Answer to question 5.

The element with atomic number 35 is bromine (Br). The nuclear charge of its atom is 35. A bromine atom contains 35 protons, 35 electrons and 45 neutrons.

§7. Changes in the composition of the nuclei of atoms of chemical elements. isotopes

Answer to question 1.

Isotopes 40 19 K and 40 18 Ar exhibit different properties, because they have different nuclear charges and different amount electrons.

Answer to question 2.

The relative atomic mass of argon is close to 40, because there are 18 protons and 22 neutrons in the nucleus of its atom, and 19 protons and 20 neutrons in the nucleus of the potassium atom, so its relative atomic mass is close to 39. Since the number of protons in the nucleus of the potassium atom is greater, it is in the table after argon.

Answer to question 3.

Isotopes are varieties of atoms of the same element that have the same number of protons and electrons and a different number of neutrons.

Answer to question 4.

Isotopes of chlorine are similar in properties, because properties are determined by the charge of the nucleus, not by its relative mass, even when the relative atomic mass chlorine isotopes by 1 or 2 units, the mass changes slightly, in contrast to hydrogen isotopes, where, with the addition of one or two neutrons, the mass of the nucleus changes by 2 or 3 times.

Answer to question 5.

Deuterium (heavy water) - a compound where 1 oxygen atom is bonded to two atoms of the hydrogen isotope 2 1 D, formula D2 O. Comparison of the properties of D2 O and H2 O

Answer to question 6.

The element with the largest relative value is placed first.

atomic mass in vapors:

Te-I (tellurium-iodine) 128 Te and 127 I.

Th-Pa (thorium-protactinium) 232 90 Th and 231 91 Pa . U-Np (uranium-neptunium) 238 92 U and 237 93 Np .

§ eight . Structure electron shells atoms

Answer to question 1.

a) Al+13			b) R				c) Oh
13 Al 2e– , 8e– , 3e–				15 Р 2e– , 8e– , 5e–			8 О 2e– , 6e–

a) - diagram of the structure of the aluminum atom; b) - diagram of the structure of the phosphorus atom; c) - diagram of the structure of the oxygen atom.

Answer to question 2.

a) compare the structure of nitrogen and phosphorus atoms.

	7 N 2e– , 5e–			15 Р 2e– , 8e– , 5e–

The structure of the electron shell of these atoms is similar, both contain 5 electrons at the last energy level. However, nitrogen has only 2 energy levels, while phosphorus has 3.

b) Let's compare the structure of phosphorus and sulfur atoms.

15 Р 2e– , 8e– , 5e–				16S 2e– , 8e– , 6e–

The atoms of phosphorus and sulfur have 3 energy levels each, and each last level is incomplete, but phosphorus has 5 electrons at the last energy level, and sulfur has 6.

Answer to question 3.

The silicon atom contains 14 protons and 14 neutrons in the nucleus. The number of electrons around the nucleus, like the number of protons, is equal to the atomic number of the element. The number of energy levels is determined by the period number and is equal to 3. The number of external electrons is determined by the group number and is equal to 4.

Answer to question 4.

The number of elements contained in the period is equal to the maximum possible number electrons in the outer energy level and this number is determined by the formula 2n2, where n is the period number.

Therefore, the first period contains only 2 elements (2 12 ), and the second period contains 8 elements (2 22 ).

Answer to question 5.

AT astronomy - The period of rotation of the Earth around its axis is 24 hours.

AT geography - Change of seasons with a period of 1 year.

AT Physics - Periodic oscillations of the pendulum.

AT biology - Each yeast cell in optimal conditions once every 20 min. is divided.

Answer to question 6.

Electrons and the structure of the atom were discovered at the beginning of the 20th century, a little later this poem was written, which reflects in many respects the nuclear, or planetary, theory of the structure of the atom, and the author also admits the possibility that electrons are also complex particles, the structure of which we simply do not yet studied.

