Genetic series of silicon. Genetic relationship between classes of inorganic substances

The material world in which we live and of which we are a tiny part is one and at the same time infinitely diverse. The unity and diversity of the chemical substances of this world is most clearly manifested in the genetic connection of substances, which is reflected in the so-called genetic series. Let's highlight the most characteristic features such rows.

1. All substances in this series must be formed by one chemical element. For example, a series written using the following formulas:

2. Substances formed by the same element must belong to different classes, i.e., reflect different shapes his existence.

3. Substances that form the genetic series of one element must be connected by mutual transformations. Based on this feature, it is possible to distinguish between complete and incomplete genetic series.

For example, the above genetic series of bromine will be incomplete, incomplete. Here's the next row:

can already be considered complete: it began with the simple substance bromine and ended with it.

Summarizing the above, we can give the following definition of the genetic series.

Genetic series- this is a series of substances - representatives of different classes, which are compounds of one chemical element, connected by mutual transformations and reflecting the common origin of these substances or their genesis.

Genetic connection- a more general concept than the genetic series, which is, albeit a vivid, but particular manifestation of this connection, which is realized during any mutual transformations of substances. Then, obviously, the first given series of substances also fits this definition.

There are three types of genetic series:

The richest series of metals exhibits different oxidation states. As an example, consider the genetic series of iron with oxidation states +2 and +3:

Let us recall that to oxidize iron into iron (II) chloride, you need to take a weaker oxidizing agent than to obtain iron (III) chloride:

Similar to the series of metals, the series of non-metals with different degrees oxidation, for example, the genetic series of sulfur with oxidation states +4 and +6:

Only the last transition can cause difficulty. Follow the rule: to obtain a simple substance from an oxidized compound of an element, you need to take for this purpose its most reduced compound, for example, a volatile hydrogen compound of a non-metal. In our case:

This reaction in nature produces sulfur from volcanic gases.

Likewise for chlorine:

3. The genetic series of the metal, which corresponds to amphoteric oxide and hydroxide,very rich in bonds, because depending on the conditions they exhibit either acidic or basic properties.

For example, consider the genetic series of zinc:

Genetic relationship between classes of inorganic substances

Characteristic are reactions between representatives of different genetic series. Substances from the same genetic series, as a rule, do not interact.

For example:
1. metal + non-metal = salt

Hg + S = HgS

2Al + 3I 2 = 2AlI 3

2. basic oxide + acidic oxide = salt

Li 2 O + CO 2 = Li 2 CO 3

CaO + SiO 2 = CaSiO 3

3. base + acid = salt

Cu(OH) 2 + 2HCl = CuCl 2 + 2H 2 O

FeCl 3 + 3HNO 3 = Fe(NO 3) 3 + 3HCl

salt acid salt acid

4. metal - main oxide

2Ca + O2 = 2CaO

4Li + O 2 =2Li 2 O

5. non-metal - acid oxide

S + O 2 = SO 2

4As + 5O 2 = 2As 2 O 5

6. basic oxide - base

BaO + H 2 O = Ba(OH) 2

Li 2 O + H 2 O = 2LiOH

7. acid oxide - acid

P 2 O 5 + 3H 2 O = 2H 3 PO 4

SO 3 + H 2 O =H 2 SO 4

This lesson is devoted to the generalization and systematization of knowledge on the topic “Classes of inorganic substances.” The teacher will tell you how you can get a substance of another class from substances of one class. The acquired knowledge and skills will be useful for drawing up reaction equations along chains of transformations.

During chemical reactions a chemical element does not disappear, atoms move from one substance to another. The atoms of a chemical element are, as it were, transferred from a simple substance to a more complex one, and vice versa. Thus, so-called genetic series arise, starting with a simple substance - a metal or non-metal - and ending with a salt.

