When studying ionic and covalent polar chemical bonds, you got acquainted with complex substances consisting of two chemical elements. Such substances are called binary (from Latin bi - two) or two-element.
Let us recall the typical binary compounds that we cited as an example to consider the mechanisms for the formation of ionic and covalent polar chemical bonds: NaCl - sodium chloride and HCl - hydrogen chloride.
In the first case, the bond is ionic: the sodium atom transferred its outer electron to the chlorine atom and turned into an ion with a charge of +1, and the chlorine atom accepted an electron and turned into an ion with a charge of -1. Schematically, the process of transformation of atoms into ions can be depicted as follows:
In the HC1 hydrogen chloride molecule, a chemical bond is formed due to the pairing of unpaired outer electrons and the formation of a common electron pair of hydrogen and chlorine atoms:
It is more correct to represent the formation of a covalent bond in a hydrogen chloride molecule as an overlap of a one-electron s-cloud of a hydrogen atom with a one-electron p-cloud of a chlorine atom:
During chemical interaction, the common electron pair is shifted towards the more electronegative chlorine atom: , i.e., the electron will not completely transfer from the hydrogen atom to the chlorine atom, but partially, thereby causing a partial charge of atoms 5 (see § 12): . If we imagine that in the HCl molecule, as well as in sodium chloride NaCl, the electron completely passed from the hydrogen atom to the chlorine atom, then they would receive charges +1 and -1: . Such conditional charges are called the oxidation state. When defining this concept, it is conditionally assumed that in covalent polar compounds, the binding electrons have completely transferred to a more electronegative atom, and therefore the compounds consist only of positively and negatively charged ions.
The oxidation state can have a negative, positive, or zero value, which is usually placed above the element symbol at the top, for example:
Those atoms that have received electrons from other atoms or to which common electron pairs are displaced, i.e., atoms of more electronegative elements, have a negative oxidation state. Fluorine always has an oxidation state of -1 in all compounds. Oxygen, the second most electronegative element after fluorine, almost always has an oxidation state of -2, except for compounds with fluorine, for example:
Those atoms that donate their electrons to other atoms or from which common electron pairs are drawn, i.e., atoms of less electronegative elements, have a positive oxidation state. Metals in compounds always have a positive oxidation state. For metals of the main subgroups: group I (group IA) in all compounds, the oxidation state is +1, group II (group IIA) is +2, group III (group IIIA) - +3, for example:
but in compounds with metals, hydrogen has an oxidation state of -1:
The zero value of the oxidation state has atoms in the molecules of simple substances and atoms in the free state, for example:
Close to the concept of "oxidation state" is the concept of "valence", which you met when considering a covalent chemical bond. However, they are not the same.
The concept of "valency" is applicable to substances that have a molecular structure. The vast majority of organic substances that you will get acquainted with in grade 10 have just such a structure. In the course of the basic school, you study inorganic chemistry, the subject of which is substances of both molecular and non-molecular, for example, ionic, structure. Therefore, it is preferable to use the concept of "oxidation state".
What is the difference between valence and oxidation state?
Often the valence and the oxidation state are numerically the same, but the valency does not have a charge sign, and the oxidation state does. For example, monovalent hydrogen has the following oxidation states in various substances:
It would seem that monovalent fluorine - the most electronegative element - should have a complete coincidence of the values of the oxidation state and valency. After all, its atom is capable of forming only one single covalent bond, since it lacks one electron until the completion of the outer electronic layer. However, here too there is a difference:
The valency and oxidation state differ even more if they do not coincide numerically. For example:
In compounds, the total oxidation state is always zero. Knowing this and the oxidation state of one of the elements, you can find the oxidation state of another element by the formula, for example, a binary compound. So, let's find the oxidation state of chlorine in the compound C1 2 O 7.
Let us denote the degree of oxidation of oxygen: . Therefore, seven oxygen atoms will have a total negative charge of (-2) × 7 = -14. Then the total charge of two chlorine atoms will be +14, and one chlorine atom: (+14) : 2 = +7. Therefore, the oxidation state of chlorine is .
