The formal charge of an atom in compounds is an auxiliary quantity; it is usually used in descriptions of the properties of elements in chemistry. This conventional electric charge is the oxidation state. Its value changes as a result of many chemical processes. Although the charge is formal, it clearly characterizes the properties and behavior of atoms in redox reactions (ORR).

Oxidation and reduction

In the past, chemists used the term "oxidation" to describe the interaction of oxygen with other elements. The name of the reactions comes from the Latin name for oxygen - Oxygenium. Later it turned out that other elements also oxidize. In this case, they are reduced - they gain electrons. Each atom, when forming a molecule, changes the structure of its valence electron shell. In this case, a formal charge appears, the magnitude of which depends on the number of conventionally given or accepted electrons. To characterize this value, the English chemical term “oxidation number” was previously used, which translated means “oxidation number”. When using it, it is based on the assumption that the bonding electrons in molecules or ions belong to an atom with a higher electronegativity (EO) value. The ability to retain their electrons and attract them from other atoms is well expressed in strong nonmetals (halogens, oxygen). Strong metals (sodium, potassium, lithium, calcium, other alkali and alkaline earth elements) have the opposite properties.

Determination of oxidation state

The oxidation state is the charge that an atom would acquire if the electrons participating in the formation of the bond were completely shifted to a more electronegative element. There are substances that do not have molecular structure(alkali metal halides and other compounds). In these cases, the oxidation state coincides with the charge of the ion. The conditional or real charge shows what process occurred before the atoms acquired their current state. The positive oxidation number is the total number of electrons that have been removed from the atoms. A negative oxidation number is equal to the number of electrons gained. By changing the oxidation state of a chemical element, one judges what happens to its atoms during the reaction (and vice versa). The color of a substance determines what changes have occurred in the oxidation state. Compounds of chromium, iron and a number of other elements, in which they exhibit different valencies, are colored differently.

Negative, zero and positive oxidation state values

Simple substances are formed by chemical elements with the same value EO. In this case, the bonding electrons belong to all structural particles equally. Consequently, in simple substances the elements are not characterized by an oxidation state (H 0 2, O 0 2, C 0). When atoms accept electrons or the general cloud shifts in their direction, charges are usually written with a minus sign. For example, F -1, O -2, C -4. By donating electrons, atoms acquire a real or formal positive charge. In the OF2 oxide, the oxygen atom gives up one electron each to two fluorine atoms and is in the O +2 oxidation state. In a molecule or polyatomic ion, the more electronegative atoms are said to receive all the bonding electrons.

Sulfur is an element exhibiting different valence and oxidation states

Chemical elements of the main subgroups often exhibit a lower valency equal to VIII. For example, the valence of sulfur in hydrogen sulfide and metal sulfides is II. An element is characterized by intermediate and highest valence in the excited state, when the atom gives up one, two, four or all six electrons and exhibits valences I, II, IV, VI, respectively. The same values, only with a minus or plus sign, have the oxidation states of sulfur:

• in fluorine sulfide donates one electron: -1;
• in hydrogen sulfide the lowest value: -2;
• in dioxide intermediate state: +4;
• in trioxide, sulfuric acid and sulfates: +6.

In his highest condition oxidation, sulfur only accepts electrons; to a lower degree, it exhibits strong reducing properties. S+4 atoms can act as reducing agents or oxidizing agents in compounds, depending on the conditions.

Transfer of electrons in chemical reactions

When a sodium chloride crystal forms, sodium donates electrons to the more electronegative chlorine. The oxidation states of elements coincide with the charges of the ions: Na +1 Cl -1. For molecules created by sharing and shifting electron pairs to a more electronegative atom, only the concept of formal charge is applicable. But we can assume that all compounds consist of ions. Then the atoms, by attracting electrons, acquire a conditional negative charge, and by giving them away, a positive charge. In reactions they indicate how many electrons are displaced. For example, in the carbon dioxide molecule C +4 O - 2 2, the index indicated in the upper right corner of the chemical symbol for carbon reflects the number of electrons removed from the atom. Oxygen in this substance is characterized by an oxidation state of -2. The corresponding index for the chemical sign O is the number of added electrons in the atom.

