The degree of oxidation is a conditional value used to record redox reactions. To determine the degree of oxidation, a 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 the 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 completion of the outer energy level. Most non-metals are oxidizing agents.

Simple substances that do not react always have a zero oxidation state.

Rice. 1. Table of oxidation states.

In the compound, a non-metal atom with a lower electronegativity has a positive oxidation state.

Definition

You can determine the maximum and minimum oxidation state (how many electrons an atom can give and take) using the periodic table of Mendeleev.

The maximum power 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 the lowest is -4. The maximum oxidation state of sulfur is +6, the minimum is -2. Most non-metals always have a variable - positive and negative - oxidation state. The exception is fluorine. 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 different oxidation states. The exception is aluminum. Despite being in group III, its oxidation state is always +3.

Rice. 3. Alkali and alkaline earth metals.

Of group VIII, only ruthenium and osmium can exhibit the highest oxidation state +8. Gold and copper, which are in group I, exhibit oxidation states of +3 and +2, respectively.

Recording

To correctly record the oxidation state, you should remember a few rules:

  • inert gases do not react, so their oxidation state is always zero;
  • in compounds, the variable oxidation state depends on the variable valency 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 has received or given away in a compound. The value depends on the number of valence electrons. Metals in compounds always have a positive oxidation state, i.e. are restorers. For alkali and alkaline earth metals, the oxidation state is always the same. Non-metals, except for fluorine, can take positive and negative oxidation states.

When defining this concept, it is conditionally assumed that the binding (valence) electrons pass to more electronegative atoms (see Electronegativity), and therefore the compounds consist, as it were, of positively and negatively charged ions. The oxidation state can have zero, negative, and positive values, which are usually placed above the element symbol at the top.

The zero value of the oxidation state is assigned to the atoms of the elements in the free state, for example: Cu, H 2 , N 2 , P 4 , S 6 . Negative meaning oxidation states are those atoms towards which the binding electron cloud (electron pair) is displaced. For fluorine in all its compounds, it is -1. Atoms that donate valence electrons to other atoms have a positive oxidation state. For example, for alkali and alkaline earth metals, it is respectively +1 and +2. In simple ions like Cl − , S 2− , K + , Cu 2+ , Al 3+ , 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. For oxygen, the oxidation state is -2, but, for example, in combination with fluorine OF 2 it will be +2, and in peroxide compounds (BaO 2, etc.) -1. In some cases, this value can also be expressed as a fractional number: 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 phosphoric 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 get 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 highest positive value. For most elements, it is equal to the group number in the periodic system and is an important quantitative characteristic of the element in its compounds. Lowest value the oxidation state of an element that occurs in its compounds is commonly 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 groups of the periodic system reflects the frequency of change in their chemical properties with increasing serial number.

The concept of the oxidation state of elements is used in the classification of substances, describing their properties, formulating 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" (cf.

The task of determining the degree of oxidation can be both a simple formality and a complex puzzle. First of all, it will depend on the formula of the chemical compound, as well as the availability of elementary knowledge in chemistry and mathematics.

Knowing the basic rules and the algorithm of sequentially logical actions, which will be discussed in this article, when solving problems of this type, everyone can easily cope with this task. And having trained and learned to determine the degree of oxidation of diverse chemical compounds, you can safely take on the equalization of complex redox reactions by the method of compiling an electronic balance.

The concept of oxidation state

To learn how to determine the degree of oxidation, first you need to figure out what this concept means?

  • The oxidation state is used when recording in redox reactions, when electrons are transferred from atom to atom.
  • The oxidation state fixes the number of electrons transferred, denoting the conditional charge of the atom.
  • The oxidation state and valency are often identical.

This designation is written above chemical element, in its right corner, and is an integer with a "+" or "-" sign. The zero value of the degree of oxidation does not carry a sign.

Rules for determining the degree of oxidation

Consider the main canons for determining the degree of oxidation:

  • Simple elemental substances, that is, those that consist of one kind of atoms, will always have a zero oxidation state. For example, Na0, H02, P04
  • There are a number of atoms that always have one, constant, oxidation state. It is better to remember the values ​​given in the table.
  • As you can see, the only exception is hydrogen in combination with metals, where it acquires an oxidation state “-1” that is not characteristic of it.
  • Oxygen also adopts the "+2" oxidation state in chemical compound with fluorine and "-1" in the compositions of peroxides, superoxides or ozonides, where oxygen atoms are connected to each other.


