Average degree of oxidation. Valency and oxidation state - preparation for the exam in chemistry

In chemistry, the terms "oxidation" and "reduction" mean reactions in which an atom or a group of atoms lose or, respectively, gain electrons. The oxidation state is a numerical value attributed to one or more atoms that characterizes the number of redistributed electrons and shows how these electrons are distributed between atoms during the reaction. Determining this quantity can be both a simple and quite complex procedure, depending on the atoms and the molecules consisting of them. Moreover, the atoms of some elements can have several oxidation states. Fortunately, there are simple unambiguous rules for determining the degree of oxidation, for the confident use of which it is enough to know the basics of chemistry and algebra.

Steps

Part 1

Determine if the substance in question is elemental. The oxidation state of atoms outside a chemical compound is zero. This rule is true both for substances formed from individual free atoms, and for those that consist of two or polyatomic molecules of one element.

• For example, Al(s) and Cl 2 have an oxidation state of 0 because both are in a chemically uncombined elemental state.
• Please note that the allotropic form of sulfur S 8, or octasulfur, despite its atypical structure, is also characterized by a zero oxidation state.

1. Determine if the substance in question consists of ions. The oxidation state of ions is equal to their charge. This is true both for free ions and for those that are part of chemical compounds.

   • For example, the oxidation state of the Cl ion is -1.
   • The oxidation state of the Cl ion in the chemical compound NaCl is also -1. Since the Na ion, by definition, has a charge of +1, we conclude that the charge of the Cl ion is -1, and thus its oxidation state is -1.

2. Note that metal ions can have several oxidation states. Atoms of many metallic elements can be ionized to different extents. For example, the charge of ions of a metal such as iron (Fe) is +2 or +3. The charge of metal ions (and their degree of oxidation) can be determined by the charges of ions of other elements with which this metal is part of a chemical compound; in the text, this charge is indicated by Roman numerals: for example, iron (III) has an oxidation state of +3.

   • As an example, consider a compound containing an aluminum ion. The total charge of the AlCl 3 compound is zero. Since we know that Cl - ions have a charge of -1, and the compound contains 3 such ions, for the total neutrality of the substance in question, the Al ion must have a charge of +3. Thus, in this case, the oxidation state of aluminum is +3.

3. The oxidation state of oxygen is -2 (with some exceptions). In almost all cases, oxygen atoms have an oxidation state of -2. There are several exceptions to this rule:

   • If oxygen is in the elemental state (O 2 ), its oxidation state is 0, as is the case for other elemental substances.
   • If oxygen is included peroxides, its oxidation state is -1. Peroxides are a group of compounds containing a single oxygen-oxygen bond (ie the peroxide anion O 2 -2). For example, in the composition of the H 2 O 2 molecule (hydrogen peroxide), oxygen has a charge and an oxidation state of -1.
   • In combination with fluorine, oxygen has an oxidation state of +2, see the rule for fluorine below.

4. Hydrogen has an oxidation state of +1, with a few exceptions. As with oxygen, there are also exceptions. As a rule, the oxidation state of hydrogen is +1 (unless it is in the elemental state H 2). However, in compounds called hydrides, the oxidation state of hydrogen is -1.

   • For example, in H 2 O, the oxidation state of hydrogen is +1, since the oxygen atom has a charge of -2, and two +1 charges are needed for overall neutrality. However, in the composition of sodium hydride, the hydrogen oxidation state is already -1, since the Na ion carries a charge of +1, and for general electroneutrality, the charge of the hydrogen atom (and therefore its oxidation state) must be -1.

5. Fluorine always has an oxidation state of -1. As already noted, the degree of oxidation of some elements (metal ions, oxygen atoms in peroxides, and so on) can vary depending on a number of factors. The oxidation state of fluorine, however, is invariably -1. This is explained by the fact that this element has the highest electronegativity - in other words, fluorine atoms are the least willing to part with their own electrons and most actively attract other people's electrons. Thus, their charge remains unchanged.

6. The sum of the oxidation states in a compound is equal to its charge. The oxidation states of all the atoms that make up a chemical compound, in total, should give the charge of this compound. For example, if a compound is neutral, the sum of the oxidation states of all its atoms must be zero; if the compound is a polyatomic ion with a charge of -1, the sum of the oxidation states is -1, and so on.

   • This is a good method of checking - if the sum of the oxidation states does not equal the total charge of the compound, then you are wrong somewhere.

   Part 2

   Determining the oxidation state without using the laws of chemistry

   1. Find atoms that do not have strict rules regarding oxidation state. In relation to some elements, there are no firmly established rules for finding the degree of oxidation. If an atom does not fit any of the rules listed above, and you do not know its charge (for example, the atom is part of a complex, and its charge is not indicated), you can determine the oxidation state of such an atom by elimination. First, determine the charge of all other atoms of the compound, and then from the known total charge of the compound, calculate the oxidation state of this atom.

