Chemistry. Examination in chemistry on the topic "Structure of substances"

The molecular structure has

1) silicon oxide (IV)

2) barium nitrate

3) sodium chloride

4) carbon monoxide (II)

Decision.

The structure of a substance is understood from which particles of molecules, ions, atoms its crystal lattice is built. Substances with ionic and metallic bonds have a non-molecular structure. Substances in the molecules of which the atoms are connected by covalent bonds can have molecular and atomic crystal lattices. Atomic crystal lattices: C (diamond, graphite), Si, Ge, B, SiO 2, SiC (carborundum), BN, Fe 3 C, TaC, red and black phosphorus. This group includes substances, as a rule, solid and refractory substances.

Silicon (IV) oxide - covalent bonds, solid, refractory substance, atomic crystal lattice. Barium nitrate and sodium chloride substances with ionic bonds - the crystal lattice is ionic. Carbon monoxide (II) is a gas in a molecule of covalent bonds, which means that this is the correct answer, the crystal lattice is molecular.

Answer: 4

A source: Demo version Unified State Exam-2012 in Chemistry.

In solid form, the molecular structure has

1) silicon oxide (IV)

2) calcium chloride

3) copper (II) sulfate

Decision.

The structure of a substance is understood from which particles of molecules, ions, atoms its crystal lattice is built. Substances with ionic and metallic bonds have a non-molecular structure. Substances in molecules of which the atoms are connected by covalent bonds can have molecular and atomic crystal lattices. Atomic crystal lattices: C (diamond, graphite), Si, Ge, B, SiO 2, SiC (carborundum), BN, Fe 3 C, TaC, red and black phosphorus. This group includes substances, as a rule, solid and refractory substances.

Substances with a molecular crystal lattice have lower boiling points than all other substances. According to the formula, it is necessary to determine the type of bond in the substance, and then determine the type crystal lattice... Silicon (IV) oxide - covalent bonds, solid, refractory substance, atomic crystal lattice. Calcium chloride and copper sulfate - substances with ionic bonds - the crystal lattice is ionic. There are covalent bonds in the iodine molecule, and it sublimes easily, which means that this is the correct answer, the crystal lattice is molecular.

Answer: 4

Source: Demo version of the Unified State Exam-2013 in Chemistry.

1) carbon monoxide (II)

3) magnesium bromide

Decision.

Substances with ionic and metallic bonds have a non-molecular structure. Substances in the molecules of which the atoms are connected by covalent bonds can have molecular and atomic crystal lattices. Atomic crystal lattices: C (diamond, graphite), Si, Ge, B, SiO2, SiC (carborundum), BN, Fe3 C, TaC, red and black phosphorus. This group includes substances, as a rule, solid and refractory substances.

Answer: 3

Source: Unified State Exam in Chemistry 06/10/2013. The main wave. Far East... Option 1.

The ionic crystal lattice has

2) carbon monoxide (II)

4) magnesium bromide

Decision.

Substances with ionic and metallic bonds have a non-molecular structure. Substances in the molecules of which the atoms are connected by covalent bonds can have molecular and atomic crystal lattices. Atomic crystal lattices: C (diamond, graphite), Si, Ge, B, SiO2, CaC2, SiC (carborundum), BN, Fe3 C, TaC, red and black phosphorus. This group includes substances, as a rule, solid and refractory substances.

Magnesium bromide has an ionic crystal lattice.

Answer: 4

Source: Unified State Exam in Chemistry 06/10/2013. The main wave. Far East. Option 2.

Sodium sulfate has a crystal lattice

1) metal

3) molecular

4) atomic

Decision.

Sodium sulfate is a salt with an ionic crystal lattice.

Answer: 2

Source: Unified State Exam in Chemistry 06/10/2013. The main wave. Far East. Option 3.

Each of two substances has a non-molecular structure:

1) nitrogen and diamond

2) potassium and copper

3) water and sodium hydroxide

4) chlorine and bromine

Decision.

Substances with ionic and metallic bonds have a non-molecular structure. Substances in the molecules of which the atoms are connected by covalent bonds can have molecular and atomic crystal lattices. Atomic crystal lattices: C (diamond, graphite), Si, Ge, B, SiO2, SiC (carborundum), BN, red and black phosphorus. This group includes substances, as a rule, solid and refractory substances.

Of these substances, only diamond, potassium, copper, and sodium hydroxide have a non-molecular structure.

Answer: 2

Source: Unified State Exam in Chemistry 06/10/2013. The main wave. Far East. Option 4.

A substance with an ionic type of crystal lattice is

3) acetic acid

4) sodium sulfate

Decision.

Substances with ionic and metallic bonds have a non-molecular structure. Substances in the molecules of which the atoms are connected by covalent bonds can have molecular and atomic crystal lattices. Atomic crystal lattices: C (diamond, graphite), Si, Ge, B, SiO2, CaC2, SiC (carborundum), BN, Fe3 C, TaC, red and black phosphorus. This group includes substances, as a rule, solid and refractory substances.

Sodium sulfate has an ionic crystal lattice.

Answer: 4

Source: Unified State Exam in Chemistry 06/10/2013. The main wave. Siberia. Option 1.

The metal crystal lattice is characteristic for

2) white phosphorus

3) aluminum oxide

4) calcium

Decision.

The metal crystal lattice is typical for metals such as calcium.

