Malyugin 14. External and internal energy levels. Completion of the energy level.

Let us briefly recall what we already know about the structure of the electron shell of atoms:

ü the number of energy levels of the atom = the number of the period in which the element is located;

ü the maximum capacity of each energy level is calculated by the formula 2n2

ü the outer energy shell cannot contain more than 2 electrons for elements of the 1st period, more than 8 electrons for elements of other periods

Once again, let us return to the analysis of the scheme for filling energy levels in elements of small periods:

Table 1. Filling of energy levels

for elements of small periods

Period number	Number of energy levels = period number	Element symbol, its ordinal number	Total electrons	Distribution of electrons by energy levels	Group number
				H +1 )1	+1 H, 1e-
		He + 2 ) 2	+2 No, 2nd
				Li + 3 ) 2 ) 1	+ 3 Li, 2e-, 1e-
		Be +4 ) 2 )2	+ 4 Be, 2e-,2 e-
		B +5 ) 2 )3	+5 B, 2e-, 3e-
		C +6 ) 2 )4	+6 C, 2e-, 4e-
		N + 7 ) 2 ) 5	+ 7 N, 2e-,5 e-
		O + 8 ) 2 ) 6	+ 8 O, 2e-,6 e-
		F + 9 ) 2 ) 7	+ 9 F, 2e-,7 e-
		Ne + 10 ) 2 ) 8	+ 10 Ne, 2e-,8 e-
				Na + 11 ) 2 ) 8 )1	+1 1 Na, 2e-, 8e-, 1e-
		mg + 12 ) 2 ) 8 )2	+1 2 mg, 2e-, 8e-, 2 e-
		Al + 13 ) 2 ) 8 )3	+1 3 Al, 2e-, 8e-, 3 e-
		Si + 14 ) 2 ) 8 )4	+1 4 Si, 2e-, 8e-, 4 e-
		P + 15 ) 2 ) 8 )5	+1 5 P, 2e-, 8e-, 5 e-
		S + 16 ) 2 ) 8 )6	+1 5 P, 2e-, 8e-, 6 e-
		Cl + 17 ) 2 ) 8 )7	+1 7 Cl, 2e-, 8e-, 7 e-
18 Ar		Ar+ 18 ) 2 ) 8 )8	+1 8 Ar, 2e-, 8e-, 8 e-

Analyze table 1. Compare the number of electrons in the last energy level and the number of the group in which the chemical element is located.

Have you noticed that the number of electrons in the outer energy level of atoms is the same as the group number, in which the element is located (the exception is helium)?

!!! This rule is true only for elements major subgroups.

Each period of the system ends with an inert element(helium He, neon Ne, argon Ar). The external energy level of these elements contains the maximum possible number of electrons: helium -2, the remaining elements - 8. These are elements of group VIII of the main subgroup. The energy level similar to the structure of the energy level of an inert gas is called completed. This is a kind of strength limit of the energy level for each element of the Periodic system. Molecules of simple substances - inert gases, consist of one atom and are distinguished by chemical inertness, i.e., they practically do not enter into chemical reactions.

For the remaining elements of the PSCE, the energy level differs from the energy level of the inert element, such levels are called unfinished. The atoms of these elements tend to complete their outer energy level by donating or accepting electrons.

Questions for self-control

1. What energy level is called external?

2. What energy level is called internal?

3. What energy level is called complete?

4. Elements of which group and subgroup have a completed energy level?

5. What is the number of electrons in the outer energy level of the elements of the main subgroups?

6. How are the elements of one main subgroup similar in the structure of the electronic level

7. How many electrons at the outer level contain the elements of a) group IIA;

b) IVA group; c) Group VII A

View answer

1. Last

2. Any but the last

3. The one that contains the maximum number of electrons. As well as the outer level, if it contains 8 electrons for period I - 2 electrons.

4. Elements of group VIIIA (inert elements)

5. The number of the group in which the element is located

6. All elements of the main subgroups on the external energy level contain as many electrons as the group number

7. a) the elements of group IIA have 2 electrons in the outer level; b) group IVA elements have 4 electrons; c) elements of group VII A have 7 electrons.

Tasks for independent solution

1. Determine the element according to the following criteria: a) it has 2 electronic levels, on the outer - 3 electrons; b) has 3 electronic levels, on the outer - 5 electrons. Write down the distribution of electrons over the energy levels of these atoms.

2. What two atoms have the same number of filled energy levels?

View answer:

1. a) Let's establish the "coordinates" of the chemical element: 2 electronic levels - II period; 3 electrons at the outer level - III A group. This is a 5B bur. Scheme of distribution of electrons by energy levels: 2e-, 3e-

b) III period, VA group, element phosphorus 15Р. Scheme of distribution of electrons by energy levels: 2e-, 8e-, 5e-

2. d) sodium and chlorine.

Explanation: a) sodium: +11 )2)8 )1 (filled 2) ←→ hydrogen: +1)1

b) helium: +2 )2 (filled 1) ←→ hydrogen: hydrogen: +1)1

c) helium: +2 )2 (filled 1) ←→ neon: +10 )2)8 (filled 2)

*G) sodium: +11 )2)8 )1 (filled 2) ←→ chlorine: +17 )2)8 )7 (filled 2)

4. Ten. Number of electrons = serial number

5 c) arsenic and phosphorus. Atoms located in the same subgroup have the same number of electrons.

