Genetic relationships between the main classes of inorganic substances. Genetic relationship of metals, non-metals and their compounds

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 us highlight the most characteristic features of such series.

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 forms of its 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 metal series, the non-metal series with different oxidation states is richer in bonds, for example, the genetic series of sulfur with oxidation states +4 and +6:

Only the last transition can cause difficulty. Follow the rule: in order 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

Genetic series of metals and their compounds

Each such row consists of a metal, its main oxide, a base and any salt of the same metal:

To move from metals to basic oxides in all these series, reactions of combination with oxygen are used, for example:

2Ca + O 2 = 2CaO; 2Mg + O 2 = 2MgO;

The transition from basic oxides to bases in the first two rows is carried out through the hydration reaction known to you, for example:

СaO + H 2 O = Сa(OH) 2.

As for the last two rows, the oxides MgO and FeO contained in them do not react with water. In such cases, to obtain bases, these oxides are first converted into salts, and then they are converted into bases. Therefore, for example, to carry out the transition from MgO oxide to Mg(OH) 2 hydroxide, successive reactions are used:

MgO + H 2 SO 4 = MgSO 4 + H 2 O; MgSO 4 + 2NaOH = Mg(OH) 2 ↓ + Na 2 SO 4.

Transitions from bases to salts are carried out by reactions already known to you. Thus, soluble bases (alkalis) located in the first two rows are converted into salts under the action of acids, acid oxides or salts. Insoluble bases from the last two rows form salts under the action of acids.

Genetic series of nonmetals and their compounds.

Each such series consists of a non-metal, an acidic oxide, a corresponding acid and a salt containing the anions of this acid:

To move from non-metals to acidic oxides in all these series, reactions of combination with oxygen are used, for example:

4P + 5O 2 = 2 P 2 O 5 ; Si + O 2 = SiO 2;

The transition from acidic oxides to acids in the first three rows is carried out through the hydration reaction known to you, for example:

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

However, you know that the oxide SiO 2 contained in the last row does not react with water. In this case, it is first converted into the corresponding salt, from which the desired acid is then obtained:

SiO 2 + 2KOH = K 2 SiO 3 + H 2 O; K 2 SiO 3 + 2HCl = 2KCl + H 2 SiO 3 ↓.

Transitions from acids to salts can be carried out by reactions known to you with basic oxides, bases or salts.

Things to remember:

· Substances of the same genetic series do not react with each other.

· Substances of different types of genetic series react with each other. The products of such reactions are always salts (Fig. 5):

Rice. 5. Diagram of the relationship between substances of different genetic series.

This diagram shows the relationships between different classes of inorganic compounds and explains the variety of chemical reactions between them.

Assignment on the topic:

Write down reaction equations that can be used to carry out the following transformations:

1. Na → Na 2 O → NaOH → Na 2 CO 3 → Na 2 SO 4 → NaOH;

2. P → P 2 O 5 → H 3 PO 4 → K 3 PO 4 → Ca 3 (PO 4) 2 → CaSO 4 ;

3. Ca → CaO → Ca(OH) 2 → CaCl 2 → CaCO 3 → CaO;

4. S → SO 2 → H 2 SO 3 → K 2 SO 3 → H 2 SO 3 → BaSO 3 ;

5. Zn → ZnO → ZnCl 2 → Zn(OH) 2 → ZnSO 4 → Zn(OH) 2;

6. C → CO 2 → H 2 CO 3 → K 2 CO 3 → H 2 CO 3 → CaCO 3 ;

7. Al → Al 2 (SO 4) 3 → Al(OH) 3 → Al 2 O 3 → AlCl 3;

8. Fe → FeCl 2 → FeSO 4 → Fe(OH) 2 → FeO → Fe 3 (PO 4) 2;

9. Si → SiO 2 → H 2 SiO 3 → Na 2 SiO 3 → H 2 SiO 3 → SiO 2;

10. Mg → MgCl 2 → Mg(OH) 2 → MgSO 4 → MgCO 3 → MgO;

11. K → KOH → K 2 CO 3 → KCl → K 2 SO 4 → KOH;

12. S → SO 2 → CaSO 3 → H 2 SO 3 → SO 2 → Na 2 SO 3;

13. S → H 2 S → Na 2 S → H 2 S → SO 2 → K 2 SO 3;

14. Cl 2 → HCl → AlCl 3 → KCl → HCl → H 2 CO 3 → CaCO 3 ;

15. FeO → Fe(OH) 2 → FeSO 4 → FeCl 2 → Fe(OH) 2 → FeO;

16. CO 2 → K 2 CO 3 → CaCO 3 → CO 2 → BaCO 3 → H 2 CO 3 ;

17. K 2 O → K 2 SO 4 → KOH → KCl → K 2 SO 4 → KNO 3;

18. P 2 O 5 → H 3 PO 4 → Na 3 PO 4 → Ca 3 (PO 4) 2 → H 3 PO 4 → H 2 SO 3;

19. Al 2 O 3 → AlCl 3 → Al(OH) 3 → Al(NO 3) 3 → Al 2 (SO 4) 3 → AlCl 3;

20. SO 3 → H 2 SO 4 → FeSO 4 → Na 2 SO 4 → NaCl → HCl;

21. KOH → KCl → K 2 SO 4 → KOH → Zn(OH) 2 → ZnO;

22. Fe(OH) 2 → FeCl 2 → Fe(OH) 2 → FeSO 4 → Fe(NO 3) 2 → Fe;

23. Mg(OH) 2 → MgO → Mg(NO 3) 2 → MgSO 4 → Mg(OH) 2 → MgCl 2;

24. Al(OH) 3 → Al 2 O 3 → Al(NO 3) 3 → Al 2 (SO 4) 3 → AlCl 3 → Al(OH) 3;

25. H 2 SO 4 → MgSO 4 → Na 2 SO 4 → NaOH → NaNO 3 → HNO 3;

26. HNO 3 → Ca(NO 3) 2 → CaCO 3 → CaCl 2 → HCl → AlCl 3;

27. CuCO 3 → Cu(NO 3) 2 → Cu(OH) 2 → CuO → CuSO 4 → Cu;

28. MgSO 4 → MgCl 2 → Mg(OH) 2 → MgO → Mg(NO 3) 2 → MgCO 3;

29. K 2 S → H 2 S → Na 2 S → H 2 S → SO 2 → K 2 SO 3;

30. ZnSO 4 → Zn(OH) 2 → ZnCl 2 → HCl → AlCl 3 → Al(OH) 3;

31. Na 2 CO 3 → Na 2 SO 4 → NaOH → Cu(OH) 2 → H 2 O → HNO 3;

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 substances, metals and non-metals, 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 for a new activity that can be algorithmized; 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 a common 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).

During the classes

I. Organizational moment.

II. Updating students' basic knowledge and methods of action.

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

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 draw up simple diagrams of 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 follows: 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 nice 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).

Annex 1

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

Task 1. “The third wheel”.

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

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 bicarbonate 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 react:

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.

Repetition. Genetic relationship of classes of inorganic compounds
Introduction

The topic of this lesson is “Repetition. Genetic relationship of classes of inorganic compounds". You will repeat how all inorganic substances are divided, and conclude how another class of inorganic compounds can be obtained from one class. Based on the information received, you will learn what the genetic connection of such classes is, the two main ways of such connections.

Topic: Introduction

Lesson: Repetition. Genetic relationship of classes of inorganic compounds

Chemistry is the science of substances, their properties and transformations into each other.

Rice. 1. Genetic relationship of classes of inorganic compounds

All inorganic substances can be divided into:

Simple substances

Complex substances.

Simple substances are divided into:

Metals

Nonmetals

Complex substances can be divided into:

Grounds

Acids

Salt. See Fig.1.

These are binary compounds consisting of two elements, one of which is oxygen in the -2 oxidation state. Fig.2.

For example, calcium oxide: Ca +2 O -2, phosphorus oxide (V) P 2 O 5., nitrogen oxide (IV) -« Fox's tail"

Rice. 2. Oxides

Are divided into:

Basic

Acidic

Basic oxides correspond grounds.

Acidic oxides correspond acids.

Salts consist of metal cations And acid residue anions.

Rice. 3. Pathways of genetic connections between substances

Thus: from one class of inorganic compounds another class can be obtained.

Therefore, everything classes of inorganic substances are interrelated.

Class Relationship inorganic compounds are often called genetic. Fig.3.

Genesis in Greek means "origin". Those. a genetic connection shows the relationship between the transformation of substances and their origin from a single substance.

There are two main ways of genetic connections between substances. One of them begins with a metal, the other with a non-metal.

Genetic series of metal shows:

Metal → Basic oxide → Salt → Base → New salt.

Genetic series of a nonmetal reflects the following transformations:

Non-metal → Acidic oxide → Acid → Salt.

For any genetic series, reaction equations can be written that show transformation of one substance into another.

First, you need to determine which class of inorganic compounds each substance of the genetic series belongs to.

Think about it how to get the substance after the arrow from the substance before the arrow.

Example No. 1. Genetic series of metal.

