Practicing tasks on chemical bonding. Basic types of chemical bond

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 a bond differ in electronegativity, the more chemical bond more polar.
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.

This material allows the teacher to conduct a general lesson on the topic "Chemical bonds. Classification of complex inorganic substances".In this lesson, the material allows students to consolidate their knowledge of identifying types chemical bond, and also learn to write electronic formulas of substances, the ability to determine classes of inorganic substances, and compose formulas based on oxidation states. This lesson can be taught when repeating 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 binary compounds by oxidation state, 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 board.

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, and determined classes of 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 due to 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 must prepare for the test. 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 of independent works in chemistry, grade 8 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

The purpose of the lesson: consolidate students' knowledge of types of chemical bonds.

Lesson objectives:

1) repeat the main types of chemical bonds, properties and mechanism of their formation;

2) develop students’ skills in drawing up schemes for the formation of various types of chemical bonds;

3) to cultivate in students organization, independence, communication skills, the ability to generalize knowledge and apply it in practice.

Lesson type: lesson to consolidate knowledge.

Technologies used: control and corrective teaching technology, information and communication technology.

Equipment: table “Types of chemical bonds”, cards with tasks for individual work (3 levels), multi-level test tasks, interactive whiteboard, multimedia projector.

Forms educational activities: frontal, pair work, individual work, work with a textbook and additional. literature.

Lesson structure:

1. Organizational moment.

2. Repetition of the topic “Types of chemical bonds” (electronic presentation prepared by students).

3. Work in pairs using cards.

4. Individual work of students' choice: oral control - conversation with a teacher or consultant, studying a topic in a textbook or additional literature, taking a test, doing independent work.

5. Summing up the lesson, homework.

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Plan - outline open lesson chemistry in 11th grade.

Topic: “Types of chemical bonds.”

The purpose of the lesson: consolidate students' knowledge of types of chemical bonds.

Lesson objectives:

1. repeat the main types of chemical bonds, properties and mechanism of their formation;
2. develop students’ skills and abilities in drawing up schemes for the formation of various types of chemical bonds;
3. to cultivate in students organization, independence, communication skills, the ability to generalize knowledge and apply it in practice.

Lesson type: lesson to consolidate knowledge.

Technologies used:control and corrective teaching technology, information and communication technology.

Equipment: table “Types of chemical bonds”, cards with tasks for individual work (3 levels), multi-level test tasks, interactive whiteboard, multimedia projector.

Forms of educational activities:frontal, pair work, individual work, work with a textbook and additional. literature.

Lesson structure:

1. Organizing time.
2. Review of the topic “Types of chemical bonds” (electronic presentation prepared by students).
3. Work in pairs using cards.
4. Individual work of students' choice: oral control - conversation with a teacher or consultant, study of a topic in a textbook or additional literature, test work, independent work.
5. Summing up the lesson, homework.

During the classes.

1 .Organizing time.Setting the lesson goal.

2. Repetition of the main types of chemical bonds. A group of students performs electronic presentation"Types of chemical bonds." A media projector and interactive whiteboard are used.

3. Work in pairs. Each pair of students receives a card with a task that they complete together, for example:

Card No. 1

1. Determine the type of chemical bond in substances and draw up bond formation schemes for these substances: MgBr 2, H 2 O, Na, H 2.

2. Determine the intermolecular chemical bond for the substance (CH 3OH)n , note the features in the properties of this substance in connection with this type of chemical bond.

4. Individual work of students of their choice.

The use of control and corrective teaching technology allows each student to develop his own educational trajectory. Students keep an activity record sheet where they mark each type of control.

After studying the topic, the student must undergo an oral interview with a teacher or consultant, complete a test and independent work. Only after this does he perform the final test. Consultants are appointed by the teacher, usually 2-3 people who mastered the topic earlier than others and passed all types of control.

Test (1st level)

1. A pair of elements between which an ionic chemical bond is formed:

A) carbon and sulfur; c) potassium and oxygen;

b) hydrogen and nitrogen; d) silicon and hydrogen.

2. Formula of a substance with a polar covalent bond:

A) NaCl; b) HCl; c) BaO; d) Ca 3 N 2.