Similarly, knowing the oxidation states of the elements, one can formulate the formula of a compound, for example, aluminum carbide (a compound of aluminum and carbon).
It is easy to see that you worked similarly with the concept of "valency" when you derived the formula of a covalent compound or determined the valency of an element by the formula of its compound.
The names of binary compounds are formed from two words - the names of their constituent chemical elements. The first word denotes the electronegative part of the compound - non-metal, its Latin name with the suffix -id is always in the nominative case. The second word denotes the electropositive part - a metal or a less electronegative element, its name is always in the genitive case:
For example: NaCl - sodium chloride, MgS - magnesium sulfide, KH - potassium hydride, CaO - calcium oxide. If the electropositive element exhibits different degrees of oxidation, then this is reflected in the name, indicating the degree of oxidation with a Roman numeral, which is placed at the end of the name, for example: - iron oxide (II) (read "iron oxide two"), - iron oxide (III) (read "iron oxide three").
If the compound consists of two non-metal elements, then the suffix -id is added to the name of the more electronegative of them, the second component is placed after that in the genitive case. For example: - oxygen fluoride (II), - sulfur oxide (IV) and - sulfur oxide (VI).
In some cases, the number of atoms of elements is indicated using the names of numerals in Greek - mono, di, three, tetra, penta, hexa, etc. For example: - carbon monoxide, or carbon monoxide (II), - carbon dioxide, or oxide carbon (IV), - lead tetrachloride, or lead (IV) chloride.
In order for chemists from different countries to understand each other, it was necessary to create a unified terminology and nomenclature of substances. The principles of chemical nomenclature were first developed by French chemists A. Lavoisier, A. Fourcroix, L. Giton de Mervaux and C. Berthollet in 1785. Currently, the International Union of Pure and Applied Chemistry (IUPAC) coordinates the activities of scientists from different countries and issues recommendations on nomenclature of substances and terminology used in chemistry.
Keywords and phrases
- Binary, or two-element, compounds.
- The degree of oxidation.
- Chemical nomenclature.
- Determination of the oxidation states of elements by the formula.
- Drawing up formulas of binary compounds according to the oxidation states of the elements.
Work with computer
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Questions and tasks
- Write down the formulas of nitrogen oxides (II), (V), (I), (III), (IV).
- Give the names of binary compounds whose formulas are: a) С1 2 0 7 , С1 2 O, С1O 2 ; b) FeCl 2, FeCl 3; c) MnS, MnO 2 , MnF 4 , MnO, MnCl 4 ; r) Cu 2 O, Mg 2 Si, SiCl 4, Na 3 N, FeS.
- Find in reference books and dictionaries all kinds of names of substances with formulas: a) CO 2 and CO; b) SO 2 and SO 3. Explain their etymology. Give two names of these substances according to the international nomenclature in accordance with the rules set out in the paragraph.
- What other name can be given to ammonia H 3 N?
- Find the volume that they have at n. y. 17 g of hydrogen sulfide.
- How many molecules are contained in this volume?
- Calculate the mass of 33.6 m3 of methane CH 2 at n. y. and determine the number of its molecules contained in this volume.
- Determine the oxidation state of carbon and write down the structural formulas of the following substances, knowing that carbon in organic compounds is always tetravalent: methane CH 4, carbon tetrachloride CC1 4, ethane C 2 H 4, acetylene C 2 H 2.
In chemical processes, the main role is played by atoms and molecules, the properties of which determine the outcome of chemical reactions. One of the important characteristics of an atom is the oxidation number, which simplifies the method of taking into account the transfer of electrons in a particle. How to determine the oxidation state or the formal charge of a particle and what rules do you need to know for this?
Definition
Any chemical reaction is due to the interaction of atoms of various substances. The reaction process and its result depend on the characteristics of the smallest particles.
The term oxidation (oxidation) in chemistry means a reaction during which a group of atoms or one of them lose electrons or gain, in the case of acquisition, the reaction is called "reduction".