The series begins with the simple metal substance copper. To make the first transition, you need to burn copper in an oxygen atmosphere.

2Cu +O 2 →2CuO

Second transition: you need to get the salt CuCl 2. It is formed by hydrochloric acid HCl, because salts of hydrochloric acid are called chlorides.

CuO +2 HCl → CuCl 2 + H 2 O

Third step: to obtain an insoluble base, you need to add alkali to the soluble salt.

CuCl 2 + 2NaOH → Cu(OH) 2 ↓ + 2NaCl

To convert copper(II) hydroxide into copper(II) sulfate, add sulfuric acid H2SO4 to it.

Cu(OH) 2 ↓ + H 2 SO 4 → CuSO 4 + 2H 2 O

Example No. 2. Genetic series of a nonmetal.

The series begins with a simple substance, the nonmetal carbon. To accomplish the first transition, carbon must be burned in an oxygen atmosphere.

C + O 2 → CO 2

If you add water to an acidic oxide, you get an acid called carbonic acid.

CO 2 + H 2 O → H 2 CO 3

To obtain the salt of carbonic acid - calcium carbonate, you need to add a calcium compound to the acid, for example calcium hydroxide Ca(OH) 2.

H 2 CO 3 + Ca (OH) 2 → CaCO 3 + 2H 2 O

The composition of any genetic series includes substances of various classes of inorganic compounds.

But these substances necessarily contain the same element. Knowing the chemical properties of classes of compounds, it is possible to select reaction equations with which these transformations can be carried out. These transformations are also used in production to select the most rational methods for obtaining certain substances.

You repeated how all inorganic substances are divided, and concluded how another class of inorganic compounds can be obtained from one class. Based on the information received, we learned what the genetic connection of such classes is, the two main ways of such connections .

1. Rudzitis G.E. Inorganic and organic chemistry. 8th grade: textbook for general education institutions: basic level / G. E. Rudzitis, F.G. Feldman.M.: Enlightenment. 2011, 176 pp.: ill.

2. Popel P.P. Chemistry: 8th grade: textbook for general education institutions / P.P. Popel, L.S. Krivlya. -K.: IC “Academy”, 2008.-240 p.: ill.

3. Gabrielyan O.S. Chemistry. 9th grade. Textbook. Publisher: Bustard: 2001. 224s.

1. No. 10-a, 10z (p. 112) Rudzitis G.E. Inorganic and organic chemistry. 8th grade: textbook for general education institutions: basic level / G. E. Rudzitis, F.G. Feldman.M.: Enlightenment. 2011, 176 pp.: ill.

2. How to obtain calcium sulfate from calcium oxide in two ways?

3. Make a genetic series for producing barium sulfate from sulfur. Write the reaction equations.

First, we present our information about the classification of substances in the form of a diagram (Scheme 1).

Scheme 1
Classification of inorganic substances

Knowing the classes of simple substances, it is possible to create two genetic series: the genetic series of metals and the genetic series of non-metals.

There are two varieties of the genetic series of metals.

1. Genetic series of metals to which alkali corresponds as a hydroxide. In general, such a series can be represented by the following chain of transformations:

For example, the genetic series of calcium:

Ca → CaO → Ca(OH) 2 → Ca 3 (PO 4) 2.

2. Genetic series of metals that correspond to an insoluble base. This series is richer in genetic connections, since it more fully reflects the idea of ​​mutual transformations (direct and reverse). In general, such a series can be represented by the following chain of transformations:

metal → basic oxide → salt →
→ base → basic oxide → metal.

For example, the genetic series of copper:

Cu → CuO → CuCl 2 → Cu(OH) 2 → CuO → Cu.

Here, too, two varieties can be distinguished.

1. The genetic series of nonmetals, to which a soluble acid corresponds as a hydroxide, can be reflected in the form of the following chain of transformations:

non-metal → acidic oxide → acid → salt.

For example, the genetic series of phosphorus:

P → P 2 O 5 → H 3 PO 4 → Ca 3 (PO 4) 2.

2. The genetic series of nonmetals, which correspond to an insoluble acid, can be represented using the following chain of transformations:

nonmetal → acid oxide → salt →
→ acid → acid oxide → non-metal.

Since of the acids we have studied, only silicic acid is insoluble, as an example of the last genetic series, consider the genetic series of silicon:

Si → SiO 2 → Na 2 SiO 3 → H 2 SiO 3 → SiO 2 → Si.

Key words and phrases

1. Genetic connection.
2. Genetic series of metals and its varieties.
3. Genetic series of nonmetals and its varieties.

Work with computer

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