Task number 1

The +2 oxidation state in all compounds exhibits

Answer: 4

Explanation:

Of all the proposed options, the +2 oxidation state in complex compounds is shown only by zinc, being an element of the secondary subgroup of the second group, where the maximum oxidation state is equal to the group number.

Tin - an element of the main subgroup of group IV, a metal, exhibits oxidation states 0 (in a simple substance), +2, +4 (group number).

Phosphorus is an element of the main subgroup of the main group, being a non-metal, it exhibits oxidation states from -3 (group number - 8) to +5 (group number).

Iron is a metal, the element is located in a secondary subgroup of the main group. For iron, oxidation states are characteristic: 0, +2, +3, +6.

Task number 2

The compound of the composition KEO 4 forms each of the two elements:

1) phosphorus and chlorine

2) fluorine and manganese

3) chlorine and manganese

4) silicon and bromine

Answer: 3

Explanation:

The salt of the KEO 4 composition contains the acid residue EO 4 - , where oxygen has an oxidation state of -2, therefore, the oxidation state of the element E in this acid residue is +7. Of the proposed options, chlorine and manganese are suitable - elements of the main and secondary subgroups of group VII, respectively.

Fluorine is also an element of the main subgroup of group VII, however, being the most electronegative element, it does not show positive oxidation states (0 and -1).

Boron, silicon and phosphorus are elements of the main subgroups of groups 3, 4 and 5, respectively, therefore, in salts, they exhibit the corresponding maximum oxidation states of +3, +4, +5.

Task number 3

• 1. Zn and Cr
• 2. Si and B
• 3. Fe and Mn
• 4.P and As

Answer: 4

Explanation:

The same highest oxidation state in the compounds, equal to the group number (+5), is shown by P and As. These elements are located in the main subgroup of group V.

Zn and Cr are elements of secondary subgroups of groups II and VI, respectively. In compounds, zinc exhibits the highest oxidation state +2, chromium - +6.

Fe and Mn are elements of the secondary subgroups of groups VIII and VII, respectively. The highest oxidation state for iron is +6, for manganese - +7.

Task number 4

The same highest oxidation state in compounds exhibit

• 1. Hg and Cr
• 2. Si and Al
• 3.F and Mn
• 4. P and N

Answer: 4

Explanation:

P and N show the same highest oxidation state in compounds, equal to the group number (+5). These elements are located in the main subgroup of group V.

Hg and Cr are elements of secondary subgroups of groups II and VI, respectively. In compounds, mercury exhibits the highest oxidation state +2, chromium - +6.

Si and Al are elements of the main subgroups of groups IV and III, respectively. Therefore, for silicon, the maximum oxidation state in complex compounds is +4 (the group number where silicon is located), for aluminum - +3 (the group number where aluminum is located).

F and Mn are elements of the main and secondary subgroups of groups VII, respectively. However, fluorine, being the most electronegative element of the Periodic Table of Chemical Elements, does not show positive oxidation states: in complex compounds, its oxidation state is -1 (group number -8). The highest oxidation state of manganese is +7.

Task number 5

The +3 oxidation state nitrogen exhibits in each of two substances:

• 1. HNO 2 and NH 3
• 2. NH 4 Cl and N 2 O 3
• 3. NaNO 2 and NF 3
• 4. HNO 3 and N 2

Answer: 3

Explanation:

In nitrous acid HNO 2, the oxidation state of oxygen in the acid residue is -2, for hydrogen - +1, therefore, in order for the molecule to remain electrically neutral, the oxidation state of nitrogen is +3. In ammonia, NH 3, nitrogen is a more electronegative element, therefore it pulls the electron pair of a covalent polar bond onto itself and has a negative oxidation state of -3, the oxidation state of hydrogen in ammonia is +1.

Ammonium chloride NH 4 Cl is an ammonium salt, so the oxidation state of nitrogen is the same as in ammonia, i.e. equals -3. In oxides, the oxidation state of oxygen is always -2, so for nitrogen it is +3.

In sodium nitrite NaNO 2 (salts of nitrous acid), the degree of oxidation of nitrogen is the same as in nitrogen in nitrous acid, because is +3. In nitrogen fluoride, the oxidation state of nitrogen is +3, since fluorine is the most electronegative element in the Periodic Table and in complex compounds it exhibits a negative oxidation state of -1. This answer option satisfies the condition of the task.

