Basic oxides correspond to bases.

For example , calcium oxide CaO corresponds to calcium hydroxide Ca(OH)2, cadmium oxide CdO corresponds to cadmium hydroxide Cd(OH)2.

Chemical properties of basic oxides

Ticket 1.

1. Basic chemical concepts (using the example of any chemical formula).

1. Complex substance - consists of different chemical elements.

2. 5 (coefficient) molecules of a complex substance.

3. Qualitative composition of a complex substance - consists of hydrogen and oxygen.

4. Quantitative composition of 1 molecule: 2 H atoms and one O atom; 5 molecules: 10 H atoms and 5 O atoms.

5. Molar mass M (H 2 O) = 1 * 2 + 16 = 18 g/mol

6. Mass of 5 molecules m (H 2 O) = 5 * 18 = 90 g

7. Mass fraction of hydrogen in the molecule: w = = = 0.3333 (33.33%)

2.

Elements of the oxygen subgroup - oxygen O, sulfur S, selenium Se, tellurium Te, polonium Ro- have a common name “chalcogens”, which means “giving birth to ores”.

Structure and properties of atoms.

Sulfur atoms, like oxygen atoms and all other elements of the main subgroup of group VI of D.I. Mendeleev’s Periodic Table, contain 6 electrons in the outer energy level, of which 2 are unpaired electrons.

Simple substances. Allotropy of oxygen is the simple substances oxygen O 2 and ozone O 3.

Sulfur, like oxygen, is characterized by allotropy. This is rhombic and plastic sulfur.

Chemical properties. Sulfur can be both an oxidizing agent and a reducing agent.

1. In relation to reducing agents - hydrogen, metals, sulfur exhibits oxidizing properties and acquires an oxidation state of -2. Under normal conditions, sulfur reacts with all alkali and alkaline earth metals, copper, mercury, silver, for example:

H 2 + S = H 2 S.

2. However, compared to oxygen and fluorine, sulfur is a reducing agent, forming compounds with an oxidation state of +4, +6.

Sulfur burns with a bluish flame, forming sulfur oxide (IV):

S + O 2 = SO 2.

This compound is commonly known as sulfur dioxide

3.

Ca + N 2 ®Ca 3 N 2

Cu + H 2 SO 4 (conc) ® CuSO 4 + SO 2 + H 2 O

Ticket 2.

1. Discovery by D.I. Mendeleev's Periodic Law. Periodic table of chemical elements.

D. I. Mendeleev arranged all the chemical elements known at the time of the discovery of the Periodic Law in a row, according to increasing atomic masses, and marked segments in it - periods , in which the properties of the elements and the substances formed by them changed in a similar way, namely (in modern terms):

1) metallic properties weakened;

2) non-metallic properties were enhanced;

3) the oxidation state of the element in higher oxides increased from +1 to +7;

4) oxides from basic through amphoteric were replaced by acidic ones;

5) hydroxides from alkalis through amphoteric hydroxides were replaced by increasingly stronger acids.

Based on these observations, D.I. Mendeleev made a conclusion in 1869 and formulated the Periodic Law:

properties of chemical elements and those formed by them substances are in a periodic depending on their atomic weights. In modern formulation atomic masses of elements replaced by nuclear charge.

2. Carbon subgroup: structure and properties of carbon atoms, simple substances formed by carbon, chemical properties of carbon.

Carbon subgroup (group 4 A) – carbon, silicon, germanium, tin, lead.

Carbon C is the first element of the main subgroup of group IV of D. I. Mendeleev’s Periodic Table. Its atoms contain 4 electrons in the outer energy level, so they can accept four electrons, acquiring an oxidation state of -4, i.e., exhibit oxidizing properties and give up their electrons to more electronegative elements, i.e., exhibit reducing properties, acquiring at This oxidation state is +4.

Carbon is a simple substance. Carbon forms allotropic modifications - diamond And graphite. They have a structure similar to graphite soot And charcoal. Coal, due to its porous surface, has the ability to absorb gases and dissolved substances. This property of some substances is called adsorption.

Chemical properties of carbon.

Diamond and graphite combine with oxygen at very high temperatures. Soot and coal interact with oxygen much more easily, burning in it. But in any case, the result of such interaction is the same - carbon dioxide is formed:

C + O 2 = CO 2

When heated, carbon forms carbides with metals, for example:

4Al + 3C = Al 4 C 3

3. Prove the presence of carbonate ion in sodium carbonate using a characteristic reaction.

CO 3 2- + H + (any acid) ® CO 2 +H 2 O

A heavy, colorless gas is released, which extinguishes the burning match.

