Oxidation states of all chemical elements in compounds. Oxidation state

In chemistry, the terms “oxidation” and “reduction” refer to reactions in which an atom or group of atoms loses or gains electrons, respectively. The oxidation state is a numerical value assigned to one or more atoms that characterizes the number of redistributed electrons and shows how these electrons are distributed between atoms during a reaction. Determining this value can be either a simple or quite complex procedure, depending on the atoms and the molecules consisting of them. Moreover, the atoms of some elements may have several oxidation states. Fortunately, there are simple, unambiguous rules for determining the oxidation state; to use them confidently, a knowledge of the basics of chemistry and algebra is sufficient.

Steps

Part 1

Determination of oxidation state according to the laws of chemistry

Determine whether the substance in question is elemental. The oxidation state of atoms outside a chemical compound is zero. This rule is true both for substances formed from individual free atoms, and for those that consist of two or polyatomic molecules of one element.

• For example, Al(s) and Cl 2 have an oxidation state of 0 because both are in a chemically unbound elemental state.
• Please note that the allotropic form of sulfur S8, or octasulfur, despite its atypical structure, is also characterized by a zero oxidation state.

1. Determine whether the substance in question consists of ions. The oxidation state of ions is equal to their charge. This is true both for free ions and for those that are part of chemical compounds.

   • For example, the oxidation state of the Cl - ion is -1.
   • The oxidation state of the Cl ion in the chemical compound NaCl is also -1. Since the Na ion, by definition, has a charge of +1, we conclude that the Cl ion has a charge of -1, and thus its oxidation state is -1.

2. Please note that metal ions can have several oxidation states. The atoms of many metallic elements can be ionized to varying degrees. For example, the charge of ions of a metal such as iron (Fe) is +2 or +3. The charge of metal ions (and their oxidation state) can be determined by the charges of ions of other elements with which the metal is part of a chemical compound; in the text this charge is indicated by Roman numerals: for example, iron (III) has an oxidation state of +3.

   • As an example, consider a compound containing an aluminum ion. The total charge of the AlCl 3 compound is zero. Since we know that Cl - ions have a charge of -1, and there are 3 such ions in the compound, for the substance in question to be overall neutral, the Al ion must have a charge of +3. Thus, in this case, the oxidation state of aluminum is +3.

3. The oxidation state of oxygen is -2 (with some exceptions). In almost all cases, oxygen atoms have an oxidation state of -2. There are a few exceptions to this rule:

   • If oxygen is in its elemental state (O2), its oxidation state is 0, as is the case for other elemental substances.
   • If oxygen is included peroxide, its oxidation state is -1. Peroxides are a group of compounds containing a simple oxygen-oxygen bond (that is, the peroxide anion O 2 -2). For example, in the composition of the H 2 O 2 (hydrogen peroxide) molecule, oxygen has a charge and oxidation state of -1.
   • When combined with fluorine, oxygen has an oxidation state of +2, read the rule for fluorine below.

4. Hydrogen has an oxidation state of +1, with some exceptions. As with oxygen, there are exceptions here too. Typically, the oxidation state of hydrogen is +1 (unless it is in the elemental state H2). However, in compounds called hydrides, the oxidation state of hydrogen is -1.

   • For example, in H2O the oxidation state of hydrogen is +1 because the oxygen atom has a -2 charge and two +1 charges are needed for overall neutrality. However, in the composition of sodium hydride, the oxidation state of hydrogen is already -1, since the Na ion carries a charge of +1, and for overall electrical neutrality, the charge of the hydrogen atom (and thus its oxidation state) must be equal to -1.

5. Fluorine Always has an oxidation state of -1. As already noted, the oxidation state of some elements (metal ions, oxygen atoms in peroxides, etc.) can vary depending on a number of factors. The oxidation state of fluorine, however, is invariably -1. This is explained by the fact that this element has the highest electronegativity - in other words, fluorine atoms are the least willing to part with their own electrons and most actively attract foreign electrons. Thus, their charge remains unchanged.

6. The sum of oxidation states in a compound is equal to its charge. The oxidation states of all atoms in a chemical compound must add up to the charge of that compound. For example, if a compound is neutral, the sum of the oxidation states of all its atoms must be zero; if the compound is a polyatomic ion with a charge of -1, the sum of the oxidation states is -1, and so on.

   • This is a good way to check - if the sum of the oxidation states does not equal the total charge of the compound, then you have made a mistake somewhere.