Explanations:

a) sodium and magnesium (in different groups); b) calcium and zinc (in the same group, but different subgroups); * c) arsenic and phosphorus (in one, main, subgroup) d) oxygen and fluorine (in different groups).

7. d) the number of electrons in the outer level

8. b) the number of energy levels

9. a) lithium (located in group IA of period II)

10. c) silicon (IVA group, III period)

11. b) boron (2 levels - IIperiod, 3 electrons in the outer level - IIIAGroup)

E.N.FRENKEL

Chemistry tutorial

A guide for those who do not know, but want to learn and understand chemistry

Part I. Elements of General Chemistry
(first level of difficulty)

Continuation. See the beginning in No. 13, 18, 23/2007

Chapter 3. Elementary information about the structure of the atom.
Periodic law of D.I. Mendeleev

Remember what an atom is, what an atom consists of, whether an atom changes in chemical reactions.

An atom is an electrically neutral particle consisting of a positively charged nucleus and negatively charged electrons.

The number of electrons during chemical processes can change, but nuclear charge always stays the same. Knowing the distribution of electrons in an atom (the structure of an atom), it is possible to predict many properties of a given atom, as well as the properties of simple and complex substances of which it is a part.

The structure of the atom, i.e. the composition of the nucleus and the distribution of electrons around the nucleus can be easily determined by the position of the element in the periodic system.

In the periodic system of D.I. Mendeleev, chemical elements are arranged in a certain sequence. This sequence is closely related to the structure of the atoms of these elements. Each chemical element in the system is assigned serial number, in addition, for it you can specify the period number, group number, subgroup type.

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Knowing the exact "address" of a chemical element - a group, subgroup and period number, one can unambiguously determine the structure of its atom.

Period is a horizontal row of chemical elements. There are seven periods in the modern periodic system. The first three periods small, because they contain 2 or 8 elements:

1st period - H, He - 2 elements;

2nd period - Li ... Ne - 8 elements;

3rd period - Na ... Ar - 8 elements.

Other periods - large. Each of them contains 2-3 rows of elements:

4th period (2 rows) - K ... Kr - 18 elements;

6th period (3 rows) - Cs ... Rn - 32 elements. This period includes a number of lanthanides.

Group is a vertical row of chemical elements. There are eight groups in total. Each group consists of two subgroups: main subgroup and secondary subgroup. For example:

The main subgroup is formed by chemical elements of small periods (for example, N, P) and large periods (for example, As, Sb, Bi).

A side subgroup is formed by chemical elements of only large periods (for example, V, Nb,
Ta).

Visually, these subgroups are easy to distinguish. The main subgroup is “high”, it starts from the 1st or 2nd period. The secondary subgroup is “low”, starting from the 4th period.

So, each chemical element of the periodic system has its own address: period, group, subgroup, ordinal number.

For example, vanadium V is a chemical element of the 4th period, group V, secondary subgroup, serial number 23.

Task 3.1. Specify the period, group and subgroup for chemical elements with serial numbers 8, 26, 31, 35, 54.

Task 3.2. Specify the serial number and name of the chemical element, if it is known that it is located:

a) in the 4th period, group VI, secondary subgroup;

b) in the 5th period, group IV, main subgroup.

How can information about the position of an element in the periodic system be related to the structure of its atom?

An atom is made up of a nucleus (positively charged) and electrons (negatively charged). In general, the atom is electrically neutral.

Positive charge of the nucleus of an atom equal to the atomic number of the chemical element.

The nucleus of an atom is a complex particle. Almost all the mass of an atom is concentrated in the nucleus. Since a chemical element is a collection of atoms with the same nuclear charge, the following coordinates are indicated near the symbol of the element:

Based on these data, the composition of the nucleus can be determined. The nucleus is made up of protons and neutrons.

Proton p has a mass of 1 (1.0073 amu) and a charge of +1. Neutron n it has no charge (neutral), and its mass is approximately equal to the mass of a proton (1.0087 amu).

The nuclear charge is determined by the protons. And the number of protons is(by size) charge of the nucleus of an atom, i.e. serial number.

Number of neutrons N determined by the difference between the quantities: "mass of the nucleus" BUT and "serial number" Z. So, for an aluminum atom:

N = BUT – Z = 27 –13 = 14n,

Task 3.3. Determine the composition of the nuclei of atoms if the chemical element is in:

a) 3rd period, group VII, main subgroup;

b) 4th period, group IV, secondary subgroup;

c) 5th period, group I, main subgroup.

Attention! When determining the mass number of the nucleus of an atom, it is necessary to round off the atomic mass indicated in the periodic system. This is done because the masses of the proton and neutron are practically integer, and the mass of electrons can be neglected.

Let us determine which of the nuclei below belong to the same chemical element:

A (20 R + 20n),

B (19 R + 20n),

IN 20 R + 19n).

Atoms of the same chemical element have nuclei A and B, since they contain the same number of protons, i.e., the charges of these nuclei are the same. Studies show that the mass of an atom does not significantly affect its chemical properties.

Isotopes are called atoms of the same chemical element (the same number of protons), differing in mass (different number of neutrons).