  • Metal ions have several values ​​of the degree of oxidation (and only positive ones), so it is determined by neighboring elements in the compound. For example, in FeCl3, chlorine has an oxidation state of "-1", it has 3 atoms, so we multiply -1 by 3, we get "-3". In order for the sum of the oxidation states of the compound to be "0", iron must have an oxidation state of "+3". In the formula FeCl2, iron, respectively, will change its degree to "+2".
  • Mathematically summing the oxidation states of all atoms in the formula (taking into account the signs), a zero value should always be obtained. For example, in hydrochloric acid H + 1Cl-1 (+1 and -1 = 0), and in sulfuric acid H2 + 1S + 4O3-2 (+1 * 2 = +2 for hydrogen, +4 for sulfur and -2 * 3 = -6 for oxygen; +6 and -6 add up to 0).
  • The oxidation state of a monatomic ion will be equal to its charge. For example: Na+, Ca+2.
  • The highest degree of oxidation, as a rule, corresponds to the group number in the periodic system of D.I. Mendeleev.


Algorithm of actions for determining the degree of oxidation

The order of finding the degree of oxidation is not complicated, but requires attention and certain actions.

Task: Arrange the oxidation states in the compound KMnO4

  • The first element, potassium, has a constant oxidation state of "+1".
    To check, you can look at periodic system, where potassium is in group 1 of elements.
  • Of the remaining two elements, oxygen tends to take on an oxidation state of "-2".
  • We get the following formula: K+1MnxO4-2. It remains to determine the oxidation state of manganese.
    So, x is the oxidation state of manganese unknown to us. Now it is important to pay attention to the number of atoms in the compound.
    The number of potassium atoms is 1, manganese - 1, oxygen - 4.
    Taking into account the electrical neutrality of the molecule, when the total (total) charge is zero,

1*(+1) + 1*(x) + 4(-2) = 0,
+1+1x+(-8) = 0,
-7+1x = 0,
(when transferring, change the sign)
1x = +7, x = +7

Thus, the oxidation state of manganese in the compound is "+7".

Task: arrange the oxidation states in the compound Fe2O3.

  • Oxygen, as you know, has an oxidation state of "-2" and acts as an oxidizing agent. Taking into account the number of atoms (3), the total value of oxygen is “-6” (-2*3= -6), i.e. multiply the oxidation state by the number of atoms.
  • To balance the formula and bring it to zero, 2 iron atoms will have an oxidation state of "+3" (2*+3=+6).
  • In sum, we get zero (-6 and +6 = 0).

Task: arrange the oxidation states in the Al(NO3)3 compound.

  • The aluminum atom is one and has a constant oxidation state of "+3".
  • There are 9 (3 * 3) oxygen atoms in the molecule, the oxidation state of oxygen, as you know, is “-2”, which means that by multiplying these values, we get “-18”.
  • It remains to equalize the negative and positive values, thus determining the degree of oxidation of nitrogen. -18 and +3, + 15 is missing. And given that there are 3 nitrogen atoms, it is easy to determine its oxidation state: divide 15 by 3 and get 5.
  • The oxidation state of nitrogen is “+5”, and the formula will look like: Al + 3 (N + 5O-23) 3
  • If it is difficult to determine the desired value in this way, you can compose and solve equations:

1*(+3) + 3x + 9*(-2) = 0.
+3+3x-18=0
3x=15
x=5


So, the degree of oxidation is a rather important concept in chemistry, symbolizing the state of atoms in a molecule.
Without knowledge of certain provisions or bases that allow you to correctly determine the degree of oxidation, it is impossible to cope with this task. Therefore, there is only one conclusion: to thoroughly familiarize yourself and study the rules for finding the degree of oxidation, clearly and concisely presented in the article, and boldly move on along the difficult path of chemical wisdom.

In school, chemistry is still one of the most difficult subjects, which, due to the fact that it hides many difficulties, causes in students (usually in the period from 8 to 9 classes) more hatred and indifference to study than interest. All this reduces the quality and quantity of knowledge on the subject, although many areas still require specialists in this field. Yes, sometimes there are even more difficult moments and incomprehensible rules in chemistry than it seems. One of the questions that concern most students is what is the oxidation state and how to determine the oxidation states of elements.