In nitric acid, nitrogen has the highest oxidation state, equal to the group number (+5). Nitrogen as a simple compound (since it consists of atoms of one chemical element) has an oxidation state of 0.

Task number 6

The highest oxide of an element of group VI corresponds to the formula

• 1. E 4 O 6
• 2.EO 4
• 3. EO 2
• 4. EO 3

Answer: 4

Explanation:

The highest oxide of an element is the oxide of the element with its highest oxidation state. In a group, the highest oxidation state of an element is equal to the group number, therefore, in group VI, the maximum oxidation state of an element is +6. In oxides, oxygen exhibits an oxidation state of -2. The numbers below the element symbol are called indices and indicate the number of atoms of this element in the molecule.

The first option is incorrect, because the element has an oxidation state of 0-(-2)⋅6/4 = +3.

In the second version, the element has an oxidation state of 0-(-2) ⋅ 4 = +8.

In the third variant, the oxidation state of the element E: 0-(-2) ⋅ 2 = +4.

In the fourth variant, the oxidation state of the element E: 0-(-2) ⋅ 3 = +6, i.e. this is the desired answer.

Task number 7

The oxidation state of chromium in ammonium dichromate (NH 4) 2 Cr 2 O 7 is

• 1. +6
• 2. +2
• 3. +3
• 4. +7

Answer: 1

Explanation:

In ammonium dichromate (NH 4) 2 Cr 2 O 7 in the ammonium cation NH 4 + nitrogen, as a more electronegative element, has a lower oxidation state of -3, hydrogen is positively charged +1. Therefore, the entire cation has a charge of +1, but since there are 2 of these cations, the total charge is +2.

In order for the molecule to remain electrically neutral, the acid residue Cr 2 O 7 2− must have a charge of -2. Oxygen in the acid residues of acids and salts always has a charge of -2, therefore, 7 oxygen atoms that make up the ammonium bichromate molecule are charged -14. Chromium atoms Cr into molecules 2, therefore, if the charge of chromium is denoted by x, then we have:

2x + 7 ⋅ (-2) = -2 where x = +6. The charge of chromium in the ammonium dichromate molecule is +6.

Task number 8

An oxidation state of +5 is possible for each of the two elements:

1) oxygen and phosphorus

2) carbon and bromine

3) chlorine and phosphorus

4) sulfur and silicon

Answer: 3

Explanation:

In the first proposed answer, only phosphorus, as an element of the main subgroup of group V, can exhibit an oxidation state of +5, which is the maximum for it. Oxygen (an element of the main subgroup of group VI), being an element with high electronegativity, in oxides exhibits an oxidation state of -2, as a simple substance - 0 and in combination with fluorine OF 2 - +1. The +5 oxidation state is not typical for it.

Carbon and bromine are elements of the main subgroups of groups IV and VII, respectively. Carbon is characterized by a maximum oxidation state of +4 (equal to the group number), and bromine exhibits oxidation states of -1, 0 (in a simple compound Br 2), +1, +3, +5 and +7.

Chlorine and phosphorus are elements of the main subgroups of groups VII and V, respectively. Phosphorus exhibits a maximum oxidation state of +5 (equal to the group number), for chlorine, similarly to bromine, oxidation states of -1, 0 (in a simple compound Cl 2), +1, +3, +5, +7 are characteristic.

Sulfur and silicon are elements of the main subgroups of groups VI and IV, respectively. Sulfur exhibits a wide range of oxidation states from -2 (group number - 8) to +6 (group number). For silicon, the maximum oxidation state is +4 (group number).

Task number 9

• 1. NaNO3
• 2. NaNO2
• 3.NH4Cl
• 4. NO

Answer: 1

Explanation:

In sodium nitrate NaNO 3, sodium has an oxidation state of +1 (group I element), there are 3 oxygen atoms in the acid residue, each of which has an oxidation state of −2, therefore, in order for the molecule to remain electrically neutral, nitrogen must have an oxidation state of: 0 − (+ 1) − (−2) 3 = +5.

In sodium nitrite NaNO 2, the sodium atom also has an oxidation state of +1 (group I element), there are 2 oxygen atoms in the acid residue, each of which has an oxidation state of −2, therefore, in order for the molecule to remain electrically neutral, nitrogen must have an oxidation state: 0 − (+1) − (−2) 2 = +3.