The oxidation state is a quantity that is measured quantitatively and characterizes the redistributed electrons during the reaction. Those. in the process of oxidation, the electrons in the atom decrease or increase, being redistributed among other interacting particles, and the level of oxidation shows exactly how they are reorganized. This concept is closely related to the electronegativity of particles - their ability to attract and repel free ions from themselves.
Determining the level of oxidation depends on the characteristics and properties of a particular substance, so the calculation procedure cannot be unequivocally called easy or complex, but its results help to conditionally record the processes of redox reactions. It should be understood that the obtained result of calculations is the result of taking into account the transfer of electrons and has no physical meaning, and is not the true charge of the nucleus.
It is important to know! Inorganic chemistry often uses the term valency instead of the oxidation state of elements, this is not a mistake, but it should be borne in mind that the second concept is more universal.
The concepts and rules for calculating the movement of electrons are the basis for classifying chemicals (nomenclature), describing their properties and compiling communication formulas. But most often this concept is used to describe and work with redox reactions.
Rules for determining the degree of oxidation
How to find out the degree of oxidation? When working with redox reactions, it is important to know that the formal charge of a particle will always be equal to the magnitude of the electron, expressed in numerical value. This feature is connected with the assumption that the electron pairs that form a bond are always completely shifted towards more negative particles. It should be understood that we are talking about ionic bonds, and in the case of a reaction at , electrons will be divided equally between identical particles.
The oxidation number can have both positive and negative values. The thing is that during the reaction, the atom must become neutral, and for this you need to either attach a certain number of electrons to the ion, if it is positive, or take them away if it is negative. To designate this concept, when writing formulas, an Arabic numeral with the corresponding sign is usually written above the designation of the element. For example, or etc.
You should know that the formal charge of metals will always be positive, and in most cases, you can use the periodic table to determine it. There are a number of features that must be taken into account in order to determine the indicators correctly.
Degree of oxidation:
Having remembered these features, it will be quite simple to determine the oxidation number of elements, regardless of the complexity and number of atomic levels.
Useful video: determining the degree of oxidation
The periodic table of Mendeleev contains almost all the necessary information for working with chemical elements. For example, schoolchildren use only it to describe chemical reactions. So, in order to determine the maximum positive and negative values of the oxidation number, it is necessary to check the designation of the chemical element in the table:
- The maximum positive is the number of the group in which the element is located.
- The maximum negative oxidation state is the difference between the maximum positive limit and the number 8.
Thus, it is enough to simply find out the extreme boundaries of the formal charge of an element. Such an action can be performed using calculations based on the periodic table.
It is important to know! One element can have several different oxidation indices at the same time.
There are two main ways to determine the level of oxidation, examples of which are presented below. The first of these is a method that requires knowledge and skills to apply the laws of chemistry. How to arrange oxidation states using this method?
The rule for determining oxidation states
For this you need:
- Determine whether a given substance is elemental and whether it is out of bond. If yes, then its oxidation number will be equal to 0, regardless of the composition of the substance (individual atoms or multilevel atomic compounds).
- Determine whether the substance in question consists of ions. If yes, then the degree of oxidation will be equal to their charge.
- If the substance in question is a metal, then look at the indicators of other substances in the formula and calculate the metal readings by arithmetic.
- If the entire compound has one charge (in fact, this is the sum of all the particles of the elements presented), then it is enough to determine the indicators of simple substances, then subtract them from the total amount and obtain metal data.
- If the relationship is neutral, then the total must be zero.
For example, consider combining with an aluminum ion whose total charge is zero. The rules of chemistry confirm the fact that the Cl ion has an oxidation number of -1, and in this case there are three of them in the compound. So the Al ion must be +3 for the entire compound to be neutral.
This method is quite good, since the correctness of the solution can always be checked by adding all the oxidation levels together.
The second method can be applied without knowledge of chemical laws:
- Find particle data for which there are no strict rules and the exact number of their electrons is unknown (possible by elimination).
- Find out the indicators of all other particles and then from the total amount by subtracting find the desired particle.
