What is a positive oxidation state. How to arrange and how to determine the oxidation state of elements

Degrees of oxidation of elements. How to find oxidation states?

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

2) It is necessary to remember the elements that are characterized by constant oxidation states. All of them are listed in the table.

3) 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.

Consider the application of this rule on simple examples.

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

Solution. 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.

Solution. 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) = 0. Solving given 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.

Solution. 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.

Solution. 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).

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. 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.

Solution. The oxidation states of hydrogen and oxygen are known, but sulfur and nitrogen are not. Classic example problems 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.

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

Solution. 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. By S-N connections there is a shift in the electron density 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.

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The formal charge of an atom in compounds is an auxiliary quantity, it is usually used in descriptions of the properties of elements in chemistry. This conditional electric charge is the degree of oxidation. Its meaning changes as a result of many chemical processes. Although the charge is formal, it vividly characterizes the properties and behavior of atoms in redox reactions (ORDs).

Oxidation and reduction

In the past, chemists used the term "oxidation" to describe the interaction of oxygen with other elements. The name of the reactions comes from the Latin name for oxygen - Oxygenium. Later it turned out that other elements also oxidize. In this case, they are restored - they attach electrons. Each atom during the formation of a molecule changes the structure of its valence electron shell. In this case, a formal charge appears, the value of which depends on the number of conditionally given or received electrons. To characterize this value, the English chemical term "oxidation number" was previously used, which means "oxidation number" in translation. Its use is based on the assumption that the bonding electrons in molecules or ions belong to the atom with the higher electronegativity (EO). The ability to retain their electrons and attract them from other atoms is well expressed in strong non-metals (halogens, oxygen). Strong metals (sodium, potassium, lithium, calcium, other alkali and alkaline earth elements) have opposite properties.

Determination of the degree of oxidation

The oxidation state is the charge that an atom would acquire if the electrons involved in the formation of the bond were completely shifted to a more electronegative element. There are substances that do not molecular structure(alkali metal halides and other compounds). In these cases, the oxidation state coincides with the charge of the ion. The conditional or real charge shows what process took place before the atoms acquired their current state. A positive oxidation state is the total number of electrons that have been removed from the atoms. The negative value of the oxidation state is equal to the number of acquired electrons. By changing the oxidation state of a chemical element, one judges what happens to its atoms during the reaction (and vice versa). The color of the substance determines what changes in the state of oxidation have occurred. Compounds of chromium, iron and a number of other elements in which they exhibit different valences are colored differently.

Negative, zero and positive oxidation state values

Simple substances are formed by chemical elements with the same EO value. In this case, the bonding electrons belong to all structural particles equally. Therefore, in simple substances ax elements are not characterized by an oxidation state (H 0 2, O 0 2, C 0). When atoms accept electrons or the general cloud shifts in their direction, it is customary to write charges with a minus sign. For example, F -1, O -2, C -4. By donating electrons, atoms acquire a real or formal positive charge. In OF 2 oxide, the oxygen atom donates one electron each to two fluorine atoms and is in the O +2 oxidation state. It is believed that in a molecule or a polyatomic ion, the more electronegative atoms receive all the binding electrons.

Sulfur is an element that exhibits different valencies and oxidation states.

Chemical elements of the main subgroups often exhibit a lower valence equal to VIII. For example, the valency of sulfur in hydrogen sulfide and metal sulfides is II. The element is characterized by intermediate and higher valencies in the excited state, when the atom gives up one, two, four or all six electrons and exhibits valences I, II, IV, VI, respectively. The same values, only with a minus or plus sign, have the oxidation states of sulfur:

• in fluorine sulfide gives one electron: -1;
• in hydrogen sulfide, the lowest value: -2;
• in dioxide intermediate state: +4;
• in trioxide, sulfuric acid and sulfates: +6.

In its highest oxidation state, sulfur only accepts electrons; in its lowest state, it exhibits strong reducing properties. The S +4 atoms can act as reducing or oxidizing agents in compounds, depending on the conditions.

