How to determine the degree of oxidation? The periodic table allows you to record a given quantitative value for any chemical element.

Definition

First, let's try to understand what this term is. The oxidation state according to the periodic table is the number of electrons that are accepted or given away by an element in the process of chemical interaction. It can take both negative and positive values.

Link to table

How is oxidation state determined? The periodic table consists of eight groups arranged vertically. Each of them has two subgroups: main and secondary. In order to set indicators for elements, certain rules must be used.

Instruction

How to calculate the oxidation states of elements? The table allows you to fully cope with a similar problem. Alkali metals, which are located in the first group (main subgroup), the oxidation state is shown in compounds, it corresponds to +, is equal to their highest valence. Metals of the second group (subgroup A) have +2 oxidation state.

The table allows you to determine this value not only for elements that exhibit metallic properties, but also for non-metals. Their maximum value will correspond to the highest valence. For example, for sulfur it will be +6, for nitrogen +5. How is their minimum (lowest) figure calculated? The table also answers this question. Subtract the group number from eight. For example, for oxygen it will be -2, for nitrogen -3.

For simple substances that did not enter into chemical interaction with other substances, the determined indicator is considered to be zero.

Let's try to identify the main actions related to the arrangement in binary compounds. How to put in them the degree of oxidation? The periodic table helps solve the problem.

For example, take calcium oxide CaO. For calcium located in the main subgroup of the second group, the value will be constant, equal to +2. For oxygen, which has non-metallic properties, this indicator will be a negative value, and it corresponds to -2. In order to check the correctness of the definition, we summarize the obtained numbers. As a result, we get zero, therefore, the calculations are correct.

Let us determine similar indicators in one more binary compound CuO. Since copper is located in a secondary subgroup (the first group), therefore, the indicator under study may show different values. Therefore, to determine it, you must first identify the indicator for oxygen.

For a non-metal located at the end of a binary formula, the oxidation state has a negative value. Since this element is located in the sixth group, when subtracting six from eight, we get that the oxidation state of oxygen corresponds to -2. Since there are no indices in the compound, therefore, the oxidation state of copper will be positive, equal to +2.

How else is the chemistry table used? The oxidation states of elements in formulas consisting of three elements are also calculated according to a certain algorithm. First, these indicators are placed at the first and last element. For the first, this indicator will have a positive value, correspond to valence. For the extreme element, which is a non-metal, this indicator has a negative value, it is determined as a difference (the group number is subtracted from eight). When calculating the oxidation state of the central element, a mathematical equation is used. The calculations take into account the indices available for each element. The sum of all oxidation states must be zero.

Example of determination in sulfuric acid

The formula of this compound is H 2 SO 4 . Hydrogen has an oxidation state of +1, oxygen has -2. To determine the oxidation state of sulfur, we compose a mathematical equation: + 1 * 2 + X + 4 * (-2) = 0. We get that the oxidation state of sulfur corresponds to +6.

Conclusion

When using the rules, you can arrange the coefficients in redox reactions. This issue is considered in the course of chemistry of the ninth grade of the school curriculum. In addition, information about the degrees of oxidation allows you to complete the tasks of the OGE and the Unified State Examination.

To place correctly oxidation states There are four rules to keep in mind.

1) In a simple substance, the oxidation state of any element is 0. Examples: Na 0, H 0 2, P 0 4.

2) You should remember the elements for which are characteristic constant oxidation states. All of them are listed in the table.

3) The highest oxidation state of an element, as a rule, coincides with the number of the group in which this element is located (for example, phosphorus is in group V, the highest SD of phosphorus is +5). Important exceptions: F, O.

4) The search for the oxidation states of the remaining elements is based on a simple rule:

In a neutral molecule, the sum of the oxidation states of all elements is equal to zero, and in an ion - the charge of the ion.

A few simple examples for determining oxidation states

Example 1. It is necessary to find the oxidation states of elements in ammonia (NH 3).