How to calculate oxidation states

Counting the number of electrons donated and gained by atoms can be time consuming. The following rules make this task easier:

1. In simple substances, the oxidation states are zero.
2. The sum of the oxidation of all atoms or ions in a neutral substance is zero.
3. In a complex ion, the sum of the oxidation states of all elements must correspond to the charge of the entire particle.
4. A more electronegative atom acquires a negative oxidation state, which is written with a minus sign.
5. Less electronegative elements receive positive oxidation states and are written with a plus sign.
6. Oxygen generally exhibits an oxidation state of -2.
7. For hydrogen, the characteristic value is: +1; in metal hydrides it is found: H-1.
8. Fluorine is the most electronegative of all the elements, and its oxidation state is always -4.
9. For most metals, the oxidation numbers and valencies are the same.

Oxidation state and valence

Most compounds are formed as a result of redox processes. The transition or displacement of electrons from one element to another leads to a change in their oxidation state and valence. Often these values ​​coincide. The phrase “electrochemical valence” can be used as a synonym for the term “oxidation state”. But there are exceptions, for example, in the ammonium ion, nitrogen is tetravalent. At the same time, the atom of this element is in the -3 oxidation state. In organic substances, carbon is always tetravalent, but the oxidation states of the C atom in methane CH 4, formic alcohol CH 3 OH and acid HCOOH have different values: -4, -2 and +2.

Redox reactions

Redox processes include many of the most important processes in industry, technology, living and inanimate nature: combustion, corrosion, fermentation, intracellular respiration, photosynthesis and other phenomena.

When compiling OVR equations, coefficients are selected using the electronic balance method, which operates with the following categories:

• oxidation states;
• the reducing agent gives up electrons and is oxidized;
• the oxidizing agent accepts electrons and is reduced;
• the number of electrons given up must be equal to the number of electrons added.

The acquisition of electrons by an atom leads to a decrease in its oxidation state (reduction). The loss of one or more electrons by an atom is accompanied by an increase in the oxidation number of the element as a result of reactions. For ORR flowing between ions of strong electrolytes in aqueous solutions, more often they use not an electronic balance, but a half-reaction method.

In chemistry, the description of various redox processes is not complete without oxidation states - special conventional quantities with which you can determine the charge of an atom of any chemical element.

If we imagine the oxidation state (do not confuse it with valency, since in many cases they do not coincide) as an entry in a notebook, then we will see simply numbers with zero signs (0 - in a simple substance), plus (+) or minus (-) above the substance of interest to us. Be that as it may, they play a huge role in chemistry, and the ability to determine CO (oxidation state) is a necessary basis in the study of this subject, without which further actions make no sense.

We use CO to describe Chemical properties a substance (or an individual element), the correct spelling of its international name (understandable for any country and nation, regardless of the language used) and formula, as well as for classification by characteristics.

The degree can be of three types: the highest (to determine it you need to know in which group the element is located), intermediate and lowest (it is necessary to subtract from the number 8 the number of the group in which the element is located; naturally, the number 8 is taken because there is only D. Mendeleev 8 groups). Determining the oxidation state and its correct placement will be discussed in detail below.

How is the oxidation state determined: constant CO

Firstly, CO can be variable or constant

Determining the constant oxidation state is not very difficult, so it is better to start the lesson with it: for this you only need the ability to use the PS (periodic table). So, there are a number of certain rules:

1. Zero degree. It was mentioned above that only simple substances have it: S, O2, Al, K, and so on.
2. If the molecules are neutral (in other words, they have no electrical charge), then their oxidation states add up to zero. However, in the case of ions, the sum must equal the charge of the ion itself.
3. In groups I, II, III of the periodic table, mainly metals are located. Elements of these groups have a positive charge, the number of which corresponds to the group number (+1, +2, or +3). Perhaps the big exception is iron (Fe) - its CO can be both +2 and +3.
4. Hydrogen CO (H) is most often +1 (when interacting with non-metals: HCl, H2S), but in some cases we set it to -1 (when forming hydrides in compounds with metals: KH, MgH2).
5. CO oxygen (O) +2. Compounds with this element form oxides (MgO, Na2O, H20 - water). However, there are also cases when oxygen has an oxidation state of -1 (in the formation of peroxides) or even acts as a reducing agent (in combination with fluorine F, because the oxidizing properties of oxygen are weaker).

Based on this information, oxidation states are assigned to a variety of complex substances, redox reactions are described, etc., but more on that later.

Variable CO

Some chemical elements differ in that they have more than one oxidation state and change it depending on what formula they are in. According to the rules, the sum of all powers must also be equal to zero, but to find it you need to do some calculations. In written form, it looks like just an algebraic equation, but over time we get better at it, and it’s not difficult to compose and quickly execute the entire algorithm of actions mentally.

