Degree oxidation 4. Proper compilation of the formulas of substances

The degree of oxidation is the conditional charge of the atoms of the chemical element in the compound calculated from the assumption that all links have an ion type. The degrees of oxidation may have a positive, negative or zero value, therefore the algebraic amount of the degrees of the oxidation of elements in the molecule, taking into account the number of their atoms, is 0, and in the ion - ion charge.

This list of oxidation degrees shows all known oxidation degrees. chemical elements Periodic table of Mendeleev. The list is based on the Greenwood Table with all additions. In lines that are highlighted by color, the inert gases are ins. The degree of oxidation of which is zero.

−1

He.

-3

BE.

−1

−4

−3

−2

−1

−3

−2

−1

−2

−1

Na.

MG.

−4

−3

−2

−1

−3

−2

−1

−2

−1

Cl.

CA.

−1

−2

−1

−3

−2

−1

MN.

−2

−1

FE.

−1

Co.

−1

Ni.

Cu.

Zn.

GA.

−4

GE.

−3

−2

−1

Br.

Kr.

RB.

Sr.

Zr.

−1

NB.

−2

−1

Mo.

−3

−1

TC.

−2

−1

Rh.

Pd.

IN.

−4

SN.

−3

SB.

−2

−1

Xe.

CS.

BA.

Pr.

ND.

PM.

SM.

GD.

DY.

HO.

ER.

TM.

YB.

Lu.

HF.

−1

TA.

−2

−1

−3

−1

Re.

−2

−1

−3

−1

IR.

Pt.

−1

AU.

Hg.

−4

PB.

−3

−2

−1

AT.

RN.

Fr.

TH.

NP.

AM.

Cm.

Bk.

CF.

100

FM.

101

MD.

102

No.

103

104

105

106

SG.

107

Bh.

108

HS.

The highest degree of oxidation of the element corresponds to the number of the periodic system number, where this element is located (the exception is: AU + 3 (I Group), Cu + 2 (II), from Group VIII The degree of oxidation is +8 can only be OSME OS and RU ruthenium.

The degree of oxidation of metals in connections

The degrees of oxidation of metals in the compounds are always positive, but to talk about nonmetallah, their degree of oxidation depends on which it is connected to the atom:

if with a nonmetal atom, the degree of oxidation can be positive and negative. It depends on the electronegativity of the atoms of the elements;
if with a metal atom, then the degree of oxidation is negative.

Negative degree of oxidation of non-metals

The highest negative degree of non-metal oxidation can be determined by subtracting from 8 group number in which this chemical element is located, i.e. The highest positive degree of oxidation is equal to the number of electrons on the outer layer, which corresponds to the number number.

Please note that the degrees of oxidation of simple substances are equal to 0, regardless of whether it is metal or nonmetall.

Sources:

Greenwood, Norman N.; Earnshaw, A. Chemistry of the Elements - 2nd ed. - Oxford: Butterworth-Heinemann, 1997
Green Stable Magnesium (i) Compounds with MG-MG Bonds / Jones C.; Stasch A .. - Journal Science, 2007. - December (Vol. 318 (No. 5857)
Science magazine, 1970. - Vol. 3929. - № 168. - P. 362.
Journal of the Chemical Society, Chemical Communications, 1975. - S. 760B-761.
Irving Langmuir The Arrangement of Electrons in Atoms and Molecules. - J. AM magazine. Chem. SOC., 1919. - Vol. 41.

Task 54.
What low degree of oxidation is hydrogen, fluorine, sulfur and nitrogen show? Why? Make a calcium compound formula with data elements to this oxidation. What are the corresponding compounds?
Decision:
The lowest degree of oxidation is determined by the conditional charge,which acquires an atom when the number of electrons is attached, which is necessary for the formation of a stable e-casing of an NS2NP6 inert gas (in the case of NS 2 hydrogen). Hydrogen, fluorine, sulfur and nitrogen are respectively in Ia-, VII, VIA- and Via 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, the hydrogen atom and the atom of the fluorine must be attached by one electron, the sulfur atom - two, the nitrogen atom - three. Hence the low degree of oxidation for hydrogen, fluorine, sulfur and nitrogen is equal to -1, -1, -2 and -3. Calcium compound formulas with data elements to this oxidation:

CAH 2 - calcium hydride;
CAF 2 - calcium fluoride;
CAS - calcium sulphide;
Ca 3 N 2 - calcium nitride.