Decision. We already know (see 2) that Art. OK. hydrogen is +1. It remains to find this characteristic for nitrogen. Let x be the desired oxidation state. We compose the simplest equation: x + 3 (+1) \u003d 0. The solution is obvious: x \u003d -3. Answer: N -3 H 3 +1.

Example 2. Specify the oxidation states of all atoms in the H 2 SO 4 molecule.

Decision. The oxidation states of hydrogen and oxygen are already known: H(+1) and O(-2). We compose an equation for determining the degree of oxidation of sulfur: 2 (+1) + x + 4 (-2) \u003d 0. Solving this equation, we find: x \u003d +6. Answer: H +1 2 S +6 O -2 4 .

Example 3. Calculate the oxidation states of all elements in the Al(NO 3) 3 molecule.

Decision. The algorithm remains unchanged. The composition of the "molecule" of aluminum nitrate includes one atom of Al (+3), 9 oxygen atoms (-2) and 3 nitrogen atoms, the oxidation state of which we have to calculate. Corresponding equation: 1 (+3) + 3x + 9 (-2) = 0. Answer: Al +3 (N +5 O -2 3) 3.

Example 4. Determine the oxidation states of all atoms in the (AsO 4) 3- ion.

Decision. In this case, the sum of the oxidation states will no longer be equal to zero, but to the charge of the ion, i.e., -3. Equation: x + 4 (-2) = -3. Answer: As(+5), O(-2).

What to do if the oxidation states of two elements are unknown

Is it possible to determine the oxidation states of several elements at once using a similar equation? If we consider this problem from the point of view of mathematics, the answer will be negative. A linear equation with two variables cannot have a unique solution. But we are not just solving an equation!

Example 5. Determine the oxidation states of all elements in (NH 4) 2 SO 4.

Decision. The oxidation states of hydrogen and oxygen are known, but sulfur and nitrogen are not. A classic example of a problem with two unknowns! We will consider ammonium sulfate not as a single "molecule", but as a combination of two ions: NH 4 + and SO 4 2-. We know the charges of ions, each of them contains only one atom with an unknown degree of oxidation. Using the experience gained in solving previous problems, we can easily find the oxidation states of nitrogen and sulfur. Answer: (N -3 H 4 +1) 2 S +6 O 4 -2.

Conclusion: if the molecule contains several atoms with unknown oxidation states, try to "split" the molecule into several parts.

How to arrange oxidation states in organic compounds

Example 6. Indicate the oxidation states of all elements in CH 3 CH 2 OH.

Decision. Finding oxidation states in organic compounds has its own specifics. In particular, it is necessary to separately find the oxidation states for each carbon atom. You can reason as follows. Consider, for example, the carbon atom in the methyl group. This C atom is connected to 3 hydrogen atoms and an adjacent carbon atom. On the C-H bond, the electron density shifts towards the carbon atom (because the electronegativity of C exceeds the EO of hydrogen). If this displacement were complete, the carbon atom would acquire a charge of -3.

The C atom in the -CH 2 OH group is bonded to two hydrogen atoms (electron density shift towards C), one oxygen atom (electron density shift towards O) and one carbon atom (we can assume that the shifts in electron density in this case not happening). The oxidation state of carbon is -2 +1 +0 = -1.

Answer: C -3 H +1 3 C -1 H +1 2 O -2 H +1.

Do not confuse the concepts of "valence" and "oxidation state"!

Oxidation state is often confused with valence. Don't make that mistake. I will list the main differences:

• the oxidation state has a sign (+ or -), valence - no;
• the degree of oxidation can be equal to zero even in a complex substance, the equality of valency to zero means, as a rule, that the atom of this element is not connected to other atoms (we will not discuss any kind of inclusion compounds and other "exotics" here);
• the degree of oxidation is a formal concept that acquires real meaning only in compounds with ionic bonds, the concept of "valency", on the contrary, is most conveniently applied to covalent compounds.

