To determine the conditional charge of atoms in redox reactions, use the table of oxidation of chemical elements. Depending on the properties of the atom, an element can exhibit a positive or negative oxidation state.

What is oxidation number

The conditional charge of the atoms of elements in complex substances is called the oxidation state. The charge value of atoms is recorded in redox reactions to understand which element is a reducing agent and which is an oxidizing agent.

The oxidation state is related to electronegativity, which shows the ability of atoms to accept or give up electrons. The higher the electronegativity value, the greater the ability of an atom to lose electrons in reactions.

Rice. 1. Electronegativity series.

The oxidation state can have three values:

• zero- the atom is at rest (all simple substances have an oxidation state of 0);
• positive- the atom gives up electrons and is a reducing agent (all metals, some non-metals);
• negative- the atom accepts electrons and is an oxidizing agent (most nonmetals).

For example, the oxidation states in the reaction of sodium with chlorine are as follows:

2Na 0 + Cl 2 0 → 2Na +1 Cl -1

In the reaction of metals with nonmetals, the metal is always the reducing agent and the nonmetal is the oxidizing agent.

How to determine

There is a table that shows all the possible oxidation states of elements.

Name	Symbol	Oxidation state
Beryllium
		1, 0, +1, +2, +3
		4, -3, -2, -1, 0, +2, +4
		3, -2, -1, 0, +1, +2, +3, +4, +5
Oxygen		2, -1, 0, +1, +2
Aluminum
		1, 0, +1, +3, +5, +7, rarely +2 and +4
Manganese		2, +3, +4, +6, +7
		2, +3, rarely +4 and +6
		2, +3, rarely +4
		2, rarely +1, +3, +4
		1, +2, rarely +3
		3, rarely +2
Germanium
		3, +3, +5, rarely +2
		2, +4, +6, rarely +2
		1, +1, +5, rarely +3, +4
Strontium
Zirconium		4, rarely +2, +3
		3, +5, rarely +2, +4
Molybdenum		3, +6, rarely +2, +3, +5
Technetium
		3, +4, +8, rarely +2, +6, +7
		4, rarely +2, +3, +6
Palladium		2, +4, rarely +6
		1, rarely +2, +3
		2, rarely +1
		3, rarely +1, +2
		3, +3, +5, rarely +4
		2, +4, +6, rare
		1, +1, +5, +7, rarely +3, +4
Praseodymium
Promethium
		3, rarely +2
		3, rarely +2
Gadolinium
Dysprosium
		3, rarely +2
Ytterbium		3, rarely +2
		5, rarely +3, +4
Tungsten		6, rarely +2, +3, +4, +5
		2, +4, +6, +7, rarely -1, +1, +3, +5
		3, +4, +6, +8, rarely +2
		3, +4, +6, rarely +1, +2
		2, +4, +6, rarely +1, +3
		1, +3, rarely +2
		1, +3, rarely +2
		3, rarely +3, +2, +4, +5
		2, +4, rarely -2, +6

Or use this version of the table in your lessons.

Rice. 2. Table of oxidation states.

In addition, the oxidation states of chemical elements can be determined from the periodic table of Mendeleev:

• the highest degree (maximum positive) coincides with the group number;
• to determine the minimum value of the oxidation state, eight is subtracted from the group number.

Rice. 3. Periodic table.

Most nonmetals have positive and negative oxidation states. For example, silicon is in group IV, which means its maximum oxidation state is +4 and minimum -4. In compounds of non-metals (SO 3 , CO 2 , SiC), the oxidizing agent is a non-metal with a negative oxidation state or with a high electronegativity value. For example, in the compound PCl 3, phosphorus has an oxidation state of +3, chlorine -1. The electronegativity of phosphorus is 2.19, chlorine is 3.16.

The second rule does not work for alkali and alkaline earth metals, which always have one positive oxidation state equal to the group number. Exceptions are magnesium and beryllium (+1, +2). Also have a constant oxidation state:

• aluminum (+3);
• zinc (+2);
• cadmium (+2).

Other metals have a variable oxidation state. In most reactions they act as a reducing agent. In rare cases, they can be oxidizing agents with a negative oxidation state.

Fluorine is the most powerful oxidizing agent. Its oxidation state is always -1.

What have we learned?