Answer: 4

Source: Unified State Exam in Chemistry 06/10/2013. The main wave. Ural. Option 1.

Maxim Avramchuk 22.04.2015 16:53

All metals except mercury have a metallic crystal lattice. Can you tell me what is the crystal lattice of mercury and amalgam?

Alexander Ivanov

Mercury in the solid state also has a metal crystal lattice.

2) calcium oxide

4) aluminum

Decision.

Substances with ionic and metallic bonds have a non-molecular structure. Substances in the molecules of which the atoms are connected by covalent bonds can have molecular and atomic crystal lattices. Atomic crystal lattices: C (diamond, graphite), Si, Ge, B, SiO2, CaC2, SiC (carborundum), BN, Fe3 C, TaC, red and black phosphorus. This group includes substances, as a rule, solid and refractory substances.

Calcium oxide has an ionic crystal lattice.

Answer: 2

Source: Unified State Exam in Chemistry 06/10/2013. The main wave. Siberia. Option 2.

Molecular crystal lattice in solid state It has

1) sodium iodide

2) sulfur (IV) oxide

3) sodium oxide

4) iron (III) chloride

Decision.

Substances with ionic and metallic bonds have a non-molecular structure. Substances in the molecules of which the atoms are connected by covalent bonds can have molecular and atomic crystal lattices. Atomic crystal lattices: C (diamond, graphite), Si, Ge, B, SiO2, CaC2, SiC (carborundum), BN, Fe3 C, TaC, red and black phosphorus. This group includes substances, as a rule, solid and refractory substances.

Among the substances listed, all except for sulfur oxide (IV) have an ionic crystal lattice, and he has a molecular one.

Answer: 2

Source: Unified State Exam in Chemistry 06/10/2013. The main wave. Siberia. Option 4.

The ionic crystal lattice has

3) sodium hydride

4) nitric oxide (II)

Decision.

Substances with ionic and metallic bonds have a non-molecular structure. Substances in the molecules of which the atoms are connected by covalent bonds can have molecular and atomic crystal lattices. Atomic crystal lattices: C (diamond, graphite), Si, Ge, B, SiO2, CaC2, SiC (carborundum), BN, Fe3 C, TaC, red and black phosphorus. This group includes substances, as a rule, solid and refractory substances.

Sodium hydride has an ionic crystal lattice.

Answer: 3

Source: Unified State Exam in Chemistry 06/10/2013. The main wave. Ural. Option 5.

For substances with a molecular crystal lattice, the characteristic property is

1) refractoriness

2) low boiling point

3) high melting point

4) electrical conductivity

Decision.

Substances with a molecular crystal lattice have lower boiling points than all other substances. Answer: 2

Answer: 2

Source: Unified State Exam in Chemistry 06/10/2013. The main wave. Centre. Option 1.

For substances with a molecular crystal lattice, the characteristic property is

1) refractoriness

2) high boiling point

3) low melting point

4) electrical conductivity

Decision.

Substances with a molecular crystal lattice have lower melting and boiling points than all other substances.

Answer: 3

Source: Unified State Exam in Chemistry 06/10/2013. The main wave. Centre. Option 2.

The molecular structure has

1) hydrogen chloride

2) potassium sulfide

3) barium oxide

4) calcium oxide

Decision.

Substances with ionic and metallic bonds have a non-molecular structure. Substances in the molecules of which the atoms are connected by covalent bonds can have molecular and atomic crystal lattices. Atomic crystal lattices: C (diamond, graphite), Si, Ge, B, SiO2, CaC2, SiC (carborundum), BN, Fe3 C, TaC, red and black phosphorus. This group includes substances, as a rule, solid and refractory substances.

All of the substances listed have an ionic crystal lattice except for hydrogen chloride.

Answer: 1

Source: Unified State Exam in Chemistry 06/10/2013. The main wave. Centre. Option 5.

The molecular structure has

1) silicon oxide (IV)

2) barium nitrate

3) sodium chloride

4) carbon monoxide (II)

Decision.

Substances with ionic and metallic bonds have a non-molecular structure. Substances in the molecules of which the atoms are connected by covalent bonds can have molecular and atomic crystal lattices. Atomic crystal lattices: C (diamond, graphite), Si, Ge, B, SiO2, CaC2, SiC (carborundum), BN, Fe3 C, TaC, red and black phosphorus. This group includes substances, as a rule, solid and refractory substances.

Carbon monoxide has a molecular structure among the listed substances.

Answer: 4

Source: Demo version of the USE-2014 in chemistry.

A substance of molecular structure is

1) ammonium chloride

2) cesium chloride

3) iron (III) chloride

4) hydrogen chloride

Decision.

The structure of a substance is understood from which particles of molecules, ions, atoms its crystal lattice is built. Substances with ionic and metallic bonds have a non-molecular structure. Substances in whose molecules the atoms are connected by covalent bonds can have molecular and atomic crystal lattices. Atomic crystal lattices: C (diamond, graphite), Si, Ge, B, SiO2, SiC (carborundum), BN, Fe3C, TaC, red and black phosphorus. This group includes substances, as a rule, solid and refractory substances.

1) ammonium chloride - ionic structure

2) cesium chloride - ionic structure

3) iron (III) chloride - ionic structure

4) hydrogen chloride - molecular structure

Answer: 4

Which chlorine compound has the highest melting point?

Answer: 3

Which oxygen compound has the highest melting point?