The oxidation state (more precisely, its modulus) is often numerically equal to the valence, but even more often these values ​​do NOT coincide. For example, the oxidation state of carbon in CO 2 is +4; valency C is also equal to IV. But in methanol (CH 3 OH), the valency of carbon remains the same, and the oxidation state of C is -1.

A small test on the topic "The degree of oxidation"

Take a few minutes to check how you have understood this topic. You need to answer five simple questions. Good luck!

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 connections

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

Chemistry preparation for ZNO and DPA
Comprehensive edition

PART AND

GENERAL CHEMISTRY

CHEMICAL BOND AND STRUCTURE OF SUBSTANCE

Oxidation state

The oxidation state is the conditional charge on an atom in a molecule or crystal that arose on it when all the polar bonds created by it were of an ionic nature.

Unlike valency, oxidation states can be positive, negative, or zero. In simple ionic compounds, the oxidation state coincides with the charges of the ions. For example, in sodium chloride NaCl (Na + Cl - ) Sodium has an oxidation state of +1, and Chlorine -1, in calcium oxide CaO (Ca +2 O -2) Calcium exhibits an oxidation state of +2, and Oxysen - -2. This rule applies to all basic oxides: the oxidation state of a metallic element is equal to the charge of the metal ion (Sodium +1, Barium +2, Aluminum +3), and the oxidation state of Oxygen is -2. The degree of oxidation is indicated by Arabic numerals, which are placed above the symbol of the element, like valence, and first indicate the sign of the charge, and then its numerical value:

If the module of the oxidation state is equal to one, then the number "1" can be omitted and only the sign can be written: Na + Cl - .

The oxidation state and valency are related concepts. In many compounds, the absolute value of the oxidation state of the elements coincides with their valency. However, there are many cases where the valency differs from the oxidation state.

In simple substances - non-metals, there is a covalent non-polar bond, a joint electron pair is shifted to one of the atoms, therefore the degree of oxidation of elements in simple substances is always zero. But the atoms are connected to each other, that is, they exhibit a certain valence, as, for example, in oxygen, the valency of Oxygen is II, and in nitrogen, the valency of Nitrogen is III:

In a hydrogen peroxide molecule, the valency of Oxygen is also II, and Hydrogen is I:

Definition of possible degrees element oxidation

The oxidation states, which elements can show in various compounds, in most cases can be determined by the structure of the external electronic level or by the place of the element in the Periodic system.

Atoms of metallic elements can only donate electrons, so in compounds they exhibit positive oxidation states. Its absolute value in many cases (with the exception of d -elements) is equal to the number of electrons in the outer level, that is, the group number in the Periodic system. atoms d -elements can also donate electrons from the front level, namely from unfilled d -orbitals. Therefore, for d -elements, it is much more difficult to determine all possible oxidation states than for s- and p-elements. It is safe to say that the majority d -elements exhibit an oxidation state of +2 due to the electrons of the outer electronic level, and the maximum oxidation state in most cases is equal to the group number.

Atoms of non-metallic elements can exhibit both positive and negative oxidation states, depending on which atom of which element they form a bond with. If the element is more electronegative, then it exhibits a negative oxidation state, and if less electronegative - positive.

The absolute value of the oxidation state of non-metallic elements can be determined from the structure of the outer electronic layer. An atom is able to accept so many electrons that eight electrons are located on its outer level: non-metallic elements of group VII take one electron and show an oxidation state of -1, group VI - two electrons and show an oxidation state of -2, etc.

Non-metallic elements are capable of giving off a different number of electrons: a maximum of as many as are located on the external energy level. In other words, the maximum oxidation state of non-metallic elements is equal to the group number. Due to electron spooling at the outer level of atoms, the number of unpaired electrons that an atom can donate in chemical reactions varies, so non-metallic elements are able to exhibit various intermediate oxidation states.

