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 conditional electric charge is the degree of oxidation. Its value changes as a result of many chemical processes. Although the charge is formal, it vividly characterizes the properties and behavior of atoms in redox reactions (ORDs).

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 restored - they attach electrons. Each atom during the formation of a molecule changes the structure of its valence electron shell. In this case, a formal charge appears, the value of which depends on the number of conditionally given or received electrons. To characterize this value, the English chemical term "oxidation number" was previously used, which means "oxidation number" in translation. Its use is based on the assumption that the bonding electrons in molecules or ions belong to the atom with the higher electronegativity (EO). The ability to retain their electrons and attract them from other atoms is well expressed in strong non-metals (halogens, oxygen). Strong metals (sodium, potassium, lithium, calcium, other alkali and alkaline earth elements) have opposite properties.

Determination of the degree of oxidation

The oxidation state is the charge that an atom would acquire if the electrons involved in the formation of the bond were completely shifted to a more electronegative element. There are substances that do not 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 took place before the atoms acquired their current state. A positive oxidation state is the total number of electrons that have been removed from the atoms. Negative meaning the oxidation state is equal to the number of acquired electrons. By change in oxidation state chemical element judge what happens to its atoms during the reaction (and vice versa). The color of the substance determines what changes in the state of oxidation have occurred. Compounds of chromium, iron and a number of other elements in which they exhibit different valences are colored differently.

Negative, zero and positive oxidation state values

Simple substances are formed by chemical elements with the same EO value. In this case, the bonding electrons belong to all structural particles equally. Therefore, in simple substances, the oxidation state (H 0 2, O 0 2, C 0) is not characteristic of the elements. When atoms accept electrons or the general cloud shifts in their direction, it is customary to write charges with a minus sign. For example, F -1, O -2, C -4. By donating electrons, atoms acquire a real or formal positive charge. In OF 2 oxide, the oxygen atom donates one electron each to two fluorine atoms and is in the O +2 oxidation state. It is believed that in a molecule or a polyatomic ion, the more electronegative atoms receive all the binding electrons.

Sulfur is an element that exhibits different valencies and oxidation states.

Chemical elements of the main subgroups often exhibit a lower valence equal to VIII. For example, the valency of sulfur in hydrogen sulfide and metal sulfides is II. The element is characterized by intermediate and higher valencies 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 gives one electron: -1;
• in hydrogen sulfide, the lowest value: -2;
• in dioxide intermediate state: +4;
• in trioxide, sulfuric acid and sulfates: +6.

In his top notch Oxidation sulfur only accepts electrons, in the lowest degree - exhibits strong reducing properties. The S +4 atoms can act as reducing or oxidizing agents in compounds, depending on the conditions.

Transfer of electrons in chemical reactions

When a crystal is formed table salt sodium donates electrons to the more electronegative chlorine. The oxidation states of the elements coincide with the charges of the ions: Na +1 Cl -1 . For molecules created by the socialization and displacement of electron pairs to a more electronegative atom, only the concept of a formal charge is applicable. But it can be assumed that all compounds are composed of ions. Then the atoms, by attracting electrons, acquire a conditional negative charge, and by giving away, they acquire a positive one. In reactions, 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 displays the number of electrons removed from the atom. Oxygen in this substance has an oxidation state of -2. The corresponding index when chemical sign O is the number of added electrons in the atom.

How to calculate oxidation states

Counting the number of electrons donated and added 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, they 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 occurs: H-1.
8. Fluorine is the most electronegative of all elements, its oxidation state is always -4.
9. For most metals, oxidation numbers and valences are the same.

Oxidation state and valency

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 valency. Often these values ​​coincide. As a synonym for the term "oxidation state", the phrase "electrochemical valence" can be used. 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 oxidation state -3. 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 HCOOH acid have different values: -4, -2 and +2.

Redox reactions

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

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

• oxidation states;
• the reducing agent donates electrons and is oxidized;
• the oxidizing agent accepts electrons and is reduced;
• the number of given electrons must be equal to the number of attached ones.

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 OVR flowing between ions of strong electrolytes in aqueous solutions, more often they use not the electronic balance, but the method of half-reactions.