Let me remind you that salts contain metals and acidic residues. So, the genetic series of a metal may look like this:

From a metal, as a result of the reaction of a compound with oxygen, a basic oxide can be obtained, a basic oxide, when interacting with water, gives a base (only if this base is an alkali), and a salt can be obtained from a base as a result of an exchange reaction with an acid, salt or acidic oxide.

Please note that this genetic series is only suitable for metals whose hydroxides are alkalis.

Let us write down the reaction equations corresponding to the transformations of lithium in its genetic series:

Li → Li 2 O → LiOH → Li 2 SO 4

As you know, metals, when interacting with oxygen, usually form oxides. When oxidized by atmospheric oxygen, lithium forms lithium oxide:

4Li + O 2 = 2Li 2 O

Lithium oxide, interacting with water, forms lithium hydroxide - a water-soluble base (alkali):

Li 2 O + H 2 O = 2LiOH

Lithium sulfate can be obtained from lithium in several ways, for example, as a result of a neutralization reaction with sulfuric acid:

2. Chemical information network ().

Homework

1. p. 130-131 No. 2.4 from Workbook in chemistry: 8th grade: to the textbook P.A. Orzhekovsky and others. “Chemistry. 8th grade” / O.V. Ushakova, P.I. Bespalov, P.A. Orzhekovsky; ed. prof. P.A. Orzhekovsky - M.: AST: Astrel: Profizdat, 2006.

2. p.204 No. 2, 4 from the textbook P.A. Orzhekovsky, L.M. Meshcheryakova, M.M. Shalashova “Chemistry: 8th grade,” 2013

9th grade Lesson No. 47 Topic: “Genetic connection of Me, NeMe and their compounds.”

Goals and objectives of the lesson:

    Get acquainted with the concept of “genetic connection”.

    Learn to compose genetic series of metals and non-metals.

    Based on students’ knowledge of the main classes of inorganic substances, bring them to the concept of “genetic connection” and the genetic series of metal and non-metal;

    To consolidate knowledge about the nomenclature and properties of substances belonging to different classes;

    Develop the ability to highlight the main thing, compare and generalize; identify and establish relationships;

    Develop ideas about the cause-and-effect relationships of phenomena.

    Restore in memory the concepts of simple and complex substance, about metals and non-metals, about the main classes of inorganic compounds;

    To develop knowledge about genetic connections and genetic series, learn to compose genetic series of metals and non-metals.

    Develop the ability to generalize facts, build analogies and draw conclusions;

    Continue to develop a culture of communication, the ability to express one’s views and judgments.

    Foster a sense of responsibility for the knowledge acquired.

Planned results:

Know definitions and classification of inorganic substances.

Be able to classify inorganic substances by composition and properties; compose genetic series of metal and non-metal;

use equations of chemical reactions to illustrate the genetic relationship between the main classes of inorganic compounds.

Competencies:

Cognitive skills : systematize and classify information from written and oral sources.

Activity skills : reflect on one’s activities, act according to an algorithm, be able to create an algorithm oneself new activity, amenable to algorithmization; understand the language of diagrams.

Communication skills : build communication with other people - conduct a dialogue in pairs, take into account the similarities and differences in positions, interact with partners to obtain total product and result.

Lesson type:

    for didactic purpose: lesson in updating knowledge;

    by method of organization: generalizing with the acquisition of new knowledge (combined lesson).

Lesson progress

I. Organizational moment.

II. Update background knowledge and ways of doing things for students.

Lesson motto:“The only way
leading to knowledge is activity” (B. Shaw). slide 1

At the first stage of the lesson, I update the background knowledge that is necessary to solve the problem. This prepares students to accept the problem. I carry out the work in an entertaining way. I conduct a brainstorming session on the topic: “Main classes of inorganic compounds” Work using cards

Task 1. “The third wheel” slide 2

Students are given cards on which three formulas are written, one of which is redundant.

Students identify an extra formula and explain why it is extra

Answers: MgO, Na 2 SO 4, H 2 S slide 3

Task 2. “Name and choose us” (“Name us”) slide 4

nonmetals

hydroxides

Anoxic acids

Give a name to the selected substance (“4-5” write answers in formulas, “3” in words).

