EntranceOxidation degree 4. Correct formulation of substance formulas
When defining this concept, it is conventionally assumed that the bonding (valence) electrons move to more electronegative atoms (see Electronegativity), and therefore compounds consist of positively and negatively charged ions. The oxidation number can have zero, negative and positive values, which are usually placed above the element symbol at the top.
A zero oxidation state is assigned to atoms of elements in a free state, for example: Cu, H2, N2, P4, S6. Those atoms towards which the connecting electron cloud (electron pair) shifts have a negative oxidation state. For fluorine in all its compounds it is equal to −1. Atoms that donate valence electrons to other atoms have a positive oxidation state. For example, for alkali and alkaline earth metals it is equal to +1 and +2, respectively. In simple ions like Cl−, S2−, K+, Cu2+, Al3+, it is equal to the charge of the ion. In most compounds, the oxidation state of hydrogen atoms is +1, but in metal hydrides (their compounds with hydrogen) - NaH, CaH 2 and others - it is −1. Oxygen is characterized by an oxidation state of −2, but, for example, in combination with fluorine OF2 it will be +2, and in peroxide compounds (BaO2, etc.) −1. In some cases, this value can be expressed as a fraction: for iron in iron oxide (II, III) Fe 3 O 4 it is equal to +8/3.
The algebraic sum of the oxidation states of atoms in a compound is zero, and in a complex ion it is the charge of the ion. Using this rule, we calculate, for example, the oxidation state of phosphorus in orthophosphoric acid H 3 PO 4. Denoting it by x and multiplying the oxidation state for hydrogen (+1) and oxygen (−2) by the number of their atoms in the compound, we obtain the equation: (+1) 3+x+(−2) 4=0, whence x=+5 . Similarly, we calculate the oxidation state of chromium in the Cr 2 O 7 2− ion: 2x+(−2) 7=−2; x=+6. In the compounds MnO, Mn 2 O 3, MnO 2, Mn 3 O 4, K 2 MnO 4, KMnO 4, the oxidation state of manganese will be +2, +3, +4, +8/3, +6, +7, respectively.
The highest oxidation state is its greatest positive value. For most elements, it is equal to the group number in the periodic table and is an important quantitative characteristic of the element in its compounds. The lowest value of the oxidation state of an element that occurs in its compounds is usually called the lowest oxidation state; all others are intermediate. So, for sulfur, the highest oxidation state is +6, the lowest is −2, and the intermediate is +4.
The change in the oxidation states of elements by group of the periodic table reflects the periodicity of changes in their chemical properties with increasing atomic number.
The concept of the oxidation state of elements is used in the classification of substances, description of their properties, compilation of formulas of compounds and their international names. But it is especially widely used in the study of redox reactions. The concept of “oxidation state” is often used in inorganic chemistry instead of the concept of “valency” (see.
The degree of oxidation is a conventional value used to record redox reactions. To determine the degree of oxidation, the 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 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 outer energy level is completed. Most nonmetals are oxidizing agents.
Simple substances that do not react always have a zero oxidation state.
Rice. 1. Table of oxidation states.
In a compound, the nonmetal atom with lower electronegativity has a positive oxidation state.
Definition
You can determine the maximum and minimum oxidation states (how many electrons an atom can give and accept) using the periodic table.
The maximum degree 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 its lowest is -4. The maximum oxidation degree of sulfur is +6, the minimum is -2. Most nonmetals always have a variable - positive and negative - oxidation state. The exception is fluoride. 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 varying degrees of oxidation. The exception is aluminum. Despite being in group III, its oxidation state is always +3.
Rice. 3. Alkali and alkaline earth metals.
From group VIII, only ruthenium and osmium can exhibit the highest oxidation state +8. Gold and copper in group I exhibit oxidation states of +3 and +2, respectively.
Record
To correctly record the oxidation state, you should remember several rules:
- inert gases do not react, so their oxidation state is always zero;
- in compounds, the variable oxidation state depends on the variable valence 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 in a compound has accepted or given up. The value depends on the number of valence electrons. Metals in compounds always have a positive oxidation state, i.e. are reducing agents. For alkali and alkaline earth metals, the oxidation state is always the same. Nonmetals, except fluorine, can take on positive and negative oxidation states.
Test on the topic
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Modern formulation of the Periodic Law, discovered by D. I. Mendeleev in 1869:
The properties of elements are periodically dependent on the ordinal number.
The periodically repeating nature of changes in the composition of the electronic shell of atoms of elements explains the periodic change in the properties of elements when moving through the periods and groups of the Periodic System.
Let us trace, for example, the change in higher and lower oxidation states of elements of groups IA – VIIA in the second – fourth periods according to Table. 3.
Positive All elements exhibit oxidation states except fluorine. Their values increase with increasing nuclear charge and coincide with the number of electrons at the last energy level (with the exception of oxygen). These oxidation states are called highest 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 is missing 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 periodicity of changes in oxidation states is reflected in the periodic changes in the composition and properties of chemical compounds of elements.
A periodic change in the electronegativity of elements in the 1st-6th periods of groups IA–VIA can be similarly traced (Table 4).
In each period of the Periodic Table, the electronegativity of elements increases with increasing atomic 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 has the lowest electronegativity among the elements of the 1st-6th periods.
Typical nonmetals have high electronegativity, while typical metals have low electronegativity.