Ticket 3.

1. Theory of atomic structure: planetary model of atomic structure, distribution of electrons across energy levels using the example of an element of the main and secondary subgroups.

Planetary model of the atom (Rutherford model)

Nucleus: protons (p +) and neutrons (n ​​0).

The concept of the electron shell of an atom (electronic layers, energy levels)

In the electron shell, there are layers on which electrons with different amounts of energy will be located, which is why they are also called energy levels.

The number of these levels in an atom of a chemical element = the corresponding period number in D.I. Mendeleev’s table:

The Al atom, an element of period 3, has three levels. Each level can accommodate a certain maximum number of electrons: 1st - 2e - , 2nd - 8e - , and although the maximum number of electrons that can fit on the 3rd level is 18, atoms of elements of this period can place on it, like atoms of elements of period 2, only 8e - .

Energy levels containing the maximum number of electrons are called completed. If they contain fewer electrons, then these levels are incomplete.

Elements of side subgroups always have 2 electrons on the outer level (with the exception of Cr and Cu, they have 1 electron). Lastly, the pre-external level is filled:

2. Subgroup of halogens: structure and properties of atoms.

Elements of the main subgroup of group VII of the Periodic Table of D. I. Mendeleev, united under the common name halogens, fluorine F, chlorine Cl, bromine Br, iodine I, astatine At (rarely found in nature) are typical non-metals. This is understandable, because their atoms contain seven electrons in the outer energy level, and they only need one electron to complete it. Halogen atoms, when interacting with metals, accept an electron from the metal atoms. In this case, salts are formed. This is where the general name of the subgroup “halogens” comes from, i.e. “giving birth to salts”.

Halogens are very strong oxidizing agents. Fluorine in chemical reactions exhibits only oxidizing properties, and is characterized only by the oxidation state of -1 in compounds. The remaining halogens can also exhibit reducing properties when interacting with more electronegative elements - fluorine, oxygen, nitrogen. In this case, their oxidation states can take the values ​​+1, +3, +5,

7. The reducing properties of halogens increase from chlorine to iodine, which is associated with an increase in the radii of their atoms: chlorine atoms are approximately one and a half times smaller than those of iodine.

Halogens are simple substances. All halogens exist in a free state in the form of diatomic molecules F 2, Cl 2, Br 2, I 2. Fluorine and chlorine are gases, bromine is a liquid, iodine is a solid. From F 2 to I 2 the color intensity of the halogens increases. Iodine crystals have a metallic sheen.

3. Prove the presence of sulfate ion in sodium sulfate using a characteristic reaction.

SO 4 2- + Ba 2+ (soluble barium salt) ® BaSO 4 ¯

White fine crystalline precipitate

Ticket 4.

1. Rules for determining oxidation states.

Elements that have a constant oxidation state:

1. Group I A: Li +, Na +, K +, Rb +, Cs +.

2. II group A: Be +2, Mg +2, Ca +2, Zn +2, Sr +2, Cd +2, Ba +2.

3. III group A: Al +3

6. H +1 (MeH -1)

7. In simple substances, s.o. = 0.

For the remaining elements, s.o. consider

H 2 +1 S X O 4 - 2 : so sulfur does not have a constant s.o., so we take it as X.

+1 *2 + X + (-2 ) * 4 = 0

Higher s.o. = Group No. (except O, F)

Lowest s.o. = Group No. – 8 (Me does not have a lower s.o.)

2. Chemical properties of halogens - simple substances.

The chemical activity of halogens, like non-metals, weakens from fluorine to iodine.

Each halogen is the strongest oxidizing agent in its period. The oxidizing properties of halogens are distinct when they interact with metals. In this case, salts are formed. Thus, fluorine already reacts under normal conditions with most metals, and when heated, it also reacts with gold, silver, and platinum, which are known for their chemical passivity. Aluminum and zinc ignite in a fluorine atmosphere:

0 0 +2 -1
Zn + F 2 = ZnF 2.

The remaining halogens react with metals mainly when heated.

The decrease in the oxidative properties and the increase in the reducing properties of halogens from fluorine to iodine can also be judged by their ability to displace each other from salt solutions.

