Electronegativity, like other properties of atoms chemical elements, changes with increasing serial number element periodically:

The graph above shows the periodicity of changes in the electronegativity of elements of the main subgroups depending on the atomic number of the element.

When moving down a subgroup of the periodic table, the electronegativity of chemical elements decreases, and when moving to the right along the period it increases.

Electronegativity reflects the nonmetallicity of elements: the higher the electronegativity value, the more nonmetallic properties the element has.

Oxidation state

How to calculate the oxidation state of an element in a compound?

1) The degree of oxidation of chemical elements in simple substances ah is always zero.

2) There are elements that manifest themselves in complex substances constant degree oxidation:

3) There are chemical elements that exhibit a constant oxidation state in the vast majority of compounds. These elements include:

Element	Oxidation state in almost all compounds	Exceptions
hydrogen H	+1	Alkali and alkaline earth metal hydrides, for example:
oxygen O	-2	Hydrogen and metal peroxides: Oxygen fluoride -

4) The algebraic sum of the oxidation states of all atoms in a molecule is always zero. The algebraic sum of the oxidation states of all atoms in an ion is equal to the charge of the ion.

5) The highest (maximum) oxidation state is equal to the group number. Exceptions that do not fall under this rule are elements of the secondary subgroup of group I, elements of the secondary subgroup of group VIII, as well as oxygen and fluorine.

Chemical elements whose group number does not coincide with their highest oxidation state (mandatory to remember)

6) The lowest oxidation state of metals is always zero, and the lowest oxidation state of non-metals is calculated by the formula:

lowest oxidation state of non-metal = group number − 8

Based on the rules presented above, you can establish the oxidation state of a chemical element in any substance.

Finding the oxidation states of elements in various compounds

Example 1

Determine the oxidation states of all elements in sulfuric acid.

Solution:

Let's write the formula of sulfuric acid:

The oxidation state of hydrogen in all complex substances is +1 (except metal hydrides).

The oxidation state of oxygen in all complex substances is -2 (except for peroxides and oxygen fluoride OF 2). Let us arrange the known oxidation states:

Let us denote the oxidation state of sulfur as x:

The sulfuric acid molecule, like the molecule of any substance, is generally electrically neutral, because the sum of the oxidation states of all atoms in a molecule is zero. Schematically this can be depicted as follows:

Those. we got the following equation:

Let's solve it:

Thus, the oxidation state of sulfur in sulfuric acid is +6.

Example 2

Determine the oxidation state of all elements in ammonium dichromate.

Solution:

Let's write the formula of ammonium dichromate:

As in the previous case, we can arrange the oxidation states of hydrogen and oxygen:

However, we see that the oxidation states of two chemical elements at once are unknown - nitrogen and chromium. Therefore, we cannot find oxidation states similarly to the previous example (one equation with two variables does not have a single solution).

Let us draw attention to the fact that this substance belongs to the class of salts and, accordingly, has an ionic structure. Then we can rightly say that the composition of ammonium dichromate includes NH 4 + cations (the charge of this cation can be seen in the solubility table). Consequently, since the formula unit of ammonium dichromate contains two positive singly charged NH 4 + cations, the charge of the dichromate ion is equal to -2, since the substance as a whole is electrically neutral. Those. the substance is formed by NH 4 + cations and Cr 2 O 7 2- anions.

We know the oxidation states of hydrogen and oxygen. Knowing that the sum of the oxidation states of the atoms of all elements in an ion is equal to the charge, and denoting the oxidation states of nitrogen and chromium as x And y accordingly, we can write:

Those. we get two independent equations:

Solving which, we find x And y:

Thus, in ammonium dichromate the oxidation states of nitrogen are -3, hydrogen +1, chromium +6, and oxygen -2.

How to determine the oxidation states of elements in organic matter you can read it.

Valence

The valence of atoms is indicated by Roman numerals: I, II, III, etc.

The valence capabilities of an atom depend on the quantity:

1) unpaired electrons

2) lone electron pairs in the orbitals of valence levels

3) empty electron orbitals of the valence level

Valence possibilities of the hydrogen atom

Let us depict the electron graphic formula of the hydrogen atom:

It has been said that three factors can influence the valence possibilities - the presence of unpaired electrons, the presence of lone electron pairs in the outer level, and the presence of vacant (empty) orbitals in the outer level. We see one unpaired electron at the outer (and only) energy level. Based on this, hydrogen can definitely have a valence of I. However, in the first energy level there is only one sublevel - s, those. The hydrogen atom at the outer level has neither lone electron pairs nor empty orbitals.

