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

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 non-metallicity of elements: the higher the electronegativity value, the more non-metallic properties the element has.

Oxidation state

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

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

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

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

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.

You can read how to determine the oxidation states of elements in organic substances.

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 the electronic structure of the 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 CO molecule 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 valence 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 lone electron pairs. This is due to the fact that a covalent chemical bond can be formed not only when two atoms provide each other with one electron, but also when one atom with a lone pair of electrons - donor () provides it to another atom with a vacant () orbital valence level (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 bonds in the ammonium cation are absolutely identical and do not differ from each other.

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 a valence of two is possible for sulfur. 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 oxidation state +2 in all compounds exhibits

Answer:4

Explanation:

Of all the proposed options, only zinc exhibits the +2 oxidation state in complex compounds, being an element of the secondary subgroup of the second group, where the maximum oxidation state is equal to the group number.

Tin is an element of the main subgroup of group IV, a metal, exhibiting oxidation states 0 (in a simple substance), +2, +4 (group number).

Phosphorus is an element of the main subgroup of the main group, being a non-metal, exhibiting oxidation states from -3 (group number - 8) to +5 (group number).

Iron is a metal, the element is located in a secondary subgroup of the main group. Iron is characterized by oxidation states: 0, +2, +3, +6.

The compound of KEO 4 composition forms each of two elements:

1) phosphorus and chlorine

2) fluorine and manganese

3) chlorine and manganese

Answer: 3

Explanation:

The salt of the composition KEO 4 contains an acid residue EO 4 -, where oxygen has an oxidation state of -2, therefore, the oxidation state of the element E in this acid residue is +7. Of the proposed options, chlorine and manganese are suitable - elements of the main and secondary subgroups of group VII, respectively.

Fluorine is also an element of the main subgroup of group VII, however, being the most electronegative element, it does not exhibit positive oxidation states (0 and -1).

Boron, silicon and phosphorus are elements of the main subgroups of groups 3, 4 and 5, respectively, therefore in salts they exhibit the corresponding maximum oxidation states of +3, +4, +5.

Answer: 4

Explanation:

The same highest oxidation state in the compounds, equal to the group number (+5), is exhibited by P and As. These elements are located in the main subgroup of group V.

Zn and Cr are elements of secondary subgroups of groups II and VI, respectively. In compounds, zinc exhibits the highest oxidation state of +2, chromium - +6.

Fe and Mn are elements of the secondary subgroups of groups VIII and VII, respectively. The highest oxidation state for iron is +6, for manganese - +7.

The compounds exhibit the same highest oxidation state

Answer: 4

Explanation:

P and N exhibit the same highest oxidation state in compounds, equal to the group number (+5). These elements are located in the main subgroup of group V.

Hg and Cr are elements of secondary subgroups of groups II and VI, respectively. In compounds, mercury exhibits the highest oxidation state of +2, chromium - +6.

Si and Al are elements of the main subgroups of groups IV and III, respectively. Consequently, for silicon the maximum oxidation state in complex compounds is +4 (the number of the group where silicon is located), for aluminum - +3 (the number of the group where aluminum is located).

F and Mn are elements of the main and secondary subgroups of group VII, respectively. However, fluorine, being the most electronegative element of the Periodic Table of Chemical Elements, does not exhibit positive oxidation states: in complex compounds its oxidation state is −1 (group number −8). The highest oxidation state of manganese is +7.

Nitrogen exhibits oxidation state +3 in each of two substances:

1) HNO 2 and NH 3

2) NH 4 Cl and N 2 O 3

Answer: 3

Explanation:

In nitrous acid HNO 2, the oxidation state of oxygen in the acid residue is -2, that of hydrogen is +1, therefore, in order for the molecule to remain electrically neutral, the oxidation state of nitrogen is +3. In ammonia NH 3, nitrogen is a more electronegative element, so it attracts an electron pair of a covalent polar bond and has a negative oxidation state of -3, the oxidation state of hydrogen in ammonia is +1.

Ammonium chloride NH 4 Cl is an ammonium salt, therefore the oxidation state of nitrogen is the same as in ammonia, i.e. is equal to -3. In oxides, the oxidation state of oxygen is always -2, so for nitrogen it is +3.

