Exercises on chemical bonding. Types of chemical bond

MUNICIPAL EDUCATIONAL INSTITUTION

"SECONDARY SCHOOL No. 63, BRYANSK"

HANDOUT ON THE TOPIC

"CHEMICAL BOND"

chemistry

8th grade

Chemistry teacher

MBOU Secondary School No. 63, Bryansk

Gaidukova Alexandra Pavlovna

Chemical bond. Basic types of chemical bonds.

What is electronegativity? How does the electronegativity of elements change within a period? How does the electronegativity of elements change within the main subgroups?

Types of Chemical Bonds

Covalent bond. Occurs between atoms of non-metal elements. There are two types of covalent bonds: a) covalent nonpolar a bond occurs between atoms of non-metal elements with the same EO value; b) covalent polar bonds arise between atoms of non-metal elements with different meaning EO. Ionic bond. Occurs between atoms of a metal element and a non-metal element, the EO values ​​of which differ sharply. Metal connection. Occurs between atoms of a given metal. Hydrogen bond. Occurs between hydrogen atom one molecule And more electronegative element another molecule .

Task 6. I option). Determine the type of chemical bond in compounds whose formulas are given: SO 3 _________________________________________

ClF 3 ________________________________________

Br 2 ________________________________________

(H 2 O) 3 ________________________________________________

CaCl 2 _______________________________________________

Cu__________________________________________

Task 7. (Complete this task if you are on II option). Determine the type of chemical bond in the compounds whose formulas are given: N 2 _________________________________________

CO 2 ________________________________________

KI__________________________________________

(NH 3) 2 ________________________________________________

HBr__________________________________________

Mg__________________________________________

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Covalent chemical bond

What is a chemical bond? List all types of chemical bonds. What chemical bond is called covalent? Name two types of covalent chemical bonds. Define them.

Number of unpaired e = 8 – N G ,

where N g is the number of the group in which the element is located

Non-metal element

Keep exploring!

Mechanism of formation of a covalent nonpolar bond

Let us consider the mechanism of formation of a covalent nonpolar bond using the example of a hydrogen molecule H2. (Explain why there is a covalent nonpolar bond in the hydrogen molecule?). The H2 molecule contains two hydrogen atoms: H and H. Draw electron graphic formulas for the structure of each atom:

N N

As can be seen from the electron-graphical formulas you constructed, the number of unpaired electrons in each hydrogen atom is ________. Connect the unpaired electrons of each atom with a wavy line. You have obtained a schematic representation of the formation of a covalent nonpolar bond in a hydrogen molecule.

Summarize! Each hydrogen atom has ______ unpaired electron located at _____ energy level. This energy level can only hold two electrons. Therefore, a hydrogen atom needs another ______ electron to complete its energy level. During the formation of a chemical bond, a common electron pair is formed between hydrogen atoms, which belongs equally to each hydrogen atom. As a result, each atom has ______ electrons. Since both hydrogen atoms have the same EO value, the shared electron pair does not shift towards either atom. That's why this type bond and is called covalent non-polar communication Electronic circuit the formation of a covalent nonpolar bond in a hydrogen molecule looks like this:

N . + . N N : N.If you replace a common electron pair with a bar, you will get the structural formula of the molecule: H – H. If there are several common electron pairs, each pair is replaced with a bar.

Do it!

Task 4. Draw the mechanism of formation of covalent nonpolar bonds in molecules Cl 2 , O 2 using electron-graphical, electronic and structural formulas. Next to the diagrams, indicate: a) the number of unpaired electrons of each atom; b) the number of electrons in the outer level of each atom; c) the number of common electron pairs in each molecule.

Complete the task on blank page 4

Keep exploring!

Mechanism of formation of a polar covalent bond

Let us consider the mechanism of formation of a covalent nonpolar bond using the example of a hydrogen chloride molecule HCl (Explain why there is a covalent polar bond in the hydrogen chloride molecule?). The HCl molecule contains two atoms: _____ and ______. Draw electron graphic formulas for the structure of each atom:

As can be seen from the formulas you constructed, the hydrogen atom has _____unpaired electron, and the chlorine atom has _____unpaired electron. Connect the unpaired electrons of each atom with a wavy line. You have obtained a schematic representation of the formation of a covalent field bond in a hydrogen chloride molecule.