Isotopes and their chemical compounds differ from each other in physical properties, but the chemical properties of isotopes of the same chemical element are the same. Thus, isotopes of carbon-14 (14 C) have the same chemical properties as carbon-12 (12 C), which enter the tissues of any living organism. The difference is manifested only in radioactivity (isotope 14 C). Therefore, isotopes are used for the diagnosis and treatment of various diseases, for scientific research.

Let us return to the description of the structure of the atom. As you know, the nucleus of an atom does not change in chemical processes. What is changing? The variable is the total number of electrons in the atom and the distribution of electrons. General number of electrons in a neutral atom it is easy to determine - it is equal to the serial number, i.e. charge of the nucleus of an atom:

Electrons have a negative charge of -1, and their mass is negligible: 1/1840 of the mass of a proton.

Negatively charged electrons repel each other and are at different distances from the nucleus. Wherein electrons having an approximately equal amount of energy are located at an approximately equal distance from the nucleus and form an energy level.

The number of energy levels in an atom is equal to the number of the period in which the chemical element is located. Energy levels are conventionally designated as follows (for example, for Al):

Task 3.4. Determine the number of energy levels in the atoms of oxygen, magnesium, calcium, lead.

Each energy level can contain a limited number of electrons:

On the first - no more than two electrons;

On the second - no more than eight electrons;

On the third - no more than eighteen electrons.

These numbers show that, for example, the second energy level can have 2, 5, or 7 electrons, but not 9 or 12 electrons.

It is important to know that regardless of the energy level number on external level(last) cannot be more than eight electrons. The outer eight-electron energy level is the most stable and is called complete. Such energy levels are found in the most inactive elements - the noble gases.

How to determine the number of electrons in the outer level of the remaining atoms? There is a simple rule for this: number of outer electrons equals:

For elements of the main subgroups - the number of the group;

For elements of secondary subgroups, it cannot be more than two.

For example (Fig. 5):

Task 3.5. Specify the number of external electrons for chemical elements with serial numbers 15, 25, 30, 53.

Task 3.6. Find chemical elements in the periodic table, in the atoms of which there is a completed external level.

It is very important to correctly determine the number of external electrons, because It is with them that the most important properties of the atom are associated. So, in chemical reactions, atoms tend to acquire a stable, completed external level (8 e). Therefore, atoms, on the outer level of which there are few electrons, prefer to give them away.

Chemical elements whose atoms can only donate electrons are called metals. Obviously, there should be few electrons at the outer level of the metal atom: 1, 2, 3.

If there are many electrons on the external energy level of an atom, then such atoms tend to accept electrons before the completion of the external energy level, i.e. up to eight electrons. Such elements are called non-metals.

Question. Do the chemical elements of the secondary subgroups belong to metals or non-metals? Why?

Answer. Metals and non-metals of the main subgroups in the periodic table are separated by a line that can be drawn from boron to astatine. Above this line (and on the line) are non-metals, below - metals. All elements of secondary subgroups are below this line.

Task 3.7. Determine whether metals or non-metals include: phosphorus, vanadium, cobalt, selenium, bismuth. Use the position of the element in the periodic table of chemical elements and the number of electrons in the outer level.

In order to compose the distribution of electrons over the remaining levels and sublevels, the following algorithm should be used.

1. Determine the total number of electrons in the atom (by serial number).

2. Determine the number of energy levels (by period number).

3. Determine the number of external electrons (according to the type of subgroup and group number).

4. Indicate the number of electrons at all levels except the penultimate one.

For example, according to points 1–4 for the manganese atom, it is determined:

Total 25 e; distributed (2 + 8 + 2) = 12 e; so, on the third level is: 25 - 12 = 13 e.

The distribution of electrons in the manganese atom was obtained:

Task 3.8. Work out the algorithm by drawing up atomic structure diagrams for elements No. 16, 26, 33, 37. Indicate whether they are metals or non-metals. Explain the answer.

When compiling the above diagrams of the structure of the atom, we did not take into account that the electrons in the atom occupy not only levels, but also certain sublevels each level. Types of sublevels are indicated by Latin letters: s, p, d.

The number of possible sublevels is equal to the level number. The first level consists of one
s-sublevel. The second level consists of two sublevels - s and R. The third level - from three sublevels - s, p and d.

Each sublevel can contain a strictly limited number of electrons:

at the s-sublevel - no more than 2e;

at the p-sublevel - no more than 6e;

at the d-sublevel - no more than 10e.

Sublevels of one level are filled in a strictly defined order: spd.

Thus, R- sublevel can't start to fill if not full s-sublevel of a given energy level, etc. Based on this rule, it is easy to compose the electronic configuration of the manganese atom:

Generally electronic configuration of an atom manganese is written like this:

25 Mn 1 s 2 2s 2 2p 6 3s 2 3p 6 3d 5 4s 2 .

Task 3.9. Make electronic configurations of atoms for chemical elements No. 16, 26, 33, 37.

Why is it necessary to make electronic configurations of atoms? To determine the properties of these chemical elements. It should be remembered that only valence electrons.

Valence electrons are in the outer energy level and incomplete
d-sublevel of the pre-outer level.

Let's determine the number of valence electrons for manganese:

or abbreviated: Mn ... 3 d 5 4s 2 .