Let us consider the second method using the Na2SO4 substance as an example, in which the sulfur atom S is not defined, it is only known that it is nonzero.
To find what all oxidation states are equal to:
- Find known elements, keeping traditional rules and exceptions in mind.
- Na ion = +1 and each oxygen = -2.
- Multiply the number of particles of each substance by their electrons and get the oxidation states of all atoms except one.
- Na2SO4 consists of 2 sodium and 4 oxygen, when multiplied it turns out: 2 X +1 \u003d 2 is the oxidizing number of all sodium particles and 4 X -2 \u003d -8 - oxygen.
- Add the results 2+(-8) = -6 - this is the total charge of the compound without a sulfur particle.
- Express the chemical notation as an equation: sum of known data + unknown number = total charge.
- Na2SO4 is represented as follows: -6 + S = 0, S = 0 + 6, S = 6.
Thus, to use the second method, it is enough to know the simple laws of arithmetic.
A chemical element in a compound, calculated from the assumption that all bonds are ionic.
The oxidation states can have a positive, negative or zero value, therefore the algebraic sum of the oxidation states of elements in a molecule, taking into account the number of their atoms, is 0, and in an ion - the charge of the ion.
1. The oxidation states of metals in compounds are always positive.
2. The highest oxidation state corresponds to the group number of the periodic system where this element is located (the exception is: Au+3(I group), Cu+2(II), from group VIII, the oxidation state +8 can only be in osmium Os and ruthenium Ru.
3. The oxidation states of non-metals depend on which atom it is connected to:
- if with a metal atom, then the oxidation state is negative;
- if with a non-metal atom, then the oxidation state can be both positive and negative. It depends on the electronegativity of the atoms of the elements.
4. The highest negative oxidation state of non-metals can be determined by subtracting from 8 the number of the group in which this element is located, i.e. the highest positive oxidation state is equal to the number of electrons on the outer layer, which corresponds to the group number.
5. The oxidation states of simple substances are 0, regardless of whether it is a metal or a non-metal.
Elements with constant oxidation states.
|
Element |
Characteristic oxidation state |
Exceptions |
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Metal hydrides: LIH-1 |
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oxidation state called the conditional charge of the particle under the assumption that the bond is completely broken (has an ionic character). H- Cl = H + + Cl - , The bond in hydrochloric acid is covalent polar. The electron pair is more biased towards the atom Cl - , because it is more electronegative whole element. How to determine the degree of oxidation?Electronegativity is the ability of atoms to attract electrons from other elements. The oxidation state is indicated above the element: Br 2 0 , Na 0 , O +2 F 2 -1 ,K + Cl - etc. It can be negative and positive. The oxidation state of a simple substance (unbound, free state) is zero. The oxidation state of oxygen in most compounds is -2 (the exception is peroxides H 2 O 2, where it is -1 and compounds with fluorine - O +2 F 2 -1 , O 2 +1 F 2 -1 ). - Oxidation state a simple monatomic ion is equal to its charge: Na + , Ca +2 . Hydrogen in its compounds has an oxidation state of +1 (exceptions are hydrides - Na + H - and type connections C +4 H 4 -1 ). In metal-non-metal bonds, the atom that has the highest electronegativity has a negative oxidation state (electronegativity data are given on the Pauling scale): H + F - , Cu + Br - , Ca +2 (NO 3 ) - etc. Rules for determining the degree of oxidation in chemical compounds.Let's take a connection KMnO 4 , it is necessary to determine the oxidation state of the manganese atom. Reasoning:
K+MnXO 4 -2 Let be X- unknown to us the degree of oxidation of manganese. The number of potassium atoms is 1, manganese - 1, oxygen - 4. It is proved that the molecule as a whole is electrically neutral, so its total charge must be equal to zero. 1*(+1) + 1*(X) + 4(-2) = 0, X = +7, Hence, the oxidation state of manganese in potassium permanganate = +7. Let's take another example of an oxide Fe2O3. It is necessary to determine the oxidation state of the iron atom. Reasoning:
2*(X) + 3*(-2) = 0, Conclusion: the oxidation state of iron in this oxide is +3. Examples. Determine the oxidation states of all atoms in the molecule. 1. K2Cr2O7. Oxidation state K+1, oxygen O -2. Given indexes: O=(-2)×7=(-14), K=(+1)×2=(+2). Because the algebraic sum of the oxidation states of elements in a molecule, taking into account the number of their atoms, is 0, then the number of positive oxidation states is equal to the number of negative ones. Oxidation states K+O=(-14)+(+2)=(-12). It follows from this that the number of positive powers of the chromium atom is 12, but there are 2 atoms in the molecule, which means that there are (+12):2=(+6) per atom. Answer: K 2 + Cr 2 +6 O 7 -2. 2.(AsO 4) 3-. In this case, the sum of the oxidation states will no longer be equal to zero, but the charge of the ion, i.e. - 3. Let's make an equation: x+4×(- 2)= - 3 . Answer: (As +5 O 4 -2) 3-. |
DEFINITION
Oxidation state is a quantitative assessment of the state of an atom of a chemical element in a compound, based on its electronegativity.