Thus, chlorine displaces bromine and iodine from solutions of their salts, for example:

Cl 2 + 2NaBr = 2NaCl + Br 2.

3. Make up molecular and ionic equations for reactions between substances: lead (II) nitrate and potassium sulfate, iron (III) chloride and silver nitrate.

Ticket 5.

1. Classification of chemical reactions according to the number of starting materials and reaction products.

2. Hydrogen halides and hydrohalic acids and their salts.

N 2 + G 2 = 2NG

(G is the conventional chemical designation for halogens).

All hydrogen halides (their general formula can be written as NG) are colorless gases with a pungent odor and are toxic. They dissolve very well in water and smoke in humid air, as they attract water vapor in the air, forming a foggy cloud.

Solutions of hydrogen halides in water are acids, these are HF - hydrofluoric, or hydrofluoric, acid, HC1 - hydrochloric, or hydrochloric acid, HBr - hydrobromic acid, HI - hydroiodic acid. The strongest of the hydrohalic acids is hydroiodic acid, and the weakest is hydrofluoric acid.

Salts of hydrohalic acids. Hydrohalic acids form salts: fluorides, chlorides, bromides and iodides. Chlorides, bromides and iodides of many metals are highly soluble in water.

To determine chloride, bromide and iodide ions in solution and distinguish them, a reaction with silver nitrate is used.

3. Calculate the mass fraction of oxygen in sodium sulfate.

Given: Na 2 SO 4	Solution: W O = = = W O = 0.451 =45.1%
W O - ? %

Answer: mass fraction of oxygen 45.1%.

Ticket 6.

1. Electrolytes and non-electrolytes.

According to the conductivity of electric current, all substances are divided into electrolytes and non-electrolytes.

Electrolytes are substances whose solutions conduct electric current. These include acids, bases, and salts. These substances conduct current because can dissociate into a cation and anion:

Acids: HAnH + + An -

Bases: MON M + + OH -

Salts: МAn→ М + + An -

The index after a simple ion or parenthesis becomes a coefficient

Ca 3 (PO 4) 2 → 3Ca 2+ + 2 (PO 4) 3-

Non-electrolytes include all others - simple substances, oxides, almost all organic substances.

2.

The physical properties of metals are determined by their structure: the presence of free electrons in the crystal lattice. Thanks to free electrons, all metals have electrical conductivity, thermal conductivity, and a metallic luster.

Electro- And thermal conductivity. Electrons moving chaotically in a metal under the influence of an applied electrical voltage acquire directional movement, resulting in the generation of an electric current. Silver, copper, as well as gold, aluminum, and iron have the highest electrical conductivity; the smallest - manganese, lead, mercury.

Most often, the thermal conductivity of metals also changes in the same sequence as electrical conductivity. It is due to the high mobility of free electrons, which, colliding with vibrating ions and atoms, exchange energy with them. Therefore, the temperature quickly equalizes throughout the entire piece of metal.

Metallic shine. The electrons that fill the interatomic space reflect light rays rather than transmit them like glass, which is why all metals in the crystalline state have a metallic luster.

The remaining properties - hardness, density, fusibility, plasticity - are different.

3. Describe one of the elements - metals (sodium, calcium, aluminum or iron) (all optional).

CHARACTERISTICS OF A METAL ELEMENT USING THE EXAMPLE OF ALUMINA

1. Position in the Periodic Table.Aluminum(serial number 13 ) is an element 3 period, main subgroups 3

2. Number of protons in an atom aluminum equals 13 , number of electrons - 13 , number of neutrons in the isotope 27 13 Al - 27-13 =14, nuclear charge +13 , electron level distribution 2, 8, 3 .

3. Simple substance.Aluminum- This amphoteric metal. Atoms aluminum show restorative properties.

4. Higher oxide, its character. Aluminum forms a higher oxide, the formula of which is Al2O3. According to its properties it is amphoteric oxide.

4. Higher hydroxide, its character. Aluminum forms a higher hydroxide, the formula of which is Al(OH)3. By properties amphoteric base.

Ticket 7.

1. The concept of strong and weak electrolytes.

Electrolytes include salts, acids, and bases.

Salts are all strong electrolytes, i.e. conduct electricity well. Therefore, in the dissociation equation they put only one arrow in the direction of disintegration into ions

МAn→ М + + An -

Strong bases are alkalis, i.e. water-soluble bases.