3. Formula of a substance with a covalent nonpolar bond:

a) Na; b) Br 2; c) HBr; d) KCl.

4. The least polar bond is:

a) C – H; b) C – Cl; c) C – F; d) C – Br.

5. The strongest molecule is:

a) H 2; b) N 2; c) F 2; d) O 2.

6. Nuclear crystal lattice It has:

a) soda; b) water; c) diamond; d) paraffin.

7. The carbon atom has an oxidation state of -3 and a valence of IV when combined with the formula:

a) CO 2; b) C 2 H 6; c) CH 3 Cl; d) CaC 2.

8. A substance between the molecules of which there is a hydrogen bond:

a) ethane; b) sodium fluoride; c) carbon monoxide (II); d) ethanol.

9. The reasons for the sharp difference in the properties of water and hydrogen sulfide lie in the following features:

a) intramolecular bond; b) intermolecular bonds.

Test (2nd level)

1. Formula of a substance with an ionic bond:

a) NH 3; b) C 2 H 4; c) KH; d) CCl 4.

2. A covalent nonpolar bond is formed between atoms:

a) hydrogen and oxygen; c) hydrogen and chlorine;

b) hydrogen and phosphorus; d) magnesium.

3. The most polar relationship is:

a) N – C; b) N – O; c) H – S; d) H – I.

4. The number of sigma and pi bonds in the substance propene, respectively:

a) 7-sigma, 2-pi; c) 6-sigma, 2-pi

b) 8-sigma, 1-pi; d) 8-sigma, 2-pi.

5. The strongest bonds in a molecule of a substance whose formula is:

a) H 2 S; b) H 2 Se; c) H 2 O; d) H 2 Te.

6. The nitrogen atom has a valence of III and an oxidation state of 0 in a molecule of a substance whose formula is:

a)) NH 3; b) N 2; c) CH 3 NO 2; d) N 2 O 3.

7. Molecular structure has a substance with the formula:

a) CH 4; b) NaOH; c) SiO 2; d) Al.

8. A hydrogen bond is formed between:

a) water molecules; c) hydrogen molecules;

b) hydrocarbon molecules; d) metal atoms and hydrogen atoms.

9. Which connection has direction:

a) ionic; b) covalent; c) metal.

Test (3rd level)

1. Chemical bonds in substances whose formulas are CH 4 and CaCl 2 respectively:

a) ionic and covalent polar;

b) covalent polar and ionic;

c) covalent nonpolar and ionic;

d) covalent polar and metallic.

2. The polarity of the bond is greater in a substance with the formula:

a) Br 2; b) LiBr; c) HBr; d) KBr.

3. Ionic nature of bonds in a series of compounds

Li 2 O - Na 2 O – K 2 O – Rb 2 O:

a) increases; c) does not change;

b) decreases; d) first decreases, then increases.

4. There is a covalent bond between atoms, formed by a donor-acceptor mechanism in a substance, the formula of which is:

a) Al(OH) 3; b) Cl; c) C 2 H 5 OH; d) C 6 H 12 O 6.

5. A pair of formulas for substances that contain only sigma bonds:

a) CH 4 and O 2; b) C 2 H 5 OH and H 2 O; c) N 2 and CO 2; d) HBr and C 2 H 4

6. The strongest connection of the given ones:

a) C – Cl; b) C – F; c) C – Br; d) C – I.

7. The valency and degree of nitrogen in ammonium chloride are respectively equal:

a) IV and +4; b) IV and -2; c) III and +2; d) IV and -3.

8. General property for substances with a molecular crystal lattice:

a) solubility in water; c) electrical conductivity of solutions;

b) high boiling point; d) volatility.

9. The formation of hydrogen bonds can be explained by:

a) solubility acetic acid in water;

b) acid properties ethanol;

V) high temperature melting of many metals;

d) insolubility of methane in water.

5. Summing up.So, today we have repeated the main types of chemical bonds, their properties and the mechanism of formation. Review what you learned and what questions you found difficult. If necessary, work through § 6 from the textbook again.

Homework:

Repeat § 6;

Execute exercise 1-3 on p.34.

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.

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 + ?