Task 55.
What low and the highest degree of oxidation show silicon, arsenic, selenium and chlorine? Why? Make the formula of data connections of elements that meet these degrees of oxidation.
Decision:
The highest degree of element oxidation determines, as a rule, the number of the periodic system
D. I. Mendeleev, in which he is. The lower degree of oxidation is determined by the conditional charge, which acquires an atom when attaching the number of electrons, which is necessary for the formation of a stable eight-electron Introduction of the NS 2 NP 6 inert gas (in the case of NS 2 hydrogen). Silicon, arsenic, selenium and chlorine are in accordance with IVA, VA-, VIA- and VIII 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. In this way, higher degree Silicon oxidation arsenic, selenium and chlorine is equal, respectively, +4, +5, +6 and +7. The formulas of data compounds of elements corresponding to these oxidation degrees: H 2 SiO 3 - flintic acid; H 3 ASO 4 - arsenic acid; H 2 SEO 4 - seeded acid; HCLO 4 - chlorine acid.

The lowest degree of oxidation of arsenic silicon, selenium and chlorine is equal to -4, -5, -6 and -7. The formulas of these elements corresponding to these oxidation degrees: H 4 Si, H 3 AS, H 2 SE, HCl.

Task 56.
Chrome forms the compounds in which it exhibits the degree of oxidation +2, +3, +6. Make the formula of its oxides and hydroxides that meet these degrees of oxidation. Write the equations of reactions proving the amphoterity of chromium hydroxide (III).
Decision:
The chrome forms the compounds in which the oxidation degrees is +2, +3, +6. The formulas of its oxides and hydroxides corresponding to these degrees of oxidation:

a) Chromium oxides:

CRO - chromium oxide (II);
CR 2 O 3 - chromium oxide (III);
CRO 3 - chromium oxide (VI).

b) chromium hydroxides:

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

CR (OH) 3 - chromium hydroxide (III) - ampholite, i.e., a substance that reacts with both acids and bases. Equations of reactions proving the amphoterity of chromium hydroxide (III):

a) CR (OH) 3 + 3HCl \u003d CRCl 3 + 3H 2 O;
b) CR (OH) 3 + 3NAOH \u003d Nacro 3 + 3H 2 O.

Task 57.
Atomic masses of elements in the periodic system are continuously increasing, while the properties of simple bodies change periodically. How can this be explained? Give a motivated 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 a natural increase in protons and neutrons in the nuclei of atoms occurs. The properties of simple bodies vary periodically, because on the outer energy level, the number of electrons changes periodically changes at atoms. In atoms of elements, periodically with an increase in the charge of the nucleus increases the number of electrons in the external energy level, which is necessary for the formation of a stable eight-electron shell (inert gas shell). For example, periodic repeatability of properties in Li, Na and K atoms is explained by the external energy level of their atoms there is one valence electron. Also periodically repeated properties in atoms not, ne, ar, kr, xe and rn - at the atoms of these elements on the outdoor energy level contained eight electrons (helium - two electrons) - all of them are chemically inert, since their atoms cannot Neither attach nor to give electrons to atoms of other elements.