What can be determined by the formula for the electronic configuration of an atom?

1. What element is it - metal or non-metal?

Manganese is a metal, because the outer (fourth) level contains two electrons.

2. What process is typical for metal?

Manganese atoms always donate electrons in reactions.

3. What electrons and how many will give a manganese atom?

In reactions, the manganese atom gives up two outer electrons (they are farthest from the nucleus and are weaker attracted by it), as well as five pre-outer d-electrons. The total number of valence electrons is seven (2 + 5). In this case, eight electrons will remain at the third level of the atom, i.e. complete outer level is formed.

All these reasoning and conclusions can be reflected using the scheme (Fig. 6):

The resulting conditional charges of an atom are called oxidation states.

Considering the structure of the atom, in a similar way it can be shown that the typical oxidation states for oxygen are -2, and for hydrogen +1.

Question. With which of the chemical elements can manganese form compounds, if we take into account the degrees of its oxidation obtained above?

Answer: Only with oxygen, tk. its atom has the opposite charge in its oxidation state. The formulas of the corresponding manganese oxides (here the oxidation states correspond to the valences of these chemical elements):

The structure of the manganese atom suggests that manganese cannot have a higher degree of oxidation, because in this case, one would have to touch upon the stable, now completed, pre-outer level. Therefore, the +7 oxidation state is the highest, and the corresponding Mn 2 O 7 oxide is the highest manganese oxide.

To consolidate all these concepts, consider the structure of the tellurium atom and some of its properties:

As a non-metal, the Te atom can accept 2 electrons before the completion of the outer level and donate "extra" 6 electrons:

Task 3.10. Draw the electronic configurations of Na, Rb, Cl, I, Si, Sn atoms. Determine the properties of these chemical elements, the formulas of their simplest compounds (with oxygen and hydrogen).

Practical Conclusions

1. Only valence electrons participate in chemical reactions, which can only be in the last two levels.

2. Metal atoms can only donate valence electrons (all or a few), taking positive oxidation states.

3. Non-metal atoms can accept electrons (missing - up to eight), while acquiring negative oxidation states, and donate valence electrons (all or a few), while they acquire positive oxidation states.

Let us now compare the properties of the chemical elements of one subgroup, for example, sodium and rubidium:
Na...3 s 1 and Rb...5 s 1 .

What is common in the structure of the atoms of these elements? At the outer level of each atom, one electron is active metals. metal activity associated with the ability to donate electrons: the easier an atom gives off electrons, the more pronounced its metallic properties.

What holds electrons in an atom? attraction to the nucleus. The closer the electrons are to the nucleus, the stronger they are attracted by the nucleus of the atom, the more difficult it is to “tear them off”.

Based on this, we will answer the question: which element - Na or Rb - gives away an external electron more easily? Which element is the more active metal? Obviously, rubidium, because its valence electrons are farther away from the nucleus (and are less strongly held by the nucleus).

Conclusion. In the main subgroups, from top to bottom, the metallic properties are enhanced, because the radius of the atom increases, and valence electrons are weaker attracted to the nucleus.

Let's compare the properties of chemical elements of group VIIa: Cl …3 s 2 3p 5 and I...5 s 2 5p 5 .

Both chemical elements are non-metals, because. one electron is missing before the completion of the outer level. These atoms will actively attract the missing electron. Moreover, the stronger the missing electron attracts a non-metal atom, the stronger its non-metallic properties (the ability to accept electrons) are manifested.

What causes the attraction of an electron? Due to the positive charge of the nucleus of the atom. In addition, the closer the electron to the nucleus, the stronger their mutual attraction, the more active the non-metal.

Question. Which element has more pronounced non-metallic properties: chlorine or iodine?

Answer: Obviously, chlorine, because. its valence electrons are closer to the nucleus.

Conclusion. The activity of non-metals in subgroups decreases from top to bottom, because the radius of the atom increases and it is more and more difficult for the nucleus to attract the missing electrons.

Let us compare the properties of silicon and tin: Si …3 s 2 3p 2 and Sn…5 s 2 5p 2 .

Both atoms have four electrons at the outer level. Nevertheless, these elements in the periodic table are on opposite sides of the line connecting boron and astatine. Therefore, for silicon, the symbol of which is above the B–At line, nonmetallic properties are more pronounced. On the contrary, tin, whose symbol is below the B–At line, has stronger metallic properties. This is due to the fact that in the tin atom, four valence electrons are removed from the nucleus. Therefore, the attachment of the missing four electrons is difficult. At the same time, the return of electrons from the fifth energy level occurs quite easily. For silicon, both processes are possible, with the first (acceptance of electrons) predominating.

Conclusions on chapter 3. The fewer external electrons in an atom and the farther they are from the nucleus, the stronger the metallic properties are manifested.

The more external electrons in an atom and the closer they are to the nucleus, the more non-metallic properties are manifested.

Based on the conclusions formulated in this chapter, for any chemical element of the periodic system, you can make a "characteristic".