An important rule is the placement rule, algorithms

There is much talk here about compounds such as oxides. To begin with, every student must learn determination of oxides- it complex connections of two elements, they contain oxygen. Oxides are classified as binary compounds because oxygen is second in line in the algorithm. When determining the indicator, it is important to know the placement rules and calculate the algorithm.

Algorithms for Acid Oxides

Oxidation states - these are numerical expressions of the valency of the elements. For example, acid oxides formed according to a certain algorithm: first come non-metals or metals (their valency is usually from 4 to 7), and then comes oxygen, as it should be, second in order, its valency is two. It is determined easily - according to the periodic table of chemical elements of Mendeleev. It is also important to know that the oxidation state of elements is an indicator that suggests either positive or negative number.

At the beginning of the algorithm, as a rule, a non-metal, and its oxidation state is positive. Non-metal oxygen in oxide compounds has a stable value, which is -2. To determine the correctness of the arrangement of all values, you need to multiply all the available numbers by the indices of one specific element, if the product, taking into account all the minuses and pluses, is 0, then the arrangement is reliable.

Arrangement in acids containing oxygen

Acids are complex substances, they are associated with some acidic residue and contain one or more hydrogen atoms. Here, to calculate the degree, skills in mathematics are required, since the indicators necessary for the calculation are digital. For hydrogen or a proton, it is always the same - +1. The negative oxygen ion has a negative oxidation state of -2.

After carrying out all these actions, you can determine the degree of oxidation and the central element of the formula. The expression for its calculation is a formula in the form of an equation. For example, for sulfuric acid, the equation will be with one unknown.

Basic terms in OVR

ORR is a reduction-oxidation reaction.

  • The oxidation state of any atom - characterizes the ability of this atom to attach or give electrons to other atoms of ions (or atoms);
  • It is customary to consider either charged atoms or uncharged ions as oxidizing agents;
  • The reducing agent in this case will be charged ions or, on the contrary, uncharged atoms that lose their electrons in the process of chemical interaction;
  • Oxidation is the donation of electrons.

How to arrange the oxidation state in salts

Salts are composed of one metal and one or more acid residues. The determination procedure is the same as in acid-containing acids.

The metal that directly forms a salt is located in the main subgroup, its degree will be equal to the number of its group, that is, it will always remain a stable, positive indicator.

As an example, consider the arrangement of oxidation states in sodium nitrate. Salt is formed using an element of the main subgroup of group 1, respectively, the oxidation state will be positive and equal to one. In nitrates, oxygen has the same value - -2. In order to get a numerical value, first an equation is drawn up with one unknown, taking into account all the minuses and pluses of the values: +1+X-6=0. By solving the equation, you can come to the fact that the numerical indicator is positive and equal to + 5. This is the indicator of nitrogen. An important key to calculate the degree of oxidation - table.

Arrangement rule in basic oxides

  • Oxides of typical metals in any compounds have a stable oxidation index, it is always no more than +1, or in other cases +2;
  • The digital indicator of the metal is calculated using the periodic table. If the element is contained in the main subgroup of group 1, then its value will be +1;
  • The value of oxides, taking into account their indices, after multiplication, summed up should be equal to zero, because the molecule in them is neutral, a particle devoid of charge;
  • Metals of the main subgroup of group 2 also have a stable positive indicator, which is +2.

Electronegativity, like other properties of atoms of chemical elements, changes periodically with an increase in the ordinal number of the element:

The graph above shows the frequency of changes in the electronegativity of the elements of the main subgroups, depending on the ordinal number of the element.

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

Electronegativity reflects the non-metallicity of elements: the higher the value of electronegativity, the more non-metallic properties of the element are expressed.

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 is always zero.

2) There are elements that exhibit a constant oxidation state in complex substances:

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

Element

The 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 match their highest oxidation state (mandatory to memorize)

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 a non-metal = group number - 8

Based on the rules presented above, it is possible to establish the degree of oxidation 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 for sulfuric acid:

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

The oxidation state of oxygen in all complex substances is -2 (except for peroxides and oxygen fluoride OF 2). Let's 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, nitrogen and chromium, are unknown. Therefore, we cannot find the oxidation states in the same way as in the previous example (one equation with two variables does not have a unique solution).

