This lesson is devoted to the generalization and systematization of knowledge about the types of chemical bonds. During the lesson, schemes for the formation of a chemical bond in various substances. The lesson will help to consolidate the ability to determine the type of chemical bond in a substance by its chemical formula.
Topic: Chemical bond. Electrolytic dissociation
Lesson: Schemes for the formation of substances with different types of bonds
Rice. 1. Scheme of bond formation in a fluorine molecule
The fluorine molecule consists of two atoms of the same non-metal chemical element with the same electronegativity, therefore, a covalent relationship is realized in this substance. non-polar bond. Let us depict the scheme of bond formation in a fluorine molecule. Rice. one.
Around each fluorine atom, using dots, we draw seven valence, that is, external, electrons. Before a steady state, each atom needs one more electron. Thus, one common electron pair is formed. Replacing it with a dash, we will depict the graphic formula of the fluorine molecule F-F.
Conclusion:a covalent non-polar bond is formed between the molecules of one chemical element-non-metal. With this type of chemical bond, common electron pairs are formed that equally belong to both atoms, that is, there is no shift in the electron density to any of the atoms of the chemical element
Rice. 2. Scheme of bond formation in a water molecule
The water molecule consists of hydrogen and oxygen atoms - two non-metal elements with different values relative electronegativity, therefore, in this substance - a covalent polar bond.
Since oxygen is a more electronegative element than hydrogen, shared electron pairs shift towards oxygen. A partial charge arises on the hydrogen atoms, and a partial negative charge on the oxygen atom. Replacing both common electron pairs with dashes, or rather arrows showing the shift in electron density, we write down the graphical formula for water Fig. 2.
Conclusion:a covalent polar bond occurs between atoms of different non-metal elements, that is, with different values of relative electronegativity. With this type of bond, common electron pairs are formed, which are shifted towards a more electronegative element..
1. No. 5,6,7 (p. 145) Rudzitis G.E. Inorganic and organic chemistry. Grade 8: textbook for educational institutions: a basic level of/ G. E. Rudzitis, F.G. Feldman. M.: Enlightenment. 2011 176 pp.: ill.
2. Indicate the particle with the largest and smallest radius: Ar atom, ions: K +, Ca 2+, Cl -. Justify your answer.
3. Name three cations and two anions that have the same electron shell as the F - ion.
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.
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 ionic bond substances 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.
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 numbers of 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, and 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 the addition of one or two neutrons changes the mass of the nucleus 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 . The structure of the electron shells of atoms
Answer to question 1.
a) Al+13 |
b) R |
c) Oh |
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13 Al 2e– , 8e– , 3e– |
15 Р 2e– , 8e– , 5e– |
8 О 2e– , 6e– |
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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– |
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Structure electron shell these atoms are similar, both on the last energy level contain 5 electrons. 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– |
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The atoms of phosphorus and sulfur have 3 energy levels, each with an incomplete last level, 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 a period is equal to the maximum possible number of electrons in the external energy level, and this number is determined by the formula 2n2, where n is the number of the period.
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 under optimal conditions 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.
§ nine . 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– |
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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– |
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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 to a greater extent non-metallic properties 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 give up 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 fill neon lamps and fluorescent lamps, which are often used to illuminate signs, posters, 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° - 3e → 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 |
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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 |
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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 |
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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
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: 
PART 1
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, taking electrons that are missing before the completion of the outer electronic layer, turn into negative ions:
3. Between oppositely charged ions arises a bond called an ionic bond.
4. Complete the table "Ionic bond".
PART 2
1. Complete the schemes for the formation of positively charged ions. From the letters corresponding to the correct answers, you will make the name of one of ancient 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
2) hydrogen and phosphorus
3) aluminum and fluorine
4) hydrogen and nitrogen
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. Make a diagram of the formation of two chemical compounds with an ionic bond according to the conditional notation:
Choose 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 ionic bond substances 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 such a 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. So, salt is life, but its consumption should be in moderation. 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.