Answer to question 7.

The quatrains given in the textbook 2 speak of V. Bryusov's enormous poetic talent and his flexible mind, since he could so easily understand and accept all the achievements of contemporary science, as well as, apparently, enlightenment and education in this area.

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

Answer to question 1.

a) Compare the structure and properties of carbon and silicon atoms

6 С 2e– , 4e–			14 Si 2e– , 8e– , 4e–

In terms of the structure of the electron shell, these elements are similar: both have 4 electrons at the last energy level, but carbon has 2 energy levels, and silicon has 3. the number of electrons in the outer level is the same, then the properties of these elements will be similar, but the radius of the silicon atom is larger, therefore, compared to carbon, it will exhibit more metallic properties.

b) Compare the structure and properties of silicon and phosphorus atoms:

14 Si 2e– , 8e– , 4e–			15 Р 2e– , 8e– , 5e–

Silicon and phosphorus atoms have 3 energy levels, each with an incomplete last level, but silicon has 4 electrons at the last energy level, and phosphorus has 5, so the radius of the phosphorus atom is smaller and it exhibits non-metallic properties to a greater extent than silicon.

Answer to question 2.

a) Consider the formation of an ionic bond between aluminum and oxygen.

1. Aluminum - an element of the main subgroup of group III, metal. It is easier for its atom to donate 3 outer electrons than to accept the missing ones.

Al0 – 3e– → Al+ 3

2. Oxygen - an element of the main subgroup of group VI, non-metal. It is easier for its atom to accept 2 electrons, which are not enough to complete the outer level, than to give 6 electrons from the outer level.

O0 + 2e– → О− 2

3. First, find the least common multiple between the charges of the formed ions, it is equal to 6(3 2). For Al atoms to give 6

electrons, they need to be taken 2 (6: 3), so that oxygen atoms can accept 6 electrons, they need to be taken 3 (6: 2).

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

2Al0 + 3O0 → Al2 +3 O3 –2 → Al2 O3

6e–

b) Consider the scheme for the formation of an ionic bond between lithium and phosphorus atoms.

1. Lithium - an element of group I of the main subgroup, a metal. It is easier for its atom to donate 1 outer electron than to accept the missing 7:

Li0 – 1e– → Li+ 1

2. Phosphorus - an element of the main subgroup of group V, non-metal. It is easier for its atom to accept 3 electrons, which are not enough to complete the outer level, than to give 5 electrons:

Р0 + 3e– → Р− 3

3. Let us find the least common multiple between the charges of the formed ions, it is equal to 3(3 1). For lithium atoms to give

3 electrons, they need to be taken 3 (3: 1), so that phosphorus atoms can accept 5 electrons, you need to take only 1 atom (3: 3).

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

3Li0 – + P0 → Li3 +1 P–3 → Li3 P

c) Consider the scheme of formation of an ionic bond between magnesium and fluorine atoms.

1. Magnesium - an element of group II of the main subgroup, a metal. It is easier for its atom to donate 2 outer electrons than to accept the missing ones.

Mg0 – 2e– → Mg+ 2

2. Fluorine - an element of the main subgroup of group VII, non-metal. It is easier for its atom to accept 1 electron, which is not enough to complete the outer level, than to give 7 electrons:

F0 + 1e– → F− 1

3. Find the least common multiple between the charges of the formed ions, it is equal to 2(2 1). For magnesium atoms to donate 2 electrons, only one atom is needed, so that fluorine atoms can accept 2 electrons, they need to be taken 2 (2: 1).

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

Mg0 +– 2F0 → Mg+2 F2 –1 → MgF2

Answer to question 3.

The most typical metals are located in the periodic table

in at the beginning of periods and at the end of groups, thus the most typical metal is francium (Fr). Typical non-metals are located

in at the end of periods and at the beginning of groups. Thus, the most typical non-metal is fluorine (F). (Helium does not show any chemical property).

Answer to question 4.

