Introduction

The textbook on the chemistry of chalcogens is the second in a series devoted to the chemistry of the elements of the main subgroups of the periodic system of D.I. Mendeleev. It was written on the basis of a course of lectures on inorganic chemistry delivered at Moscow State University over the past 10 years by Academician Yu.D. Tretyakov and Professor V.P. Zlomanov.

In contrast to previously published methodological developments, the manual presents new factual material (catenation, a variety of chalcogen oxoacids (VI), etc.), a modern explanation is given for the patterns of changes in the structure and properties of chalcogen compounds using the concepts of quantum chemistry, including the molecular orbital method, relativistic effect, etc. The material of the manual was selected for the purpose of illustrative illustration of the relationship between the theoretical course and practical training in inorganic chemistry.

[previous section] [table of contents]

§ one. General characteristics of chalcogens (E).

The elements of the VI main subgroup (or the 16th group according to the new IUPAC nomenclature) of the periodic system of elements of D.I. Mendeleev include oxygen (O), sulfur (S), selenium (Se), tellurium (Te) and polonium (Po). The group name of these elements is chalcogens(term "chalcogen" comes from the Greek words "chalkos" - copper and "genos" - born), that is, "giving birth to copper ores", due to the fact that in nature they occur most often in the form of copper compounds (sulfides, oxides, selenides, etc. ).

In the ground state, chalcogen atoms have the electronic configuration ns 2 np 4 with two unpaired p-electrons. They belong to even elements. Some properties of chalcogen atoms are presented in Table 1.

When moving from oxygen to polonium, the size of atoms and their possible coordination numbers increase, while the ionization energy (E ion) and electronegativity (EO) decrease. By electronegativity (EO), oxygen is second only to the fluorine atom, and the sulfur and selenium atoms are also inferior to nitrogen, chlorine, bromine; oxygen, sulfur and selenium are typical non-metals.

In compounds of sulfur, selenium, tellurium with oxygen and halogens, oxidation states +6, +4 and +2 are realized. With most other elements, they form chalcogenides, where they are in the -2 oxidation state.

Table 1. Properties of atoms of elements of group VI.

Properties
atomic number
Number of stable isotopes
Electronic configuration			3d 10 4s 2 4p 4	4d 10 5s 2 5p 4	4f 14 5d 10 6s 2 6p 4
Covalent radius, E
First ionization energy, E ion, kJ/mol
Electronegativity (Pauling)
Affinity of an atom to an electron, kJ/mol

The stability of compounds with the highest oxidation state decreases from tellurium to polonium, for which compounds with oxidation states 4+ and 2+ are known (for example, PoCl 4 , PoCl 2 , PoO 2). This may be due to an increase in the bond strength of 6s 2 electrons with the nucleus due to relativistic effect. Its essence is to increase the speed of movement and, accordingly, the mass of electrons in elements with a large nuclear charge (Z> 60). The "weighting" of electrons leads to a decrease in the radius and an increase in the binding energy of 6s electrons with the nucleus. This effect is more clearly manifested in compounds of bismuth, an element of group V, and is discussed in more detail in the corresponding manual.

The properties of oxygen, as well as other elements of the 2nd period, differ from the properties of their heavier counterparts. Owing to the high electron density and strong interelectron repulsion, the electron affinity and E-E bond strength of oxygen is less than that of sulfur. Metal-oxygen (M-O) bonds are more ionic than M-S, M-Se, etc. bonds. Due to the smaller radius, the oxygen atom, unlike sulfur, is able to form strong bonds (p - p) with other atoms - for example, oxygen in the ozone molecule, carbon, nitrogen, phosphorus. When moving from oxygen to sulfur, the strength of a single bond increases due to a decrease in interelectronic repulsion, and the strength of a bond decreases, which is associated with an increase in radius and a decrease in the interaction (overlap) of p-atomic orbitals. Thus, if oxygen is characterized by the formation of multiple (+) bonds, then sulfur and its analogues are characterized by the formation of single chain bonds - E-E-E (see § 2.1).

There are more analogies in the properties of sulfur, selenium and tellurium than with oxygen and polonium. So, in compounds with negative oxidation states, reducing properties increase from sulfur to tellurium, and in compounds with positive oxidation states, oxidizing properties increase.

Polonium is a radioactive element. The most stable isotope is obtained by bombarding nuclei with neutrons and subsequent -decay:

( 1/2 = 138.4 days).

The decay of polonium is accompanied by the release of a large amount of energy. Therefore, polonium and its compounds decompose solvents and vessels in which they are stored, and the study of Po compounds presents considerable difficulties.