Thus, the only valence that a hydrogen atom can exhibit is I.

Valence possibilities of the carbon atom

Let's consider electronic structure carbon atom. In the ground state, the electronic configuration of its outer level is as follows:

Those. in the ground state at the outer energy level of the unexcited carbon atom there are 2 unpaired electrons. In this state it can exhibit a valence of II. However, the carbon atom very easily goes into an excited state when energy is imparted to it, and the electronic configuration of the outer layer in this case takes the form:

Despite the fact that a certain amount of energy is spent on the process of excitation of the carbon atom, the expenditure is more than compensated for by the formation of four covalent bonds. For this reason, valency IV is much more characteristic of the carbon atom. For example, carbon has valency IV in the molecules of carbon dioxide, carbonic acid and absolutely all organic substances.

In addition to unpaired electrons and lone electron pairs, the presence of vacant ()valence level orbitals also affects the valence possibilities. The presence of such orbitals at the filled level leads to the fact that the atom can act as an electron pair acceptor, i.e. form additional covalent bonds through a donor-acceptor mechanism. For example, contrary to expectations, in the carbon monoxide molecule CO the bond is not double, but triple, as is clearly shown in the following illustration:

Valence possibilities of the nitrogen atom

Let us write the electronic graphic formula for the external energy level of the nitrogen atom:

As can be seen from the illustration above, the nitrogen atom in its normal state has 3 unpaired electrons, and therefore it is logical to assume that it is capable of exhibiting a valence of III. Indeed, a valency of three is observed in ammonia molecules (NH 3), nitrous acid(HNO 2), nitrogen trichloride (NCl 3), etc.

It was said above that the valence of an atom of a chemical element depends not only on the number of unpaired electrons, but also on the presence of lone electron pairs. This is due to the fact that covalent chemical bond can be formed not only when two atoms provide each other with one electron, but also when one atom having a lone pair of electrons - donor () provides it to another atom with a vacant () valence level orbital (acceptor). Those. For the nitrogen atom, valence IV is also possible due to an additional covalent bond formed by the donor-acceptor mechanism. For example, four covalent bonds, one of which is formed by a donor-acceptor mechanism, are observed during the formation of an ammonium cation:

Despite the fact that one of the covalent bonds is formed according to the donor-acceptor mechanism, all N-H connections in the ammonium cation are absolutely identical and do not differ from each other in any way.

The nitrogen atom is not capable of exhibiting a valency equal to V. This is due to the fact that it is impossible for a nitrogen atom to transition to an excited state, in which two electrons are paired with the transition of one of them to a free orbital that is closest in energy level. The nitrogen atom has no d-sublevel, and the transition to the 3s orbital is energetically so expensive that the energy costs are not covered by the formation of new bonds. Many may wonder, what is the valency of nitrogen, for example, in molecules of nitric acid HNO 3 or nitric oxide N 2 O 5? Oddly enough, the valency there is also IV, as can be seen from the following structural formulas:

The dotted line in the illustration shows the so-called delocalized π -connection. For this reason, terminal NO bonds can be called “one and a half bonds.” Similar one-and-a-half bonds are also present in the molecule of ozone O 3, benzene C 6 H 6, etc.

Valence possibilities of phosphorus

Let us depict the electronic graphic formula of the external energy level of the phosphorus atom:

As we see, the structure of the outer layer of the phosphorus atom in the ground state and the nitrogen atom is the same, and therefore it is logical to expect for the phosphorus atom, as well as for the nitrogen atom, possible valences equal to I, II, III and IV, as observed in practice.

However, unlike nitrogen, the phosphorus atom also has d-sublevel with 5 vacant orbitals.

In this regard, it is capable of transitioning to an excited state, steaming electrons 3 s-orbitals:

Thus, the valence V for the phosphorus atom, which is inaccessible to nitrogen, is possible. For example, the phosphorus atom has a valency of five in molecules of compounds such as phosphoric acid, phosphorus (V) halides, phosphorus (V) oxide, etc.