From the 8th grade lesson we learned about the degree of oxidation. This is a conventional value showing how many electrons an atom can give or take during a chemical reaction. The value is related to electronegativity. Oxidizing agents accept electrons and have a negative oxidation state, while reducing agents donate electrons and exhibit a positive oxidation state. Most metals are reducing agents with a constant or variable oxidation state. Nonmetals can exhibit oxidizing and reducing properties depending on the substance with which they react.

Test on the topic

Evaluation of the report

Average rating: 4.7. Total ratings received: 146.

To place correctly oxidation states, you need to keep four rules in mind.

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

2) You should remember the elements that 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 the element is located (for example, phosphorus is in group V, the highest s.d. of phosphorus is +5). Important exceptions: F, O.

4) 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.

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 create 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).

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 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.

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 a neighboring carbon atom. Along the C-H bond, 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.

Do not confuse the concepts of “valency” and “oxidation state”!

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

• the oxidation state has a sign (+ or -), the valence does not;
• the oxidation state can be zero even in a complex substance; valence equal 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; the valence of 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 equal to -1.

A short test on the topic "Oxidation state"

Take a few minutes to check your understanding of this topic. You need to answer five simple questions. Good luck!

Video tutorial 2: Oxidation state of chemical elements

Video tutorial 3: Valence. Determination of valency

Lecture: Electronegativity. Oxidation state and valence of chemical elements

Electronegativity

Electronegativity is the ability of atoms to attract electrons from other atoms to join them.

It is easy to judge the electronegativity of a particular chemical element using the table. Remember, in one of our lessons it was said that it increases when moving from left to right through periods in the periodic table and when moving from bottom to top through groups.

For example, the task was given to determine which element from the proposed series is the most electronegative: C (carbon), N (nitrogen), O (oxygen), S (sulfur)? We look at the table and find that this is O, because he is to the right and higher than the others.

What factors influence electronegativity? This:

• The radius of an atom, the smaller it is, the higher the electronegativity.
• The valence shell is filled with electrons; the more electrons there are, the higher the electronegativity.

Of all the chemical elements, fluorine is the most electronegative because it has a small atomic radius and 7 electrons in its valence shell.

Elements with low electronegativity include alkali and alkaline earth metals. They have large radii and very few electrons in the outer shell.

The electronegativity values ​​of an atom cannot be constant, because it depends on many factors, including those listed above, as well as the degree of oxidation, which can be different for the same element. Therefore, it is customary to talk about the relativity of electronegativity values. You can use the following scales:

You will need electronegativity values ​​when writing formulas for binary compounds consisting of two elements. For example, the formula of copper oxide Cu 2 O - the first element should be written down the one whose electronegativity is lower.

At the moment of formation of a chemical bond, if the electronegativity difference between the elements is greater than 2.0, a covalent polar bond is formed; if less, an ionic bond is formed.

Oxidation state

Oxidation state (CO)- this is the conditional or real charge of an atom in a compound: conditional - if the bond is polar covalent, real - if the bond is ionic.

An atom acquires a positive charge when it gives up electrons, and a negative charge when it accepts electrons.

Oxidation states are written above the symbols with a sign «+»/«-» . There are also intermediate COs. The maximum CO of an element is positive and equal to group number, and the minimum negative for metals is zero, for non-metals = (Group No. – 8). Elements with maximum CO only accept electrons, and elements with minimum CO only give up electrons. Elements that have intermediate COs can both give and receive electrons.

Let's look at some rules that should be followed to determine CO:

The CO of all simple substances is zero.

The sum of all CO atoms in a molecule is also equal to zero, since any molecule is electrically neutral.

In compounds with a covalent nonpolar bond, CO is equal to zero (O 2 0), and with an ionic bond it is equal to the charges of the ions (Na + Cl - sodium CO +1, chlorine -1). CO elements of compounds with a covalent polar bond are considered as with an ionic bond (H:Cl = H + Cl -, which means H +1 Cl -1).

Elements in a compound that have the greatest electronegativity have negative oxidation states, while those with the least electronegativity have positive oxidation states. Based on this, we can conclude that metals have only a “+” oxidation state.

Constant oxidation states:

Alkali metals +1.

All metals of the second group +2. Exception: Hg +1, +2.

Aluminum +3.

