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

If we represent the oxidation state (do not confuse with valence, since in many cases they do not match) as an entry in a notebook, then we will see just numbers with zero signs (0 - in a simple substance), plus (+) or minus (-) above 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 base in the study of this subject, without which further actions make no sense.

We use CO to describe Chemical properties substance (or 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 features.

degree can be three types: the highest (to determine it, you need to know which group the element is in), intermediate and lowest (it is necessary to subtract the number of the group in which the element is located from the number 8; naturally, the number 8 is taken because there are 8 groups in D. Mendeleev’s periodic system ). Details on determining the degree of oxidation and its correct placement will be discussed below.

How oxidation state is determined: constant CO

First, CO can be variable or constant.

Determining the constant oxidation state is not difficult, so it is better to start the lesson with it: for this, you only need the ability to use the PS (periodic system). 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 the sum of their oxidation states is zero. However, in the case of ions, the sum must equal the charge of the ion itself.
3. In I, II, III groups Periodic tables are located mainly metals. The 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 -1 (when hydrides are formed in compounds with metals: KH, MgH2).
5. CO oxygen (O) +2. Compounds with this element form oxides (MgO, Na2O, H20 - water). However, there are 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, the oxidation states are placed in a variety of complex substances, redox reactions are described, and so on, but more on that later.

CO variable

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 algebraic equation, but over time we “fill our hand”, and it is not difficult to compose and quickly execute the entire algorithm of actions mentally.

It will not be so easy to understand the words, and it is better to immediately go to practice:

HNO3 - in this formula, determine the oxidation state of nitrogen (N). In chemistry, we read the names of the elements, and we approach the arrangement of the oxidation states also from the end. So, it is known that CO2 of oxygen is -2. We must multiply the oxidation state by the coefficient on the right (if any): -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 give zero, you need to add 6. Check: +1+6-7=-0.

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

• 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).

This is just a small number of items. It takes study and practice to learn how to determine SD, but this does not mean that you need to memorize all the constants and variables of SD: just remember that the latter are much more common. Often, the coefficient and what substance is represented play a significant role - for example, sulfur (S) takes a negative degree in sulfides, oxygen (O) in oxides, and chlorine (Cl) in chlorides. Therefore, in these salts, another element takes a positive degree (and is called a reducing agent in this situation).

Solving problems for determining the degree of oxidation

Now we come to the most important thing - practice. Try 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 denote for now as an unknown variable x. The total value is 0. Therefore, we will make the equation: +1*2+2*x-2*7=0. After the decision, we get the answer 6. Let's check - everything coincided, which means that the task is solved.
2. H2SO4 - find the degree of sulfur.
   Using the same concept, we make 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 on your own, 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 they should not be forgotten: we are talking about elements with a CO variable. Also, to solve many problems and equations, it is necessary to be able to set the coefficients (and to know for what purpose this is done).

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Electronegativity, like other properties of atoms chemical elements, changes with an increase in the ordinal number of the element periodically:

The graph above shows the periodicity of the change in the electronegativity of the elements of the main subgroups, depending on the ordinal number of the element.

When moving down the subgroup of the periodic table, the electronegativity of chemical elements decreases, when moving to the right along the period, it increases.

Electronegativity reflects the non-metallicity of elements: the higher the value of electronegativity, the more non-metallic properties of the element are expressed.

Oxidation state

How to calculate the oxidation state of an element in a compound?

1) The oxidation state of chemical elements in simple substances is always zero.

2) There are elements that manifest in complex substances constant oxidation state:

3) There are chemical elements that exhibit a constant oxidation state in the vast majority of compounds. These elements include:

Element	The oxidation state in almost all compounds	Exceptions
hydrogen H	+1	Alkali and alkaline earth metal hydrides, for example:
oxygen O	-2	Hydrogen and metal peroxides: Oxygen fluoride -

4) The algebraic sum of the oxidation states of all atoms in a molecule is always zero. The algebraic sum of the oxidation states of all atoms in an ion is equal to the charge of the ion.