Summarize! The hydrogen atom has______ an unpaired electron located at the_____ energy level, and the chlorine atom has______an unpaired electron located at the _____ energy level. Therefore, a hydrogen atom and a chlorine atom need another ______ electron to complete the energy level. During the formation of a chemical bond, a common electron pair is formed between hydrogen atoms, which belongs to both the hydrogen atom and the chlorine atom. As a result, each atom has a complete electron shell. The shared electron pair in the case of a polar covalent bond is shifted towards the more electronegative element. Since from two atoms, H and Cl, the _______ atom has the greatest EO, then the common electron pair shifts towards the _______ atom. Electronic circuit the formation of a covalent nonpolar bond in a hydrogen molecule looks like this:

N . + . Cl N : Cl ( on electronic circuit the shared electron pair is depicted closer to the more EO atom). If you replace the shared electron pair with a line, you get the structural formula of the molecule: H – Cl . In the structural formula, the displacement of a shared electron pair is shown using an arrow: HCl . As a result of the displacement of the electron pair, each atom in the molecule acquires a partial charge: hydrogen - a partial positive charge (it is easier for it to “breathe” after the displacement of the electron pair), chlorine - a partial negative charge (it pulls the “extra load” onto itself), i.e. two “poles” are formed. Therefore, this type of bond is called covalentpolar communication

P.S. If the number of unpaired electrons of atom1 is greater than the number of unpaired electrons of atom2, it is necessary to take such a number of atoms2 that the number of unpaired electrons coincides.

Do it!

Task 5. Draw the mechanism for the formation of polar covalent bonds in molecules HBr, H 2 S using electron-graphical, electronic and structural formulas. Next to the diagrams, indicate: a) the number of unpaired electrons of each atom; b) the number of electrons in the outer level of each atom; c) towards which atom the common electron pairs are displaced. Explain your answer.

If there is not enough space, use reverse side leaf.

Ionic chemical bond

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An ionic bond is a chemical bond that occurs between ions due to the forces of electrostatic attraction.Ions – charged particles that are formed when an atom gives up or gains electrons. Atoms of a chemical element give up electrons only from the external energy level, and accordingly, they also accept electrons to the external energy level. If an atom of a chemical element gives up electrons, it turns into a positively charged ion (“rejoices” that it has thrown off its “burden”) For example: Na 0 – 1е Na + . Positively charged ions are calledcations . The charge of the cation is equal to the number of electrons given up.(!Atoms everyone metals always only give away electrons and always turn intocations !) If an atom of a chemical element gains electrons, it turns into a negatively charged ion (it has taken on an “extra load” and is therefore “upset”). For example: S 0 + 2 eS -2 . Negatively charged ions are calledanions . The charge of the anion is equal to the number of electrons accepted.

Do it!

Exercise 1. Write down the definitions in your notebook: a) ionic bond; b) ions. Make a diagram “Classification of ions”. Write down your explanations.

Task 2. Write down cations and anions from the proposed series of ions in the diagram: Na+; S-2; N +5 ; Cl - ; Ca+2; Al+3; P-3; O-2; S +4 ; F - .

Task 3. Draw in your notebook and fill out Table 1.

Atom of a chemical element

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Mechanism of ionic bond formation

Let us consider the mechanism of ionic bond formation using the example of lithium chloride LiCl. This compound is formed by lithium ions and chlorine ions. Let us show the formation of these ions using electron graphic formulas:

Li 0 Li +

1 s 2 2s 1 1s 2 (electronic configuration of the noble gas atom helium)

Cl 0 Cl - - 1е

Cl 0 Cl -

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

An ionic bond occurs between the resulting lithium ions Li + and chlorine Cl - . It is clear that oppositely charged particles are attracted to each other and held together due to the forces of electrostatic attraction. The entire mechanism of ionic bond formation can be shown in the form of a brief diagram:

Li 0 – 1 eLi + ionic bond

Cl 0 +1 eCl -

Do it! Complete the tasks given on the back in the notebook

Task 4.(Complete this task if you are on I option). Show the formation of an ionic bond between the Na and S atoms. Pay attention to the number of electrons that sodium will give up and the number of electrons that sulfur will accept... One sodium atom is clearly not enough... (This was a hint). After execution of this assignment answer the questions:

How many sodium atoms are needed to form an ionic bond between it and sulfur? Why?