Inert gases began to be called noble, as well as metals, because in nature they occur exclusively in free form and form chemical compounds with great difficulty.

Answer to question 5.

The expression "The streets of the night city were flooded with neon" is chemically incorrect, because. neon is an inert, rare gas, it contains very little in the air. However, neon is filled with neon lamps and fluorescent lamps, which are often used to illuminate signs, posters, and advertisements at night.

§ ten ​​. Interaction of atoms of non-metal elements among themselves

Answer to question 1.

The electronic scheme for the formation of a diatomic halogen molecule will look like this:

a + a → aa

And the structural formula

Answer to question 2.

a) Chemical bond formation scheme for AlCl3:

Aluminum is an element of group III. It is easier for its atom to donate 3 outer electrons than to accept the missing 5.

Al° - 3 e→ Al+3

Chlorine is an element of group VII. It is easier for its atom to accept 1 electron, which is not enough to complete the external level, than to give 7 electrons.

Сl° + 1 e → Сl–1

Let us find the least common multiple between the charges of the formed ions, it is equal to 3(3:1). In order for aluminum atoms to give up 3 electrons, only 1 atom (3: 3) must be taken, in order for chlorine atoms to be able to accept 3 electrons, they must be taken 3 (3: 1)

Al° + 3Сl° → Al+3 Cl–1 → AlСl3

3 e-

The bond between metal and non-metal atoms is ionic. b) Scheme of chemical bond formation for Cl2:

Chlorine is an element of the main subgroup of group VII. Its atoms have 7 electrons in their outer level. The number of unpaired electrons is

					→ClCl

The bond between atoms of the same element is covalent.

Answer to question 3.

Sulfur is an element of the main subgroup of group VI. Its atoms have 6 electrons at the outer level. The number of unpaired electrons is (8–6)2. In S2 molecules, the atoms are bonded by two shared electron pairs, so the bond is double.

The scheme for the formation of the S2 molecule will look like this:

Answer to question 4.

The S2 molecule has a double bond, the Cl molecule has a single bond, and the N2 molecule has a triple bond. Therefore, the strongest molecule will be N2, less durable S2, and even weaker Cl2.

The bond length is smallest in the N2 molecule, longer in the S2 molecule, and even longer in the Cl2 molecule.

§ eleven . covalent polar chemical bond

Answer to question 1.

Since the EO values ​​of hydrogen and phosphorus are the same, the chemical bond in the PH3 molecule will be covalent nonpolar.

Answer to question 2.

1. a) in the S2 molecule, the bond is covalent non-polar, because it is formed by atoms of the same element. The connection formation scheme will be as follows:

Sulfur is an element of the main subgroup of group VI. Its atoms have 6 electrons in the outer shell. There will be unpaired electrons: 8 - 6 = 2.

Denote the outer electrons S

b) in the K2 O molecule, the bond is ionic, because it is formed by atoms of metal and non-metal elements.

Potassium is an element of group I of the main subgroup, a metal. It is easier for its atom to give 1 electron than to accept the missing 7:

K0 – 1e– → K+ 1

Oxygen is an element of the main subgroup of group VI, a non-metal. It is easier for its atom to accept 2 electrons, which are not enough to complete the level, than to give 6 electrons:

O0 + 2e– → O− 2

Let us find the least common multiple between the charges of the formed ions, it is equal to 2(2 1). For potassium atoms to give up 2 electrons, they need to take 2, so that oxygen atoms can accept 2 electrons, only 1 atom is needed:

2K2e 0 – + O0 → K2 +1 O–2 → K2 O

c) in the H2 S molecule, the bond is covalent polar, because it is formed by atoms of elements with different EO. The connection formation scheme will be as follows:

Sulfur is an element of the main subgroup of group VI. Its atoms have 6 electrons in the outer shell. There will be unpaired electrons: 8– 6=2.

