Target: Continue studying valency. 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. To teach methods of comparison and generalization of the concepts being studied; develop skills and abilities in determining the degree of oxidation by chemical formulas; continue developing skills independent work; contribute to the development logical thinking. To form a sense of tolerance (tolerance and respect for other people's opinions) of mutual assistance; to carry out aesthetic education (through the design of the board and notebooks, when using presentations).
During the classes
I. Organizing time
Checking students for class.
II. Preparing for the lesson.
By the lesson you will need: Periodic system of D.I. Mendeleev, textbook, workbooks, pens, pencils.
III. Checking homework.
Frontal survey, some will work at the board on cards, conducting a test, and summing up this stage will be an intellectual game.
1. Work with cards.
1 card
Determine 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?
- What bond is called a covalent bond?
- What covalent bonds are isolated?
- What is valence?
- How do we define valency?
- Which elements (metals and non-metals) have variable valency?
3. Testing
1. Which molecules have non-polar covalent bonds?
2 . Which molecule forms a triple bond when a covalent-nonpolar bond is formed?
3 . What are positively charged ions called?
A) cations
B) molecules
B) anions
D) crystals
4. In which order 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 . Valency 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 a covalent-polar bond.
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 unshared electron pairs, only in the process of excitation of the atom. The highest valency of an element is usually equal to the group number. The degree of oxidation in compounds with different chemical bonds is formed unequally.
How is the oxidation state formed in molecules with different chemical bonds?
1) In compounds with an ionic bond, the oxidation state of the elements is equal to the charges of the ions.
2) In compounds with a covalent non-polar bond (in molecules of simple substances), the oxidation state of the elements is 0.
H 2 0 , CI 2 0 , F 2 0 , S 0 , AI 0
3) For molecules with a covalent-polar bond, the degree of oxidation is determined similarly to molecules with an ionic chemical bond.
The oxidation state of the element - this 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, in contrast to the valency, has a sign. It can be positive, negative or zero.
Valency is indicated by Roman numerals on top of the element symbol:
II |
I |
IV |
Fe |
Cu |
S, |
and the oxidation state is indicated by Arabic numerals with a charge above the element symbols ( Mg +2 , Ca +2 ,Na +1,CIˉ¹).
A positive oxidation state is equal to the number of electrons donated to these atoms. An atom can donate all valence electrons (for the main groups, these are electrons of the outer level) corresponding to the group number in which the element is located, while showing the highest oxidation state (with the exception of OF 2). For example: the highest oxidation state of the main subgroup of group II is +2 ( Zn +2) A positive degree is shown by both metals and non-metals, except for F, He, Ne. For example: C+4 ,Na+1 , Al+3
The negative oxidation state is equal to the number of electrons accepted by a given atom, it is shown only by non-metals. Atoms of non-metals attach as many electrons as they are not enough to complete the external level, while showing a negative degree.
For elements of the main subgroups of IV-VII groups, the minimum oxidation state is numerically equal to
For instance:
The value of the oxidation state between the highest and lowest oxidation states is called intermediate:
Higher |
Intermediate |
Inferior |
C +3, C +2, C 0, C -2 |
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In compounds with a covalent non-polar bond (in molecules of simple substances), the oxidation state of the elements is 0: H 2 0 , FROMI 2 0 , F 2 0 , S 0 , AI 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 compounds 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 a constant oxidation state 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
C) element of group III: Al - oxidation state +3
Algorithm for compiling a formula in compounds:
1 way
1 . The element with the lowest electronegativity is listed first, the element with the highest electronegativity is listed second.
2 . The element written in the first place has a positive charge "+", and in the second with a negative charge "-".
3 . Specify the oxidation state for each element.
4 . Find the total multiple of the oxidation states.
5. Divide the least common multiple by the value of the oxidation states and assign the resulting indices to the bottom right after the symbol of the corresponding element.
6. If the oxidation state is even - odd, then they become next to the symbol at the bottom right of the cross - crosswise without the sign "+" and "-":
7. If the oxidation state has an even value, then they must first be reduced to the smallest value of the oxidation state and put a cross - crosswise without the sign "+" and "-": C +4 O -2
2 way
1 . Let's denote the oxidation state of N through X, indicate the oxidation state of O: N 2 xO 3 -2
2 . Determine the sum of negative charges, for this, the oxidation state of oxygen is multiplied by the oxygen index: 3 (-2) \u003d -6
3 .For the molecule to be electrically neutral, you need to determine the sum of positive charges: X2 \u003d 2X
4 .Make an algebraic equation:
N 2 + 3 O 3 –2
V. Anchoring
1) Carrying out the fixing of the topic by the game, which is called "Snake".