Answer: 3

Alexander Ivanov

Not. This is an atomic crystal lattice

Igor Srago 22.05.2016 14:37

Since in the USE they teach that the bond between the atoms of metals and non-metals is ionic, insofar as aluminum oxide must form an ionic crystal. And substances of an ionic structure also (as well as atomic) have a melting point higher than that of a molecular substance.

Anton Golyshev

Substances with an atomic crystal lattice are best simply learned.

For substances with a metal crystal lattice, it is not typical

1) fragility

2) plasticity

3) high electrical conductivity

4) high thermal conductivity

Decision.

Metals are characterized by plasticity, high electrical and thermal conductivity, but fragility is uncharacteristic for them.

Answer: 1

Source: USE 05/05/2015. An early wave.

Decision.

Substances in whose molecules the atoms are connected by covalent bonds can have molecular and atomic crystal lattices. Atomic crystal lattices: C (diamond, graphite), Si, Ge, B, SiO2, SiC (carborundum), BN, Fe3C, TaC, red and black phosphorus. This group includes substances, as a rule, solid and refractory substances.

Answer: 1

The molecular crystal lattice has

Decision.

Substances with ionic (BaSO 4) and metallic bonds have a non-molecular structure.

Substances whose atoms are connected by covalent bonds can have molecular and atomic crystal lattices.

Atomic crystal lattices: C (diamond, graphite), Si, Ge, B, SiO 2, SiC (carborundum), B 2 O 3, Al 2 O 3.

Substances that are gaseous under normal conditions (O 2, H 2, NH 3, H 2 S, CO 2), as well as liquid (H 2 O, H 2 SO 4) and solid, but fusible (S, glucose), have a molecular structure

Therefore, carbon dioxide has a molecular crystal lattice.

Answer: 2

The atomic crystal lattice has

1) ammonium chloride

2) cesium oxide

3) silicon oxide (IV)

4) crystalline sulfur

Decision.

Substances with ionic and metallic bonds have a non-molecular structure.

Substances in whose molecules the atoms are connected by covalent bonds can have molecular and atomic crystal lattices. Atomic crystal lattices: C (diamond, graphite), Si, Ge, B, SiO2, SiC (carborundum), BN, Fe3C, TaC, red and black phosphorus. The rest belong to substances with a molecular crystal lattice.

Therefore, silicon oxide (IV) has an atomic crystal lattice.

Answer: 3

A solid, brittle substance with a high melting point, the solution of which conducts an electric current, has a crystal lattice

2) metal

3) atomic

4) molecular

Decision.

Such properties are characteristic of substances with an ionic crystal lattice.

Answer: 1

What silicon compound has a molecular crystal lattice in the solid state?

Novikova Olesya Vladimirovna

Chemistry and Biology Teacher

MOU - secondary school with. Prokudino

Atkar region

Saratov region.

Test № 1 on the topic: "The structure of substances."

Option I .

a) hydrogen chloride

b) sodium hydroxide

c) carbon monoxide (II)

d) carbon monoxide (IV)

2. A covalent polar bond is present in the molecule

a) oxygen

b) rhombic sulfur

d) hydrogen

3. Chemical bond in the carbon dioxide molecule

a) covalent non-polar

b) covalent polar

c) metal

d) ionic

4. The most durable molecule is:

and) H 2 ;

b) N 2 ;

in) F 2 ;

d) O 2 .

5. Substance, between the molecules of which there is a hydrogen bond:

b) sodium fluoride;

c) carbon monoxide (II);

d) ethanol.

6. Substances with ionic crystal lattice are characterized by:

a) poor solubility in water; c) fusibility;

b) high boiling point; d) volatility.

7. The formation of a hydrogen bond between molecules leads to:

a) to a decrease in boiling points;

b) to a decrease in the solubility of substances in water;

c) to an increase in boiling points;

d) to an increase in the volatility of substances.

8. What substance contains more oxygen in Na 2 CO 3 or in Ca (HCO 3) 2?

9. :

A) SO 2 + H 2 O͢͢ →

B) Na + H 2 O →

B) Na 2 O + H 2 O →

D) S + H 2 O →

10. Solve the problem :

How much water and sodium hydroxide do you need to prepare 180 g of a 15% solution?

11 . Solve the problem :

What is the mass of oxygen obtained by fractional distillation of 200 m 3 (n.o.) air, if the volume fraction of oxygen is 0.21?

Test number 1 on the topic "The structure of substances".

Option II .

Ionic chemical bond is realized in

a) crystalline sulfur

b) solid iodine

c) calcium iodide

d) phosphorus oxide (v)

2. There is a covalent polar bond in the molecule

a) sulfuric acid

b) plastic sulfur

d) rubidium sulfide

3. Chemical bond in the hydrogen molecule

a) covalent non-polar

b) covalent polar

c) metal

d) ionic

4. The strongest bonds in a molecule of a substance, the formula of which is:

and) H 2 S ;

b) H 2 Se ;

in) H 2 O ;

d) H 2 Te .

5. The molecular structure has substances with the formula:

and) CH 4 ;

b) NaOH ;

in) SiO 2 ;

d) Al .

6. A hydrogen bond is formed between:

a) water molecules; c) hydrocarbon molecules;

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

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

a) the solubility of acetic acid in water;

b) acidic properties of ethanol;

c) high melting point of many metals;

d) insolubility of methane in water.