Hydrogen is an element of the main subgroup of group I. Its atoms contain 1 electron per outer shell. 1 electron is unpaired (for a hydrogen atom, a two-electron level is complete). Let's denote the outer electrons:

H + S + H → H

Shared electron pairs are biased towards the sulfur atom, as it is more electronegative

H δ+→ S 2 δ−← H δ+

1. a) in the N2 molecule, the bond is covalent non-polar, because it is formed by atoms of the same element. The connection formation scheme is as follows:

Nitrogen is an element of the main subgroup of group V. Its atoms have 5 electrons in their outer shell. Unpaired electrons: 8 - 5 = 3.

Let's denote the outer electrons: N

					→ N N

N ≡ N

b) in the Li3 N molecule, the bond is ionic, because it is formed by atoms of metal and non-metal elements.

Lithium is an element of the main subgroup of group I, a metal. It is easier for its atom to give 1 electron than to accept the missing 7:

Li0 – 1e– → Li+ 1

Nitrogen is an element of the main subgroup of group V, a non-metal. It is easier for its atom to accept 3 electrons, which are not enough to complete the outer level, than to give five electrons from the outer level:

N0 + 3e– → N− 3

Let us find the least common multiple between the charges of the formed ions, it is equal to 3(3 1). For lithium atoms to donate 3 electrons, 3 atoms are needed, for nitrogen atoms to be able to accept 3 electrons, only one atom is needed:

3Li0 + N0 → Li3 +1 N–3 → Li3 N

3e–

c) in the NCl3 molecule, the bond is covalent polar, because it is formed by atoms of non-metal elements with different values ​​of EC. The connection formation scheme is as follows:

Nitrogen is an element of the main subgroup of group V. Its atoms have 5 electrons in the outer shell. There will be unpaired electrons: 8– 5=3.

Chlorine is an element of the main subgroup of group VII. Its atoms contain 7 electrons in the outer shell. Remains unpaired

Catch the answer.
a) Consider the formation of an ionic bond between sodium and
oxygen.
1. Sodium - an element of the main subgroup of group I, a metal. It is easier for its atom to give the I outer electron than to accept the missing 7:

2. Oxygen element the main subgroup of group VI, non-metal.
It is easier for its atom to accept 2 electrons, which are not enough to complete the outer level, than to give 6 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). In order for Na atoms to give up 2 electrons, they must be taken 2 (2: 1), in order for oxygen atoms to be able to accept 2 electrons, they must be taken 1.
4. Schematically, the formation of an ionic bond between sodium and oxygen atoms can be written as follows:

b) Consider the scheme for the formation of an ionic bond between lithium and phosphorus atoms.
I. Lithium - an element of group I of the main subgroup, a metal. It is easier for its atom to donate 1 outer electron than to accept the missing 7:

2. Chlorine - an element of the main subgroup of group VII, non-metal. His
It is easier for an atom to accept 1 electron than to donate 7 electrons:

2. Least common multiple of 1, i.e. in order for 1 atom of lithium to give away, and a chlorine atom to accept 1 electron, you need to take them one by one.
3. Schematically, the formation of an ionic bond between lithium and chlorine atoms can be written as follows:

c) Consider the scheme for the formation of an ionic bond between atoms
magnesium and fluorine.
1. Magnesium is an element of group II of the main subgroup, a metal. His
it is easier for an atom to donate 2 outer electrons than to accept the missing 6:

2. Fluorine - an element of the main subgroup of group VII, non-metal. His
it is easier for an atom to accept 1 electron, which is not enough to complete the outer level, than to give 7 electrons:

2. Find the least common multiple between the charges of the formed ions, it is equal to 2(2∙1). For magnesium atoms to donate 2 electrons, only one atom is needed, so that fluorine atoms can accept 2 electrons, they need to be taken 2 (2: 1).
3. Schematically, the formation of an ionic bond between lithium and phosphorus atoms can be written as follows:

Help is on the way, hold on.
a) Consider the formation of an ionic bond between sodium and
oxygen.
1. Sodium - an element of the main subgroup of group I, a metal. It is easier for its atom to give the I outer electron than to accept the missing 7:

1. Oxygen is an element of the main subgroup of group VI, non-metal.
It is easier for its atom to accept 2 electrons, which are not enough to complete the outer level, than to give 6 electrons from the outer level.