Task 58.
What is the modern formulation periodic Law? Explain that in the periodic system of elements argon, cobalt, tellurium and thorium are placed respectively before potassium, nickel, iodine and the prostactment, although they have a large atomic mass?
Decision:
Modern formulation of the periodic law: "The properties of the chemical elements and their ordinary or complex substances are in periodic dependence on the charge of the nucleus of the elements atoms. "

Since atoms K, Ni, I, PA - have a smaller relative mass, than in AR, CO, TE, TH - the charges of atomic nuclei per unit more

then Kalia, nickel, iodine and protacticity is assigned by the sequence numbers, respectively, 19, 28, 53 and 91. In order to the element in the periodic system, the sequence number is assigned not to increase its atomic mass, but by the number of protons contained in the nucleus of this atom, i.e. . At the charge of the nuclei of the atom. The number of the element indicates the charge of the nucleus (the number of protons contained in the nucleus of the atom), total number electrons contained in a given atom.

Task 59.
What low and higher degrees of oxidation are carbon, phosphorus, sulfur and iodine? Why? Make the formula of data connections of elements that meet these degrees of oxidation.
Decision:
The highest degree of oxidation of the element determines, as a rule, the number of the group of the periodic system D. I. Mendeleev, in which it is located. The lower degree of oxidation is determined by the conditional charge, which acquires an atom when the number of electrons is attached, which is necessary for the formation of a stable eight-electron Introduction of the NS2NP6 inert gas (in the case of NS2 hydrogen). 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 degree of carbon oxidation, phosphorus, sulfur and iodine is equal to +4, +5, +6 and +7, respectively. The formulas of the data compounds of elements corresponding to these oxidation degrees: CO 2 - carbon oxide (II); H 3 PO 4 - orthophosphoric acid; H 2 SO 4 - sulfuric acid; Hio 4 - iodine acid.

The lowest degree of carbon oxidation, phosphorus, sulfur and iodine is equal to -4, -5, -6 and -7. The formulas of the data compounds of the elements corresponding to these degrees of oxidation: CH 4, H 3 P, H 2 S, Hi.

Task 60.
Atoms of which elements fourth period Periodic Systems form oxide corresponding to their highest oxidation of e 2 o 5? Which one gives a gaseous connection with hydrogen? Make a formula of acids that meet these oxides and depict them graphically?
Decision:
Oxide E 2 O 5, where the element is in its highest degree of oxidation +5, is characteristic of the elements of the V group. Such oxide can form two elements of the fourth period and the V-group are element No. 23 (Vanadium) and No. 33 (arsenic). Vanadium and arsenic, like elements of the fifth group, form hydrogen compounds EN 3 composition, because they can exhibit a lower degree of oxidation -3. Since arsenic is nonmetall, it forms a gaseous compound with hydrogen - H 3 AS - ARSIN.

Acid formulas that correspond to oxides in the highest oxidation of vanadium and arsenic:

H 3 VO 4 - orthovanadium acid;
HVO 3 - metavanadium acid;
HASO 3 - Merchant Acid;
H 3 ASO 4 - arsenic (orthomyshiac) acid.

Graphic formulas acids:

In the chemistry, the terms "oxidation" and "recovery" means reactions in which an atom or a group of atoms lose or, accordingly, the electrons acquire. The degree of oxidation is attributed to one or several atoms a numerical value characterizing the amount of redistributed electrons and showing how these electrons are distributed between atoms during the reaction. The definition of this value can be both simple and quite complex procedure, depending on atoms and molecules consisting of them. Moreover, atoms of certain elements may have several oxidation degrees. Fortunately, there are simple unambiguous rules to determine the degree of oxidation, for sure the use of which is enough knowledge of the foundations of chemistry and algebra.

Steps

Part 1

Determination of the degree of oxidation according to the laws of chemistry

Determine whether the substance under consideration is elementary. The degree of oxidation of atoms outside the chemical compound is zero. This rule is valid for substances formed from individual free atoms, and for those that consist of two or multiatomic molecules of one element.

For example, Al (S) and Cl 2 have an oxidation degree of 0, since both are in a chemically unrelated elementary state.
Please note that the Allotropic shape of sulfur S 8, or octasor, despite its nonypical structure, is also characterized by a zero degree of oxidation.