Transfer of electrons in chemical reactions

In the formation of a sodium chloride crystal, sodium donates electrons to the more electronegative chlorine. The oxidation states of the elements coincide with the charges of the ions: Na +1 Cl -1 . For molecules created by the socialization and displacement of electron pairs to a more electronegative atom, only the concept of a formal charge is applicable. But it can be assumed that all compounds are composed of ions. Then the atoms, by attracting electrons, acquire a conditional negative charge, and by giving away, they acquire a positive one. In reactions, indicate how many electrons are displaced. For example, in the carbon dioxide molecule C +4 O - 2 2, the index indicated in the upper right corner of the chemical symbol for carbon displays the number of electrons removed from the atom. Oxygen in this substance has an oxidation state of -2. The corresponding index with the chemical sign O is the number of added electrons in the atom.

How to calculate oxidation states

Counting the number of electrons donated and added by atoms can be time consuming. The following rules make this task easier:

1. In simple substances, the oxidation states are zero.
2. The sum of the oxidation of all atoms or ions in a neutral substance is zero.
3. In a complex ion, the sum of the oxidation states of all elements must correspond to the charge of the entire particle.
4. A more electronegative atom acquires a negative oxidation state, which is written with a minus sign.
5. Less electronegative elements receive positive oxidation states, they are written with a plus sign.
6. Oxygen generally exhibits an oxidation state of -2.
7. For hydrogen, the characteristic value is: +1, in metal hydrides it occurs: H-1.
8. Fluorine is the most electronegative of all elements, its oxidation state is always -4.
9. For most metals, oxidation numbers and valences are the same.

Oxidation state and valency

Most compounds are formed as a result of redox processes. The transition or displacement of electrons from one element to another leads to a change in their oxidation state and valency. Often these values ​​coincide. As a synonym for the term "oxidation state", the phrase "electrochemical valency" can be used. But there are exceptions, for example, in the ammonium ion, nitrogen is tetravalent. At the same time, the atom of this element is in the oxidation state -3. In organic substances, carbon is always tetravalent, but the oxidation states of the C atom in methane CH 4, formic alcohol CH 3 OH and HCOOH acid have different values: -4, -2 and +2.

Redox reactions

Redox includes many of the most important processes in industry, technology, animate and inanimate nature: combustion, corrosion, fermentation, intracellular respiration, photosynthesis, and other phenomena.

When compiling the OVR equations, the coefficients are selected using the electronic balance method, in which the following categories are operated:

• oxidation states;
• the reducing agent donates electrons and is oxidized;
• the oxidizing agent accepts electrons and is reduced;
• the number of given electrons must be equal to the number of attached ones.

The acquisition of electrons by an atom leads to a decrease in its oxidation state (reduction). The loss of one or more electrons by an atom is accompanied by an increase in the oxidation number of the element as a result of reactions. For OVR flowing between ions of strong electrolytes in aqueous solutions, more often they use not the electronic balance, but the method of half-reactions.

When studying ionic and covalent polar chemical bonds, you got acquainted with complex substances consisting of two chemical elements. Such substances are called bi-pair (from Latin bi - “two”) or two-element.

Let us recall the typical binary compounds that we cited as an example to consider the mechanisms for the formation of ionic and covalent polar chemical bonds: NaHl - sodium chloride and HCl - hydrogen chloride. In the first case, the bond is ionic: the sodium atom transferred its outer electron to the chlorine atom and turned into an ion with a charge of -1. and the chlorine atom accepted an electron and turned into an ion with a charge of -1. Schematically, the process of transformation of atoms into ions can be depicted as follows:

In the HCl molecule, the bond is formed due to the pairing of unpaired outer electrons and the formation of a common electron pair of hydrogen and chlorine atoms.