It takes both positive and negative values. To indicate the oxidation state of an element in a compound, you need to put an Arabic numeral with the corresponding sign ("+" or "-") above its symbol.
It should be remembered that the degree of oxidation is a quantity that has no physical meaning, since it does not reflect the real charge of the atom. However, this concept is very widely used in chemistry.
Table of the oxidation state of chemical elements
The maximum positive and minimum negative oxidation state can be determined using the Periodic Table of D.I. Mendeleev. They are equal to the number of the group in which the element is located, and the difference between the value of the "highest" oxidation state and the number 8, respectively.
If we consider chemical compounds more specifically, then in substances with non-polar bonds, the oxidation state of the elements is zero (N 2, H 2, Cl 2).
The oxidation state of metals in the elementary state is zero, since the distribution of electron density in them is uniform.
In simple ionic compounds, the oxidation state of their constituent elements is equal to the electric charge, since during the formation of these compounds, an almost complete transfer of electrons from one atom to another occurs: Na +1 I -1, Mg +2 Cl -1 2, Al +3 F - 1 3 , Zr +4 Br -1 4 .
When determining the degree of oxidation of elements in compounds with polar covalent bonds, the values of their electronegativity are compared. Since, during the formation of a chemical bond, electrons are displaced to atoms of more electronegative elements, the latter have a negative oxidation state in compounds.
There are elements for which only one value of the oxidation state is characteristic (fluorine, metals of IA and IIA groups, etc.). Fluorine, which is characterized by the highest electronegativity, always has a constant negative oxidation state (-1) in compounds.
Alkaline and alkaline earth elements, which are characterized by a relatively low value of electronegativity, always have a positive oxidation state, equal to (+1) and (+2), respectively.
However, there are also such chemical elements, which are characterized by several values of the degree of oxidation (sulfur - (-2), 0, (+2), (+4), (+6), etc.).