Rules of the game: the teacher distributes cards. Each card has one question and one answer to another question.
The teacher starts the game. He reads out the question, the student who has the answer to my question 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: an Arabic numeral above the element symbol with charge "+" and "-". - What types of oxidation states are distinguished from atoms of chemical elements?
Answer: intermediate - What degree does metals exhibit?
Answer: positive, negative, zero. - What degree show simple substances or molecules with a non-polar covalent bond.
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 of substances consisting of the following elements
- N and H
- R&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. Summary of the lesson.
Rating with comments
VII. Homework
§23, p.67-72, task after §23-p. 72 No. 1-4 to complete.
DEFINITION
Oxidation state is a quantitative assessment of the state of an atom of a chemical element in a compound, based on its electronegativity.
It takes both positive and negative values. To indicate the oxidation state of an element in a compound, you need to put an Arabic numeral with the corresponding sign ("+" or "-") above its symbol.
It should be remembered that the degree of oxidation is a quantity that has no physical meaning, since it does not reflect the real charge of the atom. However, this concept is very widely used in chemistry.
Table of the oxidation state of chemical elements
The maximum positive and minimum negative oxidation states can be determined using the Periodic Table of D.I. Mendeleev. They are equal to the number of the group in which the element is located, and the difference between the value of the "highest" oxidation state and the number 8, respectively.
If we consider chemical compounds more specifically, then in substances with non-polar bonds, the oxidation state of the elements is zero (N 2, H 2, Cl 2).
The oxidation state of metals in the elementary state is zero, since the distribution of electron density in them is uniform.
In simple ionic compounds, the oxidation state of their constituent elements is equal to the electric charge, since during the formation of these compounds, an almost complete transfer of electrons from one atom to another occurs: Na +1 I -1, Mg +2 Cl -1 2, Al +3 F - 1 3 , Zr +4 Br -1 4 .
When determining the degree of oxidation of elements in compounds with polar covalent bonds, the values of their electronegativity are compared. Since, during the formation of a chemical bond, electrons are displaced to atoms of more electronegative elements, the latter have a negative oxidation state in compounds.
There are elements for which only one value of the oxidation state is characteristic (fluorine, metals of IA and IIA groups, etc.). Fluorine, which is characterized by the highest electronegativity, always has a constant negative oxidation state (-1) in compounds.
Alkaline and alkaline earth elements, which are characterized by a relatively low value of electronegativity, always have a positive oxidation state, equal to (+1) and (+2), respectively.
However, there are also such chemical elements, which are characterized by several values of the degree of oxidation (sulfur - (-2), 0, (+2), (+4), (+6), etc.).
