Objective: learn to characterize chemical elements based on their position in the Periodic Table of D.I. Mendeleev according to a certain plan.

Explanations for work:

The periodic system of Mendeleev is a natural classification of chemical elements according to the electronic structure of their atoms. The electronic structure of the atom, and hence the properties of the element, is judged by the position of the element in the corresponding period and subgroup of the per system. The patterns of filling of e-levels explain the different number of elements in the periods. The strict periodicity of the arrangement of elements in the per system of chemical elements of Mendeleev is fully explained by the consistent nature of the filling of energy levels. The theory of the structure of atoms explains the periodic change in the properties of elements. An increase in the positive charges of atomic nuclei from 1 to 107 causes a periodic repetition of the structure of the external energy level. And since the properties of the elements mainly depend on the number of electrons in the outer level, they also repeat periodically. This is the physical meaning of the periodic law. In short periods, with an increase in the positive charge of the nuclei of atoms, the number of electrons in the outer level increases (from 1 to 2 in the first period, and from 1 to 8 in the second and third periods), which explains the change in the properties of the elements: at the beginning of the period (except for the first period) there is an alkali metal, then the metallic properties gradually weaken and the properties of the non-metal increase. In large periods, as the nuclear charge increases, filling the levels with electrons is more difficult, which also explains the more complex change in the properties of elements compared to elements of small periods. So, in even rows of long periods, with increasing charge, the number of electrons in the outer level remains constant and is equal to 2 or 1. Therefore, while the electrons are filling the level following the outer (second from the outside), the properties of the elements in these rows change extremely slowly. Only in odd rows, when the number of electrons in the outer level increases with the growth of the nuclear charge (from 1 to 8), do the properties of the elements begin to change in the same way as for typical ones. In the light of the doctrine of the structure of atoms, the division of D.I. Mendeleev of all elements for 7 periods. The period number corresponds to the number of energy levels of atoms filled with electrons. Therefore, s-elements are present in all periods, p-elements in the second and subsequent periods, d-elements in the fourth and subsequent periods, and f-elements in the sixth and seventh periods. The division of groups into subgroups, based on the difference in the filling of energy levels with electrons, is also easily explained. For elements of the main subgroups, either s-sublevels (these are s-elements) or p-sublevels (these are p-elements) of the outer levels are filled. For elements of side subgroups, the (d-sublevel of the second outside level (these are d-elements) is filled. For lanthanides and actinides, the 4f- and 5f-sublevels are filled, respectively (these are f-elements). Thus, in each subgroup, elements are combined whose atoms have similar structure of the outer electronic level.At the same time, the atoms of the elements of the main subgroups contain at the outer levels the number of electrons equal to the number of the group.The secondary subgroups include elements whose atoms have two or one electron at the outer level.Differences in structure also cause differences in properties of elements of different subgroups of the same group. Thus, on the outer level of atoms of the elements of the halogen subgroup there are seven electrons of the manganese subgroup - two electrons each. The first are typical metals, and the second are metals. But the elements of these subgroups also have common properties: entering into chemical reactions, all of them (with the exception of fluorine F) can donate 7 electrons to form chemical bonds. Manganese give 2 electrons from the outer level and 5 electrons from the next level. Thus, in the elements of the secondary subgroups, the valence electrons are not only the outer, but also the penultimate (second from the outside) levels, which is the main difference in the properties of the elements of the main and secondary subgroups. It also follows that the group number, as a rule, indicates the number of electrons that can participate in the formation of chemical bonds. This is the physical meaning of the group number. So, the structure of atoms determines two regularities: 1) a change in the properties of elements horizontally - in the period from the left to the right, metallic properties are weakened and non-metallic properties are enhanced; 2) a change in the properties of elements along the vertical - in a subgroup with an increase in the serial number, metallic properties are enhanced and non-metallic properties are weakened. In this case, the element (and the cell of the system) is located at the intersection of the horizontal and vertical, which determines its properties. This helps to find and write the properties of elements whose isotopes are obtained artificially. According to the number of energy levels in the electron shell of the atom, the elements are divided into seven periods.