[previous section] [table of contents]

§ 2. Physical properties of simple substances.
Table 2. Physical properties of simple substances.

		Density	Temperatures, o C		Heat of atomization, kJ/mol	Electrical Resistance (25 ° C), Ohm. cm
			melting
S
Se	hex.
						1.3. 10 5 (liquid, 400 o C)
Those hex.	hex.
Ro

With an increase in the covalent radius in the O-S-Se-Te-Po series, the interatomic interaction and the corresponding temperatures of phase transitions, as well as atomization energy, that is, the energy of the transition of solid simple substances into the state of a monatomic gas, increases. The change in the properties of chalcogens from typical non-metals to metals is associated with a decrease in the ionization energy (Table 1) and structural features. Oxygen and sulfur are typical dielectrics, that is, substances that do not conduct electricity. Selenium and tellurium - semiconductors[substances whose electrophysical properties are intermediate between the properties of metals and non-metals (dielectrics). The electrical conductivity of metals decreases, and that of semiconductors increases with increasing temperature, which is due to the peculiarities of their electronic structure)], and polonium is a metal.

[previous section] [table of contents] [next section]

§ 2.1. Chalcogen catenation. Allotropy and polymorphism.

One of the characteristic properties of chalcogen atoms is their ability to bind to each other in rings or chains. This phenomenon is called catenation. The reason for this is related to the different strengths of single and double bonds. Consider this phenomenon on the example of sulfur (Table 3).

Table 3. Energies of single and double bonds (kJ/mol).

It follows from the given values ​​that the formation of two single -bonds for sulfur instead of one double (+) is associated with a gain in energy (530 - 421 = 109 J / mol). For oxygen, on the contrary, one double bond is energetically preferable (494-292=202 kJ/mol) than two single bonds. The decrease in the strength of the double bond upon the transition from O to S is associated with an increase in the size of the p-orbitals and a decrease in their overlap. Thus, for oxygen, catenation is limited to a small number of unstable compounds: O 3 ozone, O 4 F 2 .

cyclic polycations .

Allotropy and polymorphism of simple substances are associated with catenation. Allotropy is the ability of the same element to exist in different molecular forms. The phenomenon of allotropy is attributed to molecules containing a different number of atoms of the same element, for example, O 2 and O 3, S 2 and S 8, P 2 and P 4, etc. The concept of polymorphism applies only to solids. Polymorphism- the ability of a solid substance with the same composition to have a different spatial structure. Examples of polymorphic modifications are monoclinic sulfur and rhombic sulfur, consisting of the same S 8 cycles, but placed differently in space (see § 2.3). Let us first consider the properties of oxygen and its allotropic form - ozone, and then the polymorphism of sulfur, selenium and tellurium.

Dmitry Ivanovich Mendeleev discovered the periodic law, according to which the properties of the elements and the elements they form change periodically. This discovery was graphically displayed in the periodic table. The table shows very well and clearly how the properties of the elements change over the period, after which they are repeated in the next period.

To solve task No. 2 of the Unified State Exam in chemistry, we just need to understand and remember which properties of the elements change in which directions and how.

All this is shown in the figure below.

From left to right, electronegativity, non-metallic properties, higher oxidation states, etc. increase. And the metallic properties and radii decrease.

From top to bottom, vice versa: the metallic properties and radii of atoms increase, while the electronegativity decreases. The highest oxidation state, corresponding to the number of electrons in the outer energy level, does not change in this direction.

Let's look at examples.

Example 1 In the series of elements Na→Mg→Al→Si
A) the radii of atoms decrease;
B) the number of protons in the nuclei of atoms decreases;
C) the number of electron layers in atoms increases;
D) the highest degree of oxidation of atoms decreases;

If we look at the periodic table, we will see that all the elements of a given series are in the same period and are listed in the order in which they appear in the table from left to right. To answer this kind of question, you just need to know a few patterns of changes in properties in the periodic table. So from left to right along the period, metallic properties decrease, non-metallic ones increase, electronegativity increases, ionization energy increases, and the radius of atoms decreases. From top to bottom, metallic and reducing properties increase in a group, electronegativity decreases, ionization energy decreases, and the radius of atoms increases.

If you were attentive, you already understood that in this case the atomic radii decrease. Answer A.

Example 2 In order of increasing oxidizing properties, the elements are arranged in the following order:
A. F→O→N
B. I→Br→Cl
B. Cl→S→P
D. F→Cl→Br

As you know, in Mendeleev's periodic table, oxidizing properties increase from left to right in a period and from bottom to top in a group. Option B just shows the elements of one group in order from bottom to top. So B fits.