In order to make it easier to remember how many and what oxidation states are characteristic of a particular chemical element, tables of the oxidation states of chemical elements are used, which look like this:
|
Serial number |
Russian / English title |
chemical symbol |
Oxidation state |
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Hydrogen |
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Helium / Helium |
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Lithium / Lithium |
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Beryllium / Beryllium |
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(-1), 0, (+1), (+2), (+3) |
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Carbon / Carbon |
(-4), (-3), (-2), (-1), 0, (+2), (+4) |
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Nitrogen / Nitrogen |
(-3), (-2), (-1), 0, (+1), (+2), (+3), (+4), (+5) |
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Oxygen / Oxygen |
(-2), (-1), 0, (+1), (+2) |
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Fluorine / Fluorine |
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Sodium |
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Magnesium / Magnesium |
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Aluminum |
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Silicon / Silicon |
(-4), 0, (+2), (+4) |
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Phosphorus / Phosphorus |
(-3), 0, (+3), (+5) |
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Sulfur |
(-2), 0, (+4), (+6) |
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Chlorine / Chlorine |
(-1), 0, (+1), (+3), (+5), (+7), rarely (+2) and (+4) |
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Argon / Argon |
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Potassium / Potassium |
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Calcium / Calcium |
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Scandium / Scandium |
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Titanium / Titanium |
(+2), (+3), (+4) |
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Vanadium / Vanadium |
(+2), (+3), (+4), (+5) |
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Chromium / Chromium |
(+2), (+3), (+6) |
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Manganese / Manganese |
(+2), (+3), (+4), (+6), (+7) |
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Iron / Iron |
(+2), (+3), rarely (+4) and (+6) |
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Cobalt / Cobalt |
(+2), (+3), rarely (+4) |
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Nickel / Nickel |
(+2), rarely (+1), (+3) and (+4) |
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Copper |
+1, +2, rare (+3) |
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Gallium / Gallium |
(+3), rare (+2) |
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Germanium / Germanium |
(-4), (+2), (+4) |
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Arsenic / Arsenic |
(-3), (+3), (+5), rarely (+2) |
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Selenium / Selenium |
(-2), (+4), (+6), rarely (+2) |
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Bromine / Bromine |
(-1), (+1), (+5), rarely (+3), (+4) |
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Krypton / Krypton |
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Rubidium / Rubidium |
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Strontium / Strontium |
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Yttrium / Yttrium |
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Zirconium / Zirconium |
(+4), rarely (+2) and (+3) |
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Niobium / Niobium |
(+3), (+5), rarely (+2) and (+4) |
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Molybdenum / Molybdenum |
(+3), (+6), rarely (+2), (+3) and (+5) |
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Technetium / Technetium |
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Ruthenium / Ruthenium |
(+3), (+4), (+8), rarely (+2), (+6) and (+7) |
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Rhodium |
(+4), rarely (+2), (+3) and (+6) |
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Palladium / Palladium |
(+2), (+4), rarely (+6) |
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Silver / Silver |
(+1), rarely (+2) and (+3) |
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Cadmium / Cadmium |
(+2), rare (+1) |
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Indium / Indium |
(+3), rarely (+1) and (+2) |
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Tin / Tin |
(+2), (+4) |
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Antimony / Antimony |
(-3), (+3), (+5), rarely (+4) |
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Tellurium / Tellurium |
(-2), (+4), (+6), rarely (+2) |
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(-1), (+1), (+5), (+7), rarely (+3), (+4) |
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Xenon / Xenon |
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Cesium / Cesium |
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Barium / Barium |
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Lanthanum / Lanthanum |
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Cerium / Cerium |
(+3), (+4) |
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Praseodymium / Praseodymium |
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Neodymium / Neodymium |
(+3), (+4) |
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Promethium / Promethium |
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Samaria / Samarium |
(+3), rare (+2) |
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Europium / Europium |
(+3), rare (+2) |
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Gadolinium / Gadolinium |
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Terbium / Terbium |
(+3), (+4) |
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Dysprosium / Dysprosium |
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Holmium / Holmium |
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Erbium / Erbium |
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Thulium / Thulium |
(+3), rare (+2) |
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Ytterbium / Ytterbium |
(+3), rare (+2) |
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Lutetium / Lutetium |
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Hafnium / Hafnium |
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Tantalum / Tantalum |
(+5), rarely (+3), (+4) |
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Tungsten / Tungsten |
(+6), rare (+2), (+3), (+4) and (+5) |
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Rhenium / Rhenium |
(+2), (+4), (+6), (+7), rarely (-1), (+1), (+3), (+5) |
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Osmium / Osmium |
(+3), (+4), (+6), (+8), rarely (+2) |
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Iridium / Iridium |
(+3), (+4), (+6), rarely (+1) and (+2) |
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Platinum / Platinum |
(+2), (+4), (+6), rarely (+1) and (+3) |
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Gold / Gold |
(+1), (+3), rarely (+2) |
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Mercury / Mercury |
(+1), (+2) |
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Waist / Thallium |
(+1), (+3), rarely (+2) |
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Lead / Lead |
(+2), (+4) |
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Bismuth / Bismuth |
(+3), rarely (+3), (+2), (+4) and (+5) |
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Polonium / Polonium |
(+2), (+4), rarely (-2) and (+6) |
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Astatine / Astatine |
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Radon / Radon |
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Francium / Francium |
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Radium / Radium |
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Actinium / Actinium |
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Thorium / Thorium |
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Proactinium / Protactinium |
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Uranus / Uranium |
(+3), (+4), (+6), rarely (+2) and (+5) |
Examples of problem solving
EXAMPLE 1
- The oxidation state of phosphorus in phosphine is (-3), and in phosphoric acid - (+5). Change in the oxidation state of phosphorus: +3 → +5, i.e. the first answer.