8. Compare the sulfur content in Mg (HSO 4) 2 and CuSO 4?

9. Finish the equations possible reactions :

A) CO 2 + H 2 O͢͢ →

B) Al + H 2 O →

B) Fe + H 2 O →

D) C + H 2 O →

10. Solve the problem:

It is necessary to prepare 540 g of a 12% nitric acid solution. Calculate how much water and acid to take to prepare such a solution.

11. Solve the problem:

What is the mass of nitrogen obtained from 143.6 liters of air containing 78% nitrogen by volume?

Test work No. 2 DKR "STRUCTURE OF SUBSTANCE".

A 1... Chemical bonds in substances, the formulas of which are CH 4 and CaCl 2, respectively:

a) ionic and covalent polar, b) covalent polar and ionic,

c) covalent non-polar and ionic, d) covalent polar and metallic.

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

a) Br 2, b) LiBr, c) HBr, d) KBr

A 3. The ionic nature of the bond in the series of compounds Li 2 O - Na 2 O - K 2 O - Rb 2 O:

a) increases, b) decreases, c) does not change, d) first decreases, then increases.

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

a) Al (OH) 3, b) [CH 3 NH 3] Cl, c) C 2 H 5 OH, d) C 6 H 12 O 6.

A 5. A pair of formulas of substances, in the molecules of which there are only δ - 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.

A 6... The strongest bond of the above:

a) C - Cl, b) C - F, c) C - Br, d) C - I.

A 7... A group of formulas of compounds in which there is a similar orientation of bonds due to sp 3 - hybridization of electron orbitals:

a) CH 4, C 2 H 4, C 2 H 2, b) NH 3, CH 4, H 2 O, c) H 2 O, C 2 H 6, C 6 H 6, d) C 3 H 8, BCl 3, BeCl 2.

A 8... The valence and oxidation state of the carbon atom in the methanol molecule are, respectively, equal:

a) 4 and +4, b) 4 and -2, c) 3 and +2, d) 4 and -3.

A 9... Substances with an ionic crystal lattice are characterized by:

a) poor solubility in water, b) high boiling point, c) low melting point, d) volatility.

A 10... The formation of a hydrogen bond between molecules leads to:

a) to a decrease in the boiling points of substances, b) to a decrease in the solubility of substances in water,

c) to an increase in the boiling points of substances, d) to an increase in the volatility of substances.

A 11... Ionic Bond Formula:

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

A 12

A13.The molecular structure has a substance with the formula:

A 14. A hydrogen bond is formed between:

a) water molecules, b) hydrogen molecules,

c) hydrocarbon molecules, d) metal atoms and hydrogen atoms.

A 15. If you intensively shake a mixture of vegetable oil and water, you get:

a) suspension, b) emulsion, c) foam, d) aerosol.

A 16... Formula of a substance with a covalent polar bond:

a) Cl 2, b) KCl, c) NH 3, d) O 2.

A 17. A substance between the molecules of which there is a hydrogen bond:

a) ethanol, b) methane, c) Hydrogen, d) Benzene.

A 18. The number of common electron pairs in a hydrogen molecule:

a) one, b) two, c) three, d) four.

A 19... The polarity of the chemical bond increases in a number of compounds, the formulas of which:

a) NH 3, HI, O 2, b) CH 4, H 2 O, HF, c) PH 3, H 2 S, H 2, d) HCl, CH 4, CL 2.

A 20. Sodium chloride crystal lattice:

a) atomic, b) ionic, c) metallic, d) molecular.

A 21... The number of δ and π - bonds in the acetylene molecule:

a) 5 δ and π - no, b) 2 δ and 3 π, c) 3 δ and 2 π, d) 4 δ and 1 π.

A 22... Substances, the formulas of which are: CH 3 - CH 2 - OH and CH 3 - O - CH 3, are:

a) homologues, b) isomers, c) the same substance, d) homologues and isomers.

A 23... The homologue of a substance whose formula is CH 2 \u003d CH - CH 3 is:

a) butane, b) butene - 1, c) butene - 2, d) butyne - 1.

A 24... A covalent non-polar bond is formed between atoms:

a) hydrogen and oxygen, b) carbon and hydrogen, c) chlorine, d) magnesium.

A 25... Only δ - there is a bond in the molecule:

a) nitrogen, b) ethanol, c) ethylene, d) carbon monoxide (4).

A 26... The nitrogen atom has a valence of 3 and an oxidation state of 0 in a molecule of a substance, the formula of which is:

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

A 27... The molecular structure has a substance with the formula:

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

A28.The C - H bond is stronger than the Si - H bond, since:

a) the bond length is less, b) the bond length is longer,

c) the polarity of the bond is less, d) the polarity of the bond is greater.

A 29. There is a covalent bond between atoms, formed by the donor - acceptor mechanism in the substance, the formula of which is:

a) CH 3 NO 2, b) NH 4 NO 2, c) C 5 H 8, d) H 2 O.