(Students work in pairs at the board. (“4-5” write down answers in formulas, “3” in words).

Answers: slide 5

1. copper, magnesium;

4. phosphorus;

5. magnesium carbonate, sodium sulfate

7. salt

III. Learning new material.

1. Determine the topic of the lesson together with the students.

As a result of chemical transformations, substances of one class are transformed into substances of another: an oxide is formed from a simple substance, an acid is formed from an oxide, and a salt is formed from an acid. In other words, the classes of compounds you have studied are interrelated. Let's distribute substances into classes, according to the complexity of their composition, starting with a simple substance, according to our scheme.

Students express their versions, thanks to which we compose simple circuits 2 rows: metals and non-metals. Scheme of genetic series.

I draw students' attention to the fact that each chain has something in common - these are the chemical elements metal and non-metal, which pass from one substance to another (as if by inheritance).

(for strong students) CaO, P 2 O 5, MgO, P, H 3 PO 4, Ca, Na 3 PO 4, Ca(OH) 2, NaOH, CaCO 3, H 2 SO 4

(For weak students) CaO, CO 2, C, H 2 CO 3, Ca, Ca(OH) 2, CaCO 3 slide 6

Answers: slide 7

P P2O5 H3PO4 Na3 PO4

Ca CaO Ca(OH)2 CaCO3

What is the carrier of hereditary information called in biology? (Gene).

Which element do you think will be the “gene” for each chain? (metal and non-metal).

Therefore, such chains or series are called genetic. The topic of our lesson is “Genetic connection between Me and NeMe” slide 8. Open your notebook and write down the date and topic of the lesson. What do you think are the goals of our lesson? Get acquainted with the concept of “genetic connection”. Learn to compose genetic series of metals and non-metals.

2. Let's define a genetic connection.

Genetic connection - is the connection between substances of different classes, based on their mutual transformations and reflecting the unity of their origin. Slide 9.10

Signs that characterize the genetic series: slide 11

1. Substances of different classes;

2. Different substances formed by one chemical element, i.e. represent different forms of existence of one element;

3. Different substances of the same chemical element are connected by mutual transformations.

3. Consider examples of the genetic connection of Me.

2. Genetic series, where the base is an insoluble base, then the series can be represented by a chain of transformations: slide 12

metal→basic oxide→salt→insoluble base→basic oxide→metal

For example, Cu→CuO→CuCl2→Cu(OH)2→CuO
1. 2 Cu+O 2 → 2 CuO 2. CuO+ 2HCI→ CuCI 2 3. CuCI 2 +2NaOH→ Cu(OH) 2 +2NaCI

4.Сu(OH) 2 CuO +H 2 O

4. Consider examples of the genetic connection of NeMe.

Among non-metals, two types of series can also be distinguished: slide 13

2. Genetic series of non-metals, where a soluble acid acts as a link in the series. The chain of transformations can be represented as the following form: non-metal → acidic oxide → soluble acid → salt For example, P → P 2 O 5 → H 3 PO 4 → Ca 3 (PO 4) 2
1. 4P+5O 2 → 2P 2 O 5 2. P 2 O 5 + H 2 O → 2H 3 PO 4 3. 2H 3 PO 4 +3 Ca(OH) 2 → Ca 3 (PO 4) 2 +6 H 2 O

5. Compilation of a genetic series. Slide 14

1. Genetic series in which alkali acts as a base. This series can be represented using the following transformations: metal → basic oxide → alkali → salt

O 2 , +H 2 O, + HCI

4K+O 2 = 2K 2 O K 2 O +H 2 O= 2KOH KOH+ HCI= KCl slide 15

2. Genetic series of non-metals, where an insoluble acid acts as a link in the series:

nonmetal→acid oxide→salt→acid→acid oxide→nonmetal

For example, Si→SiO 2 →Na 2 SiO 3 →H 2 SiO 3 →SiO 2 →Si (compose equations yourself, who works “4-5”). Self-test. All equations are correct “5”, one error is “4”, two errors are “3”.