1. First, we find the least common multiple between the charges of the formed ions, it is equal to 2(2∙1). In order for Na atoms to give up 2 electrons, they must be taken 2 (2: 1), in order for oxygen atoms to be able to accept 2 electrons, they must be taken 1.
2. Schematically, the formation of an ionic bond between sodium and oxygen atoms can be written as follows:

b) Consider the scheme for the formation of an ionic bond between lithium and phosphorus atoms.
I. Lithium - an element of group I of the main subgroup, a metal. It is easier for its atom to donate 1 outer electron than to accept the missing 7:

2. Chlorine - an element of the main subgroup of group VII, non-metal. His
It is easier for an atom to accept 1 electron than to donate 7 electrons:

2. Least common multiple of 1, i.e. in order for 1 atom of lithium to give away, and a chlorine atom to accept 1 electron, you need to take them one at a time.
3. Schematically, the formation of an ionic bond between lithium and chlorine atoms can be written as follows:

c) Consider the scheme for the formation of an ionic bond between atoms
magnesium and fluorine.
1. Magnesium is an element of group II of the main subgroup, a metal. His
it is easier for an atom to donate 2 outer electrons than to accept the missing 6:

2. Fluorine - an element of the main subgroup of group VII, non-metal. His
it is easier for an atom to accept 1 electron, which is not enough to complete the outer level, than to give 7 electrons:

2. Find the least common multiple between the charges of the formed ions, it is equal to 2(2∙1). For magnesium atoms to donate 2 electrons, only one atom is needed, so that fluorine atoms can accept 2 electrons, they need to be taken 2 (2: 1).
3. Schematically, the formation of an ionic bond between lithium and phosphorus atoms can be written as follows:

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Lesson Objectives:

• To form the concept of chemical bonds using the example of an ionic bond. To achieve an understanding of the formation of an ionic bond as an extreme case of a polar one.
• During the lesson, ensure the assimilation of the following basic concepts: ions (cation, anion), ionic bond.
• To develop the mental activity of students through the creation problem situation when learning new material.

Tasks:

• learn to recognize the types of chemical bonds;
• repeat the structure of the atom;
• to investigate the mechanism of formation of ionic chemical bond;
• teach how to draw up formation schemes and electronic formulas of ionic compounds, reaction equations with the designation of the transition of electrons.

Equipment: computer, projector, multimedia resource, periodic system chemical elements D.I. Mendeleev, table "Ionic bond".

Lesson type: Formation of new knowledge.

Type of lesson: multimedia lesson.

X one lesson

I.Organizing time.

II . Checking homework.

Teacher: How can atoms take on stable electronic configurations? What are the ways of forming a covalent bond?

Student: Polar and non-polar covalent bonds formed by the exchange mechanism. The exchange mechanism includes cases when one electron is involved in the formation of an electron pair from each atom. For example, hydrogen: (slide 2)

The bond arises due to the formation of a common electron pair due to the union of unpaired electrons. Each atom has one s-electron. The H atoms are equivalent and the pairs equally belong to both atoms. Therefore, the formation of common electron pairs (overlapping p-electron clouds) occurs during the formation of the F 2 molecule. (slide 3)

H record · means that the hydrogen atom has 1 electron on the outer electron layer. The record shows that there are 7 electrons on the outer electron layer of the fluorine atom.

During the formation of the N 2 molecule. 3 common electron pairs are formed. The p-orbitals overlap. (slide 4)

The bond is called non-polar.