In sodium nitrite NaNO 2 (a salt of nitrous acid), the oxidation degree of nitrogen is the same as in nitrogen in nitrous acid, because is +3. In nitrogen fluoride, the oxidation state of nitrogen is +3, since fluorine is the most electronegative element of the Periodic Table and in complex compounds exhibits a negative oxidation state of -1. This answer option satisfies the conditions of the task.

In nitric acid, nitrogen has the highest oxidation state equal to the group number (+5). Nitrogen as a simple compound (since it consists of atoms of one chemical element) has an oxidation state of 0.

The highest oxide of a group VI element corresponds to the formula

Answer: 4

Explanation:

The highest oxide of an element is the oxide of the element with its highest oxidation state. In a group, the highest oxidation state of an element is equal to the group number, therefore, in group VI, the maximum oxidation state of an element is +6. In oxides, oxygen exhibits an oxidation state of -2. The numbers below the element symbol are called indices and indicate the number of atoms of that element in the molecule.

The first option is incorrect, because. the element has an oxidation state of 0-(-2)⋅6/4 = +3.

In the second version, the element has an oxidation state of 0-(-2) ⋅ 4 = +8.

In the third option, the oxidation state of the element E: 0-(-2) ⋅ 2 = +4.

In the fourth option, the oxidation state of the element E: 0-(-2) ⋅ 3 = +6, i.e. this is the answer you are looking for.

The oxidation state of chromium in ammonium dichromate (NH 4) 2 Cr 2 O 7 is equal to

Answer: 1

Explanation:

In ammonium dichromate (NH 4) 2 Cr 2 O 7 in the ammonium cation NH 4 +, nitrogen, as a more electronegative element, has a lower oxidation state of -3, hydrogen is positively charged +1. Therefore, the entire cation has a charge of +1, but since there are 2 of these cations, the total charge is +2.

In order for the molecule to remain electrically neutral, the acidic residue Cr 2 O 7 2− must have a charge of -2. Oxygen in acidic residues of acids and salts always has a charge of -2, so the 7 oxygen atoms that make up the ammonium bichromate molecule are charged -14. There are 2 chromium atoms in molecules, therefore, if the charge of chromium is designated as x, then we have:

2x + 7 ⋅ (-2) = -2, where x = +6. The charge of chromium in the ammonium dichromate molecule is +6.

The +5 oxidation state is possible for each of two elements:

1) oxygen and phosphorus

2) carbon and bromine

3) chlorine and phosphorus

Answer: 3

Explanation:

In the first proposed answer, only phosphorus, as an element of the main subgroup of group V, can exhibit an oxidation state of +5, which is its maximum. Oxygen (an element of the main subgroup of group VI), being an element with high electronegativity, exhibits an oxidation state of -2 in oxides, as a simple substance - 0 and in combination with fluorine OF 2 - +1. The oxidation state +5 is not typical for it.

Carbon and bromine are elements of the main subgroups of groups IV and VII, respectively. Carbon has a maximum oxidation state of +4 (equal to the group number), and bromine exhibits oxidation states of -1, 0 (in the simple compound Br 2), +1, +3, +5 and +7.

Chlorine and phosphorus are elements of the main subgroups of groups VII and V, respectively. Phosphorus exhibits a maximum oxidation state of +5 (equal to the group number); chlorine, similar to bromine, has oxidation states of -1, 0 (in a simple compound Cl 2), +1, +3, +5, +7.

Sulfur and silicon are elements of the main subgroups of groups VI and IV, respectively. Sulfur exhibits a wide range of oxidation states from -2 (group number − 8) to +6 (group number). For silicon, the maximum oxidation state is +4 (group number).

Answer: 1

Explanation:

In sodium nitrate NaNO 3, sodium has an oxidation state of +1 (group I element), there are 3 oxygen atoms in the acidic residue, each of which has an oxidation state of −2, therefore, in order for the molecule to remain electrically neutral, nitrogen must have an oxidation state of: 0 − (+ 1) − (−2)·3 = +5.

In sodium nitrite NaNO 2, the sodium atom also has an oxidation state of +1 (an element of group I), there are 2 oxygen atoms in the acid residue, each of which has an oxidation state of −2, therefore, in order for the molecule to remain electrically neutral, nitrogen must have an oxidation state of: 0 − (+1) − (−2) 2 = +3.