Property Description Algorithm
chemical element by its position
in the periodic system

1. Draw up a diagram of the structure of the atom, i.e. determine the composition of the nucleus and the distribution of electrons by energy levels and sublevels:

Determine the total number of protons, electrons and neutrons in an atom (by serial number and relative atomic mass);

Determine the number of energy levels (by period number);

Determine the number of external electrons (by type of subgroup and group number);

Indicate the number of electrons at all energy levels except the penultimate one;

2. Determine the number of valence electrons.

3. Determine which properties - metal or non-metal - are more pronounced for a given chemical element.

4. Determine the number of given (received) electrons.

5. Determine the highest and lowest oxidation states of a chemical element.

6. Compose for these oxidation states the chemical formulas of the simplest compounds with oxygen and hydrogen.

7. Determine the nature of the oxide and write an equation for its reaction with water.

8. For the substances indicated in paragraph 6, draw up equations of characteristic reactions (see Chapter 2).

Task 3.11. According to the above scheme, make descriptions of the atoms of sulfur, selenium, calcium and strontium and the properties of these chemical elements. What are the general properties of their oxides and hydroxides?

If you have completed exercises 3.10 and 3.11, then it is easy to see that not only the atoms of the elements of one subgroup, but also their compounds have common properties and a similar composition.

Periodic law of D.I. Mendeleev:the properties of chemical elements, as well as the properties of simple and complex substances formed by them, are in a periodic dependence on the charge of the nuclei of their atoms.

The physical meaning of the periodic law: the properties of chemical elements are periodically repeated because the configurations of valence electrons (the distribution of electrons of the outer and penultimate levels) are periodically repeated.

So, the chemical elements of the same subgroup have the same distribution of valence electrons and, therefore, similar properties.

For example, the chemical elements of the fifth group have five valence electrons. At the same time, in the atoms of chemical elements of the main subgroups- all valence electrons are in the outer level: ... ns 2 np 3 , where n– period number.

At atoms elements of secondary subgroups only 1 or 2 electrons are in the outer level, the rest are in d- sublevel of the pre-external level: ... ( n – 1)d 3 ns 2 , where n– period number.

Task 3.12. Make brief electronic formulas for atoms of chemical elements No. 35 and 42, and then make up the distribution of electrons in these atoms according to the algorithm. Make sure your prediction comes true.

Exercises for chapter 3

1. Formulate the definitions of the concepts "period", "group", "subgroup". What do the chemical elements that make up: a) period; b) a group; c) subgroup?

2. What are isotopes? What properties - physical or chemical - do isotopes have in common? Why?

3. Formulate the periodic law of DIMendeleev. Explain its physical meaning and illustrate with examples.

4. What are the metallic properties of chemical elements? How do they change in a group and in a period? Why?

5. What are the non-metallic properties of chemical elements? How do they change in a group and in a period? Why?

6. Make brief electronic formulas of chemical elements No. 43, 51, 38. Confirm your assumptions by describing the structure of the atoms of these elements according to the above algorithm. Specify the properties of these elements.

7. By short electronic formulas

a) ...4 s 2 4p 1 ;

b) …4 d 1 5s 2 ;

in 3 d 5 4s 1

determine the position of the corresponding chemical elements in the periodic system of D.I. Mendeleev. Name these chemical elements. Confirm your assumptions with a description of the structure of the atoms of these chemical elements according to the algorithm. Specify the properties of these chemical elements.

To be continued

Each period of the Periodic system of D. I. Mendeleev ends with an inert, or noble, gas.

The most common of the inert (noble) gases in the Earth's atmosphere is argon, which was isolated in its pure form before other analogues. What is the reason for the inertness of helium, neon, argon, krypton, xenon and radon?

The fact that atoms of inert gases have eight electrons at the outer, most distant levels from the nucleus (helium has two). Eight electrons at the outer level - the limiting number for each element of the Periodic system of D. I. Mendeleev, except for hydrogen and helium. This is a kind of ideal of strength of the energy level, to which the atoms of all other elements of the Periodic Table of D. I. Mendeleev strive.

Atoms can achieve such a position of electrons in two ways: by giving electrons from the external level (in this case, the external incomplete level disappears, and the penultimate one, which was completed in the previous period, becomes external) or by accepting electrons that are not enough to the treasured eight. Atoms that have fewer electrons on the outer level donate them to atoms that have more electrons on the outer level. It is easy to donate one electron, when it is the only one on the outer level, to the atoms of the elements of the main subgroup of group I (group IA). It is more difficult to donate two electrons, for example, to atoms of elements of the main subgroup of group II (group IIA). It is even more difficult to donate your three outer electrons to atoms of group III elements (group IIIA).

Atoms of elements-metals have a tendency to return electrons from the external level. And the easier the atoms of a metal element give up their outer electrons, the more pronounced its metallic properties are. It is clear, therefore, that the most typical metals in the Periodic system of D. I. Mendeleev are the elements of the main subgroup of group I (group IA). And vice versa, atoms of non-metal elements have a tendency to accept the missing to complete the external energy level. From what has been said, the following conclusion can be drawn. Within a period, with an increase in the charge of the atomic nucleus, and, accordingly, with an increase in the number of external electrons, the metallic properties of chemical elements weaken. The non-metallic properties of the elements, characterized by the ease of accepting electrons to the external level, are enhanced in this case.

The most typical non-metals are the elements of the main subgroup of group VII (VIIA group) of the Periodic Table of D. I. Mendeleev. There are seven electrons in the outer level of the atoms of these elements. Up to eight electrons at the outer level, that is, until the stable state of atoms, they lack one electron each. They easily attach them, showing non-metallic properties.