In order to make it easier to remember how many and what oxidation states are characteristic of a particular chemical element, tables of the oxidation states of chemical elements are used, which look like this:
|
Serial number |
Russian / English title |
chemical symbol |
Oxidation state |
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Hydrogen |
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Helium / Helium |
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Lithium / Lithium |
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Beryllium / Beryllium |
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(-1), 0, (+1), (+2), (+3) |
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Carbon / Carbon |
(-4), (-3), (-2), (-1), 0, (+2), (+4) |
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Nitrogen / Nitrogen |
(-3), (-2), (-1), 0, (+1), (+2), (+3), (+4), (+5) |
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Oxygen / Oxygen |
(-2), (-1), 0, (+1), (+2) |
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Fluorine / Fluorine |
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Sodium |
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Magnesium / Magnesium |
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Aluminum |
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Silicon / Silicon |
(-4), 0, (+2), (+4) |
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Phosphorus / Phosphorus |
(-3), 0, (+3), (+5) |
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Sulfur |
(-2), 0, (+4), (+6) |
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Chlorine / Chlorine |
(-1), 0, (+1), (+3), (+5), (+7), rarely (+2) and (+4) |
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Argon / Argon |
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Potassium / Potassium |
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Calcium / Calcium |
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Scandium / Scandium |
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Titanium / Titanium |
(+2), (+3), (+4) |
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Vanadium / Vanadium |
(+2), (+3), (+4), (+5) |
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Chromium / Chromium |
(+2), (+3), (+6) |
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Manganese / Manganese |
(+2), (+3), (+4), (+6), (+7) |
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Iron / Iron |
(+2), (+3), rarely (+4) and (+6) |
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Cobalt / Cobalt |
(+2), (+3), rarely (+4) |
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Nickel / Nickel |
(+2), rarely (+1), (+3) and (+4) |
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Copper |
+1, +2, rare (+3) |
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Gallium / Gallium |
(+3), rare (+2) |
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Germanium / Germanium |
(-4), (+2), (+4) |
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Arsenic / Arsenic |
(-3), (+3), (+5), rarely (+2) |
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Selenium / Selenium |
(-2), (+4), (+6), rarely (+2) |
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Bromine / Bromine |
(-1), (+1), (+5), rarely (+3), (+4) |
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Krypton / Krypton |
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Rubidium / Rubidium |
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Strontium / Strontium |
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Yttrium / Yttrium |
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Zirconium / Zirconium |
(+4), rarely (+2) and (+3) |
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Niobium / Niobium |
(+3), (+5), rarely (+2) and (+4) |
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Molybdenum / Molybdenum |
(+3), (+6), rarely (+2), (+3) and (+5) |
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Technetium / Technetium |
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Ruthenium / Ruthenium |
(+3), (+4), (+8), rarely (+2), (+6) and (+7) |
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Rhodium |
(+4), rarely (+2), (+3) and (+6) |
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Palladium / Palladium |
(+2), (+4), rarely (+6) |
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Silver / Silver |
(+1), rarely (+2) and (+3) |
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Cadmium / Cadmium |
(+2), rare (+1) |
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Indium / Indium |
(+3), rarely (+1) and (+2) |
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Tin / Tin |
(+2), (+4) |
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Antimony / Antimony |
(-3), (+3), (+5), rarely (+4) |
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Tellurium / Tellurium |
(-2), (+4), (+6), rarely (+2) |
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(-1), (+1), (+5), (+7), rarely (+3), (+4) |
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Xenon / Xenon |
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Cesium / Cesium |
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Barium / Barium |
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Lanthanum / Lanthanum |
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Cerium / Cerium |
(+3), (+4) |
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Praseodymium / Praseodymium |
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Neodymium / Neodymium |
(+3), (+4) |
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Promethium / Promethium |
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Samaria / Samarium |
(+3), rare (+2) |
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Europium / Europium |
(+3), rare (+2) |
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Gadolinium / Gadolinium |
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Terbium / Terbium |
(+3), (+4) |
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Dysprosium / Dysprosium |
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Holmium / Holmium |
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Erbium / Erbium |
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Thulium / Thulium |
(+3), rare (+2) |
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Ytterbium / Ytterbium |
(+3), rare (+2) |
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Lutetium / Lutetium |
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Hafnium / Hafnium |
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Tantalum / Tantalum |
(+5), rarely (+3), (+4) |
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Tungsten / Tungsten |
(+6), rare (+2), (+3), (+4) and (+5) |
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Rhenium / Rhenium |
(+2), (+4), (+6), (+7), rarely (-1), (+1), (+3), (+5) |
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Osmium / Osmium |
(+3), (+4), (+6), (+8), rarely (+2) |
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|
Iridium / Iridium |
(+3), (+4), (+6), rarely (+1) and (+2) |
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Platinum / Platinum |
(+2), (+4), (+6), rarely (+1) and (+3) |
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Gold / Gold |
(+1), (+3), rarely (+2) |
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Mercury / Mercury |
(+1), (+2) |
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Waist / Thallium |
(+1), (+3), rarely (+2) |
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Lead / Lead |
(+2), (+4) |
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Bismuth / Bismuth |
(+3), rarely (+3), (+2), (+4) and (+5) |
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Polonium / Polonium |
(+2), (+4), rarely (-2) and (+6) |
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Astatine / Astatine |
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Radon / Radon |
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Francium / Francium |
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Radium / Radium |
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Actinium / Actinium |
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Thorium / Thorium |
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Proactinium / Protactinium |
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|
Uranus / Uranium |
(+3), (+4), (+6), rarely (+2) and (+5) |
Examples of problem solving
EXAMPLE 1
- The oxidation state of phosphorus in phosphine is (-3), and in phosphoric acid - (+5). Change in the oxidation state of phosphorus: +3 → +5, i.e. the first answer.