Ca(OH) 2 → Ca 2+ +2(OH) -

Insoluble and slightly soluble are weak, therefore, when writing the dissociation equation, they put a reversibility sign (in addition to ions, there are molecules)

MON M + + OH -

Strong acids include HCl, HBr, HI, H2SO4, HNO3, HClO4, HClO3.

2. Alloys.

These are materials with characteristic properties, consisting of two or more components, at least one of which is metal.

In metallurgy, iron and all its alloys are divided into one group called black metals; other metals and their alloys have a technical name non-ferrous metals.

The vast majority of iron (or ferrous) alloys contain carbon. They are divided into cast iron and steel.

Cast iron- an iron-based alloy containing more than 2% carbon, as well as manganese, silicon, phosphorus and sulfur. Cast iron is much harder than iron, it is usually very brittle, cannot be forged, and breaks when hit. This alloy is used for the manufacture of various massive parts by casting, the so-called cast iron, and for processing into steel - pig iron.

Depending on the state of carbon in the alloy, gray and white cast iron are distinguished.

Steel is an iron-based alloy containing less than 2% carbon. Based on their chemical composition, steel is divided into two main types: carbon And alloyed.

Examples of non-ferrous alloys can be: nichrome, tertiary solder, pobedit, duralumin.

Duralumin- an alloy of aluminum (95%), magnesium, copper and manganese. Very light and durable alloy. It is equal in strength to steel, but three times lighter. Used in aircraft construction.

3. Describe one of the elements - nonmetals (chlorine, sulfur, phosphorus, nitrogen, carbon, silicon) (all optional).

CHARACTERISTICS OF A NON-METAL ELEMENT USING THE EXAMPLE OF SULPHUR

1. Position in the Periodic TableSulfur(serial number 16 ) is an element 3 period, main subgroups 6 groups of the Periodic Table.

2.The structure of the atom, its properties. The number of protons in a sulfur atom is 16 , number of electrons - 16 , number of neutrons in the isotope 32 16 S - 32-16 =16, nuclear charge +16 , distribution of electrons across levels 2, 8, 6.

3. Simple substance. Sulfur is non-metal. Sulfur atoms exhibit oxidative properties.

3.Higher oxide, its character. Sulfur forms a higher oxide, the formula of which is SO 3. According to its properties it is acid oxide.

4.Higher hydroxide, its character. Sulfur forms a higher hydroxide, the formula of which is H2SO4. By properties acid.

Ticket 8.

1. Oxides: their composition, classification and names.

Oxides- these are binary compounds, in second place is oxygen with an oxidation state of -2.

Depending on which element comes first, oxides are divided into three groups:

1) Basic. These are oxides in which the metal comes first: CaO, Na 2 O.

2) Acidic. These are oxides in which a non-metal comes first: P 2 O 5.

3) Amphoteric. These are oxides in which the first element is an amphoteric element (transition metal): Al 2 O 3, Fe 2 O 3

Basic oxides correspond to bases. For example, Na 2 O - NaOH. Acid oxides correspond to acids: P 2 O 5 - H 3 PO 4.

The names are composed of the name of oxygen (in Latin) - oxide, and the name of the first element indicating the oxidation state (if variable)

P 2 +5 O 5 phosphorus (V) oxide, Fe 2 +3 O 3 iron (III) oxide

2. Oxygen subgroup: structure and properties of atoms, simple substances, chemical properties of sulfur.

For the answer, see ticket 1, question 2.

3. Prove the presence of chloride ion in potassium chloride using a characteristic reaction.

Cl - + Ag + (soluble silver salt) ® Ag Cl ¯

White curdled sediment

Ticket 9.

1. Acids. Names and formulas of acids.

Acids are complex inorganic substances consisting of hydrogen cation and an acid residue anion.

HCl – hydrochloric

HNO 3 – nitrogen

H 2 SO 4 – sulfuric

H 2 CO 3 – coal

H 3 PO 4 – phosphoric

2. Alloys.

For the answer, see ticket 7, question 2.

3. Describe one of the elements - metals (lithium, magnesium, potassium or aluminum) (all optional).

For a sample answer, see ticket 6, question 3.

Ticket 10.

1. The position of metals in the periodic table of chemical elements D.I. Mendeleev, the structure of their atoms and crystals.

