In the series o s se te decreases. Characteristics of compounds of elements of the main subgroup of group VI

chemistry, really necessary! how they change oxidizing properties in the series of elements S---Se---Te---Po? explain the answer. and got the best answer

Answer from Yona Aleksandrovna Tkachenko[active]
In the oxygen subgroup, as the atomic number increases, the radius of the atoms increases and the ionization energy, which characterizes the metallic properties of the elements, decreases. Therefore, in the series 0--S--Se--Te--Po the properties of the elements change from non-metallic to metallic. Under normal conditions, oxygen is a typical non-metal (gas), and polonium is a metal similar to lead.
As the atomic number of elements increases, the electronegativity value of elements in a subgroup decreases. Negative oxidation states are becoming less and less common. Oxidation state oxidation becomes less and less common. Oxidative activity simple substances in the series 02--S-Se--Te decreases. So, although 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 according to their properties. belong to the group of oxidizing-reducing agents. In reactions with strong reducing agents they exhibit oxidizing properties, and when exposed to strong oxidizing agents. they oxidize, that is, they exhibit reducing properties.
Possible valencies and oxidation states of elements of the sixth group main subgroup from the point of view of atomic structure.
Oxygen, sulfur, selenium, tellurium and polonium form the main subgroup of group VI. The outer energy level of the atoms of elements of this subgroup contains 6 electrons, which have the s2p4 configuration and are distributed among the cells as follows:

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

in the series of elements O-S-Se, with increasing atomic 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.

Problem 840.
Based on the structure of the atoms of sulfur, selenium and tellurium, indicate what valence states and oxidation states are characteristic of these elements. What are the formulas of their higher hydroxides? Explain the answer.
Solution:
The S, Se and Te atoms at the outer energy level each contain six electrons (ns 2 p 4 ) – two in the s-orbital and four in the p-orbital. Graphically it can be represented like this:

Therefore, in a stationary state, these elements, having two unpaired p-electrons, exhibit covalency equal to 2. S, Se and Te have the number unpaired electrons in an atom can be increased by transferring s- and p-electrons to the d-sublevel of the outer layer. In this regard, these elements exhibit covalency, equal not only to 2, but also to 4 and 6. Thus, when excitation of S, Se and Te atoms, the p-electron can be transferred to a free d-orbital and then the atom acquires four unpaired electrons. In this regard, I show these elements covalency equal to 4. Graphically this can be represented as follows:

by transferring, upon excitation of the atom, p- and s-electrons to the d-sublevel of the outer layer, which leads to the formation of six unpaired electrons in the atom in the outer electronic layer. In this regard, these elements exhibit a covalency of 6. Graphically, this can be represented as follows:

Thus, S, Se and Te in their compounds can exhibit both negative (-2) and positive oxidation states (+2, +4 and 6). In compounds with metals and hydrogen, the oxidation state is -2, and in compounds with non-metals it can be +4 and +6. Therefore, in compounds with oxygen, these elements can exhibit an oxidation state equal to either +4 or +6, and two types of oxides with the formula are formed.

Problem 841.
Give comparative characteristics hydrogen compounds of elements of the main subgroup of group VI, indicating and explaining the nature of the change: a) thermal stability; b) melting and boiling points; c) acid-base and redox properties. Which of these compounds can be obtained by reacting hydrogen with the corresponding simple substance?
Solution:
a) Strength chemical bond in molecules of hydrogen compounds of elements of the main subgroup of group VI falls in the series
H 2 O - H 2 S - H 2 Se - H 2 Te, which is manifested by a change in the enthalpy of dissociation of molecules into atoms. The reasons for this are that with an increase in the size of the external electron clouds atoms, the degree of their overlap decreases. Therefore, when moving from H 2 O to H 2 Te, the degree of overlap of the electron clouds of hydrogen atoms and an atom of an element of the main subgroup of group VI decreases, and the overlap region is located at a greater distance from the nucleus of the element atom (O, S, Se, Te) and is more strongly screened by the increased the number of intermediate electronic layers. In addition, in the series O - S - Se - Te, the electronegativity of an element's atom decreases. Therefore, in the H 2 O molecule, the electron cloud of the hydrogen atom shifts towards the oxygen atom in to the greatest extent, and in the molecules H 2 S. H 2 Se and H 2 Te - less and less. This also leads to a decrease in the overlap of interacting electron clouds and thereby weakens the bond between atoms. With a decrease in the bond strength in the molecules of hydrogen compounds of elements of the main subgroup of group VI, the resistance to heating in the series H 2 O - H 2 S - H 2 Se - H 2 Te decreases.

b) In the series H 2 O - H 2 S - H 2 Se - H 2 Te, the boiling and melting temperatures change very regularly, they gradually increase in this series, while when moving to H 2 O they increase sharply. This is due to the association of water molecules as a result of the occurrence between them hydrogen bonds.

The natural increase in melting and boiling temperatures in the series H 2 O - H 2 S - H 2 Se - H 2 Te is explained by an increase in atomic radii with increasing atomic number of the element. In this case, the polarizability of the molecules increases, which leads to an increase in intermolecular dispersion interaction, i.e., to an increase in their attraction to each other ( Van der Waals forces), which causes an increase in melting and boiling temperatures (the exception is H 2 O).