- The oxidation state of a chemical element in a simple substance is zero. The oxidation state of phosphorus in the oxide composition P 2 O 5 is equal to (+5). Change in the oxidation state of phosphorus: 0 → +5, i.e. third answer.
- The oxidation state of phosphorus in an acid of composition HPO 3 is (+5), and H 3 PO 2 is (+1). Change in the oxidation state of phosphorus: +5 → +1, i.e. fifth answer.
EXAMPLE 2
| The task | The oxidation state (-3) carbon has in the compound: a) CH 3 Cl; b) C 2 H 2 ; c) HCOH; d) C 2 H 6 . |
| Solution | In order to give a correct answer to the question posed, we will alternately determine the degree of carbon oxidation in each of the proposed compounds. a) the oxidation state of hydrogen is (+1), and chlorine - (-1). We take for "x" the degree of oxidation of carbon: x + 3×1 + (-1) =0; The answer is incorrect. b) the oxidation state of hydrogen is (+1). We take for "y" the degree of oxidation of carbon: 2×y + 2×1 = 0; The answer is incorrect. c) the oxidation state of hydrogen is (+1), and oxygen - (-2). Let's take for "z" the oxidation state of carbon: 1 + z + (-2) +1 = 0: The answer is incorrect. d) the oxidation state of hydrogen is (+1). Let's take for "a" the oxidation state of carbon: 2×a + 6×1 = 0; Correct answer. |
| Answer | Option (d) |
(repetition)
II. Oxidation state (new material)
Oxidation state- this is the conditional charge that the atom receives as a result of the complete return (acceptance) of electrons, based on the condition that all bonds in the compound are ionic.
Consider the structure of fluorine and sodium atoms:
F +9)2)7
Na+11)2)8)1
- What can be said about the completeness of the external level of fluorine and sodium atoms?
- Which atom is easier to accept, and which is easier to give valence electrons in order to complete the external level?
Do both atoms have an incomplete outer level?
It is easier for the sodium atom to donate electrons, for fluorine to accept electrons before the completion of the external level.
F 0 + 1ē → F -1 (a neutral atom accepts one negative electron and acquires an oxidation state of "-1", turning into negatively charged ion - anion )
Na 0 – 1ē → Na +1 (a neutral atom donates one negative electron and acquires an oxidation state of "+1", turning into positively charged ion - cation )
How to determine the oxidation state of an atom in PSCE D.I. Mendeleev?
Definition rules oxidation states of an atom in PSCE D.I. Mendeleev:
1. Hydrogen usually exhibits an oxidation state (CO) +1 (exception, compounds with metals (hydrides) - hydrogen has CO equal to (-1) Me + n H n -1)
2. Oxygen usually exhibits CO -2 (exceptions: O +2 F 2, H 2 O 2 -1 - hydrogen peroxide)
3. Metals only show + n positive CO
4. Fluorine always shows CO equal -1 (F-1)
5. For items main subgroups:
Higher CO (+) = group number N groups
Inferior CO (-) = N groups – 8
Rules for determining the oxidation state of an atom in a compound:
I. Oxidation state free atoms and atoms in molecules simple substances is equal to zero - Na 0 , P 4 0 , O 2 0
II. V complex substance the algebraic sum of CO of all atoms, taking into account their indices, is equal to zero = 0 , and in complex ion its charge.
For example, H +1 N +5 O 3 -2 : (+1)*1+(+5)*1+(-2)*3 = 0
2- : (+6)*1+(-2)*4 = -2
Exercise 1 - determine the oxidation states of all atoms in the formula of sulfuric acid H 2 SO 4?
1. Let's put down the known oxidation states of hydrogen and oxygen, and take the CO of sulfur as "x"
H +1 S x O 4 -2
(+1)*1+(x)*1+(-2)*4=0
X \u003d 6 or (+6), therefore, sulfur has C O +6, i.e. S+6
Task 2 - determine the oxidation states of all atoms in the formula of phosphoric acid H 3 PO 4?