It is more correct to represent the formation of a covalent bond in a hydrogen chloride molecule as an overlap of a one-electron s-cloud of a hydrogen atom with a one-electron p-cloud of a chlorine atom:

During chemical interaction, the common electron pair is shifted towards the more electronegative chlorine atom:

Such conditional charges are called oxidation state. When defining this concept, it is conditionally assumed that in covalent polar compounds, the binding electrons have completely transferred to a more electronegative atom, and therefore the compounds consist only of positively and negatively charged ions.

is the conditional charge of the atoms of a chemical element in a compound, calculated on the basis of the assumption that all compounds (both ionic and covalently polar) consist only of ions.

The oxidation state can have a negative, positive, or zero value, which is usually placed above the element symbol at the top, for example:

Those atoms that have received electrons from other atoms or to which common electron pairs are displaced, that is, atoms of more electronegative elements, have a negative value for the degree of oxidation. Fluorine always has an oxidation state of -1 in all compounds. Oxygen, the second most electronegative element after fluorine, almost always has an oxidation state of -2, except for compounds with fluorine, for example:

Those atoms that donate their electrons to other atoms or from which common electron pairs are drawn, that is, atoms of less electronegative elements, have a positive oxidation state. Metals always have a positive oxidation state. For metals of the main subgroups:

Group I in all compounds, the oxidation state is +1,
Group II is equal to +2. Group III - +3, for example:

In compounds, the total oxidation state is always zero. Knowing this and the oxidation state of one of the elements, you can always find the oxidation state of another element using the formula of a binary compound. For example, let's find the oxidation state of chlorine in the compound Cl2O2. Let's denote the oxidation state -2
oxygen: Cl2O2. Therefore, seven oxygen atoms will have a total negative charge (-2) 7 =14. Then the total charge of two chlorine atoms will be +14, and one chlorine atom:
(+14):2 = +7.

Similarly, knowing the oxidation states of the elements, one can formulate the formula of a compound, for example, aluminum carbide (a compound of aluminum and carbon). Let's write the signs of aluminum and carbon next to AlC, and first the sign of aluminum, since it is a metal. We determine the number of external electrons from the periodic table of elements: Al has 3 electrons, C has 4. An aluminum atom will give up its 3 external electrons to carbon and receive an oxidation state of +3, equal to the charge of the ion. The carbon atom, on the contrary, will take the 4 electrons missing to the "cherished eight" and will receive an oxidation state of -4.

Let's write these values ​​in the formula: AlС, and find the least common multiple for them, it is equal to 12. Then we calculate the indices:

Knowing the oxidation states of elements is also necessary in order to be able to correctly name a chemical compound.

Names of binary compounds consist of two words - the names of the chemical elements that form them. The first word denotes the electronegative part of the compound - non-metal, its Latin name with the suffix -id is always in the nominative case. The second word denotes the electropositive part - a metal or a less electronegative element, its name is always in the genitive case. If the electropositive element exhibits different degrees of oxidation, then this is reflected in the name, indicating the degree of oxidation with a Roman numeral, which is placed at the end.

To chemists different countries understood each other, it was necessary to create a unified terminology and nomenclature of substances. Principles chemical nomenclature were first developed by French chemists A. Lavoisier, A. Fourctua, L. Giton and C. Berthollet in 1785. At present, the International Union of Pure and Applied Chemistry (IUPAC) coordinates the activities of scientists from several countries and issues recommendations on the nomenclature of substances and terminology used in chemistry.

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 periodicity of the change 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 are expressed in the element.

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 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 lowest degree oxidation 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 affect 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 of the outer 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 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, 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 offset 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 valence IV in molecules 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 molecule carbon monoxide CO 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 molecules 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 valency equal to five, the phosphorus atom has 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 electronic configuration outer level like this:

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.

Instruction

As a result, a complex compound is formed - hydrogen tetrachloraurate. The complexing agent in it is a gold ion, the ligands are chlorine ions, and the outer sphere is a hydrogen ion. How to determine the degree oxidation elements in this complex connection?

First of all, determine which of the elements that make up the molecule is the most electronegative, that is, which will pull the total electron density towards itself. This is chlorine, because it is in the upper right part of the periodic table, and second only to fluorine and oxygen. Therefore, his degree oxidation will be with a minus sign. What is the degree oxidation chlorine?