Let us pay attention to the fact that the indicated 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). Therefore, since there are two positive singly charged NH 4 + cations in the formula unit of ammonium dichromate, the charge of the dichromate ion is -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 the 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 can be read.

Valence

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

The valence possibilities of an atom depend on the quantity:

1) unpaired electrons

2) unshared electron pairs in the orbitals of valence levels

3) empty electron orbitals of the valence level

Valence possibilities of the hydrogen atom

Let's depict the electronic graphic formula of the hydrogen atom:

It was said that three factors can influence the valence possibilities - the presence of unpaired electrons, the presence of unshared electron pairs at the outer level, and the presence of vacant (empty) orbitals external level. We see one unpaired electron in the outer (and only) energy level. Based on this, hydrogen can exactly have a valency equal to I. However, at the first energy level there is only one sublevel - s, those. the hydrogen atom at the outer level does not have either unshared electron pairs or empty orbitals.

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

Valence possibilities of a carbon atom

Consider electronic structure carbon atom. In the ground state, the electronic configuration of its outer level is as follows:

Those. In the ground state, the outer energy level of an unexcited carbon atom contains 2 unpaired electrons. In this state, it can exhibit a valency equal to 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:

Although 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, valence IV is much more characteristic of the carbon atom. So, 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 () orbitals of the valence level also affects the valence possibilities. The presence of such orbitals in the filled level leads to the fact that the atom can act as an electron pair acceptor, i.e. form additional covalent bonds by the donor-acceptor mechanism. So, for example, contrary to expectations, in the carbon monoxide molecule CO, the bond is not double, but triple, which is clearly shown in the following illustration:

Valence possibilities of the nitrogen atom

Let's write down the electron-graphic formula of 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 can exhibit a valency equal to III. Indeed, a valency of three is observed in the molecules of ammonia (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 unshared electron pairs. This is due to the fact that the covalent chemical bond can be formed not only when two atoms provide each other with one electron each, but also when one atom that has an unshared pair of electrons - a donor () provides it to another atom with a vacant () orbital of the valence level (acceptor). Those. for the nitrogen atom, valency IV is also possible due to an additional covalent bond formed by the donor-acceptor mechanism. So, for example, four covalent bonds, one of which is formed by the donor-acceptor mechanism, is observed during the formation of the ammonium cation:

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

A valency equal to V, the nitrogen atom is not able to show. This is due to the fact that the transition to an excited state is impossible for the nitrogen atom, in which the pairing of two electrons occurs with the transition of one of them to a free orbital, which is the 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 then is the valency of nitrogen, for example, in the molecules of nitric acid HNO 3 or nitric oxide N 2 O 5? Oddly enough, the valence 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, NO terminal bonds can be called "one and a half". Similar one-and-a-half bonds are also found in the ozone molecule O 3 , benzene C 6 H 6 , etc.

Valence possibilities of phosphorus

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

As we can 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, which is observed in practice.

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

In this regard, it is able to pass into an excited state, steaming electrons 3 s-orbitals:

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

Valence possibilities of the oxygen atom

The electron-graphic formula of the external energy level of the 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 valency of the oxygen atom is observed in almost all compounds. Above, when considering the valence possibilities of the carbon atom, we discussed the formation of the carbon monoxide molecule. The bond in the CO molecule is triple, therefore, oxygen is trivalent there (oxygen is an electron pair donor).

Due to the fact that the oxygen atom does not have an external level d-sublevels, depairing of electrons 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 sulfur atom in the unexcited state:

The sulfur atom, like the oxygen atom, has two unpaired electrons in its normal state, so we can conclude that a valency of two is possible for sulfur. Indeed, sulfur has valency II, for example, in the hydrogen sulfide molecule H 2 S.

As we can see, the sulfur atom at the outer level has 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. So, when unpairing 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 about the possibility of sulfur atoms showing a valency equal to IV. Indeed, sulfur has valency IV in the molecules SO 2, SF 4, SOCl 2, etc.

When unpairing the second lone electron pair located on 3 s- sublevel, the external energy level acquires the following configuration:

In such a state, the manifestation of valence VI already becomes possible. An example 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.