      • For example, in the Na 2 SO 4 compound, the charge of the sulfur atom (S) is unknown - we only know that it is not zero, since sulfur is not in the elementary state. This compound serves as a good example to illustrate the algebraic method of determining the oxidation state.

   2. Find the oxidation states of the rest of the elements in the compound. Using the rules described above, determine the oxidation states of the remaining atoms of the compound. Don't forget about the exceptions to the rule in the case of O, H, and so on.

      • For Na 2 SO 4 , using our rules, we find that the charge (and hence the oxidation state) of the Na ion is +1, and for each of the oxygen atoms it is -2.

   3. Find the unknown oxidation state from the charge of the compound. Now you have all the data for a simple calculation of the desired oxidation state. Write down an equation, on the left side of which there will be the sum of the number obtained in the previous calculation step and the unknown oxidation state, and on the right side - the total charge of the compound. In other words, (Sum of known oxidation states) + (desired oxidation state) = (compound charge).

      • In our case Na 2 SO 4 the solution looks like this:
        • (Sum of known oxidation states) + (desired oxidation state) = (compound charge)
        • -6+S=0
        • S=0+6
        • S = 6. In Na 2 SO 4, sulfur has an oxidation state 6 .
   • In compounds, the sum of all oxidation states must equal the charge. For example, if the compound is a diatomic ion, the sum of the oxidation states of the atoms must be equal to the total ionic charge.
   • It is very useful to be able to use the periodic table of Mendeleev and know where the metallic and non-metallic elements are located in it.
   • The oxidation state of atoms in the elementary form is always zero. The oxidation state of a single ion is equal to its charge. Elements of group 1A of the periodic table, such as hydrogen, lithium, sodium, in elemental form have an oxidation state of +1; the oxidation state of group 2A metals, such as magnesium and calcium, in its elemental form is +2. Oxygen and hydrogen, depending on the type of chemical bond, can have 2 different oxidation states.

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

Oxidation and reduction

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

Determination of the degree of oxidation

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

Negative, zero and positive oxidation state values

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

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

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

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

In its highest oxidation state, sulfur only accepts electrons; in its lowest state, it exhibits strong reducing properties. The S +4 atoms can act as reducing or oxidizing agents in compounds, depending on the conditions.

Transfer of electrons in chemical reactions

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

How to calculate oxidation states

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

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

Oxidation state and valence

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

Redox reactions

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

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

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

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

Topics of the USE codifier: Electronegativity. The degree of oxidation and valence of chemical elements.

When atoms interact and form, the electrons between them are in most cases unevenly distributed, since the properties of the atoms differ. More electronegative the atom attracts the electron density to itself more strongly. An atom that has attracted electron density to itself acquires a partial negative charge. δ — , its "partner" is a partial positive charge δ+ . If the difference in the electronegativity of the atoms forming a bond does not exceed 1.7, we call the bond covalent polar . If the difference in electronegativity forming a chemical bond exceeds 1.7, then we call such a bond ionic .

Oxidation state is the auxiliary conditional charge of an atom of an element in a compound, calculated from the assumption that all compounds are composed of ions (all polar bonds are ionic).

What does "conditional charge" mean? We simply agree that we will simplify things a bit: we will consider any polar bonds to be completely ionic, and we will consider that an electron completely leaves or comes from one atom to another, even if in fact it is not. And conditionally, an electron leaves a less electronegative atom for a more electronegative one.

For example, in the H-Cl bond, we believe that hydrogen conditionally "gave" an electron, and its charge became +1, and chlorine "accepted" an electron, and its charge became -1. In fact, there are no such total charges on these atoms.

Surely, you have a question - why invent something that does not exist? This is not an insidious plan of chemists, everything is simple: such a model is very convenient. Ideas about the oxidation state of elements are useful in compiling classification chemicals, describing their properties, formulating compounds and nomenclature. Especially often the oxidation states are used when working with redox reactions.

The oxidation states are higher, lower and intermediate.

Higher the oxidation state is equal to the group number with a plus sign.

Inferior is defined as the group number minus 8.

And intermediate an oxidation state is almost any integer in the range from the lowest oxidation state to the highest.

For example, nitrogen is characterized by: the highest oxidation state is +5, the lowest 5 - 8 \u003d -3, and the intermediate oxidation states are from -3 to +5. For example, in hydrazine N 2 H 4, the oxidation state of nitrogen is intermediate, -2.

Most often, the oxidation state of atoms in complex substances is indicated first by a sign, then by a number, for example +1, +2, -2 etc. When it comes to the charge of an ion (assume that the ion really exists in the compound), then first indicate the number, then the sign. For example: Ca 2+ , CO 3 2- .