It will not be so easy to understand in words, and it is better to immediately move on to practice:

HNO3 - in this formula, determine the oxidation degree of nitrogen (N). In chemistry, we read the names of elements and also approach the arrangement of oxidation states from the end. So, it is known that oxygen CO is -2. We must multiply the oxidation number by the coefficient on the right (if there is one): -2*3=-6. Next we move on to hydrogen (H): its CO in the equation will be +1. This means that in order for the total CO to be zero, you need to add 6. Check: +1+6-7=-0.

More exercises will be found at the end, but first we need to determine which elements have variable oxidation states. In principle, all elements, excluding first three groups change their degrees. The most striking examples are halogens (elements of group VII, not counting fluorine F), group IV and noble gases. Below you will see a list of some metals and non-metals with variable degrees:

• H (+1, -1);
• Be (-3, +1, +2);
• B (-1, +1, +2, +3);
• C (-4, -2, +2, +4);
• N (-3, -1, +1, +3, +5);
• O(-2, -1);
• Mg (+1, +2);
• Si (-4, -3, -2, -1, +2, +4);
• P (-3, -2, -1, +1, +3, +5);
• S (-2, +2, +4, +6);
• Cl (-1, +1, +3, +5, +7).

It's just not a large number of elements. Learning to identify COs requires study and practice, but this does not mean that you need to memorize all constant and variable COs by heart: just remember that the latter are much more common. Often, a significant role is played by the coefficient and what substance is represented - for example, in sulfides, sulfur (S) takes a negative degree, in oxides - oxygen (O), in chlorides - chlorine (Cl). Consequently, in these salts another element takes on a positive degree (and is called a reducing agent in this situation).

Solving problems to determine the degree of oxidation

Now we come to the most important thing - practice. Try to complete the following tasks yourself, and then watch the breakdown of the solution and check the answers:

1. K2Cr2O7 - find the degree of chromium.
   CO for oxygen is -2, for potassium +1, and for chromium we designate it for now as an unknown variable x. The total value is 0. Therefore, we create the equation: +1*2+2*x-2*7=0. After solving it, we get the answer 6. Let's check - everything matches, which means the task is solved.
2. H2SO4 - find the degree of sulfur.
   Using the same concept, we create an equation: +2*1+x-2*4=0. Next: 2+x-8=0.x=8-2; x=6.

Brief conclusion

To learn how to determine the oxidation state yourself, you need not only to be able to write equations, but also to thoroughly study the properties of elements of various groups, remember algebra lessons, composing and solving equations with an unknown variable.
Do not forget that the rules have their exceptions and should not be forgotten: we are talking about elements with a CO variable. Also, to solve many problems and equations, you need the ability to set coefficients (and know the purpose for which this is done).

Editorial "site"

To characterize the state of elements in compounds, the concept of oxidation state was introduced. The oxidation state refers to the conditional charge of an atom in a compound, calculated based on the assumption that the compound consists of ions. The oxidation state is indicated Arabic numeral, which is placed in front of the element symbol, with a “+” or “−” sign, corresponding to the donation or acquisition of electrons. The oxidation state is merely a convenient form for accounting for electron transfer and should not be considered as either the effective charge of an atom in the molecule (for example, in the LiF molecule, the effective charges of Li and F are +0.89 and −0.89, respectively, while the degrees oxidation +1 and −1), nor as the valence of the element (for example, in the compounds CH 4, CH 3 OH, HCOOH, CO 2, the valency of carbon is 4, and the oxidation states are respectively −4, −2, +2, +4).

The numerical values ​​of valency and oxidation state can coincide in absolute value only when compounds with ionic bonds are formed. When determining the degree of oxidation, the following rules are used:

1. Atoms of elements that are in a free state or in the form of molecules of simple substances have an oxidation state equal to zero, for example Fe, Cu, H 2, N 2, etc.

2. The oxidation state of an element in the form of a monoatomic ion in a compound having an ionic structure is equal to the charge of this ion, for example,

3. Hydrogen in most compounds has an oxidation state of +1, with the exception of metal hydrides (NaH, LiH), in which the oxidation state of hydrogen is −1.

The most common oxidation state of oxygen in compounds is –2, with the exception of peroxides (Na 2 O 2, H 2 O 2 - the oxidation state of oxygen is −1) and F 2 O (the oxidation state of oxygen is +2).

For elements with a variable oxidation state, its value can be calculated by knowing the formula of the compound and taking into account that the sum of the oxidation states of all atoms in the molecule is zero. In a complex ion, this sum is equal to the charge of the ion. For example, the oxidation state of the chlorine atom in the HClO 4 molecule, calculated based on the total charge of the molecule = 0, x is the oxidation state of the chlorine atom), is +7. The oxidation state of the sulfur atom in the SO ion is +6.