Chlorine, like all other halogens, is located in the 7th group of the periodic table, there are 7 electrons in its outer electronic level. By dragging another electron to this level, it will move to a stable position. Thus, his degree oxidation will be equal to -1. And since in this complex connection four chloride ions, then the total charge will be -4.

But the sum of the powers oxidation elements that make up the molecule must be zero because any molecule is electrically neutral. Thus, -4 must be balanced with a positive charge of +4, at the expense of hydrogen and gold.

You will need

• A school textbook in chemistry for grades 8-9 of any author, the periodic table, a table of electronegativity of elements (printed in school textbooks in chemistry).

Instruction

To begin with, it is necessary to indicate that the degree is a concept that takes connections for, that is, does not go deep into the structure. If the element is in a free state, then this is the simplest case - a simple substance is formed, which means that the degree oxidation its equal to zero. For example, hydrogen, oxygen, nitrogen, fluorine, etc.

In complex substances, everything is different: electrons are distributed unevenly between atoms, and it is the degree oxidation helps to determine the number of donated or received electrons. Degree oxidation may be positive or negative. With a plus, electrons are given away, with a minus they are received. Some elements of their degree oxidation are stored in various compounds, but many do not differ in this feature. It is necessary to remember an important rule - the sum of degrees oxidation is always zero. The simplest example, CO gas: knowing that the degree oxidation oxygen in the vast majority of cases is -2 and using the above rule, you can calculate the degree oxidation for C. In sum with -2, zero gives only +2, which means the degree oxidation carbon +2. Let's complicate the problem and take CO2 gas for calculations: the degree oxidation oxygen still remains -2, but in this case there are two molecules of it. Therefore, (-2) * 2 = (-4). A number that adds up to -4 to zero, +4, that is, in this gas it has a degree oxidation+4. A more complicated example: H2SO4 - hydrogen has a degree oxidation+1, oxygen has -2. In the given compound, there are 2 hydrogens and 4 oxygens, i.e. will be, respectively, +2 and -8. In order to get a total of zero, you need to add 6 pluses. So the degree oxidation sulfur +6.

When it is difficult to determine in a compound where the plus is, where the minus is, electronegativity is needed (it is easy to find in a general textbook). Metals often have a positive degree oxidation, while non-metals are negative. But for example, PI3 - both elements are non-metals. The table indicates that the electronegativity of iodine is 2.6, and 2.2. When compared, it turns out that 2.6 is greater than 2.2, that is, electrons are pulled towards iodine (iodine has a negative degree oxidation). Following the given simple examples, it is easy to determine the degree oxidation any element in the connections.

note

Do not confuse metals and non-metals, then the oxidation state will be easier to find and not get confused.

Degree oxidation called the conditional charge of an atom in a molecule. It is assumed that all bonds are ionic. In other words, oxidation characterizes the ability of an element to form ionic bond.

You will need

• - Mendeleev table.

Instruction

In a compound, the sum of the powers of the atoms is equal to the charge of that compound. This means that in a simple substance, for example, Na or H2, the degree oxidation element is zero.

Degree oxidation oxygen in compounds is usually -2. For example, H2O water has two hydrogen atoms and one oxygen atom. Indeed, -2+1+1 = 0 - on the left side of the expression is the sum of the powers oxidation all the atoms in the compound. In CaO, calcium has a degree oxidation+2, and - -2. Exceptions to this are OF2 and H2O2 compounds.
Y degree oxidation is always -1.

Usually the maximum positive degree oxidation element matches the number of its group in Mendeleev's periodic table of elements. Max Degree oxidation is equal to the element minus eight. An example is chlorine in the seventh group. 7-8 = -1 - degree oxidation. The exception to this rule is fluorine, oxygen and iron - highest degree oxidation below their group number. The elements of the copper subgroup have the highest degree oxidation more than 1.

Sources:

• The oxidation state of elements in 2018

Degree oxidation element is the conditional charge of the atoms of a chemical element in a compound, calculated from the assumption that the compounds consist only of ions. They can have positive, negative, zero values. Metals have positive oxidation states, while non-metals can have both positive and negative oxidation states. It depends on which atom the nonmetal atom is connected to.