NH 4 Cl - ammonium chloride. In chlorides, chlorine atoms have an oxidation state of −1, hydrogen atoms, of which there are 4 in the molecule, are positively charged, therefore, in order for the molecule to remain electrically neutral, the nitrogen oxidation state is: 0 − (−1) − 4 (+1) = −3. In ammonia and cations of ammonium salts, nitrogen has a minimum oxidation state of −3 (the number of the group in which the element is located is −8).

In the nitric oxide NO molecule, oxygen exhibits a minimum oxidation state of −2, as in all oxides, therefore, the oxidation state of nitrogen is +2.

Task number 10

Nitrogen exhibits the highest oxidation state in a compound whose formula is

• 1. Fe(NO 3) 3
• 2. NaNO2
• 3. (NH 4) 2 SO 4
• 4 NO 2

Answer: 1

Explanation:

Nitrogen is an element of the main subgroup of group V, therefore, it can exhibit a maximum oxidation state equal to the group number, i.e. +5.

One structural unit of iron nitrate Fe(NO 3) 3 consists of one Fe 3+ ion and three nitrate ions. In nitrate ions, nitrogen atoms, regardless of the type of counterion, have an oxidation state of +5.

In sodium nitrite NaNO 2, sodium has an oxidation state of +1 (an element of the main subgroup of group I), there are 2 oxygen atoms in the acid residue, each of which has an oxidation state of −2, therefore, in order for the molecule to remain electrically neutral, nitrogen must have an oxidation state of 0 − ( +1) − (−2)⋅2 ​​= +3.

(NH 4) 2 SO 4 - ammonium sulfate. In sulfuric acid salts, the SO 4 2− anion has a charge of 2−, therefore, each ammonium cation is charged with 1+. On hydrogen, the charge is +1, therefore on nitrogen -3 (nitrogen is more electronegative, therefore it pulls the common electron pair of the N−H bond). In ammonia and cations of ammonium salts, nitrogen has a minimum oxidation state of −3 (the number of the group in which the element is located is −8).

In the nitric oxide NO 2 molecule, oxygen exhibits a minimum oxidation state of −2, as in all oxides, therefore, the oxidation state of nitrogen is +4.

Task number 11

28910E

In compounds of the composition Fe(NO 3) 3 and CF 4, the degree of oxidation of nitrogen and carbon is, respectively,

Answer: 4

Explanation:

One structural unit of iron (III) nitrate Fe(NO 3) 3 consists of one iron ion Fe 3+ and three nitrate ions NO 3 − . In nitrate ions, nitrogen always has an oxidation state of +5.

In carbon fluoride CF 4, fluorine is a more electronegative element and pulls the common electron pair of the C-F bond towards itself, showing an oxidation state of -1. Therefore, carbon C has an oxidation state of +4.

Task number 12

A32B0B

The oxidation state +7 chlorine exhibits in each of the two compounds:

• 1. Ca(OCl) 2 and Cl 2 O 7
• 2. KClO 3 and ClO 2
• 3. BaCl 2 and HClO 4
• 4. Mg(ClO 4) 2 and Cl 2 O 7

Answer: 4

Explanation:

In the first variant, chlorine atoms have oxidation states +1 and +7, respectively. One structural unit of calcium hypochlorite Ca(OCl) 2 consists of one calcium ion Ca 2+ (Ca is an element of the main subgroup of group II) and two hypochlorite ions OCl − , each of which has a charge of 1−. In complex compounds, except for OF 2 and various peroxides, oxygen always has an oxidation state of −2, so it is obvious that chlorine has a charge of +1. In chlorine oxide Cl 2 O 7, as in all oxides, oxygen has an oxidation state of −2, therefore, chlorine in this compound has an oxidation state of +7.

In potassium chlorate KClO 3, the potassium atom has an oxidation state of +1, and oxygen - -2. In order for the molecule to remain electrically neutral, chlorine must exhibit an oxidation state of +5. In chlorine oxide ClO 2, oxygen, as in any other oxide, has an oxidation state of −2, therefore, for chlorine, its oxidation state is +4.

In the third version, the barium cation in the complex compound is charged +2, therefore, a negative charge of −1 is concentrated on each chlorine anion in the BaCl 2 salt. In perchloric acid HClO 4, the total charge of 4 oxygen atoms is -2⋅4 = -8, on the hydrogen cation the charge is +1. For the molecule to remain electrically neutral, the charge of chlorine must be +7.