To place correctly oxidation states There are four rules to keep in mind.

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

2) You should remember the elements for which are characteristic constant oxidation states. All of them are listed in the table.

3) The highest oxidation state of an element, as a rule, coincides with the number of the group in which this element is located (for example, phosphorus is in group V, the highest SD of phosphorus is +5). Important exceptions: F, O.

4) The search for the oxidation states of the remaining elements is based on simple rule:

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

A few simple examples for determining oxidation states

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. We compose the simplest equation: x + 3 (+1) \u003d 0. The solution is obvious: x \u003d -3. Answer: N -3 H 3 +1.

Example 2. Specify 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 make an equation for determining the degree of oxidation of sulfur: 2 (+1) + x + 4 (-2) = 0. Solving given equation, we find: x \u003d +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 atom of Al (+3), 9 oxygen atoms (-2) and 3 nitrogen atoms, the oxidation state of which we have to calculate. Corresponding equation: 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 the 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).

What to do if the oxidation states of two elements are unknown

Is it possible to determine the oxidation states of several elements at once using a similar equation? If we consider this problem from the point of view of mathematics, the answer will be negative. Linear Equation with two variables cannot have a unique solution. But we are not just solving 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-. We know the charges of ions, each of them contains only one atom with an unknown degree of oxidation. 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 the molecule contains several atoms with unknown oxidation states, try to "split" the molecule into several parts.

How to arrange oxidation states in organic compounds

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 an adjacent carbon atom. By S-N connections there is a shift in the electron density towards the carbon atom (because 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 (electron density shift towards C), one oxygen atom (electron density shift towards O) and one carbon atom (we can assume that the shifts 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.

Do not confuse the concepts of "valence" and "oxidation state"!

Oxidation state is often confused with valence. Don't make that mistake. I will list the main differences:

• the oxidation state has a sign (+ or -), valence - no;
• oxidation state can be zero even in complex substance, the equality of valence to zero means, as a rule, that an atom of a given element is not connected to other atoms (we will not discuss any kind of inclusion compounds and other "exotics" here);
• oxidation state is a formal concept that acquires real meaning only in compounds with ionic bonds, the concept of "valence", on the contrary, is most conveniently applied in relation to covalent compounds.

The oxidation state (more precisely, its modulus) is often numerically equal to the valence, but even more often these values ​​do NOT coincide. For example, the oxidation state of carbon in CO 2 is +4; valency C is also equal to IV. But in methanol (CH 3 OH), the valency of carbon remains the same, and the oxidation state of C is -1.

A small test on the topic "The degree of oxidation"

Take a few minutes to check how you have understood this topic. You need to answer five simple questions. Good luck!

A chemical element in a compound, calculated from the assumption that all bonds are ionic.

Oxidation states can be positive, negative, or zero, so algebraic sum the oxidation states of elements in a molecule, taking into account the number of their atoms, is 0, and in an ion - the charge of the ion.

1. The oxidation states of metals in compounds are always positive.

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

3. The oxidation states of non-metals depend on which atom it is connected to:

• if with a metal atom, then the oxidation state is negative;
• if with a non-metal atom, then the oxidation state can be both positive and negative. It depends on the electronegativity of the atoms of the elements.

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

5. The oxidation states of simple substances are 0, regardless of whether it is a metal or a non-metal.

Elements with constant oxidation states.