Me are simple substances that easily give up electrons. For the main subgroups:

Me includes all elements of secondary subgroups. This position of Me in the periodic table is associated with their structure: a small number of electrons on the outer level (1-3), which in the main subgroups is determined by the group number, and in the side groups there are always 2 electrons. The second characteristic for Me is a large radius (increases in the table from top to bottom).

In the crystal lattice, Me has free electrons, which are responsible for the main physical properties of Me:

2. Foundations in the light of TED; their classification and chemistry. properties.

Bases are electrolytes that, when dissociated, form a metal cation and an acidic anion.

Classification:

1. Water-insoluble bases.

2. Alkalis – soluble in water.

Typical base reactions

1 . Base + acid® salt + water.

(exchange reaction)

Hl + NaOH = NaCl + H 2 O

H + + OH - = H 2 O (neutralization reaction).

2. Base + acid oxide®salt + water.

(exchange reaction)

2NaOH + N2O5 = 2NaNO3 + H2O
2OH - + N 2 O 5 = 2NO 3 - + H 2 O;

3 . Lye + salt ® new base + new salt.

(exchange reaction)

2KOH + CuSO 4 = = Cu(OH) 2 ¯+ K 2 SO 4

Cu 2+ + 2OH - = = Cu(OH) 2 ¯

4. Bases insoluble in water decompose when heated into metal oxide and water, which is not typical for alkalis, for example:

Cu(OH) 2 ¯ = CuO + H 2 O

3. Arrange the coefficients in reaction schemes using the electronic balance method. Indicate the oxidizing agent and reducing agent, oxidation and reduction processes.

Al + O 2 ® Al 2 O 3

HNO 3 + P® H 3 PO 4 + NO 2 + H 2 O

When preparing for the exam, see the solution in the laboratory journal - practical work No. 2.

Ticket 11.

1. Electronic balance method.

Al 0+ O2 0 ® Al 2 +3 O 3 -2

We write down the elements that changed the s.o.

Al 0 – 3e - → Al +3 4 Al 0 – reducing agent, oxidation process

O 2 0 +2*2e - →2O -2 3 O 2 0 – oxidizing agent, reduction process

Note. If a simple substance has an index (2), then it is transferred to the electronic balance.

We equalize the reaction using coefficients from the electronic balance (4, 3):

4Al +3O 2 ® 2 Al 2 O 3

2. General chemical properties of metals. Electrochemical voltage series of metals and the interaction of metals with solutions of acids and salts.

Metals are reducing agents. Reductive properties are exhibited in reactions with simple and complex substances.

I. With simple – non-metals

2Na + S = Na 2 S sodium sulfide

II. With complex: water, acids, salt solutions (substitution reactions). When writing all these reactions, it is necessary to take into account the activity series (electrochemical series) of metals.

K, Ca, Na, Mg, Al, Zn, Fe, Ni, Sn, Pb, (H 2), Cu, Hg, Ag, Au.

1. Metals standing in the voltage series to the left of hydrogen displace it from acid solutions, and those standing to the right, as a rule, do not displace hydrogen from acid solutions:

Zn + 2HCl = ZnCl 2 + H 2.

2. Each metal displaces from salt solutions other metals located to the right of it in the stress series, and can itself be displaced by metals located to the left, for example:

Fe + CuSO 4 = FeSO 4 + Cu,

Сu + HgCl 2 = Hg + CuCl 2.

3. Determine the mass of carbon monoxide (IV) by the amount of substance 2mmol.

Answer: 88 mg carbon monoxide (IV).

Ticket 12.

1. Hydrolysis of salts by cation.

МAn + HOH = MOH + HАn

Salt base acid

A salt undergoes hydrolysis if it is formed by at least one weak ion. If the cation is weak (from a weak base), then hydrolysis is called according to the cation.

Weak bases are insoluble in water.

For example, FeCl 3 is a salt formed by a strong acid (HCl) and a weak base (Fe(OH) 3)

FeCl3Û Fe 3+ +3Cl -

weak cation

Fe 3+ + H + OH - Û Fe OH 2+ + H+

4. Determine whether the solution is acidic

This is the case hydrolysis by cation.

2. General physical properties of metals.

See the ticket for the answer. 6 , question 2.