5. Perform differential exercises (self-test). Slide 15

Si+O 2 = SiO 2 SiO 2 +2NaOH= Na 2 SiO 3 + H 2 O Na 2 SiO 3 + 2НCI= H 2 SiO 3 +2NaCI H 2 SiO 3 = SiO 2 + H 2 O

SiO 2 +2Mg=Si+2MgO

1. Carry out transformations according to the scheme. (task “4-5”)

Task 1. In the figure, connect the formulas of the substances with lines in accordance with their location in the genetic series of aluminum. Write down reaction equations. Slide 16



Self-test.

4AI+ 3O 2 = 2AI 2 O 3 AI 2 O 3 + 6НCI= 2AICI 3 + 3Н 2 О AICI 3 + 3NaOH= AI(OH) 3 +3NaCI

AI(OH) 3 = AI 2 O 3 + H 2 O slide 17

Task 2. “Hit the target.” Select the formulas of the substances that make up the genetic series of calcium. Write down the reaction equations for these transformations. Slide 18

Self-test.

2Ca+O 2 =2CaO CaO+H 2 O =Ca(OH) 2 Ca(OH) 2 +2 HCI=CaCI 2 + 2 H 2 O CaCI 2 +2AgNO 3 =Ca(NO 3) 2 +2AgCI slide 19

2.Carry out the task according to the scheme. Write down the reaction equations for these transformations.

O 2 + H 2 O + NaOH

S SO 2 H 2 SO 3 Na 2 SO 3 or a lighter version

S+ O 2 = SO 2 + H 2 O = H 2 SO 3 + NaOH =

SO 2 + H 2 O = H 2 SO 3

H 2 SO 3 +2NaOH = Na 2 SO 3 +2H 2 O

IV. ConsolidationZUN

Option 1.

Part A.

1. The genetic series of a metal is: a) substances forming a series based on one metal

A)CO 2 b) CO c) CaO d) O 2

3.Identify substance “Y” from the transformation scheme: Na → Y→NaOH A)Na 2 O b)Na 2 O 2 c)H 2 O d)Na

4. In the transformation scheme: CuCl 2 → A → B → Cu, the formulas of intermediate products A and B are: a) CuO and Cu(OH) 2 b) CuSO 4 and Cu(OH) 2 c) CuCO 3 and Cu(OH) 2 G)Cu(OH) 2 AndCuO

5. The final product in the chain of transformations based on carbon compounds CO 2 → X 1 → X 2 → NaOH a) sodium carbonate b) sodium bicarbonate c) sodium carbide d) sodium acetate

E → E 2 O 5 → N 3 EO 4 → Na 3 EO 4 a) N b) Mn V)P d)Cl

Part B.

    Fe + Cl 2 A) FeCl 2

    Fe + HCl B) FeCl 3

    FeO + HCl B) FeCl 2 + H 2

    Fe 2 O 3 + HCl D) FeCl 3 + H 2

D) FeCl 2 + H 2 O

E) FeCl 3 + H 2 O

1 B, 2 A, 3D, 4E

a) potassium hydroxide (solution) b) iron c) barium nitrate (solution) d) aluminum oxide

e) carbon monoxide (II) e) sodium phosphate (solution)

Part C.

1. Implement the transformation scheme of substances: Fe → FeO → FeCI 2 → Fe(OH) 2 → FeSO 4

2Fe+O 2 =2FeO FeO+2HCI= FeCI 2 + H 2 O FeCI 2 + 2NaOH= Fe(OH) 2 +2NaCI

Fe(OH) 2 + H 2 SO 4= FeSO 4 +2 H 2 O

Option 2.