Element	Characteristic oxidation state	Exceptions
		Metal hydrides: LIH-1
		oxidation state called the conditional charge of the particle under the assumption that the bond is completely broken (has an ionic character). H- Cl = H + + Cl - , Communication in hydrochloric acid covalent polar. The electron pair is more biased towards the atom Cl - , because it is more electronegative whole element. How to determine the degree of oxidation? Electronegativity is the ability of atoms to attract electrons from other elements. The oxidation state is indicated above the element: Br 2 0 , Na 0 , O +2 F 2 -1 ,K + Cl - etc. It can be negative and positive. The oxidation state of a simple substance (unbound, free state) is zero. The oxidation state of oxygen in most compounds is -2 (the exception is peroxides H 2 O 2, where it is -1 and compounds with fluorine - O +2 F 2 -1 , O 2 +1 F 2 -1 ). - Oxidation state a simple monatomic ion is equal to its charge: Na + , Ca +2 . Hydrogen in its compounds has an oxidation state of +1 (exceptions are hydrides - Na + H - and type connections C +4 H 4 -1 ). In metal-non-metal bonds, the atom that has the highest electronegativity has a negative oxidation state (electronegativity data are given on the Pauling scale): H + F - , Cu + Br - , Ca +2 (NO 3 ) - etc. Rules for determining the degree of oxidation in chemical compounds. Let's take a connection KMnO 4 , it is necessary to determine the oxidation state of the manganese atom. Reasoning: Potassium - alkali metal, standing in group I of the periodic table, and therefore, has only a positive oxidation state of +1. Oxygen is known to have an oxidation state of -2 in most of its compounds. This substance is not a peroxide, which means it is no exception. Makes an equation: K+MnXO 4 -2 Let be X- unknown to us the degree of oxidation of manganese. The number of potassium atoms is 1, manganese - 1, oxygen - 4. It is proved that the molecule as a whole is electrically neutral, so its total charge must be equal to zero. 1*(+1) + 1*(X) + 4(-2) = 0, X = +7, Hence, the oxidation state of manganese in potassium permanganate = +7. Let's take another example of an oxide Fe2O3. It is necessary to determine the oxidation state of the iron atom. Reasoning: Iron is a metal, oxygen is a non-metal, which means that it is oxygen that will be an oxidizing agent and have a negative charge. We know that oxygen has an oxidation state of -2. We consider the number of atoms: iron - 2 atoms, oxygen - 3. We make an equation where X- the oxidation state of the iron atom: 2*(X) + 3*(-2) = 0, Conclusion: the oxidation state of iron in this oxide is +3. Examples. Determine the oxidation states of all atoms in the molecule. 1. K2Cr2O7. Oxidation state K+1, oxygen O -2. Given indexes: O=(-2)×7=(-14), K=(+1)×2=(+2). Because the algebraic sum of the oxidation states of elements in a molecule, taking into account the number of their atoms, is 0, then the number of positive oxidation states is equal to the number of negative ones. Oxidation states K+O=(-14)+(+2)=(-12). It follows from this that the number of positive powers of the chromium atom is 12, but there are 2 atoms in the molecule, which means that there are (+12):2=(+6) per atom. Answer: K 2 + Cr 2 +6 O 7 -2. 2.(AsO 4) 3-. In this case, the sum of the oxidation states will no longer be equal to zero, but to the charge of the ion, i.e. - 3. Let's make an equation: x+4×(- 2)= - 3 . Answer: (As +5 O 4 -2) 3-.

Chemistry preparation for ZNO and DPA
Comprehensive edition

PART AND

GENERAL CHEMISTRY

CHEMICAL BOND AND STRUCTURE OF SUBSTANCE

Oxidation state

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

Unlike valency, 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 Oxysen - -2. This rule applies to all basic oxides: the oxidation state of a metallic 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 degree of oxidation is indicated by Arabic numerals, which are placed above the symbol of the element, like valence, and first indicate the sign of the charge, and then its numerical value:

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

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

In simple substances - non-metals, there is a covalent non-polar bond, a joint electron pair is shifted to one of the atoms, therefore the degree of oxidation 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 valency of Oxygen is II, and in nitrogen, the valency of Nitrogen is III:

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

Definition of possible degrees element oxidation

The oxidation states, which elements can show in various compounds, in most cases can be determined by the structure of the external electronic level or by the place of the element in Periodic system.

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

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

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

Non-metallic elements are capable of giving different number electrons: maximum as many as located on the outer energy level. In other words, the maximum oxidation state of non-metallic elements is equal to the group number. Due to the spooling of electrons at the outer level of atoms, the number of unpaired electrons that an atom can donate to chemical reactions, is different, so non-metallic elements are able to detect different intermediate values degree of oxidation.

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, so the sum of the oxidation states of the atoms of all elements must be zero. Let us determine the degree of oxidation in sulfur(I V) oxide SO 2 tauphosphorus (V) sulfide P 2 S 5.