What noble gas configuration does the sulfur ion take?

Explain why the sodium atom gives up electrons? Why does a sulfur atom accept electrons?

Task 5.(Complete this task if you are on II option). Show the formation of an ionic bond between the Na and N atoms. Pay attention to the number of electrons that sodium will give up and the number of electrons that nitrogen will accept...One sodium atom is clearly not enough...(This was a hint). After completing this task, answer the following questions:

How many sodium atoms are needed to form an ionic bond between it and nitrogen? Why?

What noble gas configuration does the sodium ion take?

What noble gas configuration does the nitrogen ion take?

Explain why the sodium atom gives up electrons? Why does the nitrogen atom accept electrons?

Task 6. Draw the structure diagrams for the following ions: Mg +2; O-2; Ca+2; F - . Write abbreviated electronic formulas for them and indicate which noble gas configurations correspond to the configurations of these ions. Make up formulas for all possible connections, which can be formed by these ions.

Task 7. What ions can have configuration 1 s 2 2s 2 2p 6 (electronic configuration of the neon atom). Give examples of at least three cations and three anions.

Homework! Learn the topic “Ionic chemical bonding”. Prepare for s/r on the topics “Electronegativity of chemical elements”, “Covalent chemical bond”, “Ionic bond”.

LIST OF REFERENCES USED

Chemistry. Inorganic chemistry. 8th grade: textbook for general education. institutions/Rudzitis, Feldman - 13th edition-M: education, 2009- 176s

Atoms can combine with each other to form both simple and complex substances. In this case, various types of chemical bonds are formed: ionic, covalent (non-polar and polar), metallic.
One of the most essential properties of atoms of elements, which determines what kind of bond is formed between them, is electronegativity, i.e. the ability of atoms in a compound to attract electrons.
The more an atom attracts electrons to itself, the higher its electronegativity. Electronegativity depends on the size of the atom and the charge of its nucleus. The sizes of atoms of elements of the same period decrease with increasing nuclear charge. This happens because the charge of the atomic nucleus increases from element to element, but the number of electron layers remains the same. In this case, the atom becomes more compact, the size of the atom decreases towards the end of the period, and the force of attraction of electrons by the nucleus increases. Therefore, the electronegativity of elements increases in a period.
For elements of the main subgroups, as the nuclear charges increase, the number of electronic layers also increases, and therefore the size of the atoms increases. The attraction of outer electrons decreases. Therefore, the electronegativity of elements in a group decreases.
Non-metal elements have the greatest electronegativity: fluorine, oxygen, nitrogen and others. Metal elements have lower electronegativity. The lowest electronegativity is found in elements such as potassium, sodium, and calcium. In descending order of electronegativity, elements can be arranged in a row:
F, O, N, Cl, Br, S, I, C, Se, P, H, B, Si, Cu. Fe, Zn. Al, Mg, Li, Ca, Na, K
The electronegativity of fluorine is conventionally taken to be 4.0; The electronegativity of potassium is 0.8.
The type of chemical bond depends on how large the difference in electronegativity values ​​of the connecting atoms of elements is. The more the atoms of the elements forming the bond differ in electronegativity, the more polar the chemical bond.
1. An ionic bond is formed by the interaction of atoms that differ sharply from each other in electronegativity. For example, the typical metals lithium (Li), sodium (Na), potassium (K), calcium (Ca), strontium (Sr), barium (Ba) form ionic bonds with typical nonmetals. This produces a metal ion with a positive charge and a non-metal ion with a negative charge.
2. Covalent is a bond between non-metal atoms, as a result of which common electron pairs are formed.
There are non-polar and polar covalent bonds.
When atoms with the same electronegativity interact, molecules with a covalent nonpolar bond are formed. Such a bond exists in molecules simple substances: hydrogen, oxygen, nitrogen, chlorine, etc. Chemical bonds in these are formed through shared electron pairs, i.e. when the corresponding electron clouds overlap, due to electron-nuclear interaction when atoms approach each other.
When atoms whose electronegativity values ​​differ, but not sharply, interact, the common electron pair shifts to a more electronegative atom and a polar covalent bond is formed. In this case, partial charges are formed. This is the most common type of chemical bond found in both inorganic and organic compounds.
3. Metallic is a bond that is formed as a result of the interaction of relatively free electrons with metal ions. This type of bond is characteristic of simple substances - metals and their alloys. The essence of the process of metal bond formation is as follows: metal atoms easily give up valence electrons and turn into positively charged ions. Relatively free electrons detached from the atom move between positive metal ions. A metallic bond arises between them.
It is impossible to draw a sharp boundary between the types of chemical bonds. In most compounds, the type of chemical bond is intermediate; for example, a highly polar covalent chemical bond is close to an ionic bond. Depending on which of the limiting cases a chemical bond is closer in nature, it is classified as either an ionic or a covalent polar bond.