A 30. The least polar relationship is:

a) C - H, b) C - Cl, c) C - F, d) C - Br
Part B:
B 1... The number of common electron pairs between bromine atoms in the Br 2 molecule is ... ...
B 2. What bonds form a triple bond in the N 2 molecule (present the answer in the nominative case).
B 3... In the nodes of the metal crystal lattice there are …… ...
B 4. Give an example of a substance in a molecule of which there are five δ - and two π - bonds. Name the substance in the nominative case.
B 5.
B 6... The number of common electron pairs between bromine atoms in the N 2 molecule is equal to ……
B 7. What bonds form the triple bond in the C 2 H 2 molecule (present the answer in the nominative case).
B 8... At the sites of the ionic crystal lattice are …… ...
B 9. Give an example of a substance in a molecule of which there are five δ - and one π - bonds. Name the substance in the nominative case.
B 10. What is the maximum number of π - bonds that can form between two atoms in a molecule? (answer in numbers)
Part C:
From 1. Write down the structural formulas of all isomeric substances of composition C 5 H 10 O. Name each substance.
C 2 . Make up the structural formulas of substances: СHCl 3, C 2 H 2 Cl 2, F 2.

Make graphical formulas: AlN, CaSO 4, LiHCO 3.
C 3.

HNO 3, HClO 4, K 2 SO 3, KMnO 4, CH 3 F, MgOHCl 2, ClO 3 -, CrO 4 2-, NH 4 +

C 4. Write down the structural formulas of all isomeric substances of the composition C 4 H 8 O 2. Name each substance.
S 5 . Make up the structural formulas of substances: CHBr 3, C 2 H 2 Br 2, Br 2.

Make graphical formulas: Al 2 S 3, MgSO 4, Li 2 CO 3.
C 6. Determine the oxidation state in chemical compounds and ions:

CCl 4, Ba (NO 3) 2, Al 2 S 3, HClO 3, Na 2 Cr 2 O 7, K 2 O 4, SrO 2-, Cr 2 O 3 2

page 1

The chemical structure is the sequence of joining atoms in a molecule and their arrangement in space. The chemical structure is depicted using structural formulas. The dash represents a covalent chemical bond. If the connection is multiple: double, triple, then put two (not to be confused with the "equal" sign) or three dashes. The angles between the bonds are depicted as possible.

To correctly draw up structural formulas organic matter, it must be remembered that carbon atoms form 4 bonds each

(i.e., the valence of carbon by the number of bonds is four. organic chemistry it is mainly the valence by the number of bonds that is used).

Methane (also called marsh, firedamp) consists of one carbon atom covalently bonded to four hydrogen atoms. Molecular formula CH 4. Structural formula:
H
l
H - C - H
l
H

The angle between the bonds in the methane molecule is about 109 ° - the electron pairs forming covalent bonds of the carbon atom (in the center) with the hydrogen atoms are located in space at the maximum distance from each other.

In grades 10-11, it is studied that the methane molecule has the shape of a triangular pyramid - a tetrahedron, similar to the famous Egyptian pyramids.

Ethylene C 2 H 4 contains two carbon atoms linked by a double bond:

The angle between the bonds is 120 ° (electron pairs are repelled and located at the maximum distance from each other). Atoms are located in the same plane.

If you do not depict each hydrogen atom separately, then we get an abbreviated structural formula:

Acetylene C 2 H 2 contains a triple bond:
H - C ≡ C - H

The angle between the bonds is 180 °, the molecule has a linear shape.

When burninghydrocarbons, oxides of carbon (IV) and hydrogen are formed, i.e. carbon dioxide and water, while a lot of heat is released:

CH 4 + 2O 2 → CO 2 + 2H 2 O

C 2 H 4 + 3O 2 → 2CO 2 + 2H 2 O

2C 2 H 2 + 5O 2 → 4CO 2 + 2H 2 O (in the equation with acetylene, in front of the acetylene formula, we put a factor of 2 so that the number of oxygen atoms on the right side is even)

Of great practical importance is polymerization reaction ethylene - the combination of a large number of molecules to form polymer macromolecules - polyethylene... The bonds between molecules are formed by breaking one of the bonds of the double bond. In general, it can be written like this:

nCH 2 \u003d CH 2 → (- CH 2 - CH 2 -) n

where n is the number of joined molecules, called the degree of polymerization. The reaction takes place at elevated pressure and temperature, in the presence of a catalyst.

Polyethylene is used to make films for greenhouses, covers for cans, etc.

The formation of benzene from acetylene is also referred to as polymerization reactions.

4. The nature and types of chemical bonds. Covalent bond

Application. Spatial structure of molecules

Each molecule (for example, CO 2, H 2 O, NH 3) or molecular ion (for example, CO 3 2 -, H 3 O +, NH 4 +) has a certain qualitative and quantitative composition, as well as a structure (geometry). Molecule geometry formed at the expense of a fixed mutual disposition atoms and bond angles.

The bond angle is the angle between imaginary lines passing through the nuclei of chemically bound atoms. You can also say that it is the angle between two bond lines that have a common atom.

A communication line is a line connecting the nuclei of two chemically bonded atoms.

Only in the case of diatomic molecules (H 2, Cl 2, etc.), the question of their geometry does not arise - they are always linear, i.e. the nuclei of atoms are located on one straight line. The structure of more complex molecules may resemble different geometric figures, eg:

triatomic molecules and ions of the type AX 2 (H 2 O, CO 2, BeCl 2)

tetraatomic molecules and ions of the type AX 3 (NH 3, BF 3, PCl 3, H 3 O +, SO 3) or A 4 (P 4, As 4)

pentaatomic molecules and ions of the AX 4 type (CH 4, XeF 4, GeCl 4)

There are particles and more complex structure (octahedron, trigonal bipyramid, flat regular hexagon). In addition, molecules and ions can be in the form of a distorted tetrahedron, an irregular triangle; in molecules with an angular structure, the values \u200b\u200bof α can be different (90 °, 109 °, 120 °).