Instruction

note

The oxidation state can have fractional values, for example, in magnetic iron ore, Fe2O3 is +8/3.

Sources:

• "Manual in Chemistry", G.P. Khomchenko, 2005.

The degree of oxidation is a characteristic of elements often found in chemistry textbooks. There are a large number of tasks aimed at determining this degree, and many of them cause difficulties for schoolchildren and students. But by following a certain algorithm, these difficulties can be avoided.

You will need

• - periodic system chemical elements (table D.I. Mendeleev).

Instruction

Remember one thing general rule: any element in a simple substance is equal to zero (simple substances: Na, Mg, Al, - ie substances consisting of one element). To determine a substance, first simply write it down without losing the indices - the numbers in the lower right part next to the symbol of the element. An example would be sulfuric - H2SO4.

Next, open the table D.I. Mendeleev and find the degree of the leftmost element in your substance - in the case this example. By existing rule its oxidation state will always be positive, and it is written with a “+” sign, since it occupies the extreme left position in the formula of a substance. To determine the numerical value of the oxidation state, pay attention to the location of the element relative to the groups. Hydrogen is in the first group, therefore, its oxidation state is +1, but since there are two hydrogen atoms in sulfuric acid (this is shown to us by the index), write +2 above its symbol.

After that, determine the oxidation state of the rightmost element in the record - oxygen in this case. Its conditional (or oxidation state) will always be negative, since it occupies the right position in the substance notation. This rule is true in all cases. The numerical value of the right element is found by subtracting the number 8 from its group number. In this case, the oxidation state of oxygen is -2 (6-8=-2), taking into account the index - -8.

To find the conditional charge of an atom of the third element, use the rule - the sum of the oxidation states of all elements must be equal to zero. Hence, the conditional charge of the oxygen atom in the substance will be equal to +6: (+2)+(+6)+(-8)=0. After that, write +6 above the sulfur symbol.

Sources:

• as the oxidation states of chemical elements

Phosphorus - chemical element, which has the 15th serial number in the Periodic Table. It is located in her V group. A classic non-metal discovered by the alchemist Brand in 1669. There are three main modifications of phosphorus: red (which is part of the mixture for lighting matches), white and black. At very high pressures(about 8.3 * 10 ^ 10 Pa), black phosphorus passes into another allotropic state (“metallic phosphorus”) and begins to conduct current. phosphorus in various substances?

Instruction

Remember degree. This is the value corresponding to the charge of the ion in the molecule, provided that the electron pairs that carry out the bond are shifted towards the more electronegative element (located to the right and above in the Periodic Table).

You also need to know the main condition: the amount electric charges of all ions that make up the molecule, taking into account the coefficients, should always be equal to zero.

The oxidation state does not always quantitatively coincide with the valency. best example- carbon, which in organic always has equal to 4, and the oxidation state can be equal to -4, and 0, and +2, and +4.

What is the oxidation state in a phosphine PH3 molecule, for example? With all that said, this question is very easy to answer. Since hydrogen is the very first element in the Periodic Table, it, by definition, cannot be located there "more to the right and higher" than. Therefore, it is phosphorus that will attract hydrogen electrons to itself.

Each hydrogen atom, having lost an electron, will turn into a positively charged oxidation ion +1. Therefore, the total positive charge is +3. Hence, taking into account the rule that the total charge of the molecule is zero, the oxidation state of phosphorus in the phosphine molecule is -3.

Well, what is the oxidation state of phosphorus in P2O5 oxide? Take the periodic table. Oxygen is located in group VI, to the right of phosphorus, and also higher, therefore, it is definitely more electronegative. That is, the oxidation state of oxygen in this compound will be with a minus sign, and phosphorus with a plus sign. What are these degrees so that the molecule as a whole is neutral? It can be easily seen that the least common multiple of the numbers 2 and 5 is 10. Therefore, the oxidation state of oxygen is -2, and that of phosphorus is +5.