Problems for the section Chemical bonding and molecular structure are collected here.

Task 1. For sodium hydrogen sulfate, construct a graphic formula and indicate the types of chemical bonds in the molecule: ionic, covalent, polar, nonpolar covalent, coordination, metallic, hydrogen.

Task 2. Construct a graphic formula for ammonium nitrite and indicate the types of chemical bonds in this molecule. Show which (which) connections are “broken” during dissociation. Explain what it is ? Give examples of its influence on the properties of a substance.

Solution. Ammonium nitrite - ionic bond

NH 4 NO 2 = NH 4 + +NO 2 -

N–H– covalently polar bond

Between NH 4 + and NO 2 — — ionic bond

Solution. CH3Br — . Covalent bond occurs between atoms with close or equal values electronegativity. This bond can be thought of as the electrostatic attraction of the nuclei of two atoms to a common electron pair.

Unlike ionic compounds, molecules of covalent compounds are held together by "intermolecular forces", which are much weaker than chemical bonds. In this regard, covalent bonds are characterized saturability – formation of a limited number of connections.

It is known that atomic orbitals are oriented in space in a certain way, therefore, when a bond is formed, the overlap of electron clouds occurs in a certain direction. Those. such a property of a covalent bond is realized as focus.

Solution: Cloud overlap may occur different ways, in view of them various shapes. Distinguish σ-, π- and δ-bonds.

Sigma - communications are formed when clouds overlap along a line passing through the nuclei of atoms.

Pi – connections occur when clouds overlap on either side of the line connecting the nuclei of atoms.

Delta - connections are carried out by overlapping all four blades d - electron clouds located in parallel planes.

Sigma - communication more durable than Pi – connection.

C2H6 –sp 3 hybridization.

S-S— σ-bond (overlap 2sp 3 -2sp 3)

S–H— σ-bond (overlap of 2sp 3 -AO of carbon and 1s-AO of hydrogen)

C2H4 – sp 2 hybridization.

Double bond implemented by the presence of 2 types of communication - σ- and π-bonds(although it is depicted by two identical lines, their disparity should always be taken into account). σ-Bond is formed by the central overlap of sp 2 -hybridized orbitals, and π bond– with lateral overlap of the p-orbital lobes of neighboring sp 2 -hybridized carbon atoms. The formation of bonds in an ethylene molecule can be represented by the following diagram:

C=C- σ-bond (overlap 2sp 2 -2sp 2) and π-bond (2рz-2рz)

S–H— σ-bond (overlap of 2sp 2 -AO of carbon and 1s-AO of hydrogen)

C2H2 — sp hybridization

Triple bond is realized by a combination of σ- and two π-bonds formed by two sp-hybridized atoms.

σ-Bond occurs when the sp-hybridized orbitals of neighboring carbon atoms overlap centrally; π bonds are formed when the lobes overlap laterally ry-orbitals and pz-orbitals. The formation of bonds in the acetylene molecule H–C≡C–H can be depicted in the form of a diagram:

C≡C— σ-bond (2sp-2sp overlap);

π - connection (2рy-2рy);

π - connection (2рz-2рz);

S–H— σ-bond (overlap of 2sp-AO of carbon and 1s-AO of hydrogen).

Problem 5. What forces of intermolecular interaction are called dipole-dipole (orientational), inductive and dispersive? Explain the nature of these forces. What is the nature of the predominant intermolecular interaction forces in each of the following substances: H 2 O, HBr, Ar, N 2, NH 3?

Solution: Between molecules there may be electrostatic interaction. Most versatile - dispersive , because it is caused by the interaction of molecules with each other due to their instantaneous microdipoles. Their simultaneous appearance and disappearance in different molecules contributes to their attraction. In the absence of synchrony, molecules repel each other.