Example 3 The valence of elements in the higher oxide increases in the series:
A. Cl→Br→I
B. Cs→K→Li
B. Cl→S→P
D. Al→C→N

In higher oxides, the elements show their highest oxidation state, which will coincide with the valency. And the highest degree of oxidation grows from left to right in the table. We look: in the first and second versions, we are given elements that are in the same groups, where the highest degree of oxidation and, accordingly, the valence in oxides does not change. Cl → S → P - are located from right to left, that is, on the contrary, their valence in the higher oxide will fall. But in the row Al→C→N, the elements are located from left to right, the valence in the higher oxide increases in them. Answer: G

Example 4 In the series of elements S→Se→Te
A) the acidity of hydrogen compounds increases;
B) the highest degree of oxidation of elements increases;
C) the valence of elements in hydrogen compounds increases;
D) the number of electrons in the outer level decreases;

Immediately look at the location of these elements in the periodic table. Sulfur, selenium and tellurium are in the same group, one subgroup. Listed in order from top to bottom. Look again at the diagram above. From top to bottom in the periodic table, metallic properties increase, radii increase, electronegativity, ionization energy and non-metallic properties decrease, the number of electrons at the outer level does not change. Option D is ruled out immediately. If the number of external electrons does not change, then the valence possibilities and the highest oxidation state also do not change, B and C are excluded.

Option A remains. We check for order. According to the Kossel scheme, the strength of oxygen-free acids increases with a decrease in the oxidation state of an element and an increase in the radius of its ion. The oxidation state of all three elements is the same in hydrogen compounds, but the radius grows from top to bottom, which means that the strength of acids also grows.
The answer is A.

Example 5 In order of weakening of the main properties, the oxides are arranged in the following order:
A. Na 2 O → K 2 O → Rb 2 O
B. Na 2 O → MgO → Al 2 O 3
B. BeO→BaO→CaO
G. SO 3 → P 2 O 5 → SiO 2

The main properties of oxides weaken synchronously with the weakening of the metallic properties of the elements forming them. And Me-properties weaken from left to right or from bottom to top. Na, Mg and Al are just arranged from left to right. Answer B.

Chemistry is a must! how do the oxidizing properties change in the series of elements S---Se---Te---Po? explain the answer. and got the best answer

Answer from Pna Aleksandrovna Tkachenko[active]
In the oxygen subgroup, with increasing atomic number, the radius of atoms increases, and the ionization energy, which characterizes the metallic properties of elements, decreases. Therefore, in the 0--S-Se-Te-Po series, the properties of the elements change from non-metallic to metallic. Under normal conditions, oxygen is a typical non-metal (gas), while polonium is a metal similar to lead.
With an increase in the atomic number of the elements, the value of the electronegativity of the elements in the subgroup decreases. The negative oxidation state is becoming less and less characteristic. The oxidative oxidation state becomes less and less characteristic. The oxidizing activity of simple substances in the series 02--S-Se-Te decreases. So, if sulfur is much weaker, selenium directly interacts with hydrogen, then tellurium does not react with it.
In terms of electronegativity, oxygen is second only to fluorine, therefore, in reactions with all other elements, it exhibits exclusively oxidizing properties. Sulfur, selenium and tellurium in their properties. belong to the group of oxidizing-reducing agents. In reactions with strong reducing agents, they exhibit oxidizing properties, and under the action of strong oxidizing agents. they are oxidized, that is, they exhibit reducing properties.
Possible valencies and oxidation states of the elements of the sixth group of the main subgroup in terms of the structure of the atom.
Oxygen, sulfur, selenium, tellurium and polonium make up the main subgroup of group VI. The outer energy level of the atoms of the elements of this subgroup contains 6 electrons each, which have the s2p4 configuration and are distributed over the cells as follows:

Answer from 2 answers[guru]

Hey! Here is a selection of topics with answers to your question: chemistry, it is very necessary! how do the oxidizing properties change in the series of elements S---Se---Te---Po? explain the answer.

in a series of elements O-S-Se with an increase in the ordinal number of a chemical element, electronegativity 1) increases. 2) smart.
O-S-Se - decreases
C-N-O-F - increases
Fluorine is the most electronegative element.

in which each selenium atom is bonded to two other covalent bonds.