Determine whether the substance under consideration is from ions. The degree of oxidation of ions is equal to their charge. This is true for both free ions and for those that are part of chemical compounds.
- For example, the degree of oxidation of the C cl - equals -1.
- The degree of oxidation of the CL ion in the composition of the chemical compound NaCl is also equal to -1. Since Ion Na, by definition, has +1 charges, we conclude that the charge of the CL -1 ion, and thus the degree of its oxidation is -1.
Note that metal ions may have several oxidation degrees. Atoms of many metal elements can be ionized by different values. For example, the charge of the ions of such metal as iron (Fe) is +2, or +3. The charge of the metal ions (and their degree of oxidation) can be determined by charges of ions of other elements with which this metal is included in the chemical compound; In the text, this charge is denoted by Roman numbers: so, iron (III) has a degree of oxidation +3.
- As an example, consider a compound containing aluminum ion. The total charge of the ALCL 3 compound is zero. Since we know that the CL ions have a charge -1, and the connection contains 3 such ions, for the general neutrality of the substance considered, the Al Ion should have +3. So in this case The degree of aluminum oxidation is +3.
The degree of oxidation of oxygen is -2 (for some exceptions). Almost in all cases oxygen atoms have oxidation degree -2. There are several exceptions to this rule:
- If oxygen is in the elementary state (O 2), its oxidation degree is 0, as in the case of other elementary substances.
- If oxygen is part of peroxy, its degree of oxidation is -1. Packsi is a group of compounds containing a simple oxygen-oxygen bond (i.e. O 2 -2 peroxide anion). For example, in the composition of the H 2 O 2 molecule (hydrogen peroxide) oxygen has a charge and a degree of oxidation -1.
- In the compound with fluorine, oxygen has a degree of oxidation +2, read the rule for fluorine below.
Hydrogen is characterized by a degree of oxidation +1, for some exceptions. As for oxygen, there are also exceptions. As a rule, the degree of hydrogen oxidation is +1 (if it is not in the elementary state H 2). However, in compounds called hydrides, the degree of hydrogen oxidation is -1.
- For example, in H 2 o, the degree of hydrogen oxidation is +1, since the oxygen atom has a charge -2, and two charges are needed for general neutrality. However, in the composition of sodium hydride, the degree of hydrogen oxidation is already -1, since the Na ion is charged +1, and for the overall electronutrality, the charge of the hydrogen atom (and thereby its oxidation degree) should be -1.
Fluorine always It has the degree of oxidation -1. As already noted, the degree of oxidation of certain elements (metal ions, oxygen atoms in the breakdowns and so on) may vary depending on a number of factors. The degree of oxidation of fluorine, however, is constantly -1. This is explained by the fact that this element has the greatest electronenence - in other words, the fluorine atoms are least willingly part with their own electrons and their electrons are most actively attracted. Thus, their charge remains unchanged.
The sum of the oxidation degrees in the connection is equal to its charge. The degree of oxidation of all atoms included in the chemical compound in the amount should give the charge of this compound. For example, if the compound is neutral, the sum of the degrees of oxidation of all its atoms should be zero; If the compound is a polyhydric ion with charge -1, the sum of the degrees of oxidation is -1, and so on.
- This is a good test method - if the sum of the degrees of oxidation is not equal to the general charge of the compound, then you are wrong somewhere.
Part 2
Determination of the degree of oxidation without the use of chemistry laws
1. Find atoms that do not have strict rules regarding the degree of oxidation. In relation to some elements there is no firmly established rules for finding the degree of oxidation. If the atom does not fall under the same rule from those listed above, and you do not know its charge (for example, an atom is included in the complex, and its charge is not specified), you can establish the degree of oxidation of such an atom by the exclusion method. Initially, determine the charge of all other compound atoms, and then from a known total charge compound, calculate the degree of oxidation of this atom.
  - For example, in compound Na 2 SO 4, the sulfur atom (S) is unknown - we only know that it is not zero, since the sulfur is not in the elementary state. This compound serves as a good example to illustrate an algebraic method for determining the degree of oxidation.
2. Find the degree of oxidation of the other elements included in the connection. Using the above-described rules, determine the degree of oxidation of the remaining compound atoms. Do not forget about the exceptions from the rules in the case of O, H atoms and so on.
  - For Na 2 SO 4, using our rules, we find that the charge (and therefore the degree of oxidation) ion Na is +1, and for each of the oxygen atoms it is -2.
3. In the compounds, the sum of all oxidation degrees should be charged. For example, if a compound is a dioxidant ion, the sum of the degrees of oxidation of atoms should be equal to the general ion charge.
4. It is very useful to be able to use the periodic table of Mendeleev and know where metal and non-metallic elements are located in it.
5. The degree of oxidation of atoms in the elementary form is always zero. The degree of oxidation of a single ion is equal to its charge. Mendeleev table elements 1a, such as hydrogen, lithium, sodium, in elementary form have the degree of oxidation +1; The degree of oxidation of metals of group 2a, such as magnesium and calcium, in the elementary form is +2. Oxygen and hydrogen, depending on the type chemical bondmay have 2 different degrees of oxidation.