In the fourth variant, in the magnesium perchlorate molecule Mg(ClO 4) 2, the magnesium charge is +2 (in all complex compounds, magnesium exhibits an oxidation state of +2), therefore, each ClO 4 − anion has a charge of 1−. In total, 4 oxygen ions, where each exhibits an oxidation state of -2, have a charge of -8. Therefore, for the total charge of the anion to be 1−, the charge on chlorine must be +7. In chlorine oxide Cl 2 O 7 , as explained above, the charge of chlorine is +7.

Question number 5. "The highest oxidation state of nitrogen in compounds is greater than the highest oxidation state of carbon, since ..."

There are 5 electrons on the outer energy level of the nitrogen atom, the electronic formula of the outer layer of the nitrogen atom, the highest oxidation state is +5.

At the outer energy level of the carbon atom, there are 4 paired electrons in an excited state, the electronic formula of the outer layer of the carbon atom, the highest oxidation state is +4.

Answer: There are more electrons on the outer electron layer of the nitrogen atom than on the carbon atom.

Question number 6. “What volume of a 15% (by mass) solution (c=1.10 g/ml) will be required to completely dissolve 27 g of Al?”

Reaction equation:

Weight of 1 liter of 15%:

1000 H 1.10 \u003d 1100g;

1100 g of a 15% solution contains:

To dissolve 27 g of Al, you will need:

Answer: a) 890 ml.

Question number 7. “The reaction of dehydrogenation of hydrocarbons is an endothermic process.

How to shift the equilibrium of the reaction: C4H10 (g) > C4H6 (g) + 2H2 (g) towards the formation of C4H6? (give the answer as a sum of numbers corresponding to the chosen methods): C4H10 (g) > C4H6 (g) + 2H2 (g)

10) raise the temperature;

Since the butane dehydrogenation reaction is an endothermic process, it means that when the system is heated (as the temperature rises), the equilibrium shifts towards the endothermic reaction, the formation of butyne (C 4 H 6).

50) lower the pressure;

Gaseous substances take part in the butane dehydrogenation reaction. The total number of moles of the starting substances is less than the total number of moles of the resulting gaseous substances, therefore, as the pressure decreases, the equilibrium shifts towards larger volumes.

The degree of oxidation is a conditional value used to record redox reactions. To determine the degree of oxidation, a table of oxidation of chemical elements is used.

Meaning

The oxidation state of basic chemical elements is based on their electronegativity. The value is equal to the number of electrons displaced in the compounds.

The oxidation state is considered positive if the electrons are displaced from the atom, i.e. the element donates electrons in the compound and is a reducing agent. These elements include metals, their oxidation state is always positive.

When an electron is displaced towards an atom, the value is considered negative, and the element is considered an oxidizing agent. The atom accepts electrons until the completion of the outer energy level. Most non-metals are oxidizing agents.

Simple substances that do not react always have a zero oxidation state.

Rice. 1. Table of oxidation states.

In the compound, a non-metal atom with a lower electronegativity has a positive oxidation state.

Definition

You can determine the maximum and minimum oxidation state (how many electrons an atom can give and take) using the periodic table of Mendeleev.

The maximum power is equal to the number of the group in which the element is located, or the number of valence electrons. The minimum value is determined by the formula:

No. (groups) - 8.

Rice. 2. Periodic table.

Carbon is in the fourth group, therefore, its highest oxidation state is +4, and the lowest is -4. The maximum oxidation state of sulfur is +6, the minimum is -2. Most non-metals always have a variable - positive and negative - oxidation state. The exception is fluorine. Its oxidation state is always -1.

It should be remembered that this rule does not apply to alkali and alkaline earth metals of groups I and II, respectively. These metals have a constant positive oxidation state - lithium Li +1, sodium Na +1, potassium K +1, beryllium Be +2, magnesium Mg +2, calcium Ca +2, strontium Sr +2, barium Ba +2. Other metals may exhibit different oxidation states. The exception is aluminum. Despite being in group III, its oxidation state is always +3.

Rice. 3. Alkali and alkaline earth metals.

Of group VIII, only ruthenium and osmium can exhibit the highest oxidation state +8. Gold and copper, which are in group I, exhibit oxidation states of +3 and +2, respectively.

Recording

To correctly record the oxidation state, you should remember a few rules:

• inert gases do not react, so their oxidation state is always zero;
• in compounds, the variable oxidation state depends on the variable valency and interaction with other elements;
• hydrogen in compounds with metals exhibits a negative oxidation state - Ca +2 H 2 −1, Na +1 H −1;
• oxygen always has an oxidation state of -2, except for oxygen fluoride and peroxide - O +2 F 2 -1, H 2 +1 O 2 -1.

What have we learned?

The oxidation state is a conditional value showing how many electrons an atom of an element has received or given away in a compound. The value depends on the number of valence electrons. Metals in compounds always have a positive oxidation state, i.e. are restorers. For alkali and alkaline earth metals, the oxidation state is always the same. Non-metals, except for fluorine, can take positive and negative oxidation states.

Topic quiz

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Task 54.
What is the lowest oxidation state of hydrogen, fluorine, sulfur and nitrogen? Why? Write formulas for calcium compounds with these elements in this oxidation state. What are the names of the corresponding compounds?
Decision:
The lowest oxidation state is determined by the conditional charge, which an atom acquires upon addition of the number of electrons that is necessary to form a stable electron shell of an inert gas ns2np6 (in the case of hydrogen, ns 2). Hydrogen, fluorine, sulfur and nitrogen are respectively in the IA-, VIIA-, VIA- and VA-groups of the periodic system of chemical elements and have the structure of the external energy level s 1, s 2 p 5, s 2 p 4 and s 2 p 3.

Thus, to complete the external energy level, a hydrogen atom and a fluorine atom need to add one electron each, a sulfur atom - two, a nitrogen atom - three. Hence, the low oxidation state for hydrogen, fluorine, sulfur and nitrogen is -1, -1, -2 and -3, respectively. Formulas of calcium compounds with these elements in this oxidation state:

CaH 2 - calcium hydride;
CaF 2 - calcium fluoride;
CaS, calcium sulfide;
Ca 3 N 2 - calcium nitride.

Task 55.
What are the lowest and highest oxidation states of silicon, arsenic, selenium and chlorine? Why? Write formulas for compounds of these elements corresponding to these oxidation states.
Decision:
The highest oxidation state of an element is determined, as a rule, by the group number of the periodic system
D. I. Mendeleev, in which he is located. The lowest oxidation state is determined by the conditional charge that an atom acquires when attaching the number of electrons that is necessary to form a stable eight-electron shell of an inert gas ns 2 np 6 (in the case of hydrogen ns 2). Silicon, arsenic, selenium and chlorine are respectively in IVA-, VA-, VIa- and VIIA-groups and have the structure of the external energy level, respectively, s 2 p 2, s 2 p 3, s 2 p 4 and s 2 p5. Thus, the highest oxidation state of arsenic, selenium, and chlorine silicon is +4, +5, +6, and +7, respectively. Formulas of compounds of these elements corresponding to these oxidation states: H 2 SiO 3 - silicic acid; H 3 AsO 4 - arsenic acid; H 2 SeO 4 - selenic acid; HClO 4 - perchloric acid.

The lowest oxidation state of arsenic, selenium and chlorine silicon is -4, -5, -6 and -7, respectively. Formulas of compounds of these elements corresponding to these oxidation states: H 4 Si, H 3 As, H 2 Se, HCl.

Task 56.
Chromium forms compounds in which it exhibits oxidation states +2, +3, +6. Write formulas for its oxides and hydroxides corresponding to these oxidation states. Write the reaction equations that prove the amphoteric nature of chromium (III) hydroxide.
Decision:
Chromium forms compounds in which it exhibits oxidation states +2, +3, +6. The formulas of its oxides and hydroxides corresponding to these oxidation states are:

a) chromium oxides:

CrO, chromium (II) oxide;
Cr 2 O 3 - chromium oxide (III);
CrO 3 - chromium (VI) oxide.

b) chromium hydroxides:

Cr(OH) 2 - chromium (II) hydroxide;
Cr(OH) 3 - chromium (III) hydroxide;
H 2 CrO 4 - chromic acid.