Sulfur (And 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 -2. In this case Sulfur has a positive oxidation state. In different compounds, Sulfur can show different oxidation states, so in this case it must be calculated. In a molecule SO2 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, so the oxidation state of Sulfur is +4:

In the phosphorus molecule V) sulfide P 2 S 5 the more electronegative element is Sulfur, that is, it exhibits a negative oxidation state, and Phosphorus a positive one. For Sulfur, the negative oxidation state is only 2. Together, five Sulfur atoms 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 degree of oxidation in non-binary compounds - salts, bases and acids. But for this, one should also use the principle of electrical neutrality, and also remember that in most compounds the oxidation state of Oxygen is -2, Hydrogen +1.

Consider this using the example of potassium sulfate K2SO4. The oxidation state of Potassium in compounds can only be +1, and Oxygen -2:

From the principle of electroneutrality, we calculate the oxidation state of Sulfur:

2(+1) + 1(x) + 4(-2) = 0, hence 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, so the oxidation state of Fluorine in all compounds is -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 it is -1.

4. The oxidation state of Hydrogen in most compounds is +1, and in compounds with metallic 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.

To characterize the state of elements in compounds, the concept of the degree of oxidation has been 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 towards a given atom.

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

The oxidation state of metals in the elementary 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 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 degree of oxidation of elements in compounds with polar covalent bonds, the values ​​of their electronegativity are compared. Because in education chemical bond electrons are displaced to atoms of more electronegative elements, then the latter have a negative oxidation state in compounds.

Highest oxidation state

For elements that exhibit different oxidation states in their compounds, there are concepts of higher (maximum positive) and lower (minimum negative) oxidation states. The highest oxidation state of a chemical element usually numerically coincides with the group number in the Periodic system of D. I. Mendeleev. The exceptions are fluorine (the oxidation state is -1, and the element is located in group VIIA), oxygen (the oxidation state is +2, and the element is located in group VIA), helium, neon, argon (the oxidation state is 0, and the elements are located in group VIII group), as well as elements of the cobalt and nickel subgroups (the oxidation state is +2, and the elements are located in group VIII), for which the highest oxidation state is expressed by a number whose value is lower than the number of the group to which they belong. The elements of the copper subgroup, on the contrary, have the highest oxidation state more than one, although they belong to group I (the maximum positive oxidation state of copper and silver is +2, gold +3).

Examples of problem solving

EXAMPLE 1

Answer We will alternately determine the degree of sulfur oxidation in each of the proposed transformation schemes, and then choose the correct answer.

• In hydrogen sulfide, the oxidation state of sulfur is (-2), and in a simple substance - sulfur - 0:

Change in the oxidation state of sulfur: -2 → 0, i.e. sixth answer.

• In a simple substance - sulfur - the oxidation state of sulfur is 0, and in SO 3 - (+6):

Change in the oxidation state of sulfur: 0 → +6, i.e. fourth answer.

• In sulfurous acid, the oxidation state of sulfur is (+4), and in a simple substance - sulfur - 0:

1×2 +x+ 3×(-2) =0;

Change in the oxidation state of sulfur: +4 → 0, i.e. third answer.

EXAMPLE 2

The task	Valence III and oxidation state (-3) nitrogen shows in the compound: a) N 2 H 4; b) NH3; c) NH 4 Cl; d) N 2 O 5
Solution	In order to give a correct answer to the question posed, we will alternately determine the valency and oxidation state of nitrogen in the proposed compounds. a) the valency of hydrogen is always equal to I. Total number hydrogen valence units is 4th (1 × 4 = 4). Divide the value obtained by the number of nitrogen atoms in the molecule: 4/2 \u003d 2, therefore, the nitrogen valency is II. This answer is incorrect. b) the valency of hydrogen is always equal to I. The total number of hydrogen valence units is 3 (1 × 3 = 3). We divide the obtained value by the number of nitrogen atoms in the molecule: 3/1 \u003d 2, therefore, the nitrogen valency is III. The oxidation state of nitrogen in ammonia is (-3): This is the correct answer.
Answer	Option (b)