Higher oxidation states of elements. How to arrange and how to determine the oxidation state of elements

Such a school curriculum subject as chemistry causes numerous difficulties for most modern schoolchildren; few can determine the degree of oxidation in compounds. The greatest difficulties are experienced by schoolchildren who study, that is, primary school students (grades 8-9). Misunderstanding of the subject leads to the emergence of hostility among schoolchildren towards this subject.

Teachers identify a number of reasons for this “dislike” of middle and high school students for chemistry: reluctance to understand complex chemical terms, inability to use algorithms to consider a specific process, problems with mathematical knowledge. The Ministry of Education of the Russian Federation has made serious changes to the content of the subject. In addition, the number of hours for teaching chemistry was also “cut.” This had a negative impact on the quality of knowledge in the subject and decreased interest in studying the discipline.

What chemistry course topics are most difficult for schoolchildren?

According to the new program, the course of the basic school discipline “Chemistry” includes several serious topics: D.I. Mendeleev’s periodic table of elements, classes of inorganic substances, ion exchange. The most difficult thing for eighth graders is determining the degree of oxidation of oxides.

Arrangement rules

First of all, students should know that oxides are complex two-element compounds that include oxygen. A prerequisite for a binary compound to belong to the class of oxides is the location of oxygen second in this compound.

Algorithm for acid oxides

To begin with, let us note that degrees are numerical expressions of the valency of elements. Acidic oxides are formed by non-metals or metals with a valence of four to seven, the second in such oxides is always oxygen.

In oxides, the valence of oxygen always corresponds to two; it can be determined from the periodic table of elements by D.I. Mendeleev. A typical nonmetal like oxygen, being in group 6 of the main subgroup of the periodic table, accepts two electrons to completely complete its outer energy level. Nonmetals in compounds with oxygen most often exhibit a higher valence, which corresponds to the number of the group itself. It is important to remember that the oxidation state of chemical elements is an indicator that assumes a positive (negative) number.

The nonmetal at the beginning of the formula has a positive oxidation state. The nonmetal oxygen in oxides is stable, its index is -2. In order to check the reliability of the arrangement of values in acid oxides, you will have to multiply all the numbers you entered by the indices of a specific element. Calculations are considered reliable if the total sum of all the pros and cons of the given degrees is 0.

Compiling two-element formulas

The oxidation state of the atoms of elements gives the chance to create and write compounds from two elements. When creating a formula, firstly, both symbols are written side by side, and oxygen is always placed second. Above each of the recorded signs, the values of the oxidation states are written down, then between the found numbers there is a number that will be divisible by both numbers without any remainder. This indicator must be divided separately by the numerical value of the oxidation state, obtaining indices for the first and second components of the two-element substance. The highest oxidation state is numerically equal to the value of the highest valence of a typical non-metal and is identical to the number of the group where the non-metal is located in the PS.

Algorithm for setting numerical values in basic oxides

Oxides of typical metals are considered such compounds. In all compounds they have an oxidation state index of no more than +1 or +2. In order to understand what oxidation state a metal will have, you can use the periodic table. For metals of the main subgroups of the first group, this parameter is always constant, it is similar to the group number, that is, +1.

The metals of the main subgroup of the second group are also characterized by a stable oxidation state, in digital terms +2. The oxidation states of oxides in total, taking into account their indices (numbers), should give zero, since the chemical molecule is considered a neutral particle, devoid of charge.

Arrangement of oxidation states in oxygen-containing acids

Acids are complex substances consisting of one or more hydrogen atoms that are bonded to some kind of acidic moiety. Given that oxidation states are numbers, calculating them will require some math skills. This indicator for hydrogen (proton) in acids is always stable and is +1. Next, you can indicate the oxidation state for the negative oxygen ion; it is also stable, -2.

Only after these steps can the oxidation state of the central component of the formula be calculated. As a specific example, consider determining the oxidation state of elements in sulfuric acid H2SO4. Considering that the molecule of this complex substance contains two hydrogen protons and 4 oxygen atoms, we obtain an expression of the form +2+X-8=0. In order for the sum to form zero, sulfur will have an oxidation state of +6

Arrangement of oxidation states in salts

Salts are complex compounds consisting of metal ions and one or more acidic residues. The method for determining the oxidation states of each of the constituent parts in a complex salt is the same as in oxygen-containing acids. Considering that the oxidation state of elements is a digital indicator, it is important to correctly indicate the oxidation state of the metal.