The first period consists of atoms in which the electron shell consists of one energy level, in the second period - of two, in the third - of three, in the fourth - of four, etc. Each new period begins when a new energy level begins to fill level. In the periodic system, each period begins with elements whose atoms have one electron at the outer level - alkali metal atoms - and ends with elements whose atoms at the outer level have 2 (in the first period) or 8 electrons (in all subsequent ones) - noble gas atoms . The outer electron shells are similar for atoms of elements (Li, Na, K, Rb, Cs); (Be, Mg, Ca, Sr); (F, Cl, Br, I); (He, Ne, Ag, Kr, Xe), etc. That is why each of the above groups of elements is in a certain main subgroup of the periodic table: Li, Na, K, Rb, Cs in group I, F, Cl, Br, I - in VII, etc. It is due to the similarity in the structure of the electron shells of atoms that their physical and chemical properties are similar. The number of main subgroups is determined by the maximum number of elements at the energy level and is equal to 8. The number of transition elements (elements of side subgroups) is determined by the maximum number of electrons at the d-sublevel and is equal to 10 in each of the large periods. Since in the periodic system of chemical elements of Mendeleev one of the side subgroups contains three transition elements at once, similar in chemical properties (the so-called triads Fe-Co-Ni, Ru-Rh-Pd, Os-Ir-Pt), then the number of side subgroups, so as well as the main ones, it is equal to 8. By analogy with the transition elements, the number of lanthanides and actinides taken out at the bottom of the periodic system in the form of independent rows is equal to the maximum number of electrons at the f-sublevel, i.e. 14. The period begins with an element in whose atom at the outer level is one s-electron: in the first period it is hydrogen, in the rest - alkali metals. The period ends with a noble gas: the first is helium (1s2), the remaining periods are elements whose atoms at the outer level have an electronic configuration ns2np6. The first period contains two elements: hydrogen (Z=1) and helium (Z=2). The second period begins with the element lithium (Z= 3) and ends with neon (Z = 10). There are eight elements in the second period. The third period begins with sodium (Z= 11), whose electronic configuration is 1s22s22p63s1. The filling of the third energy level began with him. It ends at the inert gas argon (Z = 18), whose 3s and 3p sublevels are completely filled. Electronic formula of argon: 1s22s22p6Зs23p6. Sodium is an analogue of lithium, neon argon. In the third period, as in the second, there are eight elements. The fourth period begins with potassium (Z = 19), the electronic structure of which is expressed by the formula 1s22s22p63s23p64s1. Its 19th electron occupied the 4s sublevel, the energy of which is lower than the energy of the 3d sublevel. The outer 4s electron gives the element properties similar to those of sodium. In calcium (Z = 20), the 4s sublevel is filled with two electrons: 1s22s22p63s23p64s2. The element scandium (Z = 21) begins filling the 3d sublevel, since it is energetically more favorable than the 4p sublevel. Five orbitals of the 3d sublevel can be occupied by ten electrons, which occurs in atoms from scandium to zinc (Z = 30). Therefore, the electronic structure of Sc corresponds to the formula 1s22s22p63s23p63d14s2, and zinc - 1s22s22p63s23p63d104s2. In the atoms of subsequent elements, up to the inert gas krypton (Z=36), the 4p sublevel is filled. There are 18 elements in the fourth period. The fifth period contains elements from rubidium (Z = 37) to the inert gas xenon (Z = 54). The filling of their energy levels is the same as for the elements of the fourth period: after Rb and Sr, ten elements from yttrium (Z = 39) to cadmium (Z=48), the 4d sublevel is filled, after which the electrons occupy the 5p sublevel. In the fifth period, as in the fourth, there are 18 elements. In atoms of elements of the sixth period of cesium (Z = 55) and barium (Z = 56), the 6s sublevel is filled. In lanthanum (Z = 57), one electron enters the 5d sublevel, after which the filling of this sublevel stops, and the 4f level begins to fill, seven orbitals of which can be occupied by 14 electrons. This occurs for atoms of the lanthanide elements with Z = 58 - 71. Since the deep 4f sublevel of the third level from the outside is filled in these elements, they have very similar chemical properties. With hafnium (Z = 72), the filling of the d-sublevel resumes and ends with mercury (Z = 80), after which the electrons fill the 6p-sublevel. The filling of the level is completed at the noble gas radon (Z= 86). There are 32 elements in the sixth period. The seventh period is incomplete. The filling of electronic levels with electrons is similar to the sixth period. After filling the 7s sublevel in France (Z = 87) and radium (Z = 88), an actinium electron enters the 6d sublevel, after which the 5f sublevel begins to be filled with 14 electrons. This occurs for atoms of actinide elements with Z = 90 - 103. After the 103rd element, the b d-sublevel is filled: in kurchatovium (Z = 104), nilsborium (Z = 105), elements Z = 106 and Z = 107. The actinides, like the lanthanides, share many of the same chemical properties. Although the 3d sublevel is filled after the 4s sublevel, it is placed earlier in the formula, since all sublevels of this level are written sequentially. Depending on which sublevel is last filled with electrons, all elements are divided into four types (families). 1. s-elements: the s-sublevel of the outer level is filled with electrons. These include the first two elements of each period. 2. p-elements: the p-sublevel of the outer level is filled with electrons. These are the last 6 elements of each period (except the first and seventh). 3. d-elements: the d-sublevel of the second level from the outside is filled with electrons, and one or two electrons remain at the outer level (for Pd - zero). These include elements of intercalary decades of large periods located between s- and p-elements (they are also called transitional elements). 4. f-Elements: the f-sublevel of the third level from the outside is filled with electrons, and two electrons remain at the outer level. These are the lanthanides and actinides. There are 14 s-elements, 30 p-elements, 35 d-elements, 28 f-elements in the periodic system. Elements of the same type have a number of common chemical properties.

Consider the characteristics of the chemical element-metal by its position in the periodic system using the example of lithium.

Lithium is an element of the 2nd period of the main subgroup of group I of the periodic system of D. I. Mendeleev, an element of IA or a subgroup of alkali metals.

The structure of the lithium atom can be reflected as follows: 3Li - 2ē, 1ē. Lithium atoms will exhibit strong reducing properties: they will easily give up their only outer electron and, as a result, will receive an oxidation state (s. o.) +1. These properties of lithium atoms will be less pronounced than those of sodium atoms, which is associated with an increase in the atomic radii: Rat (Li)< Rат (Na). Восстановительные свойства атомов лития выражены сильнее, чем у бериллия, что связано и с числом внешних электронов, и с расстоянием от ядра до внешнего уровня.

Lithium is a simple substance, it is a metal, and, therefore, it has a metallic crystal lattice and a metallic chemical bond. The charge of the lithium ion: not Li + 1 (as the s. o. indicate), but Li +. General physical properties of metals arising from their crystalline structure: electrical and thermal conductivity, malleability, ductility, metallic luster, etc.

Lithium forms an oxide with the formula Li2O - it is a salt-forming, basic oxide. This compound is formed due to the ionic chemical bond Li2 + O2-, interact with water, forming an alkali.

Lithium hydroxide has the formula LiOH. This base is alkali. Chemical properties: interaction with acids, acid oxides and salts.

In the subgroup of alkali metals, there is no general formula "Volatile hydrogen compounds". These metals do not form volatile hydrogen compounds. Compounds of metals with hydrogen are binary compounds of the ionic type with the formula M+H-.

Characterization of chemical elements based on their position in the Periodic system

Practical work report 4.

Student______________________________________________________________________

Group_______

Objective:

_______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

1.Item:_____________________________________________________

2. Position in the Periodic system:

2.1. Item No.____

2.2. Period number____

2.3. Group No.____

2.4. Subgroup____

3. The composition of the atom:

3.1. Core charge_____

3.2. Number protons in the core ____

3.3. Number neutrons in the core ____

3.4. Total number electrons in the electronic shell _____

3.5. Number of Energy Levels_____

3.6. Number valence electrons _____

3.7. The number of electrons in the outer Energy Level_____

4. Distribution of electrons by Energy Levels:

4.1. Graphic scheme:

4.2. Electronic formula: ______________________________________________

5. Valence possibilities: _______________

6. Class of chemical element:______________

7. Class of a simple substance: ________________

8. Formulas and character of higher oxide and hydroxide:

8.1. Oxide:___________________________________

8.2. Hydroxide:_________________________________

Question 1.

A) Characteristics of phosphorus.

1. Phosphorus - an element of the fifth group and the third period, Z = 15,

Accordingly, the phosphorus atom contains 15 protons in the nucleus, 16

neutrons and 15 electrons. The structure of its electron shell

It can be shown using the following diagram:

15Р 2ё; 8th; 5th.

Phosphorus atoms exhibit both oxidizing properties (they accept the three electrons missing to complete the external level, while obtaining an oxidation state of -3, for example, in compounds with less electronegative elements - metals, hydrogen, etc.) and reducing properties (give 3 or 5 electrons to more electronegative elements - oxygen, halogens, etc., while acquiring oxidation states +3 and +5.)

Phosphorus is a less powerful oxidizing agent than nitrogen, but stronger than arsenic, which is associated with an increase in the atomic radii from nitrogen to arsenic. For the same reason, the restorative properties, on the contrary, are enhanced.

2. Phosphorus is a simple substance, a typical non-metal. Phosphorus is characterized by the phenomenon of allotropy. For example, there are allotropic modifications of phosphorus, such as white, red and black phosphorus, which have different chemical and physical properties. 3. The non-metallic properties of phosphorus are less pronounced than those of nitrogen, but stronger than those of arsenic (adjacent elements in the group).