The redox properties of an element depend on the degree of its oxidation. Atoms of the same element are distinguished lowest , higher And intermediate oxidation states.

Knowing the oxidation state of an element in a compound, it is possible to predict whether this compound exhibits oxidizing or reducing properties.

As an example, consider sulfur S and its compounds H 2 S, SO 2 and SO 3. The relationship between the electronic structure of the sulfur atom and its redox properties in these compounds is clearly presented in Table 7.1.

When studying ionic and covalent polar chemical bonds, you became familiar with 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.

To chemists different countries understood 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.

Oxidation states of elements. How to find oxidation states?

1) In a simple substance, the oxidation state of any element is 0. Examples: Na 0, H 0 2, P 0 4.

2) It is necessary to remember the elements that are characterized by constant oxidation states. All of them are listed in the table.

3) The search for oxidation states of other elements is based on a simple rule:

In a neutral molecule, the sum of the oxidation states of all elements is zero, and in an ion - the charge of the ion.

Let's look at the application of this rule using simple examples.

Example 1. It is necessary to find the oxidation states of elements in ammonia (NH 3).

Solution. We already know (see 2) that Art. OK. hydrogen is +1. It remains to find this characteristic for nitrogen. Let x be the desired oxidation state. Let's make the simplest equation: x + 3*(+1) = 0. The solution is obvious: x = -3. Answer: N -3 H 3 +1.

Example 2. Indicate the oxidation states of all atoms in the H 2 SO 4 molecule.

Solution. The oxidation states of hydrogen and oxygen are already known: H(+1) and O(-2). We create an equation to determine the oxidation state of sulfur: 2*(+1) + x + 4*(-2) = 0. Solving this equation, we find: x = +6. Answer: H +1 2 S +6 O -2 4.

Example 3. Calculate the oxidation states of all elements in the Al(NO 3) 3 molecule.

Solution. The algorithm remains unchanged. The composition of the “molecule” of aluminum nitrate includes one Al atom (+3), 9 oxygen atoms (-2) and 3 nitrogen atoms, the oxidation state of which we have to calculate. The corresponding equation is: 1*(+3) + 3x + 9*(-2) = 0. Answer: Al +3 (N +5 O -2 3) 3.

Example 4. Determine the oxidation states of all atoms in the (AsO 4) 3- ion.

Solution. In this case, the sum of oxidation states will no longer be equal to zero, but to the charge of the ion, i.e., -3. Equation: x + 4*(-2) = -3. Answer: As(+5), O(-2).

Is it possible to determine the oxidation states of several elements at once using a similar equation? If we consider this problem from a mathematical point of view, the answer will be negative. A linear equation with two variables cannot have a unique solution. But we are solving more than just an equation!

Example 5. Determine the oxidation states of all elements in (NH 4) 2 SO 4.

Solution. The oxidation states of hydrogen and oxygen are known, but sulfur and nitrogen are not. A classic example of a problem with two unknowns! We will consider ammonium sulfate not as a single “molecule”, but as a combination of two ions: NH 4 + and SO 4 2-. The charges of ions are known to us; each of them contains only one atom with an unknown oxidation state. Using the experience gained in solving previous problems, we can easily find the oxidation states of nitrogen and sulfur. Answer: (N -3 H 4 +1) 2 S +6 O 4 -2.

Conclusion: if a molecule contains several atoms with unknown oxidation states, try to “split” the molecule into several parts.

Example 6. Indicate the oxidation states of all elements in CH 3 CH 2 OH.

Solution. Finding oxidation states in organic compounds has its own specifics. In particular, it is necessary to separately find the oxidation states for each carbon atom. You can reason as follows. Consider, for example, the carbon atom in the methyl group. This C atom is connected to 3 hydrogen atoms and a neighboring carbon atom. By S-N connections the electron density shifts towards the carbon atom (since the electronegativity of C exceeds the EO of hydrogen). If this displacement were complete, the carbon atom would acquire a charge of -3.

The C atom in the -CH 2 OH group is bonded to two hydrogen atoms (a shift in electron density towards C), one oxygen atom (a shift in electron density towards O) and one carbon atom (it can be assumed that the shift in electron density in this case not happening). The oxidation state of carbon is -2 +1 +0 = -1.

Answer: C -3 H +1 3 C -1 H +1 2 O -2 H +1.

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