And how do the atoms of the elements of the main subgroup of the IV group (IVA group) of the Periodic Table of D. I. Mendeleev behave? After all, they have four electrons on the outer level, and it would seem that they do not care whether to give or receive four electrons. It turned out that the ability of atoms to give or receive electrons is influenced not only by the number of electrons in the outer level, but also by the radius of the atom. Within the period, the number of energy levels in the atoms of elements does not change, it is the same, but the radius decreases, as the positive charge of the nucleus (the number of protons in it) increases. As a result, the attraction of electrons to the nucleus increases, and the radius of the atom decreases, as if the atom is compressed. Therefore, it becomes more and more difficult to donate outer electrons and, conversely, it becomes easier to accept the missing up to eight electrons.

Within the same subgroup, the radius of an atom increases with an increase in the charge of the atomic nucleus, since with a constant number of electrons in the outer level (it is equal to the group number), the number of energy levels increases (it is equal to the period number). Therefore, it becomes easier for the atom to give away outer electrons.

In the Periodic system of D. I. Mendeleev, with an increase in the serial number, the properties of atoms of chemical elements change as follows.

What is the result of the acceptance or release of electrons by atoms of chemical elements?

Imagine that two atoms “meet”: a metal atom of group IA and an atom of a non-metal of group VIIA. A metal atom has a single electron in its outer energy level, while a non-metal atom lacks just one electron to complete its outer level.

A metal atom will easily give up its electron, which is the most distant from the nucleus and weakly bound to it, to a non-metal atom, which will provide it with a free place on its outer energy level.

Then the metal atom, devoid of one negative charge, will acquire a positive charge, and the non-metal atom, thanks to the received electron, will turn into a negatively charged particle - an ion.

Both atoms will fulfill their "cherished dream" - they will receive the much desired eight electrons at the external energy level. But what happens next? Oppositely charged ions, in full accordance with the law of attraction of opposite charges, will immediately unite, i.e., a chemical bond will arise between them.

A chemical bond formed between ions is called an ionic bond.

Consider the formation of this chemical bond using the well-known sodium chloride compound (table salt) as an example:

The process of transformation of atoms into ions is shown in the diagram and figure:

For example, an ionic bond is also formed during the interaction of calcium and oxygen atoms:

Such a transformation of atoms into ions always occurs during the interaction of atoms of typical metals and typical non-metals.

In conclusion, let us consider the algorithm (sequence) of reasoning when writing the scheme for the formation of an ionic bond, for example, between calcium and chlorine atoms.

1. Calcium is an element of the main subgroup of group II (HA group) of the Periodic Table of D. I. Mendeleev, metal. It is easier for its atom to donate two outer electrons than to accept the missing six:

2. Chlorine is an element of the main subgroup of group VII (VIIA group) of the Mendeleev table, non-metal. It is easier for its atom to accept one electron, which it lacks before the completion of the outer energy level, than to give up seven electrons from the outer level:

3. First, we find the least common multiple between the charges of the formed ions, it is equal to 2 (2 × 1). Then we determine how many calcium atoms need to be taken so that they donate two electrons (i.e., you need to take 1 Ca atom), and how many chlorine atoms you need to take so that they can accept two electrons (i.e., you need to take 2 Cl atoms) .

4. Schematically, the formation of an ionic bond between calcium and chlorine atoms can be written as follows:

To express the composition of ionic compounds, formula units are used - analogues of molecular formulas.

The numbers showing the number of atoms, molecules or formula units are called coefficients, and the numbers showing the number of atoms in a molecule or ions in a formula unit are called indices.

In the first part of the paragraph, we made a conclusion about the nature and causes of changes in the properties of elements. In the second part of the paragraph, we present the keywords.

Keywords and phrases

1. Atoms of metals and non-metals.
2. Ions positive and negative.
3. Ionic chemical bond.
4. Coefficients and indices.

Work with computer

1. Refer to the electronic application. Study the material of the lesson and complete the suggested tasks.
2. Search the Internet for email addresses that can serve as additional sources that reveal the content of the keywords and phrases of the paragraph. Offer the teacher your help in preparing a new lesson - make a report on the key words and phrases of the next paragraph.

Questions and tasks

1. Compare the structure and properties of atoms: a) carbon and silicon; b) silicon and phosphorus.
2. Consider the schemes for the formation of an ionic bond between the atoms of chemical elements: a) potassium and oxygen; b) lithium and chlorine; c) magnesium and fluorine.
3. Name the most typical metal and the most typical non-metal of the Periodic Table of D. I. Mendeleev.
4. Using additional sources of information, explain why inert gases began to be called noble gases.

MBOU "Gymnasium No. 1 of the city of Novopavlovsk"

Chemistry Grade 8

Subject:

"Change in the number of electrons

at the outer energy level

atoms of chemical elements"

Teacher: Tatyana Alekseevna Komarova

Novopavlovsk

The date: ___________

Lesson– 9

Lesson topic: Change in the number of electrons on the external energy

the level of atoms of chemical elements.

Lesson Objectives:

- to form the concept of metallic and non-metallic properties of elements at the atomic level;

- show the reasons for changing the properties of elements in periods and groups based on the structure of their atoms;

- to give initial ideas about the ionic bond.