If the metal forming the salt is located in the main subgroup, its oxidation state will be stable, corresponds to the group number, and is a positive value. If the salt contains a metal of a similar PS subgroup, the different metals can be revealed by the acid residue. After the oxidation state of the metal is established, set (-2), then calculate the oxidation state of the central element using a chemical equation.

As an example, consider the determination of oxidation states of elements in (average salt). NaNO3. The salt is formed by a metal of the main subgroup of group 1, therefore, the oxidation state of sodium will be +1. Oxygen in nitrates has an oxidation state of -2. To determine the numerical value of the oxidation state, the equation is +1+X-6=0. Solving this equation, we find that X should be +5, this is

Basic terms in OVR

There are special terms for the oxidation and reduction processes that schoolchildren must learn.

The oxidation state of an atom is its direct ability to attach to itself (donate to others) electrons from some ions or atoms.

An oxidizing agent is considered to be neutral atoms or charged ions that gain electrons during a chemical reaction.

The reducing agent will be uncharged atoms or charged ions that lose their own electrons in the process of chemical interaction.

Oxidation is thought of as a procedure of donating electrons.

Reduction involves the acceptance of additional electrons by an uncharged atom or ion.

The redox process is characterized by a reaction during which the oxidation state of an atom necessarily changes. This definition provides insight into how one can determine whether a reaction is ODD.

Rules for parsing OVR

Using this algorithm, you can arrange the coefficients in any chemical reaction.

The degree of oxidation is a conventional value used to record redox reactions. To determine the degree of oxidation, the table of oxidation of chemical elements is used.

Meaning

The oxidation state of basic chemical elements is based on their electronegativity. The value is equal to the number of electrons displaced in the compounds.

The oxidation state is considered positive if electrons are displaced from the atom, i.e. the element donates electrons in the compound and is a reducing agent. These elements include metals; their oxidation state is always positive.

When an electron is displaced towards an atom, the value is considered negative and the element is considered an oxidizing agent. The atom accepts electrons until the outer energy level is completed. Most nonmetals are oxidizing agents.

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

Rice. 1. Table of oxidation states.

In a compound, the nonmetal atom with lower electronegativity has a positive oxidation state.

Definition

You can determine the maximum and minimum oxidation states (how many electrons an atom can give and accept) using the periodic table.

The maximum degree is equal to the number of the group in which the element is located, or the number of valence electrons. The minimum value is determined by the formula:

No. (groups) – 8.

Rice. 2. Periodic table.

Carbon is in the fourth group, therefore, its highest oxidation state is +4, and its lowest is -4. The maximum oxidation degree of sulfur is +6, the minimum is -2. Most nonmetals always have a variable - positive and negative - oxidation state. The exception is fluoride. Its oxidation state is always -1.

It should be remembered that this rule does not apply to alkali and alkaline earth metals of groups I and II, respectively. These metals have a constant positive oxidation state - lithium Li +1, sodium Na +1, potassium K +1, beryllium Be +2, magnesium Mg +2, calcium Ca +2, strontium Sr +2, barium Ba +2. Other metals may exhibit varying degrees of oxidation. The exception is aluminum. Despite being in group III, its oxidation state is always +3.

Rice. 3. Alkali and alkaline earth metals.

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

Record

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

inert gases do not react, so their oxidation state is always zero;
in compounds, the variable oxidation state depends on the variable valence and interaction with other elements;
hydrogen in compounds with metals exhibits a negative oxidation state - Ca +2 H 2 −1, Na +1 H −1;
oxygen always has an oxidation state of -2, except for oxygen fluoride and peroxide - O +2 F 2 −1, H 2 +1 O 2 −1.

What have we learned?

The oxidation state is a conditional value showing how many electrons an atom of an element in a compound has accepted or given up. The value depends on the number of valence electrons. Metals in compounds always have a positive oxidation state, i.e. are reducing agents. For alkali and alkaline earth metals, the oxidation state is always the same. Nonmetals, except fluorine, can take on positive and negative oxidation states.

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.