5) The highest (maximum) oxidation state is equal to the group number. Exceptions that do not fall under this rule are elements of the secondary subgroup of group I, elements of the secondary subgroup of group VIII, as well as oxygen and fluorine.

Chemical elements whose group number does not match their highest oxidation state (mandatory to memorize)

6) The lowest oxidation state of metals is always zero, and the lowest oxidation state of non-metals is calculated by the formula:

lowest oxidation state of a non-metal = group number - 8

Based on the rules presented above, it is possible to establish the degree of oxidation of a chemical element in any substance.

Finding the oxidation states of elements in various compounds

Example 1

Determine the oxidation states of all elements in sulfuric acid.

Decision:

Let's write the formula for sulfuric acid:

The oxidation state of hydrogen in all complex substances is +1 (except for metal hydrides).

The oxidation state of oxygen in all complex substances is -2 (except for peroxides and oxygen fluoride OF 2). Let's arrange the known oxidation states:

Let us denote the oxidation state of sulfur as x:

The sulfuric acid molecule, like the molecule of any substance, is generally electrically neutral, because. the sum of the oxidation states of all atoms in a molecule is zero. Schematically, this can be depicted as follows:

Those. we got the following equation:

Let's solve it:

Thus, the oxidation state of sulfur in sulfuric acid is +6.

Example 2

Determine the oxidation state of all elements in ammonium dichromate.

Decision:

Let's write the formula of ammonium dichromate:

As in the previous case, we can arrange the oxidation states of hydrogen and oxygen:

However, we see that the oxidation states of two chemical elements at once, nitrogen and chromium, are unknown. Therefore, we cannot find the oxidation states in the same way as in the previous example (one equation with two variables does not have a unique solution).

Let us pay attention to the fact that the indicated substance belongs to the class of salts and, accordingly, has an ionic structure. Then we can rightly say that the composition of ammonium dichromate includes NH 4 + cations (the charge of this cation can be seen in the solubility table). Therefore, since in formula unit ammonium dichromate, two positive singly charged NH 4 + cations, the charge of the dichromate ion is -2, since the substance as a whole is electrically neutral. Those. the substance is formed by NH 4 + cations and Cr 2 O 7 2- anions.

We know the oxidation states of hydrogen and oxygen. Knowing that the sum of the oxidation states of the atoms of all elements in the ion is equal to the charge, and denoting the oxidation states of nitrogen and chromium as x and y accordingly, we can write:

Those. we get two independent equations:

Solving which, we find x and y:

Thus, in ammonium dichromate, the oxidation states of nitrogen are -3, hydrogen +1, chromium +6, and oxygen -2.

How to determine the oxidation state of elements in organic substances can be read.

Valence

The valency of atoms is indicated by Roman numerals: I, II, III, etc.

The valence possibilities of an atom depend on the quantity:

1) unpaired electrons

2) unshared electron pairs in the orbitals of valence levels

3) empty electron orbitals of the valence level

Valence possibilities of the hydrogen atom

Let's depict the electronic graphic formula of the hydrogen atom:

It was said that three factors can affect the valence possibilities - the presence of unpaired electrons, the presence of unshared electron pairs at the outer level, and the presence of vacant (empty) orbitals of the outer level. We see one unpaired electron in the outer (and only) energy level. Based on this, hydrogen can exactly have a valency equal to I. However, at the first energy level there is only one sublevel - s, those. the hydrogen atom at the outer level does not have either unshared electron pairs or empty orbitals.

Thus, the only valency that a hydrogen atom can exhibit is I.

Valence possibilities of a carbon atom

Consider electronic structure carbon atom. In the ground state, the electronic configuration of its outer level is as follows:

Those. In the ground state, the outer energy level of an unexcited carbon atom contains 2 unpaired electrons. In this state, it can exhibit a valency equal to II. However, the carbon atom very easily goes into an excited state when energy is imparted to it, and the electronic configuration of the outer layer in this case takes the form:

Although a certain amount of energy is spent on the process of excitation of the carbon atom, the expenditure is more than compensated for by the formation of four covalent bonds. For this reason, valence IV is much more characteristic of the carbon atom. So, for example, carbon has valency IV in the molecules of carbon dioxide, carbonic acid and absolutely all organic substances.