The molecular structure is reliably established experimentally using various physical methods. Various theoretical models have been developed to explain the reasons for the formation of a particular structure and predict the geometry of molecules. The easiest to understand are the model of repulsion of valence electron pairs (HVEP model) and the hybridization model of valence atomic orbitals (HVAO model).

The basis of all (including the two mentioned) theoretical models explaining the structure of molecules is the following statement: a stable state of a molecule (ion) corresponds to such a spatial arrangement of atomic nuclei, in which the mutual repulsion of electrons in the valence layer will be minimal.

This takes into account the repulsion of electrons both participating in the formation of a chemical bond (bond electrons) and not participating (lone pairs of electrons). It is taken into account that the orbital of the bonding electron pair is compactly concentrated between two atoms and therefore takes up less space than the orbital of the lone pair of electrons. For this reason, the repulsive effect of a non-bonding (lone) pair of electrons and its effect on bond angles are more pronounced than that of a bonding one.

OVEP model. This theory is based on the following basic provisions (presented in a simplified way):

the geometry of the molecule is determined only by σ-bonds (but not π-);
the angles between bonds depend on the number of lone pairs of electrons of the central atom.

These positions should be considered together, since both the electrons of the chemical bond and the lone pairs of electrons are repelled from each other, which ultimately leads to the formation of such a structure of the molecule in which this repulsion will be minimal.

Consider from the standpoint of the OVEP method the geometry of some molecules and ions; σ-bond electrons will be denoted by two dots (:), lone pairs of electrons - by a conventional symbol (or) or a dash.

Let's start with the CH 4 methane five-atom molecule. In this case, the central atom (this carbon) has completely exhausted its valence capabilities and does not contain lone pairs of valence electrons, i.e. all four valence electrons form four σ-bonds. How should the σ-bond electrons be positioned relative to each other so that the repulsion between them is minimal? Obviously, at an angle of 109 °, i.e. along the lines directed to the vertices of an imaginary tetrahedron, in the center of which is a carbon atom. In this case, the electrons participating in the formation of the bond are maximally distant from each other (for a square configuration, the distance between these bond electrons is greater and the electron-electron repulsion is smaller). For this reason, the methane molecule, as well as the CCl 4, CBr 4, CF 4 molecules, have the form of a regular tetrahedron (as they say, they have a tetrahedral structure):

The ammonium cation NH + 4 and the BF 4 - anion have the same structure, since the nitrogen and boron atoms form four σ-bonds each, and they do not have lone pairs of electrons.

Let us consider the structure of the tetraatomic ammonia molecule NH 3. In the ammonia molecule, there are three pairs of bonding electrons and one lone pair of electrons at the nitrogen atom, i.e. also four pairs of electrons. However, will the bond angle remain 109 °? No, since the lone pair of electrons, which occupies a larger volume in space, has a strong repulsive effect on the electrons of the σ-bond, which leads to a slight decrease in the bond angle, in this case this angle is approximately 107 °. The ammonia molecule has the shape of a trigonal pyramid (pyramidal structure):

The pyramidal structure also has a four-atomic hydronium ion H 3 O +: the oxygen atom forms three σ-bonds and contains one lone pair of electrons.

In the tetraatomic BF 3 molecule, the number of σ-bonds is also three, but the boron atom has no lone pairs of electrons. Obviously, the electron-electron repulsion will be minimal if the BF 3 molecule has the shape of a regular flat triangle with a bond angle of 120 °:

The molecules BCl 3, BH 3, AlH 3, AlF 3, AlCl 3, SO 3 have the same structure and for the same reasons.

What structure will a water molecule have?

There are four pairs of electrons in a triatomic water molecule, but only two of them are σ-bond electrons, the other two are lone pairs of electrons of the oxygen atom. The repulsive effect of two lone pairs of electrons in an H2O molecule is stronger than in an ammonia molecule with one lone pair, therefore the H – O – H bond angle is less than the H – N – H angle in an ammonia molecule: in a water molecule, the bond angle is approximately 105 ° :

In the CO2 molecule (O \u003d C \u003d O), there are also two pairs of bonding electrons (we consider only σ-bonds), however, unlike the water molecule, the carbon atom has no lone pairs of electrons. Obviously, the repulsion between pairs of electrons in this case will be minimal if they are located at an angle of 180 °, i.e. with a linear form of the CO 2 molecule:

Molecules BeH 2, BeF 2, BeCl 2 have a similar structure and for the same reasons. In the triatomic SO 2 molecule, the central atom (sulfur atom) also forms two σ-bonds, but has an unshared pair of electrons; therefore, the sulfur (IV) oxide molecule has an angular structure, but the bond angle in it is greater than in the water molecule (the oxygen atom two lone pairs of electrons, and the sulfur atom has only one):

Some triatomic molecules of ABC composition (for example, H – C≡N, Br – C≡N, S \u003d C \u003d Te, S \u003d C \u003d O), in which the central atom does not have lone pairs of electrons, also have a linear structure. But the HClO molecule has an angular structure (α ≈ 103 °), since the central atom, the oxygen atom, contains two lone pairs of electrons.

The OVEP model can also predict the molecular structure of organic substances. For example, in a molecule of acetylene C 2 H 2, each carbon atom forms two σ-bonds, and there are no lone pairs of electrons at carbon atoms; therefore, the molecule has a linear structure H – C≡C – H.