3. Carry out reactions confirming that sulfuric acid contains hydrogen cations and sulfate anions.

H 2 SO 4 Û 2H + + SO 4 2-

H+ - methyl orange (will turn red), or litmus (will turn red)

SO 4 2- + Ba 2+ ® Ba SO 4 ¯ (white fine-crystalline precipitate)

Ticket 13.

1. Hydrolysis of salts by anion.

Salt hydrolysis is the interaction of a soluble salt with water.

МAn + HOH = MOH + HАn

Salt base acid

A salt undergoes hydrolysis if it is formed by at least one weak ion. If the anion is weak (from a weak acid), then hydrolysis is called according to the anion.

Strong acids: H 2 SO 4, HNO 3, HClO 3, HClO 4, HCl, HBr, HI

The rest are weak.

For example, Na 2 CO 3 - a salt is formed by a weak acid and a strong base

1. Write down the salt dissociation equation. Na 2 CO 3Û 2Na + + CO 3 2-

weak anion

2. Select a weak ion: cation or anion.

3. Record its interaction with water. CO 3 2- + H + OH - Û HCO 3 - + HE -

4. Determine the solution environment: HE -- alkaline environment, H + - acidic environment, absence of H + and OH - neutral.

This is the case hydrolysis by anion.

2. General chemical properties of metals.

For the answer, see ticket 11, question 2.

3. How many grams of iodine and alcohol do you need to take to prepare 30 g of a 5% solution of iodine tincture?

When preparing for the exam, see the solution in the laboratory journal - practical work No. 1.

Ticket 14.

1 . Drawing up formulas of chemical substances by oxidation state.

1. Enter the oxidation states:

For the first element, the constant is the highest (by group number), or variable (indicated in the name of the substance)

For the second - the lowest (-(8-No. gr.)), or according to the solubility table (for a group of elements);

2. Cross the oxidation states to get the indices (reduce if necessary).

For example.

1) make aluminum oxide: Al 2 +3 O 3 -2

2) compose lead(IV) sulfide: Pb 2 +4 S 4 -2 → PbS 2

3) make calcium sulfate: Ca +2 SO 4 -2

2. Subgroup of halogens.

When preparing for the exam, see the answer in ticket 3, question 2.

3. Carry out reactions to confirm the qualitative composition of barium chloride.

BaCl 2 Û Ba 2+ + 2Cl -

Ba 2+ + SO 4 2- ® Ba SO 4 ¯ (white fine-crystalline precipitate)

Сl - + Ag + ® Ag Сl ¯ (white cheesy sediment)

Ticket 15.

1. Ion exchange reactions.

In order to record an ion exchange reaction, you must adhere to the following algorithm.

1. Write a molecular equation for the reaction

Fe(NO 3) 3 + 3NaOH = Fe(OH) 3 + 3NaNO 3

2. Check the possibility of the reaction occurring (reaction products: sediment, gas or water)

Fe(NO 3) 3 + 3NaOH = Fe(OH) 3↓ + 3NaNO 3

3. Write down the ionic equation of the reaction, and do not forget:

· We leave it in the form of a molecule - a weak electrolyte (H 2 O) and a non-electrolyte, sediment or gas;

· The coefficient in front of the formula of a substance refers to both ions!!!

· The formulas of polyatomic (complex) ions do not break: OH -, CO3 2-, PO4 3-, etc.

· The index after a simple ion or bracket goes into the coefficient in front of it in the ionic equation

Fe 3+ + 3(NO 3) - + 3Na + + 3OH - = Fe(OH) 3↓ + 3Na + + NO 3 -

4. “Reduce” similar ones

Fe 3+ + 3NO 3 - + 3Na++ 3OH - = Fe(OH) 3↓ + 3Na+ + NO 3 -

5. Rewrite the abbreviated ionic equation

Fe 3+ + 3OH - = Fe(OH) 3↓

2. General characteristics of alkali metals: atomic structure and physical properties of simple substances.

These are complex substances consisting of two chemical elements, one of which is oxygen with an oxidation state of (-2). General formula of oxides: EmABOUTn, Where m- number of atoms of the element E, A n- number of oxygen atoms. Oxides can be solid (sand SiO 2, varieties of quartz), liquid (hydrogen oxide H 2 O), gaseous (carbon oxides: carbon dioxide CO 2 and carbon dioxide).