Teacher: We have now considered the cases when molecules are formed a simple substance. But there are many substances around us, a complex structure. Let's take a hydrogen fluoride molecule. How does the formation of a connection take place in this case?

Student: When a hydrogen fluoride molecule is formed, the orbital of the s-electron of hydrogen and the orbital of the p-electron of fluorine H-F overlap. (slide 5)

The bonding electron pair is shifted to the fluorine atom, resulting in the formation dipole. Connection called polar.

III. Knowledge update.

Teacher: A chemical bond arises as a result of changes that occur with the outer electron shells of the connecting atoms. This is possible because the outer electron layers are not complete in elements other than inert gases. The chemical bond is explained by the desire of atoms to acquire a stable electronic configuration, similar to the configuration of the "nearest" inert gas to them.

Teacher: Write down the diagram electronic structure sodium atom (at the blackboard). (slide 6)

Student: To achieve the stability of the electron shell, the sodium atom must either give up one electron or accept seven. Sodium will easily give up its electron far from the nucleus and weakly bound to it.

Teacher: Make a diagram of the recoil of an electron.

Na° - 1ē → Na+ = Ne

Teacher: Write down a diagram of the electronic structure of the fluorine atom (at the blackboard).

Teacher: How to achieve the completion of the filling of the electronic layer?

Student: To achieve the stability of the electron shell, the fluorine atom must either give up seven electrons or accept one. It is energetically more favorable for fluorine to accept an electron.

Teacher: Make a scheme for receiving an electron.

F° + 1ē → F- = Ne

IV. Learning new material.

The teacher addresses a question to the class in which the task of the lesson is set:

Are there other options in which atoms can take on stable electronic configurations? What are the ways of formation of such connections?

Today we will look at one of the types of connections - ionic bond. Let us compare the structure of the electron shells of the already named atoms and inert gases.

Conversation with the class.

Teacher: What charge did the sodium and fluorine atoms have before the reaction?

Student: The atoms of sodium and fluorine are electrically neutral, because. the charges of their nuclei are balanced by electrons revolving around the nucleus.

Teacher: What happens between atoms when giving and receiving electrons?

Student: Atoms acquire charges.

The teacher gives explanations: In the formula of an ion, its charge is additionally recorded. To do this, use the superscript. In it, a number indicates the amount of charge (they do not write a unit), and then a sign (plus or minus). For example, a Sodium ion with a charge of +1 has the formula Na + (read "sodium plus"), a Fluorine ion with a charge of -1 - F - ("fluorine minus"), a hydroxide ion with a charge of -1 - OH - (" o-ash-minus"), a carbonate ion with a charge of -2 - CO 3 2- ("tse-o-three-two-minus").

In the formulas of ionic compounds, first write down, without indicating the charges, positively charged ions, and then - negatively charged. If the formula is correct, then the sum of the charges of all ions in it is equal to zero.

positively charged ion called a cation, and a negatively charged ion-anion.

Teacher: We write the definition in workbooks:

And he is a charged particle into which an atom turns into as a result of receiving or giving off electrons.

Teacher: How to determine the charge of the calcium ion Ca 2+?

Student: An ion is an electrically charged particle formed as a result of the loss or gain of one or more electrons by an atom. Calcium has two electrons in the last electronic level, the ionization of a calcium atom occurs when two electrons are given away. Ca 2+ is a doubly charged cation.

Teacher: What happens to the radii of these ions?

During the transition electrically neutral atom into an ionic state, the particle size changes greatly. An atom, giving up its valence electrons, turns into a more compact particle - a cation. For example, during the transition of a sodium atom to the Na+ cation, which, as indicated above, has a neon structure, the radius of the particle is greatly reduced. The radius of an anion is always greater than the radius of the corresponding electrically neutral atom.

Teacher: What happens to oppositely charged particles?