Examples of tasks for parts A, B1. In the 4th period the number of elements is equal to
2. Metallic properties of elements of the 3rd period from Na to Cl
1) get stronger
2) weaken
3) do not change
4) I don’t know
3. Nonmetallic properties of halogens with increasing atomic number
1) increase
2) decrease
3) remain unchanged
4) I don’t know
4. In the series of elements Zn – Hg – Co – Cd, one element not included in the group is
5. The metallic properties of elements increase in a number of ways
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) I don’t know
7. In the series of elements O – S – Se – Those sizes (radii) of an atom
1) decrease
2) increase
3) do not change
4) I don’t know
8. In the series of elements P – Si – Al – Mg, the dimensions (radii) of an atom are
1) decrease
2) increase
3) do not change
4) I don’t know
9. For phosphorus the element with less electronegativity is
10. A molecule in which the electron density is shifted towards the phosphorus atom is
11. Higher The oxidation state of elements is manifested in a set of oxides and fluorides
1) ClO 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. Lowest oxidation state of elements - in their hydrogen compounds and set fluorides
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. Valency for a multivalent atom is 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 oxidation state of carbon in it
Question No. 5. “The highest oxidation state of nitrogen in compounds is greater than the highest oxidation state of carbon, since …”
The outer energy level of the nitrogen atom contains 5 electrons, 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: The outer electron layer of the nitrogen atom has more electrons than the carbon atom.
Question No. 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 15%:
1000 H 1.10 = 1100g;
1100g of 15% solution contains:
To dissolve 27 g of Al you will need:
Answer: a) 890 ml.
Question No. 7. “The dehydrogenation reaction 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) increase the temperature;
Since the dehydrogenation reaction of butane is an endothermic process, it means that when the system is heated (with increasing temperature), the equilibrium shifts towards the endothermic reaction, the formation of butine (C 4 H 6).
50) lower the pressure;
Gaseous substances take part in the dehydrogenation reaction of butane. 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.
Task 54.
What is the lowest oxidation state of hydrogen, fluorine, sulfur and nitrogen? Why? Make up formulas for calcium compounds with these elements in this oxidation state. What are the names of the corresponding compounds?
Solution:
The lowest oxidation state is determined by the conditional charge which an atom acquires upon the addition of the number of electrons 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 outer energy level s 1, s 2 p 5, s 2 p 4 and s 2 p 3.
Thus, to complete the outer energy level, the hydrogen atom and fluorine atom need to add one electron each, the sulfur atom needs two, and the nitrogen atom needs 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? Make up formulas for compounds of these elements that correspond to these oxidation states.
Solution:
The highest oxidation state of an element is usually determined by the group number of the periodic table
D.I. Mendeleev, in which it is located. The lowest oxidation state is determined by the conventional 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 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 outer energy level s 2 p 2, s 2 p 3, s 2 p 4 and s 2 p5, respectively. Thus, the highest oxidation state of silicon for arsenic, selenium and chlorine 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 silicon, arsenic, selenium and chlorine 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. Make up formulas for its oxides and hydroxides corresponding to these oxidation states. Write reaction equations proving the amphoteric nature of chromium (III) hydroxide.
Solution:
Chromium forms compounds in which it exhibits oxidation states +2, +3, +6. Formulas of its oxides and hydroxides corresponding to these oxidation states:
a) chromium oxides:
CrO – chromium (II) oxide;
Cr 2 O 3 – chromium (III) oxide;
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 is an ampholyte, i.e. a substance that reacts with both acids and bases. Reaction equations proving the amphoteric nature 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 elements in the periodic table are continuously increasing, while the properties of simple bodies change periodically. How can this be explained? Give a reasoned answer.
Solution:
In most cases, with an increase in the charge of the nuclei of atoms of elements, their relative atomic masses naturally increase, because there is a natural 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 in atoms periodically changes at the outer energy level. In atoms of elements, periodically, as the charge of the nucleus increases, the number of electrons in 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 repeatability of the properties of Li, Na and K atoms is explained by the fact that at the outer energy level of their atoms there is one valence electron each. The properties of the atoms He, Ne, Ar, Kr, Xe and Rn are also periodically repeated - the atoms of these elements contain eight electrons at the outer energy level (helium has two electrons) - all of them are chemically inert, since their atoms cannot neither gain nor give up 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 larger atomic mass?
Solution:
The modern formulation of the periodic law: “The properties of chemical elements and the simple or complex substances they form are periodically dependent on the magnitude of the charge of the nucleus of the atoms of the elements.”
Since the K, Ni, I, Pa atoms have a lower relative mass than Ar, Co, Te, Th, respectively, 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. according to 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? Make up formulas for compounds of these elements that correspond to these oxidation states.
Solution:
The highest oxidation state of an element is determined, as a rule, by the number of the group of D.I. Mendeleev’s periodic system in which it is located. The lowest oxidation state is determined by the conventional charge that an atom acquires when adding the number of electrons that is necessary to form a stable eight-electron shell of the inert gas ns2np6 (in the case of hydrogen, ns2). Carbon, phosphorus, sulfur and iodine are located in IVA-, VA-, VIa- and VIIA-groups, respectively, and have the structure of the outer 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 – periodic acid.
The lowest oxidation states of carbon, phosphorus, sulfur and iodine are -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.
Which atoms of the fourth period of the periodic table form an oxide corresponding to their highest oxidation state E 2 O 5? Which one produces a gaseous compound with hydrogen? Make up formulas of acids corresponding to these oxides and depict them graphically?
Solution:
Oxide E 2 O 5, where the element is in its highest oxidation state +5, is characteristic of group V elements. Such an oxide can be formed by two elements of the fourth period and V-group - these are 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 a lower 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 the oxides of 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 (orthoarsenic) acid.
Graphic formulas of acids:
