6, +5
4) +2, +6, +6

8. The minimum negative oxidation state of a chemical element is usually equal to

1) period number
3) the number of electrons missing to complete the outer electron layer

9. The maximum positive oxidation state of chemical elements located in the main subgroups, as a rule, is equal to

1) period number
2) the serial number of the chemical element
3) group number
4) the total number of electrons in the element

10. Phosphorus exhibits the maximum positive oxidation state in the compound

1) HPO 3
2) H3PO3
3) Na3P
4) Ca 3 P 2

11. Phosphorus exhibits minimal oxidation state in the compound

1) HPO 3
2) H3PO3
3) Na 3 PO 4
4) Ca 3 P 2

12. The nitrogen atoms in ammonium nitrite, located in the cation and anion, exhibit oxidation states, respectively

1) –3, +3
2) –3, +5
3) +3, –3
4) +3, +5

13. The valence and oxidation state of oxygen in hydrogen peroxide are respectively equal

1) II, –2
2) II, –1
3) I, +4
4) III, –2

14. The valence and degree of oxidation of sulfur in pyrite FeS2 are respectively equal

1) IV, +5
2) II, –1
3) II, +6
4) III, +4

15. The valency and oxidation state of the nitrogen atom in ammonium bromide are respectively equal to

1) IV, –3
2) III, +3
3) IV, –2
4) III, +4

16. The carbon atom exhibits a negative oxidation state when combined with

1) oxygen
2) sodium
3) fluorine
4) chlorine

17. exhibits a constant state of oxidation in its compounds

1) strontium
2) iron
3) sulfur
4) chlorine

18. The oxidation state +3 in their compounds can exhibit

1) chlorine and fluorine
2) phosphorus and chlorine
3) carbon and sulfur
4) oxygen and hydrogen

19. The oxidation state +4 in their compounds can exhibit

1) carbon and hydrogen
2) carbon and phosphorus
3) carbon and calcium
4) nitrogen and sulfur

20. The oxidation state equal to the group number in its compounds exhibits

1) chlorine
2) iron
3) oxygen
4) fluorine