Student: Oppositely charged sodium and fluorine ions, resulting from the transition of an electron from a sodium atom to a fluorine atom, are mutually attracted and form sodium fluoride. (slide 7)

Na + + F - = NaF

The scheme of formation of ions that we have considered shows how a chemical bond is formed between the sodium atom and the fluorine atom, which is called ionic.

Ionic bond- a chemical bond formed by the electrostatic attraction of oppositely charged ions to each other.

The compounds that form in this case are called ionic compounds.

V. Consolidation of new material.

Tasks to consolidate knowledge and skills

1. Compare the structure of the electron shells of the calcium atom and the calcium cation, the chlorine atom and the chloride anion:

Comment on the formation of an ionic bond in calcium chloride:

2. To complete this task, you need to divide into groups of 3-4 people. Each member of the group considers one example and presents the results to the whole group.

Students response:

1. Calcium is an element of the main subgroup of group II, a 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, a non-metal. It is easier for its atom to accept one electron, which it lacks before the completion of the outer level, than to give up seven electrons from the outer level:

3. First, find the least common multiple between the charges of the formed ions, it is equal to 2 (2x1). Then we determine how many calcium atoms need to be taken so that they donate two electrons, that is, one Ca atom and two CI atoms must be taken.

4. Schematically, the formation of an ionic bond between calcium and chlorine atoms can be written: (slide 8)

Ca 2+ + 2CI - → CaCI 2

Tasks for self-control

1. Based on the scheme for the formation of a chemical compound, write an equation chemical reaction: (slide 9)

2. Based on the scheme for the formation of a chemical compound, make up an equation for a chemical reaction: (slide 10)

3. A scheme for the formation of a chemical compound is given: (slide 11)

Choose a pair of chemical elements whose atoms can interact in accordance with this scheme:

a) Na and O;
b) Li and F;
in) K and O;
G) Na and F

Part I

1. Atoms of metals, giving up external electrons, turn into positive ions:

where n is the number of electrons in the outer layer of the atom, corresponding to the group number of the chemical element.

2. Atoms of non-metals, accepting electrons missing before the completion of the outer electron layer, are converted into negative ions:

3. A bond arises between oppositely charged ions, which is called ionic.

4. Complete the table "Ionic bond".

Part II

1. Complete the schemes for the formation of positively charged ions. From the letters corresponding to the correct answers, you will form the name of one of the oldest natural dyes: indigo.

2. Play tic-tac-toe. Show the winning path that the formulas of substances with an ionic chemical bond make up.

3. Are the following statements true?

3) only B is true

4. Underline the pairs of chemical elements between which an ionic chemical bond is formed.
1) potassium and oxygen
3) aluminum and fluorine
Draw diagrams for the formation of a chemical bond between the selected elements.

5. Create a comic-style drawing of the formation of an ionic chemical bond.

6. Chart the formation of two chemical compounds with ionic bond by convention:

Select chemical elements"A" and "B" from the following list:
calcium, chlorine, potassium, oxygen, nitrogen, aluminum, magnesium, carbon, bromine.
Suitable for this scheme are calcium and chlorine, magnesium and chlorine, calcium and bromine, magnesium and bromine.

7. Write a short literary work(essay, short story or poem) about one of the substances with an ionic bond that a person uses in everyday life or at work. Use the Internet to complete the task.
Sodium chloride is a substance with an ionic bond, without it there is no life, although when there is a lot of it, this is also not good. There is even one folk tale, which tells that the princess loved her father the king as much as salt, for which she was expelled from the kingdom. But, when the king once tried food without salt and realized that it was impossible, he then realized that his daughter loved him very much. This means that salt is life, but its consumption should be in
measure. Because too much salt is bad for your health. Excess salt in the body leads to kidney disease, changes skin color, retains excess fluid in the body, which leads to edema and stress on the heart. Therefore, you need to control your salt intake. 0.9% sodium chloride solution is a saline solution used to infuse drugs into the body. Therefore, it is very difficult to answer the question: is salt useful or harmful? We need her in moderation.