1. Let's put down the known oxidation states of hydrogen and oxygen, and take the CO of phosphorus as "x"
H 3 +1 P x O 4 -2
2. Compose and solve the equation, according to the rule (II):
(+1)*3+(x)*1+(-2)*4=0
X \u003d 5 or (+5), therefore, phosphorus has C O +5, i.e. P+5
Task 3 - determine the oxidation states of all atoms in the formula of the ammonium ion (NH 4) + ?
1. Let's put down the known oxidation state of hydrogen, and take the CO of nitrogen as "x"
Valency (lat. valere - to have a meaning) is a measure of the "connective capacity" of a chemical element, equal to the number of individual chemical bonds that one atom can form.
Valence is determined by the number of bonds that one atom forms with others. For example, consider the molecule
To determine valency, you need to have a good idea of the graphic formulas of substances. In this article, you will see many formulas. I also inform you about chemical elements with constant valency, which are very useful to know.
In electronic theory, it is believed that the bond valency is determined by the number of unpaired (valence) electrons in the ground or excited state. We touched on the topic of valence electrons and the excited state of the atom. Using the example of phosphorus, let's combine these two topics for a complete understanding.
The vast majority of chemical elements have a variable valence value. Variable valency is characteristic of copper, iron, phosphorus, chromium, and sulfur.
Below you will see elements with variable valency and their compounds. Note that other elements help us to determine their non-permanent valency - with a constant valence.
Remember that for some simple substances, valency takes on the values: III - for nitrogen, II - for oxygen. Let's summarize the knowledge gained by writing the graphic formulas of nitrogen, oxygen, carbon dioxide and carbon monoxide, sodium carbonate, lithium phosphate, iron (II) sulfate and potassium acetate.
As you noticed, valencies are indicated by Roman numerals: I, II, III, etc. On the presented formulas, the valencies of substances are equal:
- N-III
- O-II
- H, Na, K, li - I
- S-VI
- C - II (in carbon monoxide CO), IV (in carbon dioxide CO 2 and sodium carbonate Na 2 CO 3
- Fe-II
The oxidation state (CO) is a conditional indicator that characterizes the charge of an atom in a compound and its behavior in an OVR (redox reaction). In simple substances, CO is always equal to zero, in complex substances it is determined based on the constant oxidation states of some elements.
Numerically, the oxidation state is equal to the conditional charge that can be attributed to an atom, guided by the assumption that all the electrons that form bonds have passed to a more electronegative element.
Determining the degree of oxidation, we attribute the conditional charge "+" to one element, and "-" to the other. This is due to electronegativity - the ability of an atom to attract electrons to itself. The sign "+" means a lack of electrons, and "-" - their excess. I repeat, CO is a conditional concept.
The sum of all oxidation states in a molecule is zero - this is important to remember for self-examination.
Knowing the changes in electronegativity in periods and groups of the periodic table D.I. Mendeleev, we can conclude which element takes "+" and which minus. Elements with a constant degree of oxidation also help in this matter.
Who is more electronegative, he attracts electrons to himself more strongly and "goes into the minus". Those who donate their electrons and experience a shortage of them receive the "+" sign.
Independently determine the oxidation states of atoms in the following substances: RbOH, NaCl, BaO, NaClO 3, SO 2 Cl 2, KMnO 4, Li 2 SO 3, O 2, NaH 2 PO 4. Below you will find a solution to this problem.
Compare the value of electronegativity according to the periodic table, and, of course, use your intuition :) However, as you study chemistry, accurate knowledge of oxidation states should replace even the most developed intuition ;-)
I would especially like to highlight the topic of ions. An ion is an atom or a group of atoms that, due to the loss or gain of one or more electrons, has acquired (and) a positive or negative charge.
When determining the CO of atoms in an ion, one should not strive to bring the total charge of the ion to "0", as in a molecule. Ions are given in the solubility table, they have different charges - it is necessary to bring the ion to such a charge. I'll explain with an example.
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Themes USE codifier: Electronegativity. The degree of oxidation and valence of chemical elements.