Cr (OH) 3 - chromium (III) hydroxide - ampholyte, i.e. a substance that reacts with both acids and bases. Reaction equations proving the amphotericity of chromium (III) hydroxide:

a) Cr(OH) 3 + 3HCl = CrCl 3 + 3H 2 O;
b) Cr(OH) 3 + 3NaOH = NaCrO 3 + 3H 2 O.

Task 57.
The atomic masses of the elements in the periodic system are continuously increasing, while the properties of simple bodies change periodically. How can this be explained? Give a reasoned answer.
Decision:
In most cases, with an increase in the charge of the nucleus of the atoms of elements, their relative atomic masses naturally increase, because there is a regular increase in the content of protons and neutrons in the nuclei of atoms. The properties of simple bodies change periodically, because the number of electrons periodically changes at the outer energy level of atoms. For atoms of elements, periodically with an increase in the charge of the nucleus, the number of electrons at the external energy level increases, which is necessary for the formation of a stable eight-electron shell (shell of an inert gas). For example, the periodic recurrence of the properties of the atoms of Li, Na and K is explained by the fact that at the outer energy level of their atoms there is one valence electron each. Also, the properties of He, Ne, Ar, Kr, Xe and Rn atoms are periodically repeated - the atoms of these elements contain eight electrons at the outer energy level (helium has two electrons) - they are all chemically inert, since their atoms cannot neither accept nor donate electrons to atoms of other elements.

Task 58.
What is the modern formulation of the periodic law? Explain why in the periodic table of elements argon, cobalt, tellurium and thorium are placed respectively in front of potassium, nickel, iodine and protactinium, although they have a large atomic mass?
Decision:
The modern formulation of the periodic law: "The properties of the chemical elements and the simple or complex substances they form are in a periodic dependence on the magnitude of the charge of the nucleus of the atoms of the elements."

Since the atoms K, Ni, I, Pa - having a lower relative mass than, respectively, Ar, Co, Te, Th - the charges of atomic nuclei are one more

then potassium, nickel, iodine and protactinium are assigned serial numbers 19, 28, 53 and 91, respectively. Thus, an element in the periodic system is assigned a serial number not by increasing its atomic mass, but by the number of protons contained in the nucleus of a given atom, i.e. by the charge of the atomic nucleus. The element number indicates the nuclear charge (the number of protons contained in the nucleus of an atom), the total number of electrons contained in a given atom.

Task 59.
What are the lowest and highest oxidation states of carbon, phosphorus, sulfur and iodine? Why? Write formulas for compounds of these elements corresponding to these oxidation states.
Decision:
The highest oxidation state of an element is determined, as a rule, by the group number of the periodic system of D. I. Mendeleev, in which it is located. The lowest oxidation state is determined by the conditional charge that an atom acquires when adding the number of electrons that is necessary to form a stable eight-electron shell of an inert gas ns2np6 (in the case of hydrogen, ns2). Carbon, phosphorus, sulfur and iodine are respectively in IVA-, VA-, VIa- and VIIA-groups and have the structure of the external energy level, respectively, s 2 p 2, s 2 p 3, s 2 p 4 and s 2 p 5. Thus, the highest oxidation state of carbon, phosphorus, sulfur and iodine is +4, +5, +6 and +7, respectively. Formulas of compounds of these elements corresponding to these oxidation states: CO 2 - carbon monoxide (II); H 3 PO 4 - orthophosphoric acid; H 2 SO 4 - sulfuric acid; HIO 4 - iodic acid.

The lowest oxidation state of carbon, phosphorus, sulfur and iodine is -4, -5, -6 and -7, respectively. Formulas of compounds of these elements corresponding to these oxidation states: CH 4, H 3 P, H 2 S, HI.

Task 60.
The atoms of which elements of the fourth period of the periodic system form an oxide corresponding to their highest oxidation state E 2 O 5 ? Which of them gives a gaseous combination with hydrogen? Compose the formulas of acids corresponding to these oxides and depict them graphically?
Decision:
Oxide E 2 O 5, where the element is in its highest oxidation state +5, is characteristic of the elements of group V. Such an oxide can be formed by two elements of the fourth period and the V-group - this is element No. 23 (vanadium) and No. 33 (arsenic). Vanadium and arsenic, as elements of the fifth group, form hydrogen compounds of the composition EN 3 because they can exhibit the lowest oxidation state -3. Since arsenic is a non-metal, it forms a gaseous compound with hydrogen - H 3 As - arsine.