In the C 2 H 4 ethene molecule, each carbon atom forms three σ-bonds, which, in the absence of lone pairs of electrons from carbon atoms, leads to a triangular arrangement of atoms around each carbon atom:

Table 4.2 summarizes some data on the structure of molecules and ions.

Table 4.2

Relationship between the structure of molecules (ions) and the number σ -bonds and lone pairs of electrons of the central atom

Molecule (ion) type	The number of σ-bonds formed by the central atom	Number of lone pairs of electrons	Structure, bond angle	Particle examples (central atom highlighted)
AB 2	2	0	Linear, α \u003d 180 °	C O 2, Be H 2, HC N, Be Cl 2, C 2 H 2, N 2 O, C S 2
		1	Corner, 90 °< α < 120°	Sn Cl 2, S O 2, N O 2 -
		2	Angular, α< 109°	H 2 O, O F 2, H 2 S, H 2 Se, S F 2, Xe O 2, -
AB 3	3	0	Triangular, α ≈ 120 °	B F 3, B H 3, B Cl 3, Al F 3, S O 3, C O 3 2 -, N O 3 -
AB 3	3	1	Trigonal pyramid, α< 109°	N H 3, H 3 O +, N F 3, S O 3 2 -, P F 3, P Cl 3, As H 3
AB 4	4	0	Tetrahedron, α \u003d 109 °	N H 4 +, C H 4, Si H 4, B F 4, B H 4 -, S O 4 2 -, A l H 4 -

Note. In the notation of the general formula of molecules (ions), A is the central atom, B is the terminal atoms.

GVAO model. The main position of this model is that the formation of covalent bonds involves not "pure" valence s -, p - and d - orbitals, but the so-called hybrid orbitals... Further, hybridization with the participation of 2p - and 2s-AOs is considered.

Hybridization is the phenomenon of mixing of valence orbitals, as a result of which they align in shape and energy.

The concept of hybridization is always used when in education chemical bonds electrons of different energy sublevels, not very different in energy, are involved: 2s and 2p, 4s, 4p and 3d, etc.

The hybrid orbital is not similar in shape to the original 2p - and 2s -AO. It has the shape of an irregular volumetric figure of eight:

As you can see, hybrid AOs are more elongated, so they can better overlap and form stronger covalent bonds. When the hybrid orbitals overlap, only σ-bonds are formed; because of their specific form, hybrid AOs do not participate in the formation of π-bonds (π-bonds form only non-hybrid AOs). The number of hybrid orbitals is always equal to the number of initial AOs participating in hybridization. Hybrid orbitals should be oriented in space so as to ensure their maximum distance from each other. In this case, the repulsion of electrons located on them (binding and non-binding) will be minimal; the energy of the entire molecule will also be minimal.

The GVAO model assumes that hybridization involves orbitals with close energies (i.e., valence orbitals) and a sufficiently high electron density. The electron density of the orbital decreases with an increase in its size; therefore, the role in hybridization is especially important for molecules of elements of small periods.

It should be remembered that GVAO is not a real physical phenomenon, but a convenient concept (mathematical model) that allows one to describe the structure of some molecules. The formation of hybrid AOs is not detected by any physical methods. Nevertheless, the theory of hybridization has some physical basis.

Let's consider the structure of the methane molecule. It is known that the СН 4 molecule has the shape of a regular tetrahedron with a carbon atom in the center; all four С – Н bonds are formed by the exchange mechanism and have the same energy and length; are equivalent. It is quite simple to explain the presence of four unpaired electrons in a carbon atom, assuming its transition to an excited state:

However, this process does not explain in any way the equivalence of all four C – H bonds, since, according to the above scheme, three of them are formed with the participation of the 2p-AO of the carbon atom, one with the participation of 2s-AO, and the shape and energy of 2p - and 2s-AO are different.

To explain this and other similar facts, L. Pauling developed the concept of the GVAO. It is assumed that mixing of the orbitals occurs at the time of the formation of chemical bonds. This process requires the expenditure of energy for the pairing of electrons, which, however, are compensated for by the release of energy during the formation of stronger (as compared to nonhybrid) bonds by hybrid AOs.

Based on the nature and number of AOs involved in hybridization, several types are distinguished.

In the case of sp 3 -hybridization, one s - and three p-orbitals are mixed (hence the name of the type of hybridization). For a carbon atom, the process can be represented as follows:

1 s 2 2 s 2 2 p x 1 2 p y 1 → electrons transition 1 s 2 2 s 1 2 p x 1 2 p y 1 2 p z 1 → divisation hybrid- 1 s 2 2 (s p 3) 4

or using electronic configurations:

Four sp 3 -hybrid AOs occupy an intermediate position in energy between 2p and 2s-AOs.

The sp 3 -hybridization scheme can be represented using images of the AO shape of the carbon atom:

Thus, as a result of sp 3 hybridization, four hybrid orbitals are formed, each of which contains an unpaired electron. These orbitals in space are located at an angle of 109 ° 28 ′, which ensures the minimum repulsion of the electrons located on them. If you connect the vertices of the hybrid orbitals, you get a three-dimensional figure - a tetrahedron. For this reason, the molecules of the composition AX 4 (CH 4, SiH 4, CCl 4, etc.), in which this type of hybridization is realized, have the shape of a tetrahedron.