4. The non-metallic properties of phosphorus are more pronounced than those of silicon, but weaker than those of sulfur (adjacent elements in the period). 5. The highest phosphorus oxide has the formula P205. It's an acid oxide. It exhibits all the typical properties of acidic oxides. So, for example, when it interacts with water, phosphoric acid is obtained.

P205 + ZN20 \u003d * 2H3P04.

When it interacts with basic oxides and bases, it gives salts.

Р205 + 3MgO = Mg3(P04)2; P205 + 6KOH = 2K3P04 + ZN20.

6. Higher phosphorus hydroxide - phosphoric acid H3P04,

The solution of which exhibits all the typical properties of acids:

Interaction with bases and basic oxides:

H3P04 + 3NaOH = Na3P04 + ZH20. 2H3P04 + 3CaO = Ca3(P04)2i + 3H20.

7. Phosphorus forms a volatile compound H3P - phosphine.

B) Characteristics of potassium.

1. Potassium has serial number 19, Z = 19 and relative

Atomic mass A, (K) \u003d 39. Accordingly, the charge of the nucleus of its atom is +19

(equal to the number of protons). Therefore, the number of neutrons in the nucleus

Equal to 20. Since the atom is electrically neutral, the number of electrons

The element potassium is in the fourth period of the periodic system, which means that all electrons are located in four energy levels. Thus, the structure of the potassium atom is written as follows:

19K: 2e; 8th; 8th; 1st.

Based on the structure of the atom, it is possible to predict the degree of oxidation of potassium in its compounds. Since in chemical reactions, the potassium atom gives up one external electron, exhibiting reducing properties, therefore, it acquires an oxidation state of +1.

The reducing properties of potassium are more pronounced than those of sodium, but weaker than those of rubidium, which is associated with an increase in the radii from Na to Rb.

2. Potassium is a simple substance, it is characterized by a metallic

Crystal lattice and metallic chemical bond, and

Hence - and all the properties typical of metals.

3. The metallic properties of potassium are more pronounced than those of sodium, but weaker than those of rubidium, since the potassium atom gives up an electron more easily than a sodium atom, but more difficult than a rubidium atom.

4. The metallic properties of potassium are more pronounced than those of calcium, since one electron of the potassium atom is easier to tear off than two electrons of the calcium atom.

5. Potassium oxide KrO is a basic oxide and exhibits all the typical properties of basic oxides. Interaction with acids and acid oxides.

K20 + 2HC1 = 2KS1 + H20; K20 + S03 = K2S04.

6. Potassium hydroxide corresponds to the base (alkali) KOH, which exhibits all the characteristic properties of bases: interaction with acids and acid oxides.

KOH + HNO3 = KN03 + H20; 2KOH+N205 = 2KN03+H20.

7. Potassium does not form a volatile hydrogen compound, but forms potassium hydride KH.

Question 2.

A) MgO - basic oxide, S03 - acid oxide.

1) MgO + S03 = MgS04;

2) MgO + 2HNO3 = Mg(N03)2 + H20;

3) MgO + 2H+ = Mg + + H20; 2RbOH + S03 = Rb2S04 + H20; S03+20RG=S04~+H20.

B) Mg (OH) 2 - basic hydroxide, H2SO4 - acid hydroxide.

1) Mg(OH)2 + H2SO4 = MgSO4 + 2H20; OH~ + H+ = H20;

2) Mg(OH)2 + S03 = MgS04 + H20; S03 + 20RG = H20 + S04"; I) H2b04 + Na20 = Na2b04 + H20; Na20 + 2H = 2Na + H20.

Question 3.

Magnesium is a simple substance, it is characterized by a metallic crystal lattice; it has a metallic luster, electrical conductivity.

A) 2Mg + 02 = 2MgO

6) Mg + Cl2 = MgCl2 Mg°-2e = Mg2+ 1

Question 4.

Allotropy is the phenomenon of the existence of a chemical element in

The form of several simple substances, different in structure and

Properties (so-called allotropic forms).

A) In the molecules of the composition S8, the covalent-nonpolar type is realized

Bonds (i.e., there is no displacement of the electron pair forming

B) In the molecules of the composition H2S, a covalent-polar type of bond is realized, since the electron pair is shifted to a more electronegative atom - sulfur (S).

H->S<- Н

Physical properties of rhombic sulfur (S8):

The lemon-yellow substance, stable up to t = 95.6°C, dissolves in carbon disulfide (CS2), aniline, benzene, phenol. Reaction equations:

A) 2Na + S = Na2S

Reducing agent

Ca ° -2 " = eCa2 +

B) S2Al + 3l \u003d Al2S3 A1 ° -Ze \u003d A12

E) S + 3F2 = SF6 6

1 - reducing agent 1 - oxidizing agent

Reductant 1 - oxidizer

1 - reducing agent 3 - oxidizing agent

Question 5.

The non-metallic properties of silicon are less pronounced than those of phosphorus, but stronger than those of aluminum. .

Question 6.

A) Nitrogen is more acidic than phosphorus.

Since in groups from top to bottom there is an increase in the main and

Weakening of acidic properties.

B) Sulfur has more acidic properties than phosphorus,

Since in periods from left to right there is an increase in acid and

Weakening of basic properties.

Question 7.Given: Ti(0 2) = 0.2; m(Mg) = 0.12g; Hum8 (impurities) = 2%. Find: V (air)

R^nie: 1. Find the mass of magnesium without impurities: pure (Me) \u003d m (Me) - m (Me) - 10m0 (impurities); pure (Mg) \u003d 0.121-0.12g-0.02 \u003d 0.1176g.

2. Let's write down the reaction equation for burning magnesium. OD 176

2Mg + 02 = 2MgO,

Y=2mol; y=1mol;

M = 24 g/mol; Vv = 22.4 l/mol;

We will make up the proportion to the reaction equation: 48 g -22.4 l 0.1176 g -chl

X \u003d ° "1176" 22 "4 \u003d 0.05488 l. 48

Therefore, 0.05488 liters of pure oxygen is required for combustion

0.1176 g of magnesium.

3. Find the volume of air required to burn magnesium:

Y(air) = X(°ll = 0,05488 = 0.2744 l.

Answer: U (air) \u003d 0.2744 l.

Question 8.

Given: m(S) = 1.6 kg -1600 g.

Find: V(S02)

Solution: 1. Let's write the reaction equation for the combustion of sulfur in oxygen.

1 £(\(\ Г VTT

slide 2

1. Characteristics of a non-metal using nitrogen as an example

Position of N in the Periodic system and the structure of its atom a) Position of N in the Periodic system N serial number - 7 2 (small) period, V group, main subgroup

slide 3

b) The composition of the atom P+ = 7 (serial number) ē = P+ = 7 n0 = Ar - № = 14-7=7

slide 4

c) the structure of the atom N: The number of energy levels = the number of the period = 2 The number ē at the last level = the number of the group in which the element is located, i.e. 5. N+7)) 1s2 2s2 2p3 2 5 2 2 3

slide 5

The nitrogen atom has 5 electrons on the outer electron layer, 3 electrons (8-5) are missing until completion, the nitrogen atom can both accept and donate electrons in chemical reactions, showing both oxidizing and reducing properties. N0 + 3 ē → N-3(reduction, oxidizer) N0 - 5ē→N+5(oxidation, reducing agent)

slide 6

Electronegativity - the ability of atoms of chemical elements to pull the electrons of atoms towards themselves. The most electronegative element is F, then O, then N. Nitrogen is the third most electronegative element.