To find the oxidation states use the following regulations :

1. The oxidation state of atoms in simple substances is equal to zero;
2. AT neutral molecules the algebraic sum of the oxidation states is zero, for ions this sum is equal to the charge of the ion;
3. Oxidation state alkali metals (elements of group I of the main subgroup) in compounds is +1, the oxidation state alkaline earth metals (elements of group II of the main subgroup) in compounds is +2; oxidation state aluminum in compounds it is +3;
4. Oxidation state hydrogen in compounds with metals (- NaH, CaH 2, etc.) is equal to -1 ; in compounds with non-metals () +1 ;
5. Oxidation state oxygen is equal to -2 . Exception constitute peroxides- compounds containing the -О-О- group, where the oxidation state of oxygen is -1 , and some other compounds ( superoxides, ozonides, oxygen fluorides OF 2 and etc.);
6. Oxidation state fluorine in all complex substances is equal to -1 .

The above are the situations when we consider the degree of oxidation constant . For all other chemical elements, the oxidation state — variable, and depends on the order and type of atoms in the compound.

Examples:

Exercise: determine the oxidation states of the elements in the potassium dichromate molecule: K 2 Cr 2 O 7.

Solution: the oxidation state of potassium is +1, the oxidation state of chromium is denoted as X, oxygen oxidation state -2. The sum of all oxidation states of all atoms in a molecule is 0. We get the equation: +1*2+2*x-2*7=0. We solve it, we get the oxidation state of chromium +6.

In binary compounds, a more electronegative element is characterized by a negative oxidation state, a less electronegative element is characterized by a positive one.

note that the concept of oxidation state is very conditional! The oxidation state does not show the real charge of the atom and has no real physical meaning.. This is a simplified model that works effectively when we need, for example, to equalize the coefficients in a chemical reaction equation, or to algorithmize the classification of substances.

Oxidation state is not valence! The oxidation state and valence in many cases do not match. For example, the valency of hydrogen in a simple substance H 2 is I, and the oxidation state, according to rule 1, is 0.

These are the basic rules that will help you determine the oxidation state of atoms in compounds in most cases.

In some situations, you may find it difficult to determine the oxidation state of an atom. Let's take a look at some of these situations and see how to resolve them:

1. In double (salt-like) oxides, the degree at the atom, as a rule, is two oxidation states. For example, in iron oxide Fe 3 O 4 iron has two oxidation states: +2 and +3. Which one to indicate? Both. To simplify, this compound can be represented as a salt: Fe (FeO 2) 2. In this case, the acid residue forms an atom with an oxidation state of +3. Or a double oxide can be represented as follows: FeO * Fe 2 O 3.
2. In peroxo compounds, the degree of oxidation of oxygen atoms connected by covalent nonpolar bonds, as a rule, changes. For example, in hydrogen peroxide H 2 O 2, and alkali metal peroxides, the oxidation state of oxygen is -1, because one of the bonds is covalent non-polar (H-O-O-H). Another example is peroxomonosulfuric acid (Caro's acid) H 2 SO 5 (see figure) contains two oxygen atoms with an oxidation state of -1, the remaining atoms with an oxidation state of -2, so the following entry will be more understandable: H 2 SO 3 (O2). Chromium peroxo compounds are also known - for example, chromium (VI) peroxide CrO (O 2) 2 or CrO 5, and many others.
3. Another example of compounds with ambiguous oxidation states are superoxides (NaO 2) and salt-like ozonides KO 3 . In this case, it is more appropriate to talk about the molecular ion O 2 with a charge of -1 and O 3 with a charge of -1. The structure of such particles is described by some models that are used in the Russian curriculum in the first courses of chemical universities: MO LCAO, the method of superposition of valence schemes, etc.
4. In organic compounds, the concept of oxidation state is not very convenient to use, because there are a large number of covalent non-polar bonds between carbon atoms. However, if you draw the structural formula of a molecule, then the oxidation state of each atom can also be determined by the type and number of atoms with which this atom is directly bonded. For example, for primary carbon atoms in hydrocarbons, the oxidation state is -3, for secondary -2, for tertiary atoms -1, for quaternary - 0.

Let's practice determining the oxidation state of atoms in organic compounds. To do this, you need to draw the complete structural formula of the atom, and select the carbon atom with its immediate environment - the atoms with which it is directly connected.

• To simplify the calculations, you can use the solubility table - the charges of the most common ions are indicated there. In most Russian chemistry exams (USE, GIA, DVI), the use of a solubility table is allowed. This is a ready-made cheat sheet, which in many cases can save a lot of time.
• When calculating the oxidation state of elements in complex substances, we first indicate the oxidation states of the elements that we know for sure (elements with a constant oxidation state), and the oxidation state of elements with a variable oxidation state is denoted as x. The sum of all charges of all particles is equal to zero in a molecule or equal to the charge of an ion in an ion. It is easy to form and solve an equation from these data.

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

The graph above shows the frequency of changes 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 exhibit a constant oxidation state 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 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.

Solution:

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 the atoms in the 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, 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 there are two positive singly charged NH 4 + cations in the formula unit of ammonium dichromate, 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 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, 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 some energy is expended in 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 lone 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 valency of nitrogen, for example, in the molecules of 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 the electronic configuration of the outer level of the following form:

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.