V) Acid properties in the series H 2 O - H 2 S - H 2 Se - H 2 Te naturally increase. Water has a very low degree of dissociation, because its molecules form associates due to hydrogen bonds, so the dissociation of an H 2 O molecule requires a significant expenditure of energy. Therefore the water is typically weak electrolyte-ampholyte, whose molecules disintegrate into ions:

H 2 O ↔ H + + OH -

H 2 O, H 2 S, H 2 Se, H 2 Te are typical acids. The acidic properties in the series naturally increase. This is explained by the fact that within one subgroup the radii of ions of the same charge increase with increasing nuclear charge. This pattern is explained by an increase in the number of electronic layers and an increasing distance outer electrons from the core. When going from H 2 S to H 2 Te, the degree of overlap of the electron clouds of hydrogen atoms with S, Se and Te atoms decreases, and the region of overlap of the electron clouds itself is located at a greater distance from the nucleus of the element’s atom and is more strongly screened by an increased number of intermediate layers. In addition, in the series S - Se - Te, the electronegativity of atoms decreases. This also leads to a decrease in the overlap of interacting electron clouds, and thus to a weakening of the bonds between atoms. Thus, in the series H 2 S - H 2 Se - H 2 Te, the distance between the atoms of the element and the hydrogen atoms increases, therefore, when dissolved in water, their acid-type dissociation increases. Oxidative activity in the series H 2 O - H 2 S - H 2 Se - H 2 Te consistently decreases, and the reducing properties increase. This is explained by a natural increase in the radii of the ions in a row, which facilitates the release of electrons. Therefore, oxygen is a very strong oxidizing agent, while tellurium is a strong reducing agent. In the series H 2 O - H 2 S - H 2 Se - H 2 Te, the reducing properties increase. Under the action of strong oxidizing agents, they are oxidized to dioxides or to the corresponding acids of the composition H 2 RO 3 or even to H 2 RO 4.

Water and hydrogen sulfide can be obtained by direct interaction of hydrogen with oxygen or sulfur with sufficient high temperatures. H 2 Se and H 2 Te are obtained indirectly, for example, by treating selenides or tellurides with strong acids:

Na 2 Se + H 2 SO 4 = Na 2 SO 4 + H 2 Se;
Na 2 Te + H 2 SO 4 = Na 2 SO 4 + H 2 Te.

Problem 842.
Which substance is more easily oxidized: sodium sulfide or sodium telluride? Explain the answer.
Solution:
Within one subgroup, the radii of ions of the same charge increase with increasing nuclear charge. This pattern is explained by an increase in the number of electronic layers and the growing distance of outer electrons from the nucleus. Therefore, the Te 2- ion will give up electrons more easily than the S 2- ion. The process of losing electrons, accompanied by an increase in the oxidation state of an element, is called oxidation. A substance that contains an oxidizing element is called a reducing agent. Thus, sodium telluride will oxidize more easily than sodium sulfide.

Dmitry Ivanovich Mendeleev discovered periodic law, according to which the properties of elements and those formed by them change periodically. This discovery was graphically displayed in the periodic table. The table very clearly and clearly shows how the properties of elements change over a period, and then repeat 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 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 metallic properties and radii decrease.

From top to bottom, it’s the other way around: metallic properties and atomic radii increase, and 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) atomic radii decrease;
B) the number of protons in the nuclei of atoms decreases;
C) the number of electronic layers in atoms increases;
D) decreases highest degree oxidation of atoms;

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 they appear in the table from left to right. To answer a question of this kind, you just need to know several patterns of changes in properties in the periodic table. So from left to right across the period, metallic properties decrease, non-metallic properties increase, electronegativity increases, ionization energy increases, and the radius of atoms decreases. In the group from top to bottom, metallic and reducing properties increase, electronegativity decreases, ionization energy decreases, and the radius of the atoms increases.

If you were careful, you already realized that in this case the radii of the atoms decrease. Answer A.

Example 2. In order to enhance their oxidizing properties, the elements are arranged in the following order:
A. F→O→N
B. I→Br→Cl
B. Cl→S→P
G. F→Cl→Br

As you know, in the periodic table of Mendeleev, oxidizing properties increase from left to right across the period and from bottom to top across the group. In option B, the elements of one group are shown in order from bottom to top. So B is suitable.

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
G. Al→C→N

In higher oxides, elements exhibit their highest oxidation state, which will coincide with the valence. And the highest oxidation state increases from left to right in the table. Let's look: in the first and second options we are given elements that are in the same groups, there the highest oxidation state and, accordingly, the valency in the oxides does not change. Cl→S→P – located from right to left, that is, on the contrary, their valency in the higher oxide will decrease. But in the series Al→C→N the elements are located from left to right, and their valence in the higher oxide increases. Answer: G

Example 4. In the series of elements S→Se→Te
A) the acidity of hydrogen compounds increases;
B) the highest oxidation state of elements increases;
C) the valency of elements in hydrogen compounds;
D) the number of electrons at the external level decreases;

We immediately look at the location of these elements in the periodic table. Sulfur, selenium and tellurium are in one group, one subgroup. Listed in order from top to bottom. Let's 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 immediately excluded. 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.

That leaves option A. Let's check for order. According to the Kossel scheme, the strength of oxygen-free acids increases with a decrease in the oxidation state of the element and an increase in the radius of its ion. All oxidation states three elements is the same in hydrogen compounds, but the radius increases from top to bottom, which means the strength of the acids increases.
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 the oxides weaken synchronously with the weakening metallic properties their constituent elements. A 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.