• Hydrogen +1. Exception: hydrides of active metals NaH, CaH 2, etc., where the oxidation state of hydrogen is –1.

  Oxygen –2. Exception: F 2 -1 O +2 and peroxides that contain the –O–O– group, in which the oxidation state of oxygen is –1.

When an ionic bond is formed, a certain transfer of electron occurs, from a less electronegative atom to an atom of greater electronegativity. Also, in this process, atoms always lose electrical neutrality and subsequently turn into ions. Integer charges are also formed. When a polar covalent bond is formed, the electron is transferred only partially, so partial charges arise.

Valence

Valenceis the ability of atoms to form n - the number of chemical bonds with atoms of other elements.

Valence is also the ability of an atom to hold other atoms near itself. As you know from your school chemistry course, different atoms are bonded to each other by electrons from the outer energy level. An unpaired electron seeks a pair from another atom. These outer level electrons are called valence electrons. This means that valence can also be defined as the number of electron pairs connecting atoms to each other. Look at the structural formula of water: H – O – H. Each dash is an electron pair, which means it shows the valency, i.e. oxygen here has two lines, which means it is divalent, hydrogen molecules come from one line each, which means hydrogen is monovalent. When writing, valency is indicated by Roman numerals: O (II), H (I). Can also be indicated above the element.

Valence can be constant or variable. For example, in metal alkalis it is constant and equals I. But chlorine in various compounds exhibits valencies I, III, V, VII.

How to determine the valence of an element?

Let's look again at the Periodic Table. Metals of the main subgroups have a constant valency, so metals of the first group have valency I, the second - II. And metals of side subgroups have variable valency. It is also variable for non-metals. The highest valence of an atom is equal to group number, the lowest is equal to = group number - 8. A familiar formulation. Doesn't this mean that the valency coincides with the oxidation state? Remember, valence may coincide with the oxidation state, but these indicators are not identical to each other. Valency cannot have a =/- sign, and also cannot be zero.

The second method is to determine valency using a chemical formula, if the constant valency of one of the elements is known. For example, take the formula of copper oxide: CuO. Oxygen valence II. We see that for one oxygen atom in this formula there is one copper atom, which means that the valence of copper is equal to II. Now let's take a more complicated formula: Fe 2 O 3. The valency of the oxygen atom is II. There are three such atoms here, multiply 2*3 =6. We found that there are 6 valences per two iron atoms. Let's find out the valence of one iron atom: 6:2=3. This means that the valence of iron is III.

In addition, when it is necessary to estimate the "maximum valence", one should always start from the electronic configuration that is present in the "excited" state.

Electronegativity is the ability of an atom of a chemical element in a compound to attract electrons from associated atoms of other chemical elements.

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

The oxidation state is the conditional charge of an atom of a chemical element in a compound, calculated based on the assumption that all bonds in its molecule are ionic, i.e. all bonding electron pairs are shifted to atoms with higher electronegativity.

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:

Element	Oxidation state in almost all compounds	Exceptions
hydrogen H	+1	Hydrides of alkali and alkaline earth metals, 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 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

Valence - the number of chemical bonds that an atom of an element forms in a chemical compound.

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:

Summarizing the information on the valence capabilities of the carbon atom:

1) Valences II, III, IV are possible for carbon

2) The most common valence of carbon in compounds IV

3) In the carbon monoxide CO molecule there is a triple bond (!), with one of the three bonds formed according to the donor-acceptor mechanism

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.

i>Summarizing the information on the valence capabilities of the nitrogen atom:

1) For nitrogen, valences I, II, III and IV are possible

2) Valence V nitrogen doesn't!

3) In the molecules of nitric acid and nitrogen oxide N 2 O 5, nitrogen has a valency IV+5 (!) .

4) In compounds in which the nitrogen atom is tetravalent, one of the covalent bonds is formed according to the donor-acceptor mechanism (ammonium salts NH 4 +, nitric acid, 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.

Thus, oxygen almost always has a valence of II, but in some particles it is trivalent, in particular in the carbon monoxide molecule C≡O. In the case when oxygen has valency III, one of the covalent bonds is formed according to the donor-acceptor mechanism.

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.

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

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

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

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

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

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

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

Covalent polar O-H bonds are indicated in blue.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

A hydrogen atoms: 1 - 0 = +1

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

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

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

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

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

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

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