In addition to unpaired electrons and unshared electron pairs, the presence of vacant () orbitals of the valence level also affects the valence possibilities. The presence of such orbitals in the filled level leads to the fact that the atom can act as an electron pair acceptor, i.e. form additional covalent bonds by the donor-acceptor mechanism. So, for example, contrary to expectations, in the carbon monoxide molecule CO, the bond is not double, but triple, which is clearly shown in the following illustration:

Valence possibilities of the nitrogen atom

Let's write down the electron-graphic formula of the external energy level of the nitrogen atom:

As can be seen from the illustration above, the nitrogen atom in its normal state has 3 unpaired electrons, and therefore it is logical to assume that it can exhibit a valence equal to III. Indeed, a valency of three is observed in the molecules of ammonia (NH 3), nitrous acid (HNO 2), nitrogen trichloride (NCl 3), etc.

It was said above that the valence of an atom of a chemical element depends not only on the number of unpaired electrons, but also on the presence of unshared electron pairs. This is due to the fact that a covalent chemical bond can form not only when two atoms provide each other with one electron each, but also when one atom that has an unshared pair of electrons - donor () provides it to another atom with a vacant () orbital valence level (acceptor). Those. for the nitrogen atom, valency IV is also possible due to an additional covalent bond formed by the donor-acceptor mechanism. So, for example, four covalent bonds, one of which is formed by the donor-acceptor mechanism, is observed during the formation of the ammonium cation:

Despite the fact that one of the covalent bonds is formed by the donor-acceptor mechanism, all N-H bonds in the ammonium cation are absolutely identical and do not differ from each other.

A valency equal to V, the nitrogen atom is not able to show. This is due to the fact that the transition to an excited state is impossible for the nitrogen atom, in which the pairing of two electrons occurs with the transition of one of them to a free orbital, which is the closest in energy level. The nitrogen atom has no d-sublevel, and the transition to the 3s-orbital is energetically so expensive that the energy costs are not covered by the formation of new bonds. Many may wonder, what then is the valence of nitrogen, for example, in molecules nitric acid HNO 3 or nitric oxide N 2 O 5? Oddly enough, the valence there is also IV, as can be seen from the following structural formulas:

The dotted line in the illustration shows the so-called delocalized π -connection. For this reason, NO terminal bonds can be called "one and a half". Similar one-and-a-half bonds are also found in the ozone molecule O 3 , benzene C 6 H 6 , etc.

Valence possibilities of phosphorus

Let us depict the electron-graphic formula of the external energy level of the phosphorus atom:

As we can see, the structure of the outer layer of the phosphorus atom in the ground state and the nitrogen atom is the same, and therefore it is logical to expect for the phosphorus atom, as well as for the nitrogen atom, possible valences equal to I, II, III and IV, which is observed in practice.

However, unlike nitrogen, the phosphorus atom also has d-sublevel with 5 vacant orbitals.

In this regard, it is able to pass into an excited state, steaming electrons 3 s-orbitals:

Thus, the valency V for the phosphorus atom, which is inaccessible to nitrogen, is possible. So, for example, a phosphorus atom has a valence of five in the molecules of such compounds as phosphoric acid, phosphorus (V) halides, phosphorus (V) oxide, etc.

Valence possibilities of the oxygen atom

The electron-graphic formula of the external energy level of the oxygen atom has the form:

We see two unpaired electrons at the 2nd level, and therefore valency II is possible for oxygen. It should be noted that this valency of the oxygen atom is observed in almost all compounds. Above, when considering the valence possibilities of the carbon atom, we discussed the formation of the carbon monoxide molecule. The bond in the CO molecule is triple, therefore, oxygen is trivalent there (oxygen is an electron pair donor).