Target: Continue studying valence. Give the concept of oxidation state. Consider the types of oxidation states: positive, negative, zero value. Learn to correctly determine the oxidation state of an atom in a compound. Teach techniques for comparing and generalizing the concepts being studied; develop skills in determining the degree of oxidation using chemical formulas; continue to develop independent work skills; promote the development of logical thinking. To develop a sense of tolerance (tolerance and respect for other people’s opinions) and mutual assistance; carry out aesthetic education (through the design of boards and notebooks, when using presentations).

During the classes

I. Organizing time

Checking students for the lesson.

II. Preparing for the lesson.

For the lesson you will need: D.I. Mendeleev's periodic table, textbook, workbooks, pens, pencils.

III. Checking homework.

A frontal survey, some will work at the board using cards, a test, and the conclusion of this stage will be an intellectual game.

1. Working with cards.

1 card

Determine the mass fractions (%) of carbon and oxygen in carbon dioxide (CO 2 ) .

2 card

Determine the type of bond in the H 2 S molecule. Write the structural and electronic formulas of the molecule.

2. Frontal survey

What is a chemical bond?
What types of chemical bonds do you know?
Which bond is called a covalent bond?
What covalent bonds are distinguished?
What is valence?
How do we define valence?
Which elements (metals and non-metals) have variable valence?

3. Testing

1. In which molecules does a nonpolar covalent bond exist?

2 . Which molecule forms a triple bond when a covalently nonpolar bond is formed?

3 . What are positively charged ions called?

A) cations

B) molecules

B) anions

D) crystals

4. In which row are the substances of an ionic compound located?

A) CH 4, NH 3, Mg

B) CI 2, MgO, NaCI

B) MgF 2, NaCI, CaCI 2

D) H 2 S, HCI, H 2 O

5 . Valence is determined by:

A) by group number

B) by the number of unpaired electrons

B) by type of chemical bond

D) by period number.

4. Intellectual game “Tic-tac-toe” »

Find substances with covalently polar bonds.

IV. Learning new material

The oxidation state is an important characteristic of the state of an atom in a molecule. Valence is determined by the number of unpaired electrons in an atom, orbitals with lone electron pairs, only in the process of excitation of the atom. The highest valence of an element is usually equal to the group number. The degree of oxidation in compounds with different chemical bonds is formed differently.

How is the oxidation state formed for molecules with different chemical bonds?

1) In compounds with ionic bonds, the oxidation states of the elements are equal to the charges of the ions.

2) In compounds with a covalent nonpolar bond (in molecules of simple substances), the oxidation state of the elements is 0.

N 2 0 , CI 2 0 , F 2 0 , S 0 , A.I. 0

3) For molecules with a covalently polar bond, the oxidation state is determined similarly to molecules with an ionic chemical bond.

Element oxidation state is the conditional charge of its atom in a molecule, if we assume that the molecule consists of ions.

The oxidation state of an atom, unlike its valency, has a sign. It can be positive, negative and zero.

Valency is indicated by Roman numerals above the element symbol:

II	I	IV
Fe	Cu	S,

and the oxidation state is indicated by Arabic numerals with the charge above the element symbols ( Mg +2 , Ca +2 ,Na +1,C.I.ˉ¹).

A positive oxidation state is equal to the number of electrons given to these atoms. An atom can give up all the valence electrons (for main groups these are electrons of the outer level) corresponding to the number of the group in which the element is located, while exhibiting the highest oxidation state (with the exception of ОF 2). For example: the highest oxidation state of the main subgroup of group II is +2 ( Zn +2) A positive degree is exhibited by both metals and non-metals, except F, He, Ne. For example: C+4,Na+1 , Al+3

A negative oxidation state is equal to the number of electrons accepted by a given atom; it is exhibited only by non-metals. Nonmetal atoms add as many electrons as they lack to complete the outer level, thus exhibiting a negative degree.

For elements of the main subgroups of groups IV-VII, the minimum oxidation state is numerically equal to

For example:

The value of the oxidation state between the highest and lowest oxidation states is called intermediate:

*Higher*	*Intermediate*	*Lowest*
	C +3, C +2, C 0, C -2

In compounds with a covalent nonpolar bond (in molecules of simple substances), the oxidation state of the elements is 0: N 2 0 , WITHI 2 0 , F 2 0 , S 0 , A.I. 0

To determine the oxidation state of an atom in a compound, a number of provisions should be taken into account:

1. Oxidation stateFin all connections is equal to “-1”.Na +1 F -1 , H +1 F -1

2. The oxidation state of oxygen in most compounds is (-2) exception: OF 2 , where the oxidation state is O +2F -1

3. Hydrogen in most compounds has an oxidation state of +1, except for compounds with active metals, where the oxidation state is (-1): Na +1 H -1

4. The degree of oxidation of metals of the main subgroupsI, II, IIIgroups in all compounds is +1,+2,+3.

Elements with constant oxidation states are:

A) alkali metals (Li, Na, K, Pb, Si, Fr) - oxidation state +1

B) elements of the II main subgroup of the group except (Hg): Be, Mg, Ca, Sr, Ra, Zn, Cd - oxidation state +2

B) element of group III: Al - oxidation state +3

Algorithm for composing formulas in compounds:

1 way

1 . The element with lower electronegativity is written in first place, and in second place with higher electronegativity.

2 . The element written in the first place has a positive charge “+”, and the element written in the second place has a negative charge “-”.

3 . Indicate the oxidation state for each element.

4 . Find the common multiple of the oxidation states.

5. Divide the least common multiple by the value of oxidation states and assign the resulting indices to the lower right after the symbol of the corresponding element.

6. If the oxidation state is even - odd, then they appear next to the symbol at the bottom right - a cross - crisscross without the "+" and "-" signs:

7. If the oxidation state has an even value, then they must first be reduced to the lowest value of the oxidation state and put a cross without the “+” and “-” signs: C +4 O -2

Method 2

1 . Let us denote the oxidation state of N by X, indicate the oxidation state of O: N 2 xO 3 -2

2 . Determine the sum of negative charges; to do this, multiply the oxidation state of oxygen by the oxygen index: 3· (-2) = -6

3 For a molecule to be electrically neutral, you need to determine the sum of positive charges: X2 = 2X

4 .Make up an algebraic equation:

N 2 + 3 O 3 –2

V. Consolidation

1) Reinforcing the topic with a game called “Snake”.

Rules of the game: the teacher distributes cards. Each card contains one question and one answer to another question.

The teacher starts the game. When the question is read out, the student who has the answer to my question on the card raises his hand and says the answer. If the answer is correct, then he reads his question and the student who has the answer to this question raises his hand and answers, etc. A snake of correct answers is formed.

How and where is the oxidation state of an atom of a chemical element indicated?
Answer: Arabic numeral above the symbol of the element with charge "+" and "-".
What types of oxidation states are distinguished in atoms of chemical elements?
Answer: intermediate
What degree does metal exhibit?
Answer: positive, negative, zero.
What degree do simple substances or molecules with non-polar covalent bonds exhibit?
Answer: positive
What charge do cations and anions have?
Answer: null.
What is the name of the oxidation state that stands between the positive and negative oxidation states.
Answer: positive, negative

2) Write formulas for substances consisting of the following elements

N and H
R and O
Zn and Cl

3) Find and cross out substances that do not have a variable oxidation state.

Na, Cr, Fe, K, N, Hg, S, Al, C

VI. Lesson summary.

Rating with comments

VII. Homework

§23, pp.67-72, complete the task after §23-page 72 No. 1-4.

Video tutorial 2: Oxidation state of chemical elements

Video tutorial 3: Valence. Determination of valency

Lecture: Electronegativity. Oxidation state and valence of chemical elements

Electronegativity

Electronegativity is the ability of atoms to attract electrons from other atoms to join them.

It is easy to judge the electronegativity of a particular chemical element using the table. Remember, in one of our lessons it was said that it increases when moving from left to right through periods in the periodic table and when moving from bottom to top through groups.

For example, the task was given to determine which element from the proposed series is the most electronegative: C (carbon), N (nitrogen), O (oxygen), S (sulfur)? We look at the table and find that this is O, because he is to the right and higher than the others.

What factors influence electronegativity? This:

The radius of an atom, the smaller it is, the higher the electronegativity.
The valence shell is filled with electrons; the more electrons there are, the higher the electronegativity.

Of all the chemical elements, fluorine is the most electronegative because it has a small atomic radius and 7 electrons in its valence shell.

Elements with low electronegativity include alkali and alkaline earth metals. They have large radii and very few electrons in the outer shell.

The electronegativity values of an atom cannot be constant, because it depends on many factors, including those listed above, as well as the degree of oxidation, which can be different for the same element. Therefore, it is customary to talk about the relativity of electronegativity values. You can use the following scales:

You will need electronegativity values when writing formulas for binary compounds consisting of two elements. For example, the formula of copper oxide Cu 2 O - the first element should be written down the one whose electronegativity is lower.