The chains are parallel to each other. Intermolecular interaction takes place between atoms of the same type in neighboring chains. The melting and boiling points of gray Se are respectively 219o C and 685o C. Photo-

the conductivity of gray selenium can be explained by the fact that under the action of the incident

of light, electrons acquire energy that allows them to overcome certain

a large barrier between the valence band and the conduction band, which is used

etsya in photocells. The electrical conductivity of selenium in the dark is very low, but it greatly increases in the light. Less stable modifications of selenium are

are: red selenium, which has eight-membered rings in its structure

ca, like sulfur, and black vitreous selenium, in which helical chains are not

reputations.

Tellurium has two modifications: amorphous dark brown and silver.

crystal gray, with a structure similar to that of gray selenium. The melting and boiling point of Te is 450o C and 990o C.

Simple substances are capable of exhibiting reducing and oxidizing

casting properties.

In the series S, Se, Te, the reducing ability of simple substances increases, while the oxidative activity decreases.

The reaction S (t.) + H2 Se (g.) \u003d H2 S (g.) + Se (gray) shows that sulfur is more

A stronger oxidizing agent than selenium.

Selenium and tellurium react with metals when heated, forming selenium.

dy and tellurides.

2Cu + Se = Cu2Se,

2Ag + Te = Ag2Te.

Selenium and tellurium are oxidized by oxygen to form dioxides

EO 2 only when heated. Both nonmetals are stable in air.

When Se and Te are oxidized with concentrated nitric and sulfuric acids, selenous and tellurous acids are obtained.

E + 2H2 SO4 = H2 EO3 + 2 SO2 + H2 O

When boiling in alkali solutions, selenium and tellurium disproportionate.

3Se + 6KOH = 2K2Se + K2SeO3 +3H2O

Selenium and tellurium compounds

Selenides and tellurides

Alkali metals, copper and silver form selenides and tellurides of normal stoichiometry, and they can be considered as salts of seleno- and tel-

hydrochloric acids. known natural selenides and tellurides:

Cu2 Se, PbSe, Cu2 Te, Ag2 Te, PbTe.

Selenium and tellurium compounds with hydrogen: H2 Se and H2 Te are colorless toxic gases with a very unpleasant odor. Dissolve in water to form

weak acids. In the series H2 S, H2 Se, H2 Te, the strength of acids increases due to the weakening of the H–E bond due to an increase in the size of the atom. In the same series, the restorative properties are enhanced. In aqueous solutions of H2 Se and

H2 Te are rapidly oxidized by atmospheric oxygen.

2H2Se + O2 = 2Se + 2H2O.

Oxides and oxygen acids of selenium and tellurium

Dioxides of selenium and tellurium- crystalline substances.

Oxide SeO2 - dissolves well in water, forming selenous acid

H2 SeO3 . TeO2 oxide is poorly soluble in water. Both oxides are highly soluble

are in alkali, for example:

SeO2 + 2NaOH = Na2 SeO3 + H2 O

Acid H 2 SeO 3 is a white solid.

tellurous acid describe the formula TeO 2 . xH 2 O, indicating-

on its variable composition.

Selenous and tellurous acids are weak , telluric exhibits amphotericity. Selenic acid is highly soluble, while telluric acid is

only in dilute solution.

selenites and tellurites similar to sulfites. When exposed to strong acids, selenous and telluric acids.

The oxidation state (+4) of selenium and tellurium is stable , but strong oxidizing agents can oxidize Se (+4) and Te (+4) compounds to the oxidation state

5H2 SeO3 + 2KMnO4 + 3H2 SO4 = 5H2 SeO4 + 2MnSO4 + K2 SO4 + 3H2 O

The reducing properties of the compounds Se (+4) and Te (+4) are expressed in terms of

noticeably weaker than that of sulfur (+4). Therefore, reactions of the type are possible: H2 EO3 + 2SO2 + H2 O \u003d E + 2H2 SO4

This method can be used to isolate red selenium and black selenium deposits.

Selenic acid H 2 SeO 4 in its pure form is a colorless solid

substance highly soluble in water. Selenic acid is close in strength to

sulfuric. and telluric is a weak acid.

Telluric acid has the formula H6 TeO6 . All six hydrogen

atoms can be replaced by metal atoms, as, for example, in salts:

Ag6 TeO6 , Hg3 TeO6 . This is a weak acid.

Selenic and telluric acids are slow-acting, but strong

nye oxidizing agents, stronger than sulfuric acid.