The nomenclature of chemical compounds developed as factual material accumulated. At first, while the number of known compounds was small, they were widely used trivial names, not reflecting the composition, structure and properties of the substance, - red lead Pb 3 O 4, litharge PHO, magnesia MgO, iron scale Fe 3 O 4, laughing gas N 2 O, white arsenic As 2 O 3 The trivial nomenclature was replaced by semi-systematic nomenclature - the name included an indication of the number of oxygen atoms in the compound: nitrous- for lower ones, oxide- for higher oxidation states; anhydride- for acidic oxides.

Currently, the transition to modern nomenclature is almost complete. According to international nomenclature, in the title oxide, the valency of the element should be indicated; for example, SO 2 - sulfur(IV) oxide, SO 3 - sulfur(VI) oxide, CrO - chromium(II) oxide, Cr 2 O 3 - chromium(III) oxide, CrO 3 - chromium(VI) oxide.

Based on their chemical properties, oxides are divided into salt-forming and non-salt-forming.

Non-salt-forming These are oxides that do not react with alkalis or acids and do not form salts. There are few of them and they contain non-metals.

Salt-forming These are oxides that react with acids or bases to form salt and water.

Among salt-forming oxides distinguish between oxides basic, acidic, amphoteric.

Basic oxides- these are oxides that correspond to bases. For example: CuO corresponds to the base Cu(OH) 2, Na 2 O - the base NaOH, Cu 2 O - CuOH, etc.

Typical reactions of basic oxides

1. Basic oxide + acid = salt + water (exchange reaction):

2. Basic oxide + acidic oxide = salt (compound reaction):

3. Basic oxide + water = alkali (compound reaction):

Acidic oxides are those oxides that correspond to acids. These are non-metal oxides: N 2 O 5 corresponds to HNO 3, SO 3 - H 2 SO 4, CO 2 - H 2 CO 3, P 2 O 5 - H 4 PO 4 as well as metal oxides with high oxidation states: Cr 2 + 6 O 3 corresponds to H 2 CrO 4, Mn 2 +7 O 7 - HMnO 4.

Typical acid oxide reactions

1. Acid oxide + base = salt + water (exchange reaction):

2. Acid oxide + basic oxide salt (compound reaction):

3. Acidic oxide + water = acid (compound reaction):

Such a reaction is possible only if the acid oxide is soluble in water.

Amphoteric are called oxides, which, depending on conditions, exhibit basic or acidic properties. These are ZnO, Al 2 O 3, Cr 2 O 3, V 2 O 5.

Amphoteric oxides do not directly combine with water.

Typical reactions of amphoteric oxides

1. Amphoteric oxide + acid = salt + water (exchange reaction):

2. Amphoteric oxide + base = salt + water or complex compound:

Basic oxides. TO main include oxides of typical metals, They correspond to hydroxides that have the properties of bases.

Preparation of basic oxides

Oxidation of metals when heated in an oxygen atmosphere.

2Mg + O2 = 2MgO

2Cu + O 2 = 2CuO

The method is not applicable for the production of alkali metal oxides. In reaction with oxygen, alkali metals usually produce peroxides, so the oxides Na 2 O, K 2 O are difficult to obtain.

Sulphide roasting

2CuS + 3O 2 = 2CuO + 2SO 2

4FeS 2 + 110 2 = 2Fe 2 O 3 + 8SO 2

The method is not applicable for sulfides of active metals that oxidize to sulfates.

Hydroxide decomposition

Cu(OH) 2 = CuO + H 2 O

ThisThis method cannot produce alkali metal oxides.

Decomposition of salts of oxygen-containing acids.

BaCO 3 = BaO + CO 2

2Pb(NO 3) 2 = 2PbO + 4N0 2 + O 2

4FeSO 4 = 2Fe 2 O 3 + 4SO 2 + O 2

Decomposition is easily carried out for nitrates and carbonates, including basic salts.

2 CO 3 = 2ZnO + CO 2 + H 2 O

Preparation of acid oxides

Acidic oxides are represented by oxides of nonmetals or transition metals in high oxidation states. They can be obtained by methods similar to those of basic oxides, for example:

1. 4P + 5O 2 = 2P 2 O 5
2. 2ZnS + 3O 2 = 2ZnO + 2SO 2
3. K 2 Cr 2 O 7 + H 2 SO 4 = 2CrO 3 ↓ + K 2 SO 4 + H 2 O
4. Na 2 SiO 3 + 2HCl = 2NaCl + SiO 2 ↓ + H 2 O