NH 4 Cl – ammonium chloride. In chlorides, chlorine atoms have an oxidation state of −1, hydrogen atoms, of which there are 4 in the molecule, are positively charged, therefore, in order for the molecule to remain electrically neutral, the oxidation state of nitrogen is: 0 − (−1) − 4 · (+1) = −3. In ammonia and ammonium salt cations, nitrogen has a minimum oxidation state of −3 (the number of the group in which the element is located is 8).

In the molecule of nitric oxide NO, oxygen exhibits a minimum oxidation state of −2, as in all oxides, therefore, the oxidation state of nitrogen is +2.

0EB205

Nitrogen exhibits its highest oxidation state in a compound whose formula is

Answer: 1

Explanation:

Nitrogen is an element of the main subgroup of group V, therefore, it can exhibit a maximum oxidation state equal to the group number, i.e. +5.

One structural unit of iron nitrate Fe(NO 3) 3 consists of one Fe 3+ ion and three nitrate ions. In nitrate ions, nitrogen atoms, regardless of the type of counterion, have an oxidation state of +5.

In sodium nitrite NaNO2, sodium has an oxidation state of +1 (an element of the main subgroup of group I), there are 2 oxygen atoms in the acid residue, each of which has an oxidation state of −2, therefore, in order for the molecule to remain electrically neutral, nitrogen must have an oxidation state of 0 − ( +1) − (−2)⋅2 ​​= +3.

(NH 4) 2 SO 4 – ammonium sulfate. In sulfuric acid salts, the SO 4 2− anion has a charge of 2−, therefore, each ammonium cation has a charge of 1+. Hydrogen has a charge of +1, so nitrogen has a charge of −3 (nitrogen is more electronegative, so it attracts the common electron pair of the N–H bond). In ammonia and ammonium salt cations, nitrogen has a minimum oxidation state of −3 (the number of the group in which the element is located is 8).

In the molecule of nitrogen oxide NO2, oxygen exhibits a minimum oxidation state of −2, as in all oxides, therefore, the oxidation state of nitrogen is +4.

28910E

In compounds of the composition Fe(NO 3) 3 and CF 4, the oxidation states of nitrogen and carbon are equal, respectively

Answer: 4

Explanation:

One structural unit of iron (III) nitrate Fe(NO 3) 3 consists of one iron ion Fe 3+ and three nitrate ions NO 3 −. In nitrate ions, nitrogen always has an oxidation state of +5.

In carbon fluoride CF 4, fluorine is a more electronegative element and attracts the common electron pair of the C-F bond, exhibiting an oxidation state of -1. Therefore, carbon C has an oxidation state of +4.

A32B0B

Chlorine exhibits an oxidation state of +7 in each of two compounds:

1) Ca(OCl) 2 and Cl 2 O 7

2) KClO 3 and ClO 2

3) BaCl 2 and HClO 4

Answer: 4

Explanation:

In the first variant, chlorine atoms have oxidation states +1 and +7, respectively. One structural unit of calcium hypochlorite Ca(OCl) 2 consists of one calcium ion Ca 2+ (Ca is an element of the main subgroup of group II) and two hypochlorite ions OCl −, each of which has a charge of 1−. In complex compounds, except OF 2 and various peroxides, oxygen always has an oxidation state of −2, so it is obvious that chlorine has a charge of +1. In chlorine oxide Cl 2 O 7, as in all oxides, oxygen has an oxidation state of −2, therefore, the chlorine in this compound has an oxidation state of +7.

In potassium chlorate KClO 3, the potassium atom has an oxidation state of +1, and oxygen - -2. In order for the molecule to remain electrically neutral, chlorine must exhibit an oxidation state of +5. In chlorine oxide ClO 2, oxygen, as in any other oxide, has an oxidation state of −2; therefore, for chlorine its oxidation state is +4.

In the third option, the barium cation in the complex compound is charged +2, therefore, a negative charge of −1 is concentrated on each chlorine anion in the BaCl 2 salt. In perchloric acid HClO 4 the total charge of 4 oxygen atoms is −2⋅4 = −8, the charge on the hydrogen cation is +1. For the molecule to remain electrically neutral, the charge of chlorine must be +7.