Part A. (tasks with one correct answer)

b) substances forming a series based on one non-metal c) substances forming a series based on a metal or non-metal d) substances from different classes of substances related by transformations

2. Identify substance “X” from the transformation scheme: P → X → Ca 3 (PO 4) 2 A)P 2 O 5 b) P 2 O 3 c) CaO d) O 2

a) Ca b)CaO c)CO 2 d)H 2 O

4. In the transformation scheme: MgCl 2 → A → B → Mg, the formulas of intermediate products A and B are: a) MgO and Mg(OH) 2 b) MgSO 4 and Mg(OH) 2 c) MgCO 3 and Mg(OH) 2 G)Mg(OH) 2 AndMgO

CO 2 → X 1 → X 2 → NaOH a) sodium carbonate b) sodium bicarbonate

6. Element “E” participating in the chain of transformations:

Part B. (tasks with 2 or more correct answer options)

1. Establish a correspondence between the formulas of the starting substances and the reaction products:

Formulas of starting substances Formulas of products

    NaOH+ CO 2 A) NaOH + H 2

    Na + H 2 O B) NaHCO 3

    NaOH + HCl D) NaCl + H 2 O

1B, 2B, 3 A, 4G

a) sodium hydroxide (solution) b) oxygen c) sodium chloride (solution) d) calcium oxide

e) potassium permanganate (crystalline) e) sulfuric acid

Part C. (with a detailed answer option)

S+ O 2 = SO 2 2SO 2 + O 2 = 2 SO 3 SO 3 +H 2 O= H 2 SO 4 H 2 SO 4 +Ca(OH) 2 = CaSO 4 +2 H 2 O

CaSO 4 + BaCI 2 = BaSO 4 + CaCI 2

V.Resultslesson. Grading.

VI.D/Z pp. 215-216 prepare for exercise No. 3 Option 1 assignments No. 2,4, 6, Option 2 assignments No. 2,3, 6. slide 20

VII. Reflection.

Students write on pieces of paper what they did well in the lesson and what they didn’t. What were the difficulties? And a wish to the teacher.

The lesson is over. Thank you all and Have a good day. Slide 21

If there is time left.

Task
Yuh once conducted experiments to measure the electrical conductivity of solutions of different salts. On his laboratory table were beakers with solutions. KCl, BaCl 2 , K 2 CO 3 ,Na 2 SO 4 and AgNO 3 . Each glass had a label carefully affixed to it. In the laboratory there lived a parrot whose cage did not lock very well. When Yukh, absorbed in the experiment, looked back at the suspicious rustling, he was horrified to discover that the parrot, in gross violation of safety regulations, was trying to drink from a glass with a BaCl 2 solution. Knowing that all soluble barium salts are extremely poisonous, Yuh quickly grabbed a glass with a different label from the table and forcibly poured the solution into the parrot's beak. The parrot was saved. A glass with what solution was used to save the parrot?
Answer:
BaCl 2 + Na 2 SO 4 = BaSO 4 (precipitate) + 2NaCl (barium sulfate is so slightly soluble that it cannot be poisonous, like some other barium salts).

Appendix 1

9"B" class F.I._______________________ (for weak students)

Task 1. “The third wheel”.

(4 correct ones – “5”, 3-“4”, 2-“3”, 1-“2”)

nonmetals

hydroxides

Anoxic acids

Students define their chosen class and select appropriate substances from the handout provided.

copper, silicon oxide, hydrochloric acid, barium hydroxide, coal, magnesium, phosphoric, barium hydroxide, magnesium oxide, iron (III) hydroxide, magnesium carbonate, sodium sulfate.

(“4-5” write answers in formulas, “3” in words).

12 answers “5”, 11-10- “4”, 9-8- “3”, 7 or less – “2”

Task 3.

O 2 , +H 2 O, + HCI

For example, K→ K 2 O →KOH→ KCl (compose equations yourself, who works “3”, one error “3”, two errors “2”).