Valence possibilities of the oxygen atom

The electron graphic formula for the external energy level of an oxygen atom has the form:

We see two unpaired electrons at the 2nd level, and therefore valence II is possible for oxygen. It should be noted that this valence of the oxygen atom is observed in almost all compounds. Above, when considering the valency capabilities of the carbon atom, we discussed the formation of the carbon monoxide molecule. The bond in the CO molecule is triple, therefore, the oxygen there is trivalent (oxygen is an electron pair donor).

Due to the fact that the oxygen atom does not have an external d-sublevel, electron pairing s And p- orbitals is impossible, which is why the valence capabilities of the oxygen atom are limited compared to other elements of its subgroup, for example, sulfur.

Valence possibilities of the sulfur atom

External energy level of a sulfur atom in an unexcited state:

The sulfur atom, like the oxygen atom, normally has two unpaired electrons, so we can conclude that sulfur can have a valence of two. Indeed, sulfur has valency II, for example, in the hydrogen sulfide molecule H 2 S.

As we see, the sulfur atom appears at the external level d-sublevel with vacant orbitals. For this reason, the sulfur atom is able to expand its valence capabilities, unlike oxygen, due to the transition to excited states. Thus, when pairing a lone electron pair 3 p-sublevel, the sulfur atom acquires the electronic configuration of the outer level of the following form:

In this state, the sulfur atom has 4 unpaired electrons, which tells us that sulfur atoms can exhibit a valence of IV. Indeed, sulfur has valency IV in molecules SO 2, SF 4, SOCl 2, etc.

When pairing the second lone electron pair located at 3 s-sublevel, the external energy level acquires the configuration:

In this state, the manifestation of valency VI becomes possible. Examples of compounds with VI-valent sulfur are SO 3, H 2 SO 4, SO 2 Cl 2, etc.

Similarly, we can consider the valence possibilities of other chemical elements.

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 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.

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Table. Oxidation states of chemical elements.

Table. Oxidation states of chemical elements.

Oxidation state is the conditional charge of the atoms of a chemical element in a compound, calculated on the assumption that all bonds have ion type. Oxidation states can be positive, negative or zero, so algebraic sum the oxidation states of elements in a molecule, taking into account the number of their atoms, is equal to 0, and in an ion - the charge of the ion. The oxidation states of metals in compounds are always positive. The highest oxidation state corresponds to the group number periodic table, where is this element(exceptions are: Au +3(I group), Cu +2(II), from group VIII the oxidation state +8 can only be found in osmium Os and ruthenium Ru. The oxidation states of nonmetals 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 either positive or negative. It depends on the electronegativity of the atoms of the elements. The highest negative oxidation state of non-metals can be determined by subtracting from 8 the number of the group in which the element is located, i.e. the highest positive oxidation state is equal to the number of electrons in the outer layer, which corresponds to the group number. The oxidation states of simple substances are 0, regardless of whether it is a metal or a non-metal.

Table: Elements with constant oxidation states.

Table. Oxidation states of chemical elements in alphabetical order. Element Name Oxidation state 7 N -III, 0, +I, II, III, IV, V 89 Ace 13 Al Aluminum 95 Am Americium 0, + II, III, IV 18 Ar 85 At -I, 0, +I, V 56 Ba 4 Be Beryllium 97 Bk 5 B -III, 0, +III 107 Bh 35 Br -I, 0, +I, V, VII 23 V 0, + II, III, IV, V 83 Bi 1 H -I, 0, +I 74 W Tungsten 64 Gd Gadolinium 31 Ga 72 Hf 2 He 32 Ge Germanium 67 Ho 66 Dy Dysprosium 105 Db 63 Eu 26 Fe 79 Au 49 In 77 Ir 39 Y 70 Yb Ytterbium 53 I -I, 0, +I, V, VII 48 Cd 19 TO 98 Cf Californium 20 Ca 54 Xe 0, + II, IV, VI, VIII 8 O Oxygen -II, I, 0, +II 27 Co 36 Kr 14 Si -IV, 0, +11, IV 96 Cm 57 La 3 Li 103 Lr Lawrence 71 Lu 12 Mg 25 Mn Manganese 0, +II, IV, VI, VIII 29 Cu 109 Mt Meitnerium 101 MD Mendelevium 42 Mo Molybdenum 33 As - III, 0, +III, V 11 Na 60 Nd 10 Ne 93 Np Neptunium 0, +III, IV, VI, VII 28 Ni 41 Nb 102 No 50 Sn 76 Os 0, +IV, VI, VIII 46 Pd Palladium 91 Pa. Protactinium 61 Pm Promethium 84 Po 59 Rg Praseodymium 78 Pt 94 P.U. Plutonium 0, +III, IV, V, VI 88 Ra 37 Rb 75 Re 104 Rf Rutherfordium 45 Rh 86 Rn 0, + II, IV, VI, VIII 44 Ru 0, +II, IV, VI, VIII 80 Hg 16 S -II, 0, +IV, VI 47 Ag 51 Sb 21 Sc 34 Se -II, 0,+IV, VI 106 Sg Seaborgium 62 Sm 38 Sr Strontium 82 Pb 81 Тl 73 Ta 52 Te -II, 0, +IV, VI 65 Tb 43 Tc Technetium 22 Ti 0, + II, III, IV 90 Th 69 Tm 6 C -IV, I, 0, +II, IV 92 U 100 Fm 15 P -III, 0, +I, III, V 87 Fr 9 F -I, 0 108 Hs 17 Cl 24 Cr 0, + II, III, VI 55 Cs 58 Ce 30 Zn 40 Zr Zirconium 99 ES Einsteinium 68 Er