   Part 2

   Determination of oxidation state without using the laws of chemistry

   1. Find atoms that do not have strict rules regarding oxidation numbers. For some elements there are no firmly established rules for finding the oxidation state. If an atom does not fall under any of the rules listed above and you do not know its charge (for example, the atom is part of a complex and its charge is not specified), you can determine the oxidation number of such an atom by elimination. First, determine the charge of all other atoms of the compound, and then, from the known total charge of the compound, calculate the oxidation state of a given atom.

      • For example, in the compound Na 2 SO 4 the charge of the sulfur atom (S) is unknown - we only know that it is not zero, since sulfur is not in an elemental state. This compound serves as a good example to illustrate the algebraic method of determining oxidation state.

   2. Find the oxidation states of the remaining elements in the compound. Using the rules described above, determine the oxidation states of the remaining atoms of the compound. Don't forget about the exceptions to the rules in the case of O, H atoms, and so on.

      • For Na 2 SO 4, using our rules, we find that the charge (and therefore the oxidation state) of the Na ion is +1, and for each of the oxygen atoms it is -2.

   3. Find the unknown oxidation number from the charge of the compound. Now you have all the data to easily calculate the desired oxidation state. Write down an equation, on the left side of which there will be the sum of the number obtained in the previous step of calculations and the unknown oxidation state, and on the right side - the total charge of the compound. In other words, (Sum of known oxidation states) + (desired oxidation state) = (charge of compound).

      • In our case, the Na 2 SO 4 solution looks like this:
        • (Sum of known oxidation states) + (desired oxidation state) = (charge of compound)
        • -6 + S = 0
        • S = 0 + 6
        • S = 6. In Na 2 SO 4 sulfur has an oxidation state 6 .
   • In compounds, the sum of all oxidation states must equal the charge. For example, if the compound is a diatomic ion, the sum of the oxidation states of the atoms must equal the total ionic charge.
   • It is very useful to be able to use the periodic table and know where metallic and non-metallic elements are located in it.
   • The oxidation state of atoms in elemental form is always zero. The oxidation state of a single ion is equal to its charge. Elements of group 1A of the periodic table, such as hydrogen, lithium, sodium, in their elemental form have an oxidation state of +1; Group 2A metals such as magnesium and calcium have an oxidation state of +2 in their elemental form. Oxygen and hydrogen, depending on the type of chemical bond, can have 2 different oxidation states.

At school, chemistry still occupies the place of one of the most difficult subjects, which, due to the fact that it hides many difficulties, causes students (usually in the period from 8th to 9th grades) 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, there are sometimes even more difficult moments and unclear rules in chemistry than it seems. One of the questions that worries most students is what is oxidation number and how to determine the oxidation numbers of elements.

Important rule – placement rule, algorithms

There is a lot of talk here about compounds such as oxides. To begin with, any student must learn determination of oxides- these are complex compounds of two elements, they contain oxygen. Oxides are classified as binary compounds because oxygen comes second in the algorithm. When determining an 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 elements. For example, acid oxides are 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 valence is equal to two. It can be easily determined using Mendeleev’s periodic table of chemical elements. It is also important to know that the oxidation state of elements is an indicator that suggests either a positive or negative number.

At the beginning of the algorithm, as a rule, the metal is a non-metal, and its oxidation state is positive. The nonmetal oxygen in oxide compounds has a stable value of -2. To determine the correctness of the arrangement of all values, you need to multiply all available numbers by the indices of one specific element; if the product, taking into account all the minuses and pluses, equals 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 required for the calculation are digital. For hydrogen or proton it is always the same – +1. The negative oxygen ion has a negative oxidation state of -2.

After all these steps, you can determine the oxidation state of the central element of the formula. The expression for calculating it is a formula in the form of an equation. For example, for sulfuric acid the equation will have one unknown.

Basic terms in OVR

ORR are reduction-oxidation reactions.

• The oxidation state of any atom characterizes the ability of this atom to attach or give away electrons of ions (or atoms) to other atoms;
• It is generally accepted that oxidizing agents are either charged atoms or uncharged ions;
• 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 involves the loss of electrons.

How to assign oxidation numbers in salts

Salts consist of one metal and one or more acidic residues. The determination procedure is the same as for acid-containing acids.