Task 4. Carry out the task according to the scheme. Write down the reaction equations for these transformations.

O 2 + H 2 O + NaOH

S SO 2 H 2 SO 3 Na 2 SO 3

or a lighter version

H 2 SO 3 + NaOH =

Option 1.

Part A. (tasks with one correct answer)

1. The genetic series of a metal is: a) substances that form a series based on one metal

b) substances forming a series based on one non-metal c) substances forming a series based on a metal or non-metal d) substances from different classes of substances related by transformations

2. Identify substance “X” from the transformation scheme: C → X → CaCO 3

a) CO 2 b) CO c) CaO d) O 2

3. Identify substance “Y” from the transformation scheme: Na → Y→NaOH a)Na 2 O b)Na 2 O 2 c)H 2 O d)Na

4. In the transformation scheme: CuCl 2 → A → B → Cu, the formulas of intermediate products A and B are: a) CuO and Cu(OH) 2 b) CuSO 4 and Cu(OH) 2 c) CuCO 3 and Cu(OH) 2 g)Cu(OH) 2 and CuO

5. The final product in the chain of transformations based on carbon compounds CO 2 → X 1 → X 2 → NaOH a) sodium carbonate b) sodium hydrogen carbonate c) sodium carbide d) sodium acetate

6. Element “E” participating in the chain of transformations: E → E 2 O 5 → H 3 EO 4 → Na 3 EO 4 a)N b) Mn c)P d)Cl

Part B. (tasks with 2 or more correct answer options)

1. Establish a correspondence between the formulas of the starting substances and the reaction products:

Formulas of starting substances Formulas of products

    Fe + Cl 2 A) FeCl 2

    Fe + HCl B) FeCl 3

    FeO + HCl B) FeCl 2 + H 2

    Fe 2 O 3 + HCl D) FeCl 3 + H 2

D) FeCl 2 + H 2 O

E) FeCl 3 + H 2 O

2. A solution of copper (II) sulfate reacts:

a) potassium hydroxide (solution) b) iron c) barium nitrate (solution) d) aluminum oxide

e) carbon monoxide (II) f) sodium phosphate (solution)

Part C. (with a detailed answer option)

1. Implement a scheme for the transformation of substances:

Fe →FeO → FeCI 2 → Fe(OH) 2 → FeSO 4

Appendix 2

9"B" class F.I._______________________ (for strong students)

Task 1. “The third wheel”. Identify the redundant formula and explain why it is redundant.

(4 correct ones – “5”, 3-“4”, 2-“3”, 1-“2”)

Task 2. “Name and choose us” (“Name us”). Give the name of the selected substance and fill out the table.

Students define their chosen class and select appropriate substances from the handout provided.

copper, silicon oxide, hydrochloric acid, barium hydroxide, coal, magnesium, phosphoric, barium hydroxide, magnesium oxide, iron (III) hydroxide, magnesium carbonate, sodium sulfate. (“4-5” write answers in formulas, “3” in words).

12 answers “5”, 11-10- “4”, 9-8- “3”, 7 or less – “2”

Task 3.

Si→SiO 2 →Na 2 SiO 3 →H 2 SiO 3 →SiO 2 →Si (compose equations yourself, who works “4-5”). Self-test. All equations are correct “5”, one error is “4”, two errors are “3”.

Task 4. In the figure, connect the formulas of the substances with lines in accordance with their location in the genetic series of aluminum. Write down reaction equations. All equations are correct “5”, one error is “4”, two errors are “3”.



Task 5. “Hit the target.” Select the formulas of the substances that make up the genetic series of calcium. Write down the reaction equations for these transformations. All equations are correct “5”, one error is “4”, two errors are “3”.

Option 2.