Due to the fact that the oxygen atom does not have an external level d-sublevels, depairing of electrons s and p- orbitals is impossible, which is why the valence capabilities of the oxygen atom are limited compared to other elements of its subgroup, for example, sulfur.

Valence possibilities of the sulfur atom

The external energy level of the sulfur atom in the unexcited state:

The sulfur atom, like the oxygen atom, has two unpaired electrons in its normal state, so we can conclude that a valency of two is possible for sulfur. Indeed, sulfur has valency II, for example, in the hydrogen sulfide molecule H 2 S.

As we can see, the sulfur atom at the outer level has d sublevel with vacant orbitals. For this reason, the sulfur atom is able to expand its valence capabilities, unlike oxygen, due to the transition to excited states. So, when unpairing a lone electron pair 3 p-sublevel the sulfur atom acquires electronic configuration outer level like this:

In this state, the sulfur atom has 4 unpaired electrons, which tells us about the possibility of sulfur atoms showing a valency equal to IV. Indeed, sulfur has valency IV in the molecules SO 2, SF 4, SOCl 2, etc.

When unpairing the second lone electron pair located on 3 s- sublevel, the external energy level acquires the following configuration:

In such a state, the manifestation of valence VI already becomes possible. An example of compounds with VI-valent sulfur are SO 3 , H 2 SO 4 , SO 2 Cl 2 etc.

Similarly, we can consider the valence possibilities of other chemical elements.

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 oxidation state of elements in compounds with polar covalent bonds compare their electronegativity values. 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. Highest Degree oxidation 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 a higher oxidation state of 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

Exercise	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
Decision	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). We divide the obtained value 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)

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Table. Degrees of oxidation of chemical elements.

Table. Degrees of oxidation of chemical elements.

Oxidation state is the conditional charge of the atoms of a chemical element in a compound, calculated from the assumption that all bonds are of the ionic type. Oxidation states can be positive, negative, or zero value, therefore, the algebraic sum of the oxidation states of elements in a molecule, taking into account the number of their atoms, is 0, and in an ion - the charge of the ion. The oxidation states of metals in compounds are always positive. The highest oxidation state corresponds to the group number 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. 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. 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. The oxidation states of simple substances are 0, regardless of whether it is a metal or a non-metal.

Table: Elements with constant oxidation states.

Table. The oxidation states of chemical elements in alphabetical order. Element Name Oxidation state 7 N -III, 0, +I, II, III, IV, V 89 ace 13 Al Aluminum 95 Am Americium 0, + II , III, IV 18 Ar 85 At -I, 0, +I, V 56 Ba 4 Be Beryllium 97 bk 5 B -III, 0, +III 107 bh 35 Br -I, 0, +I, V, VII 23 V 0, + II , III, IV, V 83 Bi 1 H -I, 0, +I 74 W Tungsten 64 Gd Gadolinium 31 Ga 72 hf 2 He 32 Ge Germanium 67 Ho 66 Dy Dysprosium 105 Db 63 EU 26 Fe 79 Au 49 In 77 Ir 39 Y 70 Yb Ytterbium 53 I -I, 0, +I, V, VII 48 CD 19 To 98 cf Californium 20 Ca 54 Xe 0, + II , IV, VI, VIII 8 O Oxygen -II, I, 0, +II 27 co 36 Kr 14 Si -IV, 0, +11, IV 96 cm 57 La 3 Li 103 lr Laurence 71 Lu 12 mg 25 Mn Manganese 0, +II, IV, VI, VIII 29 Cu 109 Mt Meitnerius 101 md Mendelevium 42 Mo Molybdenum 33 As -III, 0, +III, V 11 Na 60 Nd 10 Ne 93 Np Neptunium 0, +III, IV, VI, VII 28 Ni 41 Nb 102 no 50 sn 76 Os 0, +IV, VI, VIII 46 Pd Palladium 91 Pa. Protactinium 61 Pm Promethium 84 Ro 59 Rg Praseodymium 78 Pt 94 PU Plutonium 0, +III, IV, V, VI 88 Ra 37 Rb 75 Re 104 RF Rutherfordium 45 Rh 86 Rn 0, + II , IV, VI, VIII 44 Ru 0, +II, IV, VI, VIII 80 hg 16 S -II, 0, +IV, VI 47 Ag 51 Sb 21 sc 34 Se -II, 0,+IV, VI 106 Sg Seaborgium 62 sm 38 Sr Strontium 82 Pb 81 Tl 73 Ta 52 Te -II, 0, +IV, VI 65 Tb 43 Tc Technetium 22 Ti 0, + II , III, IV 90 Th 69 Tm 6 C -IV, I, 0, + II, IV 92 U 100 fm 15 P -III, 0, +I, III, V 87 Fr 9 F -I, 0 108 hs 17 Cl 24 Cr 0, + II , III , VI 55 Cs 58 Ce 30 Zn 40 Zr Zirconium 99 ES Einsteinium 68 Er