Slide 7

Slide 8

2. Comparison of the properties of the nitrogen atom with the properties of atoms - neighbors in the group and period

R at (N) R at (N) > R at (O) Nitrogen atoms exhibit stronger oxidizing properties, tk. have: a) less R at than C atoms b) and a large number of ē But nitrogen is a less powerful oxidizing agent than oxygen.

Slide 9

3. Simple substance nitrogen - N2 - non-metal

N2- k.n.p., gas. The non-metallic properties of the simple substance nitrogen are more pronounced than those of phosphorus. The non-metallic properties of the simple substance nitrogen are more pronounced than those of carbon, but weaker than those of the simple substance oxygen.

Slide 10

4. Highest oxide - N2O5

Acid. Reacts with bases, basic oxides, and water

slide 11

N2O5 + 2NaOH = 2NaNO3 + H2O – exchange r. N2O5 + 2Na+ + 2OH- = 2Na+ + 2NO3- + H2O compounds N2O5 + H2O = 2HNO3 - r. connections

slide 12

5. Higher hydroxide - HNO3 - acid

Reacts with Bases Basic oxides Metal salts

slide 13

2HNO3 + Cu(OH)2 = Cu(NO3)2 + 2H2O - r. exchange, 2HNO3 + СaO = Ca(NO3)2 + H2O - r. exchange 2HNO3 + Na2SiO3 = 2NaNO3 + H2SiO3 ↓ - р. exchange

Slide 14

6. NH3 - volatile hydrogen compound

slide 15

Genetic series of nitrogen

N2→ N2O5 → HNO3 → NaNO3

slide 16

Consolidation of knowledge. Testing

1. The charge of the nucleus of a nitrogen atom is equal to the number of a) protons b) electrons in the outer electron layer c) neutrons d) energy levels

Patterns of changes in some properties of chemical elements in PS. Characteristic Within a period Within one group (for elements of the main subgroups) The charge of the atomic nucleus Increases The number of energy levels Does not change Increases The number of electrons at the external energy level Increases Does not change The radius of the atom Decreases Increases Electronegativity Increases Decreases Reducing properties Decrease Increase Metallic properties Decrease Increase

Sodium Chlorine Nuclear charge Number of nucleonsp=11, n=12p=17,n=18 Number of electronse=11E=17 Number of energy levels 33 Electronic formula 1s 2 2s 2 2p 6 3s 1 1s 2 2s 2 2p 6 3s 2 3p 5 Highest degree oxidation + 1 + 7 Redox properties Reducing agent Oxidizing agent 1. The position of the element in PS and the structure of its atom

Sodium Chlorine Sodium oxide Na2O exhibits basic properties. It corresponds to the base NaOH. Na 2 O + H 2 O \u003d 2NaOH Na 2 O + 2HCl \u003d 2NaCl + H 2 O Na 2 O + SO 3 \u003d Na 2 SO 4 Higher chlorine oxide Cl2O7 is an acid oxide. It corresponds to the acid HClO4. Cl 2 O 7 + H 2 O \u003d 2HClO 4 Cl 2 O 7 + Na 2 O \u003d 2NaClO 4 Cl 2 O 7 + 2NaOH \u003d 2NaClO 4 + H 2 O

Sodium Chlorine Sodium hydroxide, NaOH, is a strong base and exhibits the properties of a base. NaOH + HCl = NaCl + H2O 2NaOH + CO2 = Na2CO3 + H2O 2NaOH + CuCl2 = Cu(OH)2 + 2NaCl Perchloric acid HClO4 exhibits the properties of a strong acid. HClO2 + KOH = KClO4 + H2O

(from other Greek αλλος “another”, τροπος “turn, property”) the existence of the same chemical element in the form of two or more simple substances, different in structure and properties of the so-called allotropic modifications or allotropic forms. Greek chemical element of simple substances

First level

Option 1

1. The reaction equation for the neutralization of sodium hydroxide with hydrochloric acid is given:
NaOH + HCl = NaCl + H20 + Q.

thermal effect;
participation of a catalyst;
direction.
Consider this chemical reaction from the point of view of the theory of electrolytic dissociation. Write down the full and abbreviated ionic equations.
NaOH + HCl = NaCl + H2O + Q
Starting materials: 1 mol of sodium hydroxide solid (1 sodium atom, 1 hydrogen atom, 1 oxygen atom), 1 mol of hydrochloric acid (1 hydrogen atom, 1 chlorine atom).
Reaction products: 1 mol of sodium chloride solid (1 sodium atom, 1 chlorine atom), 1 mol of water (1 oxygen atom, 2 hydrogen atom).
The reaction is exothermic
The starting materials and products are in solution.
without catalyst

irreversible reaction
Na+ + OH- + H+ + Cl- = Na+ + Cl- + H2O
OH- + H+ = H2O

2. Give a description of the chemical element magnesium according to the plan:
the position of the element in the PSCE;
the structure of the atom;

Magnesium -- Mg
Ordinal number Z=12; mass number A = 24, nuclear charge + 12, number of protons = 12, neutrons (N = A-Z = 12) 24 - 12 = 12 neutrons, electrons = 12, period - 3, energy levels - 3,
The structure of the electron shell: 12 M g 2e; 8e; 2e.
12 M g)))
2 8 2
+2 oxidation state;
The reduction properties of magnesium are more pronounced than those of beryllium, but weaker than those of calcium, which is associated with an increase in the radii of the atoms Be - M g - Ca;
Magnesium ion M g 2+
MgO - magnesium oxide is the main oxide and exhibits all the characteristic properties of oxides. Magnesium forms hydroxide Mg (OH) 2, which exhibits all the characteristic properties of bases.

3. Write the equations for the reactions of magnesium oxide and hydroxide with hydrochloric acid in molecular and ionic form.
MgO+2HCl=MgCl₂ + H₂O
MgO+2H+=Mg2+ + H₂O
Mg(OH)2+2HCl= MgCl₂ + 2H₂O
Mg(OH)2+2H+= Mg2+ + 2H₂O

Option 2

1. The scheme of aluminum combustion reaction is given.
Al + 02 → A1203 + Q.

Describe the following responses:
the number and composition of the starting materials and reaction products;
thermal effect;
aggregate state of substances;
participation of a catalyst;
change in the oxidation states of elements;
direction.
0 0 +3 –2
Al + O2 = Al2O3+Q
4Al + 3O2 = 2Al2O3
Aluminum is a reducing agent and oxygen is an oxidizing agent.
Starting materials: 4 moles of aluminum, 3 moles of oxygen (3 molecules of 2 oxygen atoms). Reaction product: 2 moles of aluminum oxide (2 aluminum atoms, 3 oxygen atoms in one molecule).
The reaction is exothermic.
Aluminum - solid, oxygen - g., aluminum oxide - solid.
Without the participation of a catalyst

Irreversible.