Equipment: PSCE, table "Ionic bond".

During the classes

Organizing time.

Knowledge check

Characteristics of chemical elements according to the table (3 people)

The structure of atoms (2 people)

Learning new material

Consider the following questions:

1 . Atoms of which chemical elements have completed energy levels?

- these are atoms of inert gases, which are located in the main subgroup of the 8th group.

Completed electronic layers have increased resistance and stability.

atoms Group VIII (He Ne Ar Kr Xe Rn) contain 8e - on the external level, which is why they are inert, i.e. . chemically inactive, do not interact with other substances, i.e. their atoms have increased resistance and stability. That is, all chemical elements (having a different electronic structure) tend to obtain completed outer energy level ,8e - .

Example:

N a Mg F Cl

11 +12 +9 +17

2 8 1 2 8 2 2 7 2 8 7

1s 2 2s 2 p 6 3 s 1 1s 2 2s 2 p 6 3 s 2 1s 2 2s 2 p 5 1s 2 2s 2 p 6 3 s 2 p 5

How do you think the atoms of these elements can reach eight electrons at the outer level?

If (suppose) to close the last level of Na and Mg by hand, then complete levels are obtained. Therefore, these electrons must be given away from the external electronic level! Then, when electrons are donated, the pre-outer layer of 8e - , becomes outer.

And for the elements F and Cl, you should take 1 missing electron to your energy level than give 7e -. And so, there are 2 ways to achieve the completed energy level:

A) Recoil ("extra") electrons from the outer layer.

B) Admission to the external level ("missing") electrons.

2. The concept of metallicity and non-metallicity at the atomic level:

Metals are elements whose atoms donate their outer electrons.

Nonmetals - These are elements whose atoms accept electrons to the external energy level.

The easier the Me atom gives up its electrons, the more pronounced are its metallic properties.

The easier the HeMe atom accepts the missing electrons to the outer layer, the more pronounced are its non-metallic properties.

3. Changes in the Me and NeMe properties of atoms ch.e. in periods and groups in the PSCE.

In periods:

Example: Na (1e -) Mg (2e -) - write down the structure of the atom.

- What do you think, which element has more pronounced metallic properties, Na or Mg? What is easier to give 1st - or 2nd -? (Of course, 1e -, therefore, Na has more pronounced metallic properties).

Example: Al (3e -) Si (4e -), etc.

Over the period, the number of electrons in the outer level increases from left to right.

(brighter metallic properties are expressed in Al).

Of course, the ability to donate electrons over the period will decrease, i.e. metallic properties will be weakened.

Thus, the strongest Me are located at the beginning of the periods.

- And how will the ability to attach electrons change? (will increase)

Example:

SiCl

14 r +17 r

2 8 4 2 8 7

It is easier to accept 1 missing electron (from Cl) than 4e from Si.

Conclusion:

The non-metallic properties over the period will increase from left to right, and the metallic properties will weaken.

Another reason for the enhancement of non-Me properties is a decrease in the radius of the atom with the same number of levels.

Because within the 1st period, the number of energy levels for atoms does not change, but the number of external electrons e - and the number of protons p - in the nucleus increase. As a result of this, the attraction of electrons to the nucleus increases (Coulomb's law), and the radius (r) of the atom decreases, the atom, as it were, contracts.

General conclusion:

Within one period, with an increase in the atomic number (N) of the element, the metallic properties of the elements weaken, and the non-metallic properties increase, because:

- The number e is growing - at the external level it is equal to the number of the group and the number of protons in the nucleus.

- The radius of the atom decreases

— The number of energy levels is constant.

4. Consider the vertical dependence of the change in the properties of elements (within the main subgroups) in groups.

Example: VII group main subgroup (halogens)

FCl

9 +17

2 7 2 8 7

1s 2 2s 2 p 5 1s 2 2s 2 p 6 3s 2 p 5

The number e is the same on the outer levels of these elements, but the number of energy levels is different,

at F -2e - , and Cl - 3e - /

Which atom has the larger radius? (- chlorine, because 3 energy levels).

The closer the e are located to the nucleus, the stronger they are attracted to it.

- An atom of which element will be easier to attach e - at F or Cl?

(F - it is easier to attach 1 missing electron), because it has a smaller radius, which means that the force of attraction of an electron to the nucleus is greater than that of Cl.

Coulomb's law

The strength of the interaction of two electric charges is inversely proportional to the square

distances between them, i.e. the greater the distance between atoms, the smaller the force

attraction of two opposite charges (in this case, electrons and protons).

F is stronger than Cl ˃Br ˃J, etc.

Conclusion:

In groups (main subgroups), non-metallic properties decrease, and metallic properties increase, because:

one). The number of electrons at the outer level of atoms is the same (and is equal to the group number).

2). The number of energy levels in atoms is growing.

3). The radius of the atom increases.

Orally, according to the PSCE table, consider I - the group of the main subgroup. Conclude that the strongest metal is Fr francium, and the strongest non-metal is F fluorine.

Ionic bond.