In determining this concept, it is conventionally believed that the binders (valence) electrons go to more electronegative atoms (see electronegativity), and therefore compounds consist of as it were from positive and negatively charged ions. The degree of oxidation may have zero, negative and positive values, which are usually set over the symbol of the element from above.

The zero value of the degree of oxidation is attributed to atoms of elements in a free state, for example: Cu, H 2, N 2, P 4, S 6. The negative value of the degree of oxidation have those atoms in the direction of which the binder electron cloud (electron pair) is shifted. Fluorine in all its connections it is equal to -1. A positive degree of oxidation has atoms that give valence electrons to other atoms. For example, alkaline and alkaline earth metals, it is respectively +1 and +2. In ordinary ions like Cl -, S 2-, K +, Cu 2+, Al 3+, it is equal to the charge of the ion. In most compounds, the degree of oxidation of hydrogen atoms is +1, but in metal hydrides (compounds with hydrogen) - NAH, CAH 2 and others - it is equal to -1. For oxygen, the degree of oxidation -2 is characteristic, but, for example, in a compound with fluorine of 2 it will be +2, and in peroxidation compounds (Bao 2, etc.) -1. In some cases, this value can be expressed and fractional number: for iron in iron oxide (II, III) Fe 3 O 4 it is +8/3.

The algebraic sum of the oxidation of atoms in the compound is zero, and in the complex ion - the charge of the ion. With this rule, calculate, for example, the degree of oxidation of phosphorus in orthophosphoric acid H 3 PO 4. Recalling it through X and multiplying the degree of oxidation for hydrogen (+1) and oxygen (-2) by the number of their atoms in the compound, we obtain the equation: (+1) 3 + x + (- 2) 4 \u003d 0, where x \u003d + 5 . Similarly, calculate the degree of chromium oxidation in the CR 2 O 7 2-: 2x + (- 2) 7 \u003d -2; x \u003d + 6. In MnO, Mn 2 O 3, MnO 2, Mn 2 O 3, Mno 2, Mn 3 O 4, K 2 MNO 4, KMNO 4, the degree of oxidation of manganese will be +2, +3, +4, +8/3, +6, +7.

The highest degree of oxidation is the greatest positive value. For most elements, it is equal to the group number in the periodic system and is an important quantitative characteristic of the element in its connections. The smallest value of the degree of oxidation of the element, which occurs in its compounds, is customary to be called a lower degree of oxidation; All others are intermediate. So, for sulfur, the highest degree of oxidation is +6, lower -2, intermediate +4.

The change in the degrees of the oxidation of elements by groups of the periodic system reflects the frequency of change chemical properties With the growth of the sequence number.

The concept of the degree of oxidation of the elements is used in the classification of substances, describing their properties, drawing up the formulas of the compounds and their international names. But it is particularly widely used in the study of redox reactions. The concept of "degree of oxidation" is often used in inorganic chemistry Instead of the notion of "valence" (see