Orientation interaction appears between polar molecules. The greater the polarity of the molecule, the stronger the force of their attraction to each other, and thus the greater the orientational interaction.

Inductive interaction molecules arises due to their induced dipoles. When two molecules—polar and nonpolar—meet, the nonpolar molecule deforms, which contributes to the formation of a dipole in it. An induced dipole is capable of attraction to the permanent dipole of a polar molecule. Inductive interaction the greater, the greater the electrical moment and polarizability of the molecule.

The relative contribution of each type of interaction depends on the polarity and polarizability of the molecules. Thus, the higher the polarity of the molecule, the role is more important orientation forces; the greater the polarizability, the greater the influence of dispersion forces. Inductive forces depend on both factors, but themselves usually play a secondary role.

From these substances orientational and inductive interaction occurs in polar molecules - H 2 O and NH 3. Dispersion interaction- in non-polar and low-polar molecules - HBr, Ar, N2

Problem 6. Give two schemes for filling MOs during the interaction of two AOs with populations: a) electron + electron (1+1) and b) electron + vacant orbital (1+0). Determine the covalency of each atom and the bond order. What are the limits of binding energy? Which of the following bonds are in the hydrogen molecule H 2 and the molecular ion?

Solution :

A) Consider, for example, K 2 and Li 2. Participate in the formation of connections s – orbitals:

Contact order:

b) Consider, for example, K 2 + and Li 2 +. Participate in the formation of connections s – orbitals:

Contact order:

Covalency each atom is equal to 1.

Communication energy depends on the number of valence electrons: the fewer electrons, the lower the binding energy. In K 2 and Li 2 and K 2 + and Li 2 + the binding energy lies in the range of 200-1000 kJ/mol.

In the molecule H 2 a connection of the type is implemented electron + electron, A in the molecular ion H 2 + — electron + vacant orbital.

Task 7. Give electronic configuration NO molecules using the MO method. How do the magnetic properties and bond strength change when moving from an NO molecule to molecular ion NO + ?

Example 2.1. Write the electronic formula Cr in stable oxidation states. Give examples of chromium compounds in these oxidation states.

Solution

The following oxidation states are characteristic of chromium: 0, +2, +3, +6.

The electronic formulas of chromium in these oxidation states are as follows:

Cr 0 1 s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 5 ,

Cr +2 1 s 2 2s 2 2p 6 3s 2 3p 6 3d 4 ,

Cr +3 1 s 2 2s 2 2p 6 3s 2 3p 6 3d 3 ,

Cr +6 1 s 2 2s 2 2p 6 3s 2 3p 6 .

The zero oxidation state of chromium appears in the simple substance as well as in the carbonyl.

Chromium has an oxidation state of +2 in Cr(OH)2 hydroxide, salts such as CrCl2, etc.

An example of a chromium compound in the +3 oxidation state is Cr 2 O 3 oxide. This oxidation state is most characteristic of chromium.

The oxidation state +6 is manifested in CrO 3 oxide, K 2 CrO 4 type chromates, etc.

Example 2.2. From the standpoint of the valence bond (VB) method, show the formation of the VH 3 molecule. Which orbitals of connecting atoms are involved in the formation of bonds? What and how many - or -bonds does the molecule contain? How many bonds are there in a molecule?

What is the spatial structure of the molecule? What is the type of hybridization of the central atom in the indicated compound (if any)? Note the polarity of the bonds and the polarity of the molecule as a whole.

Solution

Boron and hydrogen have the following electronic formulas:

1 N: 1 s 1

5V: 1 s 2 2s 2 2p 1

In the unexcited state, the boron atom has one unpaired electron. To form three bonds, pairing 2 is necessary s-electrons with the transition of one of them to 2 R-orbital:

5V*: 1 s 2 2s 1 2p 2

2R

To form three identical B–H bonds, hybridization of one 2 s and two 2 R-orbitals – sp 2 -hybridization with the formation of three hybrid orbitals located in the same plane at an angle of 120° relative to each other:

The hybrid orbitals formed overlap with s-orbitals of the hydrogen atom with the formation of three -bonds:

The BH 3 molecule has a flat triangular structure.