Table. The oxidation states of chemical elements by number. Element Name Oxidation state 1 H -I, 0, +I 2 He 3 Li 4 Be Beryllium 5 B -III, 0, +III 6 C -IV, I, 0, + II, IV 7 N -III, 0, +I, II, III, IV, V 8 O Oxygen -II, I, 0, +II 9 F -I, 0 10 Ne 11 Na 12 mg 13 Al Aluminum 14 Si -IV, 0, +11, IV 15 P -III, 0, +I, III, V 16 S -II, 0, +IV, VI 17 Cl -I, 0, +I, III, IV, V, VI, VII 18 Ar 19 To 20 Ca 21 sc 22 Ti 0, + II , III, IV 23 V 0, + II , III, IV, V 24 Cr 0, + II , III , VI 25 Mn Manganese 0, +II, IV, VI, VIII 26 Fe 27 co 28 Ni 29 Cu 30 Zn 31 Ga 32 Ge Germanium 33 As -III, 0, +III, V 34 Se -II, 0,+IV, VI 35 Br -I, 0, +I, V, VII 36 Kr 37 Rb 38 Sr Strontium 39 Y 40 Zr Zirconium 41 Nb 42 Mo Molybdenum 43 Tc Technetium 44 Ru 0, +II, IV, VI, VIII 45 Rh 46 Pd Palladium 47 Ag 48 CD 49 In 50 sn 51 Sb 52 Te -II, 0, +IV, VI 53 I -I, 0, +I, V, VII 54 Xe 0, + II , IV, VI, VIII 55 Cs 56 Ba 57 La 58 Ce 59 Rg Praseodymium 60 Nd 61 Pm Promethium 62 sm 63 EU 64 Gd Gadolinium 65 Tb 66 Dy Dysprosium 67 Ho 68 Er 69 Tm 70 Yb Ytterbium 71 Lu 72 hf 73 Ta 74 W Tungsten 75 Re 76 Os 0, +IV, VI, VIII 77 Ir 78 Pt 79 Au 80 hg 81 Tl 82 Pb 83 Bi 84 Ro 85 At -I, 0, +I, V 86 Rn 0, + II , IV, VI, VIII 87 Fr 88 Ra 89 ace 90 Th 91 Pa. Protactinium 92 U 93 Np Neptunium 0, +III, IV, VI, VII 94 PU Plutonium 0, +III, IV, V, VI 95 Am Americium 0, + II , III, IV 96 cm 97 bk 98 cf Californium 99 ES Einsteinium 100 fm 101 md Mendelevium 102 no 103 lr Laurence 104 RF Rutherfordium 105 Db 106 Sg Seaborgium 107 bh 108 hs 109 Mt Meitnerius

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

The oxidation states can have a positive, negative or zero value, therefore the algebraic sum of the oxidation states of elements in a molecule, taking into account the number of their atoms, is 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. Oxidation state 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 is an alkali metal 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 given oxide equals +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-. AT 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-.