Possible oxidation states s - and p-elements

PS Group
Highest oxidation state
Intermediate oxidation state
Lower oxidation state

Determination of oxidation states in compounds

Any electrically neutral molecule, so the sum of the oxidation states of the atoms of all elements must be zero. Let us determine the degree of oxidation in sulfur(I V) oxide SO 2 tauphosphorus (V) sulfide P 2 S 5.

Sulfur (And V) oxide SO 2 formed by atoms of two elements. Of these, Oxygen has the largest electronegativity, so Oxygen atoms will have a negative oxidation state. For Oxygen it is -2. In this case Sulfur has a positive oxidation state. In different compounds, Sulfur can show different oxidation states, so in this case it must be calculated. In a molecule SO2 two oxygen atoms with an oxidation state of -2, so the total charge of the oxygen atoms is -4. In order for the molecule to be electrically neutral, the Sulfur atom has to completely neutralize the charge of both Oxygen atoms, so the oxidation state of Sulfur is +4:

In the phosphorus molecule V) sulfide P 2 S 5 the more electronegative element is Sulfur, that is, it exhibits a negative oxidation state, and Phosphorus a positive one. For Sulfur, the negative oxidation state is only 2. Together, five Sulfur atoms carry a negative charge of -10. Therefore, two Phosphorus atoms have to neutralize this charge with a total charge of +10. Since there are two Phosphorus atoms in the molecule, each must have an oxidation state of +5:

It is more difficult to calculate the degree of oxidation in non-binary compounds - salts, bases and acids. But for this, one should also use the principle of electrical neutrality, and also remember that in most compounds the oxidation state of Oxygen is -2, Hydrogen +1.

Consider this using the example of potassium sulfate K2SO4. The oxidation state of Potassium in compounds can only be +1, and Oxygen -2:

From the principle of electroneutrality, we calculate the oxidation state of Sulfur:

2(+1) + 1(x) + 4(-2) = 0, hence x = +6.

When determining the oxidation states of elements in compounds, the following rules should be followed:

1. The oxidation state of an element in a simple substance is zero.

2. Fluorine is the most electronegative chemical element, so the oxidation state of Fluorine in all compounds is -1.

3. Oxygen is the most electronegative element after Fluorine, therefore the oxidation state of Oxygen in all compounds except fluorides is negative: in most cases it is -2, and in peroxides it is -1.