The metal that directly forms the 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, we can consider the arrangement of oxidation states in sodium nitrate. The salt is formed using an element of the main subgroup of group 1; accordingly, the oxidation state will be positive and equal to one. In nitrates, oxygen has one value – -2. In order to obtain a numerical value, first an equation with one unknown is drawn up, taking into account all the pros and cons of the values: +1+X-6=0. Having solved the equation, you can come to the fact that the numerical indicator is positive and equal to + 5. This is an indicator of nitrogen. An important key to calculate the oxidation state is the 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 an element is contained in the main subgroup of group 1, then its value will be +1;
• The value of the oxides, taking into account their indices, after multiplication must be summed up and 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 equal to +2.

Many school textbooks and manuals teach how to create formulas based on valencies, even for compounds with ionic bonds. To simplify the procedure for drawing up formulas, this, in our opinion, is acceptable. But you need to understand that this is not entirely correct due to the above reasons.

A more universal concept is the concept of oxidation state. Using the values ​​of the oxidation states of atoms, as well as the valency values, you can compose chemical formulas and write down formula units.

Oxidation state- this is the conditional charge of an atom in a particle (molecule, ion, radical), calculated in the approximation that all bonds in the particle are ionic.

Before determining oxidation states, it is necessary to compare the electronegativity of the bonded atoms. An atom with a higher electronegativity value has a negative oxidation state, and an atom with a lower electronegativity has a positive oxidation state.

In order to objectively compare the electronegativity values ​​of atoms when calculating oxidation states, in 2013 IUPAC recommended using the Allen scale.

* So, for example, according to the Allen scale, the electronegativity of nitrogen is 3.066, and chlorine is 2.869.

Let us illustrate the above definition with examples. Let's compose the structural formula of a water molecule.

Covalent polar O-H bonds are indicated in blue.

Let's imagine that both bonds are not covalent, but ionic. If they were ionic, then one electron would transfer from each hydrogen atom to the more electronegative oxygen atom. Let's denote these transitions with blue arrows.

*In thatexample, the arrow serves to visually illustrate the complete transfer of electrons, and not to illustrate the inductive effect.

It is easy to notice that the number of arrows shows the number of electrons transferred, and their direction indicates the direction of electron transfer.

There are two arrows directed at the oxygen atom, which means that two electrons are transferred to the oxygen atom: 0 + (-2) = -2. A charge of -2 is formed on the oxygen atom. This is the oxidation state of oxygen in a water molecule.

Each hydrogen atom loses one electron: 0 - (-1) = +1. This means that hydrogen atoms have an oxidation state of +1.

The sum of oxidation states always equals the total charge of the particle.

For example, the sum of oxidation states in a water molecule is equal to: +1(2) + (-2) = 0. The molecule is an electrically neutral particle.

If we calculate the oxidation states in an ion, then the sum of the oxidation states is, accordingly, equal to its charge.

The oxidation state value is usually indicated in the upper right corner of the element symbol. Moreover, the sign is written in front of the number. If the sign comes after the number, then this is the charge of the ion.

For example, S -2 is a sulfur atom in the oxidation state -2, S 2- is a sulfur anion with a charge of -2.

S +6 O -2 4 2- - values ​​of the oxidation states of atoms in the sulfate anion (the charge of the ion is highlighted in green).

Now consider the case when the compound has mixed bonds: Na 2 SO 4. The bond between the sulfate anion and sodium cations is ionic, the bonds between the sulfur atom and the oxygen atoms in the sulfate ion are covalent polar. Let's write down the graphic formula of sodium sulfate, and use arrows to indicate the direction of electron transition.

*Structural formula displays the order of covalent bonds in a particle (molecule, ion, radical). Structural formulas are used only for particles with covalent bonds. For particles with ionic bonds, the concept of a structural formula has no meaning. If the particle contains ionic bonds, then a graphical formula is used.

We see that six electrons leave the central sulfur atom, which means the oxidation state of sulfur is 0 - (-6) = +6.

The terminal oxygen atoms each take two electrons, which means their oxidation states are 0 + (-2) = -2

The bridging oxygen atoms each accept two electrons and have an oxidation state of -2.

It is also possible to determine the degree of oxidation using a structural-graphical formula, where covalent bonds are indicated by dashes, and the charge of ions is indicated.

In this formula, the bridging oxygen atoms already have single negative charges and an additional electron comes to them from the sulfur atom -1 + (-1) = -2, which means their oxidation states are equal to -2.

The degree of oxidation of sodium ions is equal to their charge, i.e. +1.