Table. Oxidation states of chemical elements by number. Element Name Oxidation state 1 H -I, 0, +I 2 He 3 Li 4 Be Beryllium 5 B -III, 0, +III 6 C -IV, I, 0, +II, IV 7 N -III, 0, +I, II, III, IV, V 8 O Oxygen -II, I, 0, +II 9 F -I, 0 10 Ne 11 Na 12 Mg 13 Al Aluminum 14 Si -IV, 0, +11, IV 15 P -III, 0, +I, III, V 16 S -II, 0, +IV, VI 17 Cl -I, 0, +I, III, IV, V, VI, VII 18 Ar 19 TO 20 Ca 21 Sc 22 Ti 0, + II, III, IV 23 V 0, + II, III, IV, V 24 Cr 0, + II, III, VI 25 Mn Manganese 0, +II, IV, VI, VIII 26 Fe 27 Co 28 Ni 29 Cu 30 Zn 31 Ga 32 Ge Germanium 33 As - III, 0, +III, V 34 Se -II, 0,+IV, VI 35 Br -I, 0, +I, V, VII 36 Kr 37 Rb 38 Sr Strontium 39 Y 40 Zr Zirconium 41 Nb 42 Mo Molybdenum 43 Tc Technetium 44 Ru 0, +II, IV, VI, VIII 45 Rh 46 Pd Palladium 47 Ag 48 Cd 49 In 50 Sn 51 Sb 52 Te -II, 0, +IV, VI 53 I -I, 0, +I, V, VII 54 Xe 0, + II, IV, VI, VIII 55 Cs 56 Ba 57 La 58 Ce 59 Rg Praseodymium 60 Nd 61 Pm Promethium 62 Sm 63 Eu 64 Gd Gadolinium 65 Tb 66 Dy Dysprosium 67 Ho 68 Er 69 Tm 70 Yb Ytterbium 71 Lu 72 Hf 73 Ta 74 W Tungsten 75 Re 76 Os 0, +IV, VI, VIII 77 Ir 78 Pt 79 Au 80 Hg 81 Тl 82 Pb 83 Bi 84 Po 85 At -I, 0, +I, V 86 Rn 0, + II, IV, VI, VIII 87 Fr 88 Ra 89 Ace 90 Th 91 Pa. Protactinium 92 U 93 Np Neptunium 0, +III, IV, VI, VII 94 P.U. Plutonium 0, +III, IV, V, VI 95 Am Americium 0, + II, III, IV 96 Cm 97 Bk 98 Cf Californium 99 ES Einsteinium 100 Fm 101 MD Mendelevium 102 No 103 Lr Lawrence 104 Rf Rutherfordium 105 Db 106 Sg Seaborgium 107 Bh 108 Hs 109 Mt Meitnerium

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The chemical element in a compound, calculated from the assumption that all bonds are ionic.