In the fourth variant, in the magnesium perchlorate molecule Mg(ClO 4) 2 the charge of magnesium is +2 (in all complex compounds, magnesium exhibits an oxidation state of +2), therefore, for each ClO 4 − anion there is a charge of 1−. In total, 4 oxygen ions, each exhibiting an oxidation state of −2, are charged −8. Therefore, for the total charge of the anion to be 1−, the chlorine must have a charge of +7. In chlorine oxide Cl 2 O 7, as explained above, the charge of chlorine is +7.

Knowledge and ability to determine the oxidation state of elements in molecules allows one to solve very complex reaction equations and, accordingly, correctly calculate the amounts of selected substances for reactions, experiments and technological processes. The oxidation state is one of the most important, key concepts in chemistry. This table helps in determining the oxidation state of elements, exceptions to the rule are also indicated, and an algorithm for performing tasks of this type is given

Download:

Preview:

conclusions : The highest positive oxidation state of most elements is numerically equal to the group number of the table of elements in which it is found. The lowest negative oxidation state of a non-metal element is determined by the number of electrons that are missing to fill the valence layer

Consolidation: determine the oxidation states of elements in the given formulas of binary compounds. SiF 4, P 2 O 5, As 2 O 5, CaH 2, Li 3 N, OsF 8, SiCl 4, H 3 P, SCl 4, PCL 3, H 4 C, H 3 As, SF 6, AlN, CuO , Fe

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. It can take on a negative 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 valency. 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 values. Therefore, to determine it, you must first identify the indicator for oxygen.

The nonmetal located at the end of the binary formula has a negative oxidation number. 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 a chemistry table used? 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

The formula of this compound is 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 issue is discussed in the ninth grade chemistry course of the school curriculum. In addition, information about oxidation states allows you to complete OGE and USE tasks.

The ability to find the oxidation state of chemical elements is a prerequisite for successfully solving chemical equations that describe redox reactions. Without it, you will not be able to create an exact formula for a substance resulting from a reaction between various chemical elements. As a result, solving chemical problems based on such equations will be either impossible or erroneous.

In redox reactions, the concept of oxidation state is used to determine the chemical formulas of compounds of elements resulting from the interaction of several substances.

At first glance, it may seem that the oxidation number is equivalent to the concept of valence of a chemical element, but this is not so. Concept valence used to quantify electronic interactions in covalent compounds, that is, compounds formed by the formation of shared electron pairs. Oxidation number is used to describe reactions that lose or gain electrons.

Unlike valency, which is a neutral characteristic, the oxidation state can have a positive, negative, or zero value. A positive value corresponds to the number of electrons given up, and a negative value to the number of electrons added. A value of zero means that the element is either in its elemental form, has been reduced to 0 after oxidation, or has been oxidized to zero after a previous reduction.

How to determine the oxidation state of a specific chemical element
Determining the oxidation state for a specific chemical element is subject to the following rules:

1. The oxidation state of simple substances is always zero.
   
2. Alkali metals, which are in the first group of the periodic table, have an oxidation state of +1.
   
3. Alkaline earth metals, which occupy the second group in the periodic table, have an oxidation state of +2.
   
4. Hydrogen in compounds with various non-metals always exhibits an oxidation state of +1, and in compounds with metals +1.
   
5. The oxidation state of molecular oxygen in all compounds considered in the school course of inorganic chemistry is -2. Fluorine -1.
   
6. When determining the degree of oxidation in the products of chemical reactions, they proceed from the rule of electrical neutrality, according to which the sum of the oxidation states of the various elements that make up the substance must be equal to zero.
   
7. Aluminum in all compounds exhibits an oxidation state of +3.
   

There are higher, lower and intermediate oxidation states. The highest oxidation state, like valency, corresponds to the group number of a chemical element in the periodic table, but has a positive value. The lowest oxidation state is numerically equal to the difference between the number 8 group of the element. An intermediate oxidation state will be any number ranging from the lowest oxidation state to the highest.

To help you navigate the variety of oxidation states of chemical elements, we bring to your attention the following auxiliary table. Select the element you are interested in and you will receive the values ​​of its possible oxidation states. Rarely occurring values ​​will be indicated in parentheses.