The concept of sp 3 -hybridization of AO also explains well the structure of H 2 O and NH 3 molecules. It is assumed that the 2s and 2p AOs of nitrogen and oxygen atoms are involved in hybridization. In these atoms, the number of valence electrons (5 and 6, respectively) exceeds the number of sp 3 -hybrid AOs (4); therefore, some hybrid AOs contain unpaired electrons, and some contain lone pairs of electrons:

We see that in the nitrogen atom the lone pair of electrons is on one hybrid AO, and in the oxygen atom - on two. Only AOs with unpaired electrons participate in the formation of bonds with hydrogen atoms, and lone pairs of electrons will have a repulsive effect (Fig. 4.5) on each other (in the case of oxygen) and on the bonding electrons (for oxygen and nitrogen).

Fig. 4.5. Scheme of the repulsive action of bonding and non-bonding orbitals in a molecule of ammonia (a) and water (b)

Stronger repulsion is expressed in the case of a water molecule. Since the oxygen atom has two lone pairs of electrons, the deviation from the ideal bond angle for this type of hybridization (109 ° 28 ′) in the water molecule is greater than in the ammonia molecule (in the H2O and NH 3 molecules, the bond angle is 104 , 5 ° and 107 °).

The sp 3 -hybridization model is used to explain the structure of diamond, silicon, NH 4 + and H 3 O + ions, alkanes, cycloalkanes, etc. In the case of carbon, this type of hybridization is always used when an atom of this element forms only σ-bonds.

In the case of sp 2 hybridization, one s and two p orbitals are mixed. Let us consider this type of hybridization using the boron atom as an example. The process is represented using energy diagrams

Thus, as a result of sp 2 -hybridization of the valence orbitals of the boron atom, three hybrid AOs are formed, directed at an angle of 120 °, and one of the 2p-orbitals does not participate in hybridization. Hybrid orbitals contain one unpaired electron, are located in the same plane, and if you connect their vertices, you get a regular triangle. For this reason, the AX 3 molecules with sp 2 hybridization of the orbitals of the A atom have a triangular structure, as shown for the BF 3 molecule:

The nonhybrid 2p-AO of the boron atom is free (not occupied) and is oriented perpendicular to the plane of the B – F bonds; therefore, the BF3 molecule is an electron acceptor during the formation of a covalent bond by the donor – acceptor mechanism when interacting with an ammonia molecule.

The concept of sp 2 hybridization is used to explain the nature of the carbon - carbon double bond in alkenes, the structure of benzene and graphite, i.e. in cases where a carbon atom forms three σ- and one π-bond.

The spatial arrangement of the orbitals of the carbon atom for sp 2 -hybridization looks like this: the non-hybrid 2p-AO is oriented perpendicular to the plane in which the hybrid orbitals are located (both hybrid and non-hybrid AO contain an unpaired electron).

Consider the formation of chemical bonds in the ethylene molecule H 2 C \u003d CH 2. In it, hybrid AOs overlap with each other and with the 1s-AO of the hydrogen atom, forming five σ-bonds: one C – C and four C – H. Non-hybrid 2p-AOs overlap laterally and form a π-bond between carbon atoms (Fig. 4.6).

Fig. 4.6. Diagram of the formation of σ-bonds (a) and π-bonds (b) in an ethylene molecule

In the case of sp-hybridization, one s - and one p-orbital are mixed. Let us consider this type of hybridization using the beryllium atom as an example. Let's imagine the hybridization process using the energy scheme:

and depicting the shape of the orbitals

Thus, as a result of sp-hybridization, two hybrid AOs are formed, each containing one unpaired electron. Two 2p-AOs do not take part in hybridization and, in the case of beryllium, remain vacant. The hybrid orbitals are oriented at an angle of 180 °, therefore, molecules of the AX 2 type with sp-hybridization of the orbitals of the A atom have a linear structure (Fig. 4.7).

Fig. 4.7. Spatial structure of the BeCl 2 molecule

Using the model of sp-hybridization of the orbitals of the carbon atom, the nature of the triple bond in the molecules of alkynes is explained. In this case, on two hybrid and two non-hybrid 2p-AOs (shown by horizontal arrows →, ←) contains an unpaired electron:

In the acetylene molecule HC≡CH due to hybrid AO, σ-bonds C – H and C – C are formed:

Hybrid 2p-AOs overlap in two perpendicular planes and form two π-bonds between carbon atoms (Fig. 4.8).

Fig. 4.8. Schematic representation of π-bonds (a) and planes of π-bonds (b) in the acetylene molecule (the wavy line shows the lateral overlap of the 2p-AO carbon atom)

The concept of sp-hybridization of carbon atom orbitals makes it possible to explain the formation of chemical bonds in carbyne, CO and CO 2 molecules, propadiene (CH 2 \u003d C \u003d CH 2), i.e. in all cases when a carbon atom forms two σ- and two π-bonds.

The main characteristics of the considered types of hybridization and the geometric configurations of molecules corresponding to some types of hybridization of the orbitals of the central atom A (taking into account the influence of non-bonding electron pairs) are presented in Table. 4.3 and 4.4.

Table 4.3

Main characteristics of different types of hybridization

Comparing the data in the table. 4.2 and 4.4, we can conclude that both models - OVEP and HVAO - lead to the same results regarding the structure of molecules.

Table 4.4

Types of spatial configuration of molecules corresponding to some types of hybridization