Gold dissolves in concentrated selenic acid: 2Au + 6 H2 SeO4 = Au2 (SeO4) 3 + 3 SeO2 + 6 H2 O

A mixture of concentrated selenic and hydrochloric acids dissolves the plate

Pt + 2 H2 SeO4 + 6HCl = H2 + 2 SeO2 +4 H2 O

TeO 3 trioxide is a yellow solid, insoluble in water, diluted

added acids and bases. TeO3 is obtained by decomposition of orthotelluric

howling acid when heated.

SeO 3 trioxide is a white solid formed by molecules

trimer (SeO3 )3 . Selenium trioxide is highly soluble in water, has a strong

nye oxidizing properties. SeO3 is obtained by displacing it from selenic acid with sulfur trioxide.

Selenium and tellurium halides. Many selenium and tellurium halides are known (EF6, EF4, SeF2, TeCl2), they are obtained by direct synthesis from simple elements

Conclusion

The VIA subgroup is formed by p-elements: O, S, Se, Te, Po.

All of them are non-metals, except for Po.

The general formula for valence electrons is ns 2 np 4 .

Elements of the VIA subgroup are often combined under the general name "hal-

cohens", which means "forming ores".

The most characteristic oxidation states for S, Se, Te: -2, +4, +6.

The minimum oxidation state (–2) is stable for all elements

Sulfur from positive oxidation states is more stable +6.

For Se, Te - the most stable oxidation state is +4.

Sulfur occurs in nature in the form of a simple substance, in the form of sulfide and sulfate minerals. Sulfide ores contain small amounts of selenides and tellurides.

Simple substances are capable of exhibiting both oxidative and reductive

beneficial properties.

In the series S, Se, Te, the reducing properties of simple substances are enhanced,

and oxidative activity is reduced.

Sulfur, selenium and tellurium react with metals to form sulfides, se-

lenides and tellurides, acting as oxidizers.

Sulfur, selenium and tellurium are oxidized by oxygen to form dioxides EO2.

In oxidation state(–2) all elements form weak acids of the type

H2 E.

In the series H2 S, H2 Se, H2 Te, the strength of acids increases.

Chalcogen compounds in the oxidation state (–2) show

innovative properties. They intensify when going from S to Te.

All oxides and hydroxides of chalcogens exhibit acidic properties.

The strength of acids increases with an increase in the degree of oxidation and decreases with over-

move from S to Te.

H2 SO4 and H2 SeO4 are strong acids, H2 TeO6 acid is weak.

The acids of the elements in the oxidation state (+4) are weak, and the oxide Te (+4)

exhibits amphotericity.

Oxides SO2 and SeO2 dissolve in water. TeO2 oxide is poorly soluble in water. All oxides are highly soluble in alkali.

The trioxides SO3 and SeO3 are highly soluble in water, while TeO3 is insoluble.

Sulfuric acid is the most used acid, as in chemical practice.

tick, and in industry.

The world production of H2 SO4 is 136 million tons/year.

Compounds in the +4 oxidation state can be both oxidized and reduced.

S(+4) compounds are more characteristic of reducing properties.

The reducing properties of Se (+4) and Te (+4) compounds are expressed

noticeably weaker than that of sulfur (+4).

The oxidation state (+4) of selenium and tellurium is stable, but strong oxidizing agents can oxidize Se (+4) and Te (+4) to the oxidation state (+6).

Sulfuric acid contains two oxidizing agents: hydrogen ion and

sulfate ion.

In dilute sulfuric acid, the oxidation of metals is carried out by hydrogen ions.

In concentrated sulfuric acid, the sulfate ion acts as an oxidizing agent.

which can be restored to SO2, S, H2 S, depending on the strength of the recovery

builder.

Selenic and telluric acids are slow acting but strong

oxidizing agents stronger than sulfuric acid.

1. Stepin B.D., Tsvetkov A.A. Inorganic Chemistry: Textbook for High Schools / B.D.

Stepin, A.A. Tsvetkov. - M .: Higher. school, 1994.- 608 p.: ill.

2. Karapetyants M.Kh. General and inorganic chemistry: Textbook for university students / M.Kh. Karapetyants, S.I. Drakin. - 4th ed., ster. - M.: Chemistry, 2000. -

3. Ugay Ya.A. General and inorganic chemistry: Textbook for university students,

students in the direction and specialty "Chemistry" / Ya.A. Wow. - 3rd

ed., rev. - M.: Higher. school, 2007. - 527 p.: ill.

4. Nikolsky A.B., Suvorov A.V. Chemistry. Textbook for universities /

A.B. Nikolsky, A.V. Suvorov. - St. Petersburg: Himizdat, 2001. - 512 p.: ill.