2. Give a description of the chemical element sodium according to the plan:
the position of the element in the PSCE;
the structure of the atom;
formulas of oxide and hydroxide, their character.
Sodium -- Na

11 Na)))
2 8 1
+1 oxidation state;

Sodium ion Na+

3. Write the equations for the reactions of sodium oxide and hydroxide with a solution of sulfuric acid in molecular and ionic form.
2NaOH+H2SO4=2H2O+Na2SO4
2OH-+2H+=2H2O
Na2O+H2SO4=H2O+Na2SO4
Na2O+2H+=H2O+2Na+

Option 3

1. The reaction scheme for obtaining sulfur oxide (VI) from sulfur oxide (IV) is given.
S02 + 02  S03 + Q.
Write an equation for this reaction by placing the coefficients in it using the electronic balance method. Specify the oxidizing agent and reducing agent.
Describe the following responses:
the number and composition of the starting materials and reaction products;
thermal effect;
aggregate state of substances;
participation of a catalyst;
change in the oxidation states of elements;
direction.
2S+4O2 + O02 = 2S+6O-23+ Q
S+4 -2e →S+6 reducing agent
O02 +4e→2O-2 oxidant
The initial substances are 2 mol of sulfur oxide 4 (in one molecule 1 sulfur atom, 2 oxygen atoms) and 1 mol of oxygen (in one molecule 2 oxygen atoms).
The reaction product is 2 mol of sulfur oxide 6 (one molecule contains 1 sulfur atom, 3 oxygen atoms)
The reaction is exothermic.
Sulfur oxide 4 and oxygen - gases, Sulfur oxide (VI) liquid
with catalyst

Reversible.

2. Give a description of the chemical element lithium according to the plan:
the structure of the atom;
formulas of oxide and hydroxide, their character.
lithium Li
Ordinal number Z=3; mass number A \u003d 7, nuclear charge + 3, number of protons \u003d 3, neutrons (N \u003d A-Z \u003d 4) 7 - 3 \u003d 4 neutrons, electrons \u003d 3, period - 2, energy levels - 2
The structure of the electron shell: 3 Li 2е; 1e.
3Li))
2 1
+1 oxidation state;
The reducing properties of lithium are less pronounced than those of sodium and potassium, which is associated with an increase in the atomic radii;
Lithium ion Li+
Li 2O - lithium oxide is the main oxide and exhibits all the characteristic properties of oxides. Lithium Li forms hydroxide Li OH (alkali), which exhibits all the characteristic properties of bases.

3. Write the equations for the reactions of lithium oxide and hydroxide with sulfuric acid in molecular and ionic form.
2 LiOH+H2SO4=2H2O+ Li2SO4
2OH-+2H+=2H2O
Li2O+H2SO4=H2O+ Li2SO4
Li2O+2H+=H2O+2Li +

Option 4

1. The equation for the reaction of zinc with hydrochloric acid is given:
Zn + 2HCl = ZnCl2 + H2 + Q.
Describe the following responses:
the number and composition of the starting materials and reaction products;
thermal effect;
aggregate state of the substances involved in the reaction;
participation of a catalyst;
change in the oxidation states of chemical elements;
direction.
Consider this chemical reaction from the point of view of the theory of electrolytic dissociation: write down the full and reduced ionic equations.
2HCl+Zn=ZnCl2+H2 + Q
Starting materials: 1 mol zinc, 2 mol hydrochloric acid (1 hydrogen atom, 1 chlorine atom per molecule). Reaction products: 1 mol of zinc chloride (1 zinc atom, 2 chlorine atoms in FE), 1 mol of hydrogen (2 hydrogen atoms).
exothermic reaction
Zinc - TV., Hydrochloric acid - well., Zinc chloride TV. (solution), hydrogen - g.
without catalyst
With a change in oxidation states
irreversible
2H++2Cl-+Zn0=Zn2++2Cl-+H20
2H++Zn0=Zn2++H20

2. Give a description of the chemical element calcium according to the plan:
the position of the element in the Periodic system;
the structure of the atom;
formulas of higher oxide and hydroxide, their character.
Calcium Ca
Ordinal number Z=20; mass number A \u003d 40, nuclear charge + 20, number of protons \u003d 20, neutrons (N \u003d A-Z \u003d 20) 40 - 20 \u003d 20 neutrons, electrons \u003d 20, period - 4, energy levels - 4,
The structure of the electron shell: 20 M g 2e; 8e; 8e; 2e.
20 Ca))))
2 8 8 2
+2 oxidation state;
The reducing properties of calcium are more pronounced than those of magnesium, but weaker than those of strontium, which is associated with an increase in the atomic radii
Calcium ion Ca 2+
CaO - calcium oxide is the main oxide and exhibits all the characteristic properties of oxides. Calcium forms hydroxide Ca (OH) 2, which exhibits all the characteristic properties of bases.

3. Write the equations for the reactions of calcium oxide and hydroxide with nitric acid in molecular and ionic form.
CaO + 2HNO3 \u003d Ca (NO3) ₂ + H ₂ O
CaO + 2H + \u003d Ca 2+ + H₂O
Ca(OH)2+2HNO3= Ca(NO3)₂ + 2H₂O
Ca (OH) 2 + 2H + \u003d Ca 2+ + 2H₂O

Second level

Option 1

1. The reaction equation for the production of nitric oxide (II) is given:
N2 + 02 2NO - Q.


N20 + O20 2N+2O-2 - Q
N20 - 2 * 2e \u003d 2N + 2 reducing agent
O20 + 2 * 2e \u003d 2O-2 oxidizing agent
Starting materials: nitrogen 1 mol, 2 N atoms, oxygen 1 mol (2 O atoms).
Reaction product: 2 mol of nitric oxide 2 (in the molecule 1 nitrogen atom and 1 oxygen atom).
The starting materials and reaction products are gases.
The reaction is endothermic.
Reversible.
Without catalyst.
With a change in oxidation states.