4. The oxidation state of Hydrogen in most compounds is +1, and in compounds with metallic elements (hydrides) - -1.

5. The oxidation state of metals in compounds is always positive.

6. A more electronegative element always has a negative oxidation state.

7. The sum of the oxidation states of all atoms in a molecule is zero.

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 states 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
	Hydrogen
	Helium / Helium
	Lithium / Lithium
	Beryllium / Beryllium
			(-1), 0, (+1), (+2), (+3)
	Carbon / Carbon		(-4), (-3), (-2), (-1), 0, (+2), (+4)
	Nitrogen / Nitrogen		(-3), (-2), (-1), 0, (+1), (+2), (+3), (+4), (+5)
	Oxygen / Oxygen		(-2), (-1), 0, (+1), (+2)
	Fluorine / Fluorine
	Sodium
	Magnesium / Magnesium
	Aluminum
	Silicon / Silicon		(-4), 0, (+2), (+4)
	Phosphorus / Phosphorus		(-3), 0, (+3), (+5)
	Sulfur		(-2), 0, (+4), (+6)
	Chlorine / Chlorine		(-1), 0, (+1), (+3), (+5), (+7), rarely (+2) and (+4)
	Argon / Argon
	Potassium / Potassium
	Calcium / Calcium
	Scandium / Scandium
	Titanium / Titanium		(+2), (+3), (+4)
	Vanadium / Vanadium		(+2), (+3), (+4), (+5)
	Chromium / Chromium		(+2), (+3), (+6)
	Manganese / Manganese		(+2), (+3), (+4), (+6), (+7)
	Iron / Iron		(+2), (+3), rarely (+4) and (+6)
	Cobalt / Cobalt		(+2), (+3), rarely (+4)
	Nickel / Nickel		(+2), rarely (+1), (+3) and (+4)
	Copper		+1, +2, rare (+3)
	Gallium / Gallium		(+3), rare (+2)
	Germanium / Germanium		(-4), (+2), (+4)
	Arsenic / Arsenic		(-3), (+3), (+5), rarely (+2)
	Selenium / Selenium		(-2), (+4), (+6), rarely (+2)
	Bromine / Bromine		(-1), (+1), (+5), rarely (+3), (+4)
	Krypton / Krypton
	Rubidium / Rubidium
	Strontium / Strontium
	Yttrium / Yttrium
	Zirconium / Zirconium		(+4), rarely (+2) and (+3)
	Niobium / Niobium		(+3), (+5), rarely (+2) and (+4)
	Molybdenum / Molybdenum		(+3), (+6), rarely (+2), (+3) and (+5)
	Technetium / Technetium
	Ruthenium / Ruthenium		(+3), (+4), (+8), rarely (+2), (+6) and (+7)
	Rhodium		(+4), rarely (+2), (+3) and (+6)
	Palladium / Palladium		(+2), (+4), rarely (+6)
	Silver / Silver		(+1), rarely (+2) and (+3)
	Cadmium / Cadmium		(+2), rare (+1)
	Indium / Indium		(+3), rarely (+1) and (+2)
	Tin / Tin		(+2), (+4)
	Antimony / Antimony		(-3), (+3), (+5), rarely (+4)
	Tellurium / Tellurium		(-2), (+4), (+6), rarely (+2)
			(-1), (+1), (+5), (+7), rarely (+3), (+4)
	Xenon / Xenon
	Cesium / Cesium
	Barium / Barium
	Lanthanum / Lanthanum
	Cerium / Cerium		(+3), (+4)
	Praseodymium / Praseodymium
	Neodymium / Neodymium		(+3), (+4)
	Promethium / Promethium
	Samaria / Samarium		(+3), rare (+2)
	Europium / Europium		(+3), rare (+2)
	Gadolinium / Gadolinium
	Terbium / Terbium		(+3), (+4)
	Dysprosium / Dysprosium
	Holmium / Holmium
	Erbium / Erbium
	Thulium / Thulium		(+3), rare (+2)
	Ytterbium / Ytterbium		(+3), rare (+2)
	Lutetium / Lutetium
	Hafnium / Hafnium
	Tantalum / Tantalum		(+5), rarely (+3), (+4)
	Tungsten / Tungsten		(+6), rare (+2), (+3), (+4) and (+5)
	Rhenium / Rhenium		(+2), (+4), (+6), (+7), rarely (-1), (+1), (+3), (+5)
	Osmium / Osmium		(+3), (+4), (+6), (+8), rarely (+2)
	Iridium / Iridium		(+3), (+4), (+6), rarely (+1) and (+2)
	Platinum / Platinum		(+2), (+4), (+6), rarely (+1) and (+3)
	Gold / Gold		(+1), (+3), rarely (+2)
	Mercury / Mercury		(+1), (+2)
	Waist / Thallium		(+1), (+3), rarely (+2)
	Lead / Lead		(+2), (+4)
	Bismuth / Bismuth		(+3), rarely (+3), (+2), (+4) and (+5)
	Polonium / Polonium		(+2), (+4), rarely (-2) and (+6)
	Astatine / Astatine
	Radon / Radon
	Francium / Francium
	Radium / Radium
	Actinium / Actinium
	Thorium / Thorium
	Proactinium / Protactinium
	Uranus / Uranium		(+3), (+4), (+6), rarely (+2) and (+5)

Examples of problem solving

EXAMPLE 1

Answer We will alternately determine the degree of phosphorus oxidation in each of the proposed transformation schemes, and then choose the correct answer.

• 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)