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 equal to 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 number of the group of the periodic system where the element is located (exceptions are: Au +3(I group), Cu +2(II), from group VIII the oxidation state +8 can only be found 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 either positive or 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 the element is located, i.e. the highest positive oxidation state is equal to the number of electrons in 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
		Metal hydrides: LIH -1
		Oxidation state called the conditional charge of a particle under the assumption that the bond is completely broken (has an ionic character). H- Cl = H + + Cl - , Contact in hydrochloric acid covalent polar. Electron pair in to a greater extent shifted towards the atom Cl - , because it is a more electronegative element. How to determine the oxidation state? Electronegativity is the ability of atoms to attract electrons from other elements. The oxidation number is indicated above the element: Br 2 0 , Na 0 , O +2 F 2 -1 ,K + Cl - etc. It can be negative and positive. Oxidation state of a simple substance (unbound, free state) is equal to zero. The oxidation state of oxygen for most compounds is -2 (the exception is peroxides H 2 O 2, where it is equal to -1 and compounds with fluorine - O +2 F 2 -1 , O 2 +1 F 2 -1 ). - Oxidation state of 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-nonmetal bonds, the negative oxidation state is that atom that has greater electronegativity (data on electronegativity are given in 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 the connection KMnO 4 , it is necessary to determine the oxidation state of the manganese atom. Reasoning: Potassium is an alkali metal in Group I of the periodic table, and therefore has only a positive oxidation state of +1. Oxygen, as is known, in most of its compounds has an oxidation state of -2. This substance is not a peroxide, which means it is no exception. Makes up the equation: K+Mn X O 4 -2 Let X- unknown to us oxidation state of manganese. The number of potassium atoms is 1, manganese - 1, oxygen - 4. It has been proven that the molecule as a whole is electrically neutral, so its total charge must be zero. 1*(+1) + 1*(X) + 4(-2) = 0, X = +7, This means that 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: Iron is a metal, oxygen is a non-metal, which means that oxygen will be an oxidizing agent and have a negative charge. We know that oxygen has an oxidation state of -2. We count the number of atoms: iron - 2 atoms, oxygen - 3. We create an equation where X- oxidation state of the iron atom: 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 equal to 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 chromium atom has 12 positive powers, but there are 2 atoms in the molecule, which means there are (+12) per atom: 2 = (+6). Answer: K 2 + Cr 2 +6 O 7 -2. 2.(AsO 4) 3- . In this case, the sum of oxidation states will no longer be equal to zero, but to the charge of the ion, i.e. - 3. Let's make an equation: x+4×(- 2)= - 3 . Answer: (As +5 O 4 -2) 3- .

How to determine the oxidation state? The periodic table allows you to record this quantitative value for any chemical element.

Definition

First, let's try to understand what this term represents. The oxidation state according to the periodic table represents the number of electrons that are accepted or given up by an element in the process of chemical interaction. She can accept negativity and positive value.

Linking to a table

How is the 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 metrics for elements, you must use certain rules.

Instructions

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

The table allows you to determine this value not only for elements exhibiting 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 answers this question as well. You need to subtract the group number from eight. For example, for oxygen it will be -2, for nitrogen -3.

For simple substances that have not entered into chemical interaction with other substances, the determined indicator is considered equal to zero.

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

For example, let's 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 figures. As a result, we get zero, therefore, the calculations are correct.

Let us determine similar indicators in another binary compound CuO. Since copper is located in a secondary subgroup (first group), therefore, the studied indicator may exhibit different meanings. Therefore, to determine it, you must first identify the indicator for oxygen.

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

How else is it used? chemical table? The oxidation states of elements in formulas consisting of three elements are also calculated using a specific algorithm. First, these indicators are placed at the first and last element. For the first, this indicator will have a positive value, corresponding to valence. For the outermost 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 a central element, a mathematical equation is used. When calculating, the indices available for each element are taken into account. The sum of all oxidation states must be zero.

Example of determination in sulfuric acid

Formula of this connection has the form H 2 SO 4. Hydrogen has an oxidation state of +1, and oxygen has an oxidation state of -2. To determine the oxidation state of sulfur, we create a mathematical equation: + 1 * 2 + X + 4 * (-2) = 0. We find that the oxidation state of sulfur corresponds to +6.

Conclusion

Using the rules, you can assign coefficients in redox reactions. This question covered in a ninth grade chemistry course school curriculum. In addition, information about oxidation states allows you to perform OGE assignments and Unified State Exam.