Possible oxidation states of atoms. Highest oxidation state

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

DEFINITION

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

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

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

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

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

When determining the oxidation state of elements in compounds with polar covalent bonds, their electronegativity values ​​are compared. Since during the formation of a chemical bond, electrons are displaced to the atoms of more electronegative elements, 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 highest (maximum positive) and lowest (minimum negative) oxidation states. The highest oxidation state of a chemical element usually numerically coincides with the group number in D.I. Mendeleev’s Periodic Table. Exceptions are fluorine (oxidation state is -1, and the element is located in group VIIA), oxygen (oxidation state is +2, and the element is located in group VIA), helium, neon, argon (oxidation state is 0, and the elements are located in VIII group), as well as elements of the cobalt and nickel subgroup (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. Elements of the copper subgroup, on the contrary, have a highest oxidation state greater 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

• 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 option.

• 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 option.

EXAMPLE 2

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

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

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

2. The highest oxidation state corresponds to the number of the group of the periodic table where the element is located (exceptions are: Au +3(I group), Cu +2(II), from group VIII the oxidation state +8 can only be found 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 either positive or 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 the element is located, i.e. the highest positive oxidation state is equal to the number of electrons in the outer layer, which corresponds to the group number.

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.

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 valence 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 valence 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 valency 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 valency 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

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Electronegativity, like other properties of atoms of chemical elements, changes periodically with increasing atomic number of the element:

The graph above shows the periodicity of changes in the electronegativity of elements of the main subgroups depending on the atomic number of the element.

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

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

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 exhibit a constant state of oxidation in complex substances:

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

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 coincide with their highest oxidation state (mandatory to remember)

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 non-metal = group number − 8

Based on the rules presented above, you can establish the oxidation state 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.

Solution:

Let's write the formula of sulfuric acid:

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

The oxidation state of oxygen in all complex substances is -2 (except for peroxides and oxygen fluoride OF 2). Let us 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.

Solution:

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 are unknown - nitrogen and chromium. Therefore, we cannot find oxidation states similarly to the previous example (one equation with two variables does not have a single solution).

Let us draw attention to the fact that this 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). Consequently, since the formula unit of ammonium dichromate contains two positive singly charged NH 4 + cations, the charge of the dichromate ion is equal to -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 an 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.

You can read how to determine the oxidation states of elements in organic substances.

Valence

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

The valence capabilities of an atom depend on the quantity:

1) unpaired electrons

2) lone electron pairs in the orbitals of valence levels

3) empty electron orbitals of the valence level

Valence possibilities of the hydrogen atom

Let us depict the electron graphic formula of the hydrogen atom:

It has been said that three factors can influence the valence possibilities - the presence of unpaired electrons, the presence of lone electron pairs in the outer level, and the presence of vacant (empty) orbitals in the outer level. We see one unpaired electron at the outer (and only) energy level. Based on this, hydrogen can definitely have a valence of I. However, in the first energy level there is only one sublevel - s, those. The hydrogen atom at the outer level has neither lone electron pairs nor empty orbitals.

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

Valence possibilities of the carbon atom

Let's consider the electronic structure of the carbon atom. In the ground state, the electronic configuration of its outer level is as follows:

Those. in the ground state at the outer energy level of the unexcited carbon atom there are 2 unpaired electrons. In this state it can exhibit a valence of 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:

Despite the fact that 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, valency IV is much more characteristic of the carbon atom. 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 lone electron pairs, the presence of vacant ()valence level orbitals also affects the valence possibilities. The presence of such orbitals at the filled level leads to the fact that the atom can act as an electron pair acceptor, i.e. form additional covalent bonds through a donor-acceptor mechanism. For example, contrary to expectations, in the carbon monoxide CO molecule the bond is not double, but triple, as is clearly shown in the following illustration:

Valence possibilities of the nitrogen atom

Let us write the electronic graphic formula for 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 is capable of exhibiting a valence of III. Indeed, a valence 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 lone electron pairs. This is due to the fact that a covalent chemical bond can be formed not only when two atoms provide each other with one electron, but also when one atom with a lone pair of electrons - donor () provides it to another atom with a vacant () orbital valence level (acceptor). Those. For the nitrogen atom, valence IV is also possible due to an additional covalent bond formed by the donor-acceptor mechanism. For example, four covalent bonds, one of which is formed by a donor-acceptor mechanism, are observed during the formation of an ammonium cation:

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

The nitrogen atom is not capable of exhibiting a valency equal to V. This is due to the fact that it is impossible for a nitrogen atom to transition to an excited state, in which two electrons are paired with the transition of one of them to a free orbital that is 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 is the valency of nitrogen, for example, in molecules of nitric acid HNO 3 or nitric oxide N 2 O 5? Oddly enough, the valency 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, terminal NO bonds can be called “one and a half bonds.” Similar one-and-a-half bonds are also present in the molecule of ozone O 3, benzene C 6 H 6, etc.

Valence possibilities of phosphorus

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

As we 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, as observed in practice.

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

In this regard, it is capable of transitioning to an excited state, steaming electrons 3 s-orbitals:

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

Valence possibilities of the oxygen atom

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

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

Due to the fact that the oxygen atom does not have an external d-sublevel, electron pairing 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

External energy level of a sulfur atom in an unexcited state:

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

As we see, the sulfur atom appears at the external level 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. Thus, when pairing a lone electron pair 3 p-sublevel, the sulfur atom acquires the electronic configuration of the outer level of the following form:

In this state, the sulfur atom has 4 unpaired electrons, which tells us that sulfur atoms can exhibit a valence of IV. Indeed, sulfur has valency IV in molecules SO 2, SF 4, SOCl 2, etc.

When pairing the second lone electron pair located at 3 s-sublevel, the external energy level acquires the configuration:

In this state, the manifestation of valency VI becomes possible. Examples 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.