6 C))
2 4
+4 oxidation state;

3. Make formulas for higher carbon monoxide and hydroxide, indicate their nature.
CO2 + H2O ↔ H2CO3
CO2 + H2O ↔ 2H+ + CO32-
Na2O + CO2 → Na2CO3
Na2O + CO2 → 2Na+ + CO32-
2NaOH + CO2 → Na2CO3 + H2O
OH- + CO2 → CO32- + H2O
Ca(OH)2 + CO2 → CaCO3 ↓+ H2O

H2CO3 + Ca = CaCO3 + H2
2H+ +CO32- + Ca = CaCO3 ↓+ H2
H2CO3 + CaO = CaCO3 ↓+ H2O

H2CO3 + 2NaOH = Na2CO3 + 2H2O

2H+ +OH- = 2H2O

Option 2

1. The reaction equation for the synthesis of ammonia is given:
N2 + 3H2  2NH3 + Q.
Give a description of the reaction according to all the classification features you have studied.
Consider this reaction in terms of OVR. Specify the oxidizing agent and reducing agent.
3H2 + N2 2NH3 + Q
N20 +2*3е→2N-3 oxidizer
H20 -2*1e→2H+1 reducing agent
Starting substances: 1 mol of nitrogen (a molecule of 2 nitrogen atoms), 3 mol of hydrogen (a molecule of 2 hydrogen atoms). The reaction product is ammonia, 2 mol. Molecule of 1 nitrogen atom and 2 hydrogen atoms. The starting materials are the products of the reaction - gases.
Reaction:
exothermic.
Redox.
Straight.
catalytic.
Reversible.

2. Give a description of the chemical element sulfur according to its position in the Periodic system.
Sulfur - S
Serial number Z=16 and mass number A= 32, nuclear charge + 16, number of protons = 16, neutrons (N= A-Z= 12) 32 - 16=16 neutrons, electrons = 16, period - 3, energy levels - 3
16S)))
The structure of the electron shell: 16 S 2e; 8e; 6e.
16S)))
2 8 6
Oxidation state - (-2) and (+ 2; +4; +6)
The oxidizing properties of sulfur are more pronounced than those of selenium, but weaker than those of oxygen, which is associated with an increase in the atomic radii from oxygen to selenium
SO 3 - sulfur oxide is an acidic oxide and exhibits all the characteristic properties of oxides.
Sulfur forms hydroxide H2SO4, which exhibits all the characteristic properties of acids.
Sulfur from hydrogen compounds forms H2S.

3. Make formulas for higher oxide and sulfur hydroxide, indicate their nature. Write the equations of all reactions characteristic of these substances in ionic and molecular forms.
SO3 + H2O → H2SO4
2NaOH + SO3 → Na2SO4 + H2O
2OH- + SO3 → SO42- + H2O
Na2O + SO3 → Na2SO4
Na2O + SO3 → 2Na+ +SO42-
Zn0 + H2+1SO4(razb) → Zn+2SO4 + H20
Zn0 + 2H+ → Zn2+ + H20
CuO + H2SO4 → CuSO4 + H2O
CuO + 2H+ → Cu2+ + H2O
H2SO4 + 2NaOH → Na2SO4 + 2H2O (neutralization reaction)
H+ + OH- → H2O
H2SO4 + Cu(OH)2 → CuSO4 + 2H2O
2H+ + Cu(OH)2 → Cu2+ + 2H2O
BaCl2 + H2SO4 → BaSO4↓ + 2HCl
Ba2+ + SO42- → BaSO4↓
MgCO3 + H2SO4 → MgSO4 + H2O + CO2
MgCO3 + 2H+ → Mg2+ + H2O + CO2¬

Option 3

1. The equation for the reaction of copper (II) chloride with sodium hydroxide is given:
CuCl2 + 2NaOH = Cu(OH)2↓ + 2NaCl.
Give a description of the reaction according to all the classification features you have studied.
Consider the reaction from the point of view of TED: write down the full and reduced ionic equations.
CuCl2 + 2NaOH = Cu(OH)2↓ + 2NaCl
Cu2+ + 2OH- = Cu(OH)2↓
Starting materials: 1 mol of copper chloride (1 copper atom, 2 chlorine atoms), 2 mol of sodium hydroxide (1 sodium atom, 1 oxygen atom, 1 hydrogen atom in FE).
Reaction products: 1 mol of copper hydroxide (1 copper atom, 2 oxygen atoms, 2 hydrogen atoms), 2 mol sodium chloride (1 sodium atom, 1 chlorine atom in FE).
The reaction products and starting materials are solid dissolved. Cu(OH)2 is a solid precipitate.
Reaction:
exothermic
No change in oxidation states
Straight
Without the participation of a catalyst
Irreversible.

2. Give a description of the chemical element phosphorus according to its position in the Periodic system of D. I. Mendeleev.
Characteristic P (phosphorus)
Atomic mass \u003d 31. The charge of the nucleus of the atom is P + 15, t. There are 15 protons in the nucleus. Scheme:
15R 2e) 8e) 5e)

3. Make formulas for higher oxide and phosphorus hydroxide, indicate their nature. Write the equations of all reactions characteristic of these substances in ionic and molecular forms.
P2O5 + 3H2O = 2H3PO4
P2O5 + 3H2O = 6H+ +2PO43-
3CaO + P2O5 = Ca3(PO4)2




6H++ 3CO3 2-= 3H2O + 3CO2
3NaOH + H3PO4 = Na3PO4 + 3H2O
3OH- + 3H+= 3H2O

Option 4

1. The equation for the reaction of potassium carbonate with hydrochloric acid is given:
K2CO3 + 2HCl = 2KCl + CO2 + H20.
Give a description of the reaction according to all the classification features you have studied.
Consider this reaction from the point of view of TED: write down the full and reduced ionic equations.
K2CO3 + 2HCl = 2KCl + H2O + CO2
2K+ +CO32- + 2H+ + 2Cl-= 2K+ 2Cl-+ H2O + CO2
CO32- + 2H+= H2O + CO2
Starting materials: 1 mol of potassium carbonate (2 potassium atoms, 1 carbon atom, 3 oxygen atoms) solid, 2 mol of hydrochloric acid (1 hydrogen atom, 1 chlorine atom in a molecule) liquid.
Reaction products: 2 mol of potassium chloride (in FE 1 potassium atom, 1 chlorine atom) solid, 1 mol of water (2 volumes of hydrogen, 1 oxygen atom) liquid, 1 mol of carbon dioxide (1 carbon atom, 2 oxygen atoms) - gas.
Reaction:
exothermic.
No change in oxidation states.
Straight.
Without the participation of a catalyst.
Irreversible.