Let us determine the oxidation states of elements in potassium superoxide (superoxide). To do this, let’s create a graphical formula for potassium superoxide and show the redistribution of electrons with an arrow. The O-O bond is a covalent non-polar bond, so it does not indicate the redistribution of electrons.

* Superoxide anion is a radical ion. The formal charge of one oxygen atom is -1, and the other, with an unpaired electron, is 0.

We see that the oxidation state of potassium is +1. The oxidation state of the oxygen atom written opposite potassium in the formula is -1. The oxidation state of the second oxygen atom is 0.

In the same way, you can determine the degree of oxidation using the structural-graphic formula.

The circles indicate the formal charges of the potassium ion and one of the oxygen atoms. In this case, the values ​​of formal charges coincide with the values ​​of oxidation states.

Since both oxygen atoms in the superoxide anion have different oxidation states, we can calculate arithmetic mean oxidation state oxygen.

It will be equal to / 2 = - 1/2 = -0.5.

Values ​​for arithmetic mean oxidation states are usually indicated in gross formulas or formula units to show that the sum of the oxidation states is equal to the total charge of the system.

For the case with superoxide: +1 + 2(-0.5) = 0

It is easy to determine oxidation states using electron-dot formulas, in which lone electron pairs and electrons of covalent bonds are indicated by dots.

Oxygen is an element of group VIA, therefore its atom has 6 valence electrons. Let's imagine that the bonds in a water molecule are ionic, in this case the oxygen atom would receive an octet of electrons.

The oxidation state of oxygen is correspondingly equal to: 6 - 8 = -2.

A hydrogen atoms: 1 - 0 = +1

The ability to determine oxidation states using graphic formulas is invaluable for understanding the essence of this concept; this skill will also be required in a course in organic chemistry. If we are dealing with inorganic substances, then it is necessary to be able to determine oxidation states using molecular formulas and formula units.

To do this, first of all you need to understand that oxidation states can be constant and variable. Elements exhibiting constant oxidation states must be remembered.

Any chemical element is characterized by higher and lower oxidation states.

Lowest oxidation state- this is the charge that an atom acquires as a result of receiving the maximum number of electrons on the outer electron layer.

In view of this, the lowest oxidation state has a negative value, with the exception of metals, whose atoms never accept electrons due to low electronegativity values. Metals have a lowest oxidation state of 0.

Most nonmetals of the main subgroups try to fill their outer electron layer with up to eight electrons, after which the atom acquires a stable configuration ( octet rule). Therefore, in order to determine the lowest oxidation state, it is necessary to understand how many valence electrons an atom lacks to reach the octet.

For example, nitrogen is a group VA element, which means that the nitrogen atom has five valence electrons. The nitrogen atom is three electrons short of the octet. This means the lowest oxidation state of nitrogen is: 0 + (-3) = -3

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:

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

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

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

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

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

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

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

Finding the oxidation states of elements in various compounds

Example 1

Determine the oxidation states of all elements in sulfuric acid.

Solution:

Let's write the formula of sulfuric acid:

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

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

Let us denote the oxidation state of sulfur as x:

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

Those. we got the following equation:

Let's solve it:

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

Example 2

Determine the oxidation state of all elements in ammonium dichromate.

Solution:

Let's write the formula of ammonium dichromate:

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

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

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

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

Those. we get two independent equations:

Solving which, we find x And y:

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

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.

Oxidation states 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 oxidation states of other elements is based on a simple rule:

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

Let's look at the application of this rule using 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. Let's make the simplest equation: x + 3*(+1) = 0. The solution is obvious: x = -3. Answer: N -3 H 3 +1.

Example 2. Indicate 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 create an equation to determine the oxidation state of sulfur: 2*(+1) + x + 4*(-2) = 0. Solving this equation, we find: x = +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 Al atom (+3), 9 oxygen atoms (-2) and 3 nitrogen atoms, the oxidation state of which we have to calculate. The corresponding equation is: 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 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 a mathematical point of view, the answer will be negative. A linear equation with two variables cannot have a unique solution. But we are solving more than just 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. A classic example of a problem with two unknowns! We will consider ammonium sulfate not as a single “molecule”, but as a combination of two ions: NH 4 + and SO 4 2-. The charges of ions are known to us; each of them contains only one atom with an unknown oxidation state. 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 a 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 a neighboring carbon atom. Along the C-H bond, the electron density shifts towards the carbon atom (since 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 (a shift in electron density towards C), one oxygen atom (a shift in electron density towards O) and one carbon atom (it can be assumed that the shift 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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