Part A. (tasks with one correct answer)

1. The genetic series of a non-metal is: a) substances that form a series based on one metal

b) substances forming a series based on one non-metal c) substances forming a series based on a metal or non-metal d) substances from different classes of substances related by transformations

2. Identify substance “X” from the transformation scheme: P → X → Ca 3 (PO 4) 2 a) P 2 O 5 b) P 2 O 3 c) CaO d) O 2

3. Identify substance “Y” from the transformation scheme: Ca → Y→Ca(OH) 2

a) Ca b) CaO c) CO 2 d) H 2 O

4. In the transformation scheme: MgCl 2 → A → B → Mg, the formulas of intermediate products A and B are: a) MgO and Mg(OH) 2 b) MgSO 4 and Mg(OH) 2 c) MgCO 3 and Mg(OH) 2 g)Mg(OH) 2 and MgO

5.The final product in the chain of transformations based on carbon compounds:

CO 2 → X 1 → X 2 → NaOH a) sodium carbonate b) sodium bicarbonate

c) sodium carbide d) sodium acetate

6. Element “E” participating in the chain of transformations:

E → EO 2 → EO 3 → N 2 EO 4 → Na 2 EO 4 a)N b) S c)P d)Mg

Part B. (tasks with 2 or more correct answer options)

1. Establish a correspondence between the formulas of the starting substances and the reaction products:

Formulas of starting substances Formulas of products

    NaOH+ CO 2 A) NaOH + H 2

    NaOH +CO 2 B) Na 2 CO 3 + H 2 O

    Na + H 2 O B) NaHCO 3

    NaOH + HCl D) NaCl + H 2 O

2. Hydrochloric acid does not interact:

a) sodium hydroxide (solution) b) oxygen c) sodium chloride (solution) d) calcium oxide

e) potassium permanganate (crystalline) f) sulfuric acid

Part C. (with a detailed answer option)

    Implement the transformation scheme of substances: S →SO 2 → SO 3 → H 2 SO 4 → CaSO 4 → BaSO 4

Appendix 3

Answer sheet "4-5":

Task 1. MgO, Na 2 SO 4, H 2 S

Task 2.

1. copper, magnesium;

3. silicon oxide, magnesium oxide;

4. phosphorus,

5. magnesium carbonate, sulfate;

6. barium hydroxide, iron (III) hydroxide;

7. sodium hydrochloride

Task 3.

SiO 2 + 2NaOH = Na 2 SiO 3 + H 2 O

Na 2 SiO 3 + 2НCI = H 2 SiO 3 + 2NaCI

H 2 SiO 3 = SiO 2 + H 2 O

SiO 2 +2Mg=Si+2MgO

Task 4.

4AI+ 3O 2 = 2AI 2 O 3

AI 2 O 3 + 6НCI = 2AICI 3 + 3Н 2 О

AICI 3 + 3NaOH= AI(OH) 3 + 3NaCI

AI(OH) 3 = AI 2 O 3 + H 2 O

Task 5.

CaO+H 2 O =Ca(OH) 2

Ca(OH) 2 +2 HCI=CaCI 2 + 2H2O

CaCI 2 +2AgNO 3 =Ca(NO 3) 2 +2AgCI

Self-assessment sheet.

Student's full name

Job No.

In this article we will talk about genetic series of metals. Individual chemicals usually divided into 2 groups: simple substances And complex.

This diagram gives a simplified idea of genetic sequence of metals.

On top are a group of metals and hydrogen, the structure of which differs from the structure of atoms of other elements. There is 1 electron in the outer level, like alkali metals, but at the same time there is not enough 1st electron to fill the outer level.

Based on genetic line metals form basic oxides. Hydrogen forms a specific amphoteric oxide - water H2O, which, when interacting with the main oxide, gives a base (alkali). Such reactions proceed, as a rule, without changing the oxidation state. With a change, only those reactions occur in which simple substances complex ones are formed:

2 Cu + O 2 = 2 CuO,

Basic oxides react with nonmetals, acidic oxides, acids, acid salts.

Depending on the acid, metal or non-metal are formed various salts. For example:

Cu(HE) 2 + H 2 SO 4 = SO 4 .