2. Give a description of the chemical element nitrogen according to its position in the Periodic system.
Nitrogen N - non-metal, period II (small), group V, main subgroup.
Atomic mass = 14, nuclear charge - +7, number of energy levels = 2
p=7, e=7,n=Ar-p=14-7=7.
The structure of the electron shell: 7 N 2e; 5e
7 N))
2 5
+5 oxidation state;
The oxidizing properties are more pronounced than those of carbon, but weaker than those of oxygen, which is associated with an increase in the charge of the nucleus.
N2O5 nitric oxide is an acidic oxide and exhibits all the characteristic properties of oxides. Nitrogen forms the acid HNO3, which exhibits all the characteristic properties of acids.
Volatile hydrogen compound - NH3

3. Make the formulas of the higher nitrogen oxide and hydroxide, indicate their nature.
Write the equations of all reactions characteristic of these substances in ionic and molecular forms.
N2O5 + H2O = 2НNO3
N2O5 + H2O = 2H+ + NO3-
N2O5 + BaO = Ba(NO3)2
N2O5 + BaO = Ba2+ +2NO3-
N2O5 + 2KOH (solution) = 2KNO3 + H2O
N2O5 + 2K+ +2OH- = 2K+ +NO32- + H2O
N2O5 + 2OH- = NO32- + H2O
K2O + 2HNO3 → 2KNO3 + H2O
K2O + 2H+ + 2NO3- → 2K+ + 2NO3- + H2O
K2O + 2H+ → 2K+ + H2O
HNO3 + NaOH → NaNO3 + H2O
H+ + NO3- + Na+ + OH- → Na+ + NO3- + H2O
H+ + OH- → H2O
2HNO3 + Na2CO3 → 2NaNO3 + H2O + CO2¬
2H+ + 2NO3- + 2Na+ + СO32- → 2Na+ + 2NO3- + H2O + CO2¬
2H+ + CO32- → H2O + CO2¬
S0 + 6HNO3(conc) → H2S+6O4 + 6NO2 + 2H2O
B0 + 3HNO3 → H3B+3O3 + 3NO2
3P0 + 5HNO3 + 2H2O → 5NO + 3H3P+5O4
From razb.
4Zn + 9HNO3 = NH3 + 4Zn(NO3)2 + 3H2O
4Zn + 9H+ + 9NO3- = NH3 + 4Zn2+ + 8NO3- + 3H2O
3Cu + 8HNO3 = 2NO + 3Cu(NO3)2+ 4H2O
3Cu + 8H+ +8NO3-= 2NO + 3Cu2+ +6NO3-+ 4H2O
conc.
Zn + 4HNO3 = 2NO2 + 2H2O + Zn(NO3)2
Zn + 4H+ +4NO3-= 2NO2 + 2H2O + Zn2+ +2NO3-
Cu + 4HNO3 = 2NO2 + 2H2O + Cu(NO3)2
Cu + 4H+ +4NO3- = 2NO2 + 2H2O + Cu2+ +2NO3-

Third level

Option 1

1. The equation for the reaction of obtaining nitric acid is given:
4N02 + 02 + 2H20 = 4HN03 + Q.
Give a description of the reaction according to all the classification features you have studied.

4N+4O2 + О02 + 2H2O ↔ 4HN+5O-23
N+4 -1e = N+5 reducing agent
O20 +4e = 2O-2 oxidant
Reaction:
exothermic.
With a change in the oxidation state (OVR).
Without the participation of a catalyst.
Straight.
Reversible.
Source substances: 4 mol of nitric oxide 4 (1 nitrogen atom, 2 oxygen atoms in the molecule) - gas, 1 mol of oxygen (2 oxygen atoms in the molecule) - gas, 2 mol of water (1 oxygen atom, 2 hydrogen atoms in the molecule) - liquid
The reaction product - 4 moles of nitric acid (1 nitrogen atom, 1 hydrogen atom, 3 oxygen atoms in a molecule) - is a liquid.

2. Give a description of the chemical element magnesium according to its position in the Periodic system.
Magnesium - serial number in the Periodic system Z = 12 and mass number A = 24. Nuclear charge +12 (number of protons). The number of neutrons in the nucleus N \u003d A - Z \u003d 12. The number of electrons \u003d 12.
The element magnesium is in the 3rd period of the Periodic Table. The structure of the electron shell:
12 Mg)))
2 8 2

+2 oxidation state.
The reduction properties of magnesium are more pronounced than those of beryllium, but weaker than those of calcium (elements of the IIA group), which is associated with an increase in the radii of atoms upon transition from Be to Mg and Ca.
Magnesium oxide MgO is a basic oxide and exhibits all the typical properties of basic oxides. Magnesium hydroxide corresponds to the base Mg(OH)2, which exhibits all the characteristic properties of bases.

3. Make formulas for magnesium oxide and hydroxide, indicate their nature.
Write the equations of all reactions characteristic of these substances in ionic and molecular forms.
Magnesium oxide MgO is the basic oxide, the base Mg(OH)2 exhibits all the characteristic properties of bases.
MgO + H2O = Mg(OH)2
MgO + CO2 = MgCO3
MgO + CO2 = Mg2+ + CO32-
MgO + H2SO4 = MgSO4 + H2O
MgO + 2H+ = Mg2+ + H2O
Mg(OH)2 + 2HCl = MgCl2 + 2H2O
Mg(OH)2 + 2H+ = Mg2+ + 2H2O
Mg(OH)2 + CO2 = Mg2+ +CO32- + H2O
3Mg(OH)2 + 2FeCl3 = 2Fe(OH)3 + 3MgCl2
3Mg(OH)2 + 2Fe3+ = 2Fe(OH)3 + 3Mg2+
Mg(OH)2 + 2NH4Cl = MgCl2 + 2NH3 + 2H2O
Mg(OH)2 + 2NH4+= Mg2+ + 2NH3 + 2H2O
MgSO4 + 2NaOH = Mg(OH)2 + Na2SO4
Mg2+ + 2OH- = Mg(OH)2

Option 2

1. The equation for the reaction of iron with chlorine is given:
2Fe + 3Cl2 = 2FeCl3 + Q.
Give a description of the chemical reaction according to all the classification features you have studied.
Consider the reaction in terms of redox processes. Specify the oxidizing agent and reducing agent.
2Fe + 3Cl2 = 2FeCl3 + Q
2
3 Fe - 3e - = Fe + III,
Cl2 + 2e– = 2Cl–I
2Fe – 6e– = 2Fe+III,
3Cl2 + 6e– = 6Cl–I.
Fe – 3e– = Fe+III reducing agent
Cl2 + 2e– = 2Cl–I oxidant
exothermic
OVR
Straight
irreversible
Non-catalytic
Starting substances: 2 mol of iron - solid, 2 mol of chlorine (a molecule of 2 atoms) - gas
Product: 2 mol of iron chloride (from 1 iron atom, 2 chlorine atoms in FE) - tv.

2. Give a description of the chemical element sodium according to its position in the Periodic system of D. I. Mendeleev.
Sodium -- Na
Ordinal number Z=11; mass number A \u003d 23, nuclear charge + 11, number of protons \u003d 11, neutrons (N \u003d A-Z \u003d 11) 23 - 11 \u003d 12 neutrons, electrons \u003d 11, period - 3, energy levels - 3,
The structure of the electron shell: 11 Na 2е; 8e; 1e.
11 Na)))
2 8 1
+1 oxidation state;
The reducing properties of sodium are more pronounced than those of lithium, but weaker than those of potassium, which is associated with an increase in the atomic radii;
Sodium ion Na+
Na 2O - sodium oxide is the main oxide and exhibits all the characteristic properties of oxides. Sodium forms hydroxide NaOH (alkali), which exhibits all the characteristic properties of bases.