- The oxidation state of a chemical element in a simple substance is zero. The oxidation state of phosphorus in the oxide composition P 2 O 5 is equal to (+5). Change in the oxidation state of phosphorus: 0 → +5, i.e. third answer.
- The oxidation state of phosphorus in an acid of composition HPO 3 is (+5), and H 3 PO 2 is (+1). Change in the oxidation state of phosphorus: +5 → +1, i.e. fifth answer.
EXAMPLE 2
| Exercise | The oxidation state (-3) carbon has in the compound: a) CH 3 Cl; b) C 2 H 2 ; c) HCOH; d) C 2 H 6 . |
| Decision | In order to give a correct answer to the question posed, we will alternately determine the degree of carbon oxidation in each of the proposed compounds. a) the oxidation state of hydrogen is (+1), and chlorine - (-1). We take for "x" the degree of oxidation of carbon: x + 3×1 + (-1) =0; The answer is incorrect. b) the oxidation state of hydrogen is (+1). We take for "y" the degree of oxidation of carbon: 2×y + 2×1 = 0; The answer is incorrect. c) the oxidation state of hydrogen is (+1), and oxygen - (-2). Let's take for "z" the oxidation state of carbon: 1 + z + (-2) +1 = 0: The answer is incorrect. d) the oxidation state of hydrogen is (+1). Let's take for "a" the oxidation state of carbon: 2×a + 6×1 = 0; Correct answer. |
| Answer | Option (d) |
To characterize the state of elements in compounds, the concept of the degree of oxidation has been introduced.
DEFINITION
The number of electrons displaced from an atom of a given element or to an atom of a given element in a compound is called oxidation state.
A positive oxidation state indicates the number of electrons that are displaced from a given atom, and a negative oxidation state indicates the number of electrons that are displaced towards a given atom.
From this definition it follows that in compounds with non-polar bonds, the oxidation state of the elements is zero. Molecules consisting of identical atoms (N 2 , H 2 , Cl 2) can serve as examples of such compounds.
The oxidation state of metals in the elementary state is zero, since the distribution of electron density in them is uniform.
In simple ionic compounds, the oxidation state of their constituent elements is equal to the electric charge, since during the formation of these compounds, an almost complete transfer of electrons from one atom to another occurs: Na +1 I -1, Mg +2 Cl -1 2, Al +3 F - 1 3 , Zr +4 Br -1 4 .
When determining the degree of oxidation of elements in compounds with polar covalent bonds, the values of their electronegativity are compared. Since, during the formation of a chemical bond, electrons are displaced to atoms of more electronegative elements, the latter have a negative oxidation state in compounds.
Lowest oxidation state
For elements that exhibit different oxidation states in their compounds, there are concepts of higher (maximum positive) and lower (minimum negative) oxidation states. The lowest oxidation state of a chemical element is usually numerically equal to the difference between the group number in the Periodic system of D. I. Mendeleev, in which the chemical element is located, and the number 8. For example, nitrogen is in the VA group, which means its lowest oxidation state is (-3): V-VIII = -3; sulfur is in group VIA, so its lowest oxidation state is (-2): VI-VIII = -2, etc.
Examples of problem solving
EXAMPLE 1