Formulas of acids corresponding to oxides in the highest oxidation state of vanadium and arsenic:

H 3 VO 4 - orthovanadic acid;
HVO 3 - metavanadic acid;
HAsO 3 - metaarsenic acid;
H 3 AsO 4 - arsenic (ortho-arsenic) acid.

Graphic formulas of acids:

The modern formulation of the Periodic Law, discovered by D. I. Mendeleev in 1869:

The properties of the elements are in a periodic dependence on the ordinal number.

The periodically recurring nature of the change in the composition of the electron shell of the atoms of the elements explains the periodic change in the properties of the elements when moving through the periods and groups of the Periodic system.

Let us trace, for example, the change in the higher and lower oxidation states of the elements of the IA - VIIA groups in the second - fourth periods according to Table. 3.

Positive oxidation states are exhibited by all elements, with the exception of fluorine. Their values ​​increase with increasing nuclear charge and coincide with the number of electrons at the last energy level (except for oxygen). These oxidation states are called higher oxidation states. For example, the highest oxidation state of phosphorus P is +V.

Negative oxidation states are exhibited by elements starting with carbon C, silicon Si and germanium Ge. Their values ​​are equal to the number of electrons missing up to eight. These oxidation states are called inferior oxidation states. For example, the phosphorus atom P at the last energy level lacks three electrons to eight, which means that the lowest oxidation state of phosphorus P is -III.

The values ​​of higher and lower oxidation states are repeated periodically, coinciding in groups; for example, in the IVA group, carbon C, silicon Si and germanium Ge have the highest oxidation state +IV, and the lowest oxidation state - IV.

This frequency of changes in oxidation states is reflected in the periodic change in the composition and properties of chemical compounds of elements.

Similarly, a periodic change in the electronegativity of elements in the 1st-6th periods of the IA–VIIA groups can be traced (Table 4).

In each period of the Periodic Table, the electronegativity of the elements increases with increasing serial number (from left to right).

In each group In the periodic table, electronegativity decreases as the atomic number increases (from top to bottom). Fluorine F has the highest, and cesium Cs the lowest electronegativity among the elements of the 1st-6th periods.

Typical non-metals have high electronegativity, while typical metals have low electronegativity.

Examples of tasks of parts A, B

1. In the 4th period, the number of elements is

2. Metallic properties of elements of the 3rd period from Na to Cl

1) force

2) weaken

3) do not change

4) don't know

3. Non-metallic properties of halogens with increasing atomic number

1) increase

2) go down

3) remain unchanged

4) don't know

4. In the series of elements Zn - Hg - Co - Cd, one element that is not included in the group is

5. The metallic properties of the elements increase in a row

1) In-Ga-Al

2) K - Rb - Sr

3) Ge-Ga-Tl

4) Li - Be - Mg

6. Non-metallic properties in the series of elements Al - Si - C - N

1) increase

2) decrease

3) do not change

4) don't know

7. In the series of elements O - S - Se - Te, the dimensions (radii) of the atom

1) decrease

2) increase

3) do not change

4) don't know

8. In the series of elements P - Si - Al - Mg, the dimensions (radii) of the atom

1) decrease

2) increase

3) do not change

4) don't know

9. For phosphorus, the element with lesser electronegativity is

10. A molecule in which the electron density is shifted to the phosphorus atom is

11. Supreme the oxidation state of the elements is manifested in a set of oxides and fluorides

1) СlO 2, PCl 5, SeCl 4, SO 3

2) PCl, Al 2 O 3, KCl, CO

3) SeO 3, BCl 3, N 2 O 5, CaCl 2

4) AsCl 5 , SeO 2 , SCl 2 , Cl 2 O 7

12. Inferior the degree of oxidation of elements - in their hydrogen compounds and fluorides of the set

1) ClF 3 , NH 3 , NaH, OF 2

2) H 3 S +, NH+, SiH 4, H 2 Se

3) CH 4 , BF 4 , H 3 O + , PF 3

4) PH 3 , NF+, HF 2 , CF 4

13. Valence for a polyvalent atom the same in a series of compounds

1) SiH 4 - AsH 3 - CF 4

2) PH 3 - BF 3 - ClF 3

3) AsF 3 - SiCl 4 - IF 7

4) H 2 O - BClg - NF 3

14. Indicate the correspondence between the formula of a substance or ion and the degree of oxidation of carbon in them