3. Make formulas for sodium oxide and hydroxide, indicate their nature. Write the equations of all reactions characteristic of these substances in ionic and molecular forms.
2NaOH+H2SO4=2H2O+Na2SO4
2OH-+2H+=2H2O
2NaOH + CO2 ---> Na2CO3 + H2O
2OH(-) + CO2 ---> CO3(2-) + H2O
2NaOH + SO2 ---> Na2SO3 + H2O
2OH(-) + SO2 ---> SO3(2-) + H2O
NaOH+ Al(OH)3 ---> Na
OH(-) + Al(OH)3 ---> Al(OH)4 (-)
Na2O+H2SO4=H2O+Na2SO4
Na2O+2H+=H2O+2Na+
Na2O + H2O ---> 2NaOH
Na2O + H2O ---> 2Na+ +2OH-
Na2O + 2HCl ----> 2NaCl + H2O
Na2O + 2H+ ----> 2Na+ + H2O
Na2O + CO2 ---> Na2CO3
Na2O + CO2 ---> 2Na++CO32-
Na2O + SO2 ---> Na2SO3
Na2O + SO2 ---> 2Na++SO32-

Option 3

1. The reaction equation for the decomposition of potassium nitrate is given:
2KN03 = 2KN02 + O2 - Q.
Give a description of the reaction according to all the classification features you have studied.
Consider the reaction in terms of redox processes. Specify the oxidizing agent and reducing agent.
2KNO3 = 2KNO2 + O2-Q
oxidizing agent: N5+ + 2e− = N=3+|2| recovery
reducing agent: O2− − 4e− = O20 |1| oxidation
Starting substances: 2 mol of potassium nitrate (in FE 1 potassium atom, 1 nitrogen atom, 3 oxygen atoms) - TV.
The reaction products - 2 mol of potassium nitrite (in FE 1 potassium atom, 1 nitrogen atom, 2 oxygen atoms) - solid, 1 mol of oxygen (2 oxygen atoms) - gas.
Endothermic
OVR
Straight
irreversible
Non-catalytic

2. Give a description of the chemical element carbon by its position in the Periodic system.
Carbon C is a chemical element of group IV of the periodic system of Mendeleev: atomic number 6, atomic mass 12.011.
Ordinal number Z=6; mass number A \u003d 12, nuclear charge + 6 number of protons \u003d 6, neutrons (N \u003d A-Z \u003d 6) 12 - 6 \u003d 6 neutrons, electrons \u003d 6, period - 2, energy levels - 2,
The structure of the electron shell: 6 C 2e; 4th
6 C))
2 4
+4 oxidation state;
The oxidizing properties of carbon are more pronounced than those of boron, but weaker than those of nitrogen, which is associated with an increase in the nuclear charge.
CO2 acid oxide, H2CO3 acid.

3. Make formulas for carbon monoxide and hydroxide, indicate their nature.
Write the equations of all reactions characteristic of these substances in ionic and molecular forms.
CO2 carbon monoxide is an acidic oxide and exhibits all the characteristic properties of oxides. Carbon forms the acid H2CO3, which exhibits all the characteristic properties of acids.
CO2 + H2O ↔ H2CO3
CO2 + H2O ↔ 2H+ + CO32-
Na2O + CO2 → Na2CO3
Na2O + CO2 → 2Na+ + CO32-
2NaOH + CO2 → Na2CO3 + H2O
OH- + CO2 → CO32- + H2O
Ca(OH)2 + CO2 → CaCO3 ↓+ H2O
Ca2+ +2OH- + CO2 → CaCO3 ↓+ H2O
H2CO3 + Ca = CaCO3 + H2
2H+ +CO32- + Ca = CaCO3 ↓+ H2
H2CO3 + CaO = CaCO3 ↓+ H2O
2H+ +CO32- + CaO = CaCO3 ↓+ H2O
H2CO3 + 2NaOH = Na2CO3 + 2H2O
2H+ + CO32- + 2Na+ +OH- = 2Na++CO32- + 2H2O
2H+ +OH- = 2H2O
Ca(OH)2 + H2CO3 → CaCO3 ↓+ 2H2O
Ca2+ +2OH- + 2H+ +CO32- → CaCO3 ↓+ 2H2O

Option 4

1. The reaction equation for the formation of iron hydroxide (III) is given:
4Fe(OH)2 + 2H20 + 02 = 4Fe(OH)3.
Give a description of the reaction according to all the classification features you have studied.
Consider the reaction in terms of redox processes. Specify the oxidizing agent and reducing agent.
4Fe(OH)2 + O2 + 2H2O = 4Fe(OH)3↓
Fe2+ ​​-1е→ Fe3+ reducing agent
O20 + 4е → 2O2- oxidant
Starting substances: 4 mol of iron hydroxide 2 (in FE 1 iron atom, 2 oxygen atoms, 2 hydrogen atoms) - solid, 1 mol of oxygen (2 oxygen atoms) - gas, 2 mol of water (2 hydrogen atoms, 1 oxygen atom in molecule) - f.
The reaction product is 4 mol of iron hydroxide 3 (in FE 1 iron atom, 3 oxygen atoms, 3 hydrogen atoms) - TV.
exothermic
OVR
Straight
irreversible
Non-catalytic.

2. Give a description of the chemical element phosphorus according to its position in the Periodic system.
Characteristic P (phosphorus)
The element with serial number 15 is in the 3rd period of the 5th group, the main subgroup.
Atomic mass \u003d 31. The charge of the nucleus of the atom is P + 15, t. There are 15 protons in the nucleus.
Scheme 15P 2e) 8e) 5e)
There are 16 neutrons in the nucleus of an atom. There are 15 electrons in an atom, since their number is equal to the number of protons and the serial number. There are 3 electron layers in the phosphorus atom, since P is in the 3rd period. There are 5 electrons on the last layer, since phosphorus is in group 5. The last layer is not completed. P-non-metal, because in the chemical. reactions with metals takes 3 electrons to complete the layer. Its oxide is Р2О5-acid. He is mutual. with H2O, bases and basic oxides. Its hydroxide is H3PO4-acid. She interacts. with metals standing up to H (hydrogen), with basic oxides, bases.

3. Make formulas for phosphorus oxide and hydroxide, indicate their nature.
Write the equations of all reactions characteristic of these substances in ionic and molecular forms.
P2O5 + 3H2O = 2H3PO4
P2O5 + 3H2O = 6H+ +2PO43-
3CaO + P2O5 = Ca3(PO4)2
3Ca(OH)2 + P2O5 = Ca3(PO4)2 + 3H2O.
3Mg + 2H3PO4 = Mg3(PO4)2↓ + 3H2
3Mg + 6H++ 2PO43- = Mg3(PO4)2↓ + 3H2
2H3PO4+3Na2CO3 = 2Na3PO4 + 3H2O + 3CO2
6H++ 3CO3 2-= 3H2O + 3CO2
3NaOH + H3PO4 = Na3PO4 + 3H2O
3OH- + 3H+= 3H2O