To determine the polarity of B-H bonds, it is necessary to compare the OEO values ​​of the B and H atoms; OEO(B) = 2.0; OEO(N) = 2.1. Since the electronegativity of hydrogen is greater, the B–H bond will be polar. However, in general, the BH 3 molecule does not have polarity, since the polarity of the B–H bonds directed to the vertices of a regular triangle are mutually compensated.

Thus, the formation of the BH 3 molecule takes part s- orbitals of the H atom and sp 2-hybrid orbitals of boron. The BH 3 molecule is not polar, although it contains three polar -bonds, and has a flat triangular structure. Atom B is in a state sp 2-hybridization.

Example 2.3. Using the relative electronegativity values ​​of the atoms, arrange the compounds HF, HCl, HBr, HI in order of increasing bond ionicity. To which of the connecting atoms is the electron cloud shifted and why?

Solution

The degree of ionicity of a bond can be judged based on the difference in the relative electronegativities of the atoms:

OEO: N – 2.1; F – 4; Cl – 3.0; Br – 2.8; I – 2.5.

Bond: HF HCl HBr HI

EOE: 1.9 0.9 0.7 0.4

Consequently, in order of increasing ionicity of bonds, these molecules can be arranged in a row: HI – HBr – HCl – HF; When a chemical bond is formed, the electron density shifts to a more electronegative atom. Therefore, in HF the electron density is shifted towards F; in HCl – to Cl; in HBr – to Br; in HI – to I.

Example 2.4. In the indicated complex compound, determine the oxidation states of all components, indicate the complexing agent, ligands, ions of the outer and inner spheres and the coordination number, charge of the complexing agent.

Write down the dissociation equation for this complex compound. Give this connection a name.

This material allows the teacher to conduct a general lesson on the topic "Chemical bonding. Classification of complex inorganic substances." In this lesson, the material allows students to consolidate knowledge of determining the types of chemical bonding, as well as learn to write electronic formulas of substances, the ability to determine classes of inorganic substances, drawing up formulas according to oxidation states. This lesson can be used while reviewing these topics in 9th grade.

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Final chemistry lesson in 8th grade on the topic "Chemical bonding. Classification of complex inorganic substances"

Goals:

• Educational:summarize knowledge on the topic covered, repeat fundamental questions and concepts, develop the ability to determine types of bonds, draw up diagrams of the formation of covalent and ionic bonds, electronic and structural formulas, chemical formulas of binary compounds by oxidation states, identification of the main classes of inorganic substances and compilation of formulas by name.
• Developmental: development of attention, logical thinking through relationships: the ability to identify valence electrons and the formation of various types of bonds; development of creative thinking; development of group work skills.
• Educational: nurturing perseverance in achieving knowledge, a sense of camaraderie and collectivism when working in a group.

Lesson type: generalization.

Equipment: task cards, interactive whiteboard.

DURING THE CLASSES

I. Actualization and goal setting

– We devoted several lessons to studying something very important, the formation of substances with different types bonds, the ability to calculate oxidation states, determined classes inorganic compounds. We got acquainted with how to determine the types of bonds, and most importantly, we learned how to determine and compose the chemical formulas of oxides, acids, bases, salts, and then use this knowledge in performing chemical reactions.

The goal of our lesson today is to recall and summarize the knowledge gained on the basic concepts and fundamental issues of the topic, to develop skills in determining the types of bonds and drawing up formulas of substances. You are required to be active in class, because... The assessment will consist of oral answers and test work.

II. Examination homework(3 minutes)

Message to students:

1. Those who completed the task using only a sample of execution and those who needed a textbook or additional literature to complete the task.
2. Those who completed the task halfway tried to do it, but are not sure of the correctness of execution.
3. Who did not start the task.

– Who wants to complete the task on the board: one person for a nitrogen molecule, another for hydrogen sulfide, a third for potassium chloride?

While 3 students are completing the task on the board, the rest are in the next stage.

III. Warm-up (5 minutes)

Repetition of theoretical material on the topic “Chemical bonding”.

The teacher distributes warm-up questions to each table and then the students answer them. He says that it is necessary to repeat the material in order to consolidate it using examples from the theoretical part and then apply knowledge in the practical part of the lesson.