Consider what happens to the atoms of elements if they reach an octet (i.e. 8e -) at the outer level:

Let's write out the formulas of the elements:

Na 0 +11 2e - 8e - 1e - Mg 0 +12 2e - 8e - 2e - F 0 +9 2e - 7e - Cl 0 +17 2e - 8e - 7e -

Na x +11 2e - 8e - 0e - Mg x +12 2e - 8e - 0e - F x +9 2e - 8e - Cl x +17 2e - 8e - 8e -

The top row of formulas contains the same number of protons and electrons, because these are the formulas of neutral atoms (there is a zero charge "0" - this is the degree of oxidation).

The bottom row is a different number of p + and e -, i.e. These are the formulas for charged particles.

Let us calculate the charge of these particles.

Na +1 +11 2e - 8e - 0e - 2 + 8 \u003d 10, 11-10 \u003d 1, oxidation state +1

F - +9 2e - 8e - 2 + 8 \u003d 10, 9-10 \u003d -1, oxidation state -1

mg +2 +12 2e — 8e — 0e — 2+8=10, 12-10=-2, oxidation state -2

As a result of the attachment - recoil of electrons, charged particles are obtained, which are called ions.

The atoms of Me upon recoil e - acquires "+" (positive charge)

Heme atoms accepting "foreign" electrons are charged "-" (negative charge)

A chemical bond formed between ions is called an ionic bond.

An ionic bond occurs between strong Me and strong non-Me.

Examples.

a) the formation of an ionic bond. Na + Cl

N a Cl + —

11 + +17 +11 +17

2 8 1 2 8 7 2 8 2 8 8

1e-

The process of converting atoms into ions:

1 e -

N a 0 + Cl 0 Na + + Cl - Na + Cl -

atom atom ion ion ionic compound

2e -

b) Ca O 2+ 2-

Ca 0 + 2 C l 0 Ca 2+ Cl 2 -

2 e -

Consolidation of knowledge, skills, abilities.

Atoms Me and NeMe

Ions "+" and "-"

Ionic chemical bond

Coefficients and indices.

D/Z§ 9, #1, #2, p.58

Lesson summary

Literature:

1. Chemistry grade 8. textbook for general education

institutions/O.S. Gabrielyan. Bustard 2009

2. Gabrielyan O.S. Handbook of the teacher.

Chemistry Grade 8, Bustard, 2003

Chemistry lesson in 8th grade. "_____" ___________________ 20_____

Change in the number of electrons at the external energy level of atoms of chemical elements.

Target. Consider changes in the properties of atoms of chemical elements in PSCE D.I. Mendeleev.

Educational. Explain the patterns of changes in the properties of elements within small periods and main subgroups; determine the causes of changes in metallic and non-metallic properties in periods and groups.

Developing. To develop the ability to compare and find patterns of changes in properties in PSCE D.I. Mendeleev.

Educational. Foster a culture of learning in the classroom.

During the classes.

1. Org. moment.

2. Repetition of the studied material.

Independent work.

1 option.

		Answer options
	Aluminum

6-10. Specify the number of energy levels in the atoms of the following elements.

		Answer options

	Electronic formula	Answer options

Option 2.

1-5. Specify the number of neutrons in the nucleus of an atom.

		Answer options

6-10. Specify the number of electrons in the outer energy level.

		Answer options
	Aluminum

11-15. The indicated electronic formula of the atom corresponds to the element.

		Answer options
	1s22s22p63s23p6 4s1

3. Learning a new topic.

Exercise. Distribute the electrons according to the energy levels of the following elements: Mg, S, Ar.

Completed electronic layers have increased resistance and stability. Atoms that have 8 electrons in their outer energy level - inert gases - have stability.

An atom will always be stable if it has 8ē on its outer energy level.

How can the atoms of these elements reach the 8-electron outer level?

2 ways to complete:

donate electrons

Accept electrons.

Metals are elements that donate electrons; they have 1-3 ē on the external energy level.

Non-metals are elements that accept electrons; they have 4-7 ē at the external energy level.

Changing properties in the PSCE.

Within one period, with an increase in the ordinal number of the element, the metallic properties weaken, and the non-metallic properties increase.

1. The number of electrons in the outer energy level is growing.

2. The radius of the atom decreases

3. The number of energy levels is constant

In the main subgroups, non-metallic properties decrease, and metallic properties increase.

1. The number of electrons in the external energy level is constant;

2. The number of energy levels increases;

3. The radius of the atom increases.

Thus, francium is the strongest metal, fluorine is the strongest non-metal.

4. Fixing.

Exercises.

1. Arrange these chemical elements in order of increasing metallic properties:

A) Al, Na, Cl, Si, P

B) Mg, Ba, Ca, Be

C) N, Sb, Bi, As

D) Cs, Li, K, Na, Rb

2. Arrange these chemical elements in order of increasing non-metallic properties:

B) C, Sn, Ge, Si

C) Li, O, N, B, C

D) Br, F, I, Cl

3. Underline the symbols of chemical metals:

A) Cl, Al, S, Na, P, Mg, Ar, Si

B) Sn, Si, Pb, Ge, C

Arrange in order of decreasing metallic properties.

4. Underline the symbols of the chemical elements of non-metals:

A) Li, F, N, Be, O, B, C

B) Bi, As, N, Sb, P

Arrange in order of decreasing non-metallic properties.

Homework. Page 61-63. Ex. 4 page 66