At the moment of formation of a chemical bond, if the electronegativity difference between the elements is greater than 2.0, a covalent polar bond is formed; if less, an ionic bond is formed.

Oxidation state

Oxidation state (CO)- this is the conditional or real charge of an atom in a compound: conditional - if the bond is polar covalent, real - if the bond is ionic.

An atom acquires a positive charge when it gives up electrons, and a negative charge when it accepts electrons.

Oxidation states are written above the symbols with a sign «+»/«-» . There are also intermediate COs. The maximum CO of an element is positive and equal to group number, and the minimum negative for metals is zero, for non-metals = (Group No. – 8). Elements with maximum CO only accept electrons, and elements with minimum CO only give up electrons. Elements that have intermediate COs can both give and receive electrons.

Let's look at some rules that should be followed to determine CO:

The CO of all simple substances is zero.

The sum of all CO atoms in a molecule is also equal to zero, since any molecule is electrically neutral.

In compounds with a covalent nonpolar bond, CO is equal to zero (O 2 0), and with an ionic bond it is equal to the charges of the ions (Na + Cl - sodium CO +1, chlorine -1). CO elements of compounds with a covalent polar bond are considered as with an ionic bond (H:Cl = H + Cl -, which means H +1 Cl -1).

Elements in a compound that have the greatest electronegativity have negative oxidation states, while those with the least electronegativity have positive oxidation states. Based on this, we can conclude that metals have only a “+” oxidation state.

Constant oxidation states:

Alkali metals +1.

All metals of the second group +2. Exception: Hg +1, +2.

Aluminum +3.

Hydrogen +1. Exception: hydrides of active metals NaH, CaH 2, etc., where the oxidation state of hydrogen is –1.
Oxygen –2. Exception: F 2 -1 O +2 and peroxides that contain the –O–O– group, in which the oxidation state of oxygen is –1.

When an ionic bond is formed, a certain transfer of electron occurs, from a less electronegative atom to an atom of greater electronegativity. Also, in this process, atoms always lose electrical neutrality and subsequently turn into ions. Integer charges are also formed. When a polar covalent bond is formed, the electron is transferred only partially, so partial charges arise.

Valence

Valenceis the ability of atoms to form n - the number of chemical bonds with atoms of other elements.

Valence is also the ability of an atom to hold other atoms near itself. As you know from your school chemistry course, different atoms are bonded to each other by electrons from the outer energy level. An unpaired electron seeks a pair from another atom. These outer level electrons are called valence electrons. This means that valence can also be defined as the number of electron pairs connecting atoms to each other. Look at the structural formula of water: H – O – H. Each dash is an electron pair, which means it shows the valency, i.e. oxygen here has two lines, which means it is divalent, hydrogen molecules come from one line each, which means hydrogen is monovalent. When writing, valence is indicated by Roman numerals: O (II), H (I). Can also be indicated above the element.

Valence can be constant or variable. For example, in metal alkalis it is constant and equals I. But chlorine in various compounds exhibits valencies I, III, V, VII.

How to determine the valence of an element?

Let's look again at the Periodic Table. Metals of the main subgroups have a constant valency, so metals of the first group have valency I, the second - II. And metals of side subgroups have variable valency. It is also variable for non-metals. The highest valence of an atom is equal to group number, the lowest is equal to = group number - 8. A familiar formulation. Doesn't this mean that the valency coincides with the oxidation state? Remember, valence may coincide with the oxidation state, but these indicators are not identical to each other. Valency cannot have a =/- sign, and also cannot be zero.

The second method is to determine valency using a chemical formula, if the constant valency of one of the elements is known. For example, take the formula of copper oxide: CuO. Oxygen valency II. We see that for one oxygen atom in this formula there is one copper atom, which means that the valence of copper is equal to II. Now let's take a more complicated formula: Fe 2 O 3. The valency of the oxygen atom is II. There are three such atoms here, multiply 2*3 =6. We found that there are 6 valences per two iron atoms. Let's find out the valence of one iron atom: 6:2=3. This means that the valence of iron is III.

In addition, when it is necessary to estimate the "maximum valence", one should always start from the electronic configuration that is present in the "excited" state.