Warm-up questions:
1. What is a chemical bond?
2. Electrons, due to which atoms enter into chemical bonds, are called ………
3.Valence electrons are located on ………
4. What does the group number indicate?
5. How many valence electrons are there in a Ca and N atom?
6. What is the reason for the formation of chemical bonds?
7. How many types of chemical bonds have we studied?
8. What types of chemical bonds do you know?
9. A chemical bond formed due to the formation of shared electron pairs is called ………
10. Ionic bonding is … … …
11. What is EO?
12. EO in the period from left to right ………, in the group from top to bottom … … …
Student answers:

1. Chemical bonds are interaction forces that connect individual atoms into molecules, ions, and crystals.

2. Electrons through which atoms enter into chemical bonds are called valence electrons.

3.Valence electrons are located at the outer energy level.

4.The group number shows the number of electrons in the outer level.

5.Calcium has two valence electrons, and nitrogen has five, which corresponds to the group number of each chemical element.

6.The reason for the formation of chemical bonds is a stable external energy level. If this is the first level, then there are two electrons, and all subsequent eight are like noble gases.

7.We know 4 types of connections.

8. Four types: covalent polar, non-polar, ionic and metallic.

9. A chemical bond formed through the formation of shared electron pairs is called covalent.

10. Ionic bonding is a bond between ions carried out by electrostatic attraction.

11.Electronegativity is the ability of the atoms of an element to attract electrons that bind them to other atoms.

12. EO increases from left to right in a period, and decreases in a group from top to bottom.

IV. Completing individual tasks using cards (10 minutes):

Practical assignment on the topic “Chemical bonding”

The teacher gives assignments to students on the topic “Types of chemical bonds.” In your notebook, compose electronic and structural formulas of substances with different types of bonds. Determine what type they are.

5 students work using cards. Cards with tasks are given.

Determine the types of chemical bonds in the molecules of the following substances:

1) Write the mechanism of formation of molecules No. 1 OF ₂, HCl, O 2;

2) Write the mechanism of formation of molecules No. 2 NaCl, N 2;

3) Write the mechanism of formation of molecules No. 3 NH 3 , BaO;

Answers to task No. 1 are given by students

4) Write the mechanism of formation of molecules No. 4, CH 4 , CaF 2 .

5) Write the mechanism of formation of molecules No. 5 O₂, Li₃N.

The assignments for the class are given on the board. ABOUT₂ , CH ₄ , NaF .

Students work in notebooks. Then we check the same work on the interactive board.

V. Work in teams (10 minutes):

The teacher sets the task of consolidating knowledge, skills and abilities in identifying substances, naming them and drawing up formulas for substances. Work in groups. Total 5 groups. The teacher gives a task to determine classes of inorganic substances. Children work in groups of 4-5 people, the group completes tasks together using a card. The lead advisor in the group then gives one student the opportunity to defend their work.

Task No. 1

From the list of substances, select separately and insert into the table oxides, acids, bases and salts. Determine the oxidation states of the elements included in the compounds.

H 2 O, CaSO 4, HNO 3, HCI, CaS, NO 2, CuO, MgO, Pb(OH) 2, PbO, NaOH, NaCI

Answers to task No. 1 are given by students

Task No. 2. From the list of elements, create formulas for oxides, name them and indicate their oxidation states:

S(IV) , Mg ,C(II) ,Na ,Al ,Fe(III) , Ag(I), N(V)

Answers to task No. 2 are given by students

SO ₂ , MgO, CO, Na ₂ O, Al ₂ O ₃ , Fe ₂ O ₃ , Ag ₂ O, N ₂ O₅

Task No. 3. From the list of substances, write down the formulas of the oxides, name them and indicate the oxidation states:

Na 2 O H 2 SO 4 CO 2 NaOH K 2 O H 2 O NaCI SO 2 CaSO 4 CaO HNO 3

Answers to task No. 3 are given by students

Na 2 O – sodium oxide; CO2 - carbon(IV) monoxide;K 2 O – potassium oxide; H 2 O - hydrogen oxide; SO 2 sulfur(IV) oxide; CaO – calcium oxide;

Task No. 4. Find and name the substances acid and base:

P 2 O 5 , Cu(OH) 2 , H ₂ SiO ₃ , NO 2 , H ₃ PO ₄ Zn(OH) 2 , HCl, AgNO 3 , Fe 2 (SO 4 ) 3 , H 2 SO 4 , AlBr 3 KOH , H 2 S , NaOH , Сa(OH) ₂ НNO ₃

Answers to task No. 4 are given by students

Acids:

H₂SiO₃ - silicic acid; H₃PO ₄ -orthophosphoric acid; HCl-hydrochloric acid;

H2SO4 -sulfuric acid; H2S- hydrosulfide acid.

Reasons:

Cu(OH)2 - copper(II) hydroxide; Zn(OH) 2 -zinc hydroxide; KOH- potassium hydroxide;

NaOH - sodium hydroxide;Сa(OH) ₂ - calcium hydroxide;

Task No. 5. Find and name the salts:

MgO, Fe(OH) 2, HBr, CaSiO 3, HNO 3, K 2 CO 3, I 2 O 7, Ag 2 S, KOH, MnCl 2

Answers to task No. 5 are given by students

CaSiO3 - calcium silicate; K 2 CO 3 - potassium carbonate; Ag 2 S-silver sulfide;

MnCl2 - manganese(II) chloride;

The assignments for the class are given on the board.

Compose the structural formulas NaOH, Ca(OH)₂ ,HCl, H ₂ SO ₃ , Fe ₂ O ₃ , H ₂ O, NaCl, K N O ₃

Students work in notebooks. Then we check the same work on the interactive whiteboard

VI. Consolidation of knowledge with simultaneous self-testing. (3 minutes)

At this stage of the lesson we work with supporting outline on this topic. The teacher and students consolidate knowledge on the topic “Classes of inorganic substances.” There are questions and supporting diagrams on the tables.

Basic diagram for the topics “Main classes of inorganic substances”

Questions:

1.What is an oxide called?

2.What is an acid?

3.What are grounds?

4.What are salts?

Student answers:

1) Oxide is a compound consisting of two elements, one of which is oxygen in the oxidation state -2.

2) Acids are complex substances whose molecules consist of hydrogen atoms that can be replaced by metal atoms and acidic residues.

3) Bases are complex substances consisting of metals and hydroxide ions.

4) Salts are complex substances consisting of metal ions and an acid residue.

We check the correct answers on the interactive board.

Experience: We pour indicators into three test tubes. We determine the nature of the environment.

VII. Homework: Repetition § 14-27, Novoshinskaya, Work II, options 6 (1-5), 8 1-5).

19(1-5).In your homework you should prepare for test work. Assignments are given on the topics “Chemical Bonding” and “Classes of Inorganic Substances”. Complete tasks on drawing up formulas of substances, determine the types of bonds. Today we have summarized the material on these topics, which will make it easier to complete your homework, but on your own.

VIII. Summing up the lesson

– So, today we summarized our knowledge, repeated the main questions related to the types of chemical bonds and the most important classes of compounds.

Grading.

Main features of using digital educational resources:

While studying this topic, illustrative materials from Internet resources will be used, containing reference data on the type of chemical bonds and classes of inorganic compounds; CD resources containing information and illustrative material on this topic; computer software - Microsoft Word, Microsoft Power Point, and others for preparing materials for the lesson and students’ independent work.

Expected learning outcomes:

As a result of studying this topic, students:

· Gain knowledge about the structure of substances with different types of bonds, the ability to identify substances of different classes of compounds using formulas, create formulas and give them names.

· Acquire knowledge physical properties substances, based on the study of the topic “Chemical Bonding”.

· Become familiar with the most important applications of various compounds.

· Acquire the ability to explain the reason for the danger of acid and alkali.

· Acquire skills in writing formulas for substances.

· Able to use Microsoft Word, Microsoft Power Point, Microsoft Office to prepare presentations, abstracts, reports, design work on this topic.

Used Books:

1.I.I.Novoshinsky, N.S.Novoshinskaya Textbook for educational institutions 7th edition. Moscow.: " Russian word» 2012.

2.V.V.Eremin N.E. Kuzmenko. Collection of problems and exercises in chemistry. School course. Moscow "ONICS21 century" Peace and Education 2007.

4.I.I.Novoshinsky, N.S.Novoshinskaya. Collection independent work in chemistry 8th grade Moscow: “Russian Word” 2008

5.N.P.Tregubov Control and measuring materials. M: VAKO, 2010.

6.E.V.Savinkina Chemistry express diagnostics grade 8 52 diagnostic options; national education, Moscow 2012