When atoms interact and form, the electrons between them are in most cases unevenly distributed, since the properties of the atoms differ. More electronegative the atom attracts the electron density to itself more strongly. An atom that has attracted electron density to itself acquires a partial negative charge. δ — , its "partner" is a partial positive charge δ+ . If the difference in the electronegativity of the atoms forming a bond does not exceed 1.7, we call the bond covalent polar . If the difference in electronegativity forming a chemical bond exceeds 1.7, then we call such a bond ionic .
Oxidation state is the auxiliary conditional charge of an atom of an element in a compound, calculated from the assumption that all compounds are composed of ions (all polar bonds are ionic).
What does "conditional charge" mean? We simply agree that we will simplify things a bit: we will consider any polar bonds to be completely ionic, and we will consider that an electron completely leaves or comes from one atom to another, even if in fact it is not. And conditionally, an electron leaves a less electronegative atom for a more electronegative one.
For example, in the H-Cl bond, we believe that hydrogen conditionally "gave" an electron, and its charge became +1, and chlorine "accepted" an electron, and its charge became -1. In fact, there are no such total charges on these atoms.
Surely, you have a question - why invent something that does not exist? This is not an insidious plan of chemists, everything is simple: such a model is very convenient. Ideas about the oxidation state of elements are useful in compiling classification chemicals, describing their properties, formulating compounds and nomenclature. Especially often the oxidation states are used when working with redox reactions.
The oxidation states are higher, lower And intermediate.
Higher the oxidation state is equal to the group number with a plus sign.
Inferior is defined as the group number minus 8.
AND intermediate an oxidation state is almost any integer in the range from the lowest oxidation state to the highest.
For example, nitrogen is characterized by: the highest oxidation state is +5, the lowest 5 - 8 \u003d -3, and the intermediate oxidation states are from -3 to +5. For example, in hydrazine N 2 H 4, the oxidation state of nitrogen is intermediate, -2.
Most often, the oxidation state of atoms in complex substances is indicated first by a sign, then by a number, for example +1, +2, -2 etc. When we are talking about the charge of the ion (assume that the ion really exists in the compound), then first indicate the number, then the sign. For example: Ca 2+ , CO 3 2- .
To find the oxidation states use the following regulations :
- The oxidation state of atoms in simple substances is equal to zero;
- V neutral molecules the algebraic sum of the oxidation states is zero, for ions this sum is equal to the charge of the ion;
- Oxidation state alkali metals (elements of group I of the main subgroup) in compounds is +1, the oxidation state alkaline earth metals (elements of group II of the main subgroup) in compounds is +2; oxidation state aluminum in compounds it is +3;
- Oxidation state hydrogen in compounds with metals (- NaH, CaH 2, etc.) is equal to -1 ; in compounds with non-metals () +1 ;
- Oxidation state oxygen is equal to -2 . An exception constitute peroxides- compounds containing the -О-О- group, where the oxidation state of oxygen is -1 , and some other compounds ( superoxides, ozonides, oxygen fluorides OF 2 and etc.);
- Oxidation state fluorine in all complex substances is equal to -1 .
The above are the situations when we consider the degree of oxidation permanent . For all other chemical elements, the oxidation state — variable, and depends on the order and type of atoms in the compound.
Examples:
The task: determine the oxidation states of the elements in the potassium dichromate molecule: K 2 Cr 2 O 7.
Solution: the oxidation state of potassium is +1, the oxidation state of chromium is denoted as X, oxygen oxidation state -2. The sum of all oxidation states of all atoms in a molecule is 0. We get the equation: +1*2+2*x-2*7=0. We solve it, we get the oxidation state of chromium +6.
In binary compounds, a more electronegative element is characterized by a negative oxidation state, a less electronegative element is characterized by a positive one.
note that the concept of oxidation state is very conditional! The oxidation state does not show the real charge of the atom and has no real physical meaning.. This is a simplified model that works effectively when we need, for example, to equalize the coefficients in a chemical reaction equation, or to algorithmize the classification of substances.
Oxidation state is not valence! The oxidation state and valence in many cases do not match. For example, the valency of hydrogen in a simple substance H 2 is I, and the oxidation state, according to rule 1, is 0.
These are the basic rules that will help you determine the oxidation state of atoms in compounds in most cases.
In some situations, you may find it difficult to determine the oxidation state of an atom. Let's take a look at some of these situations and how to resolve them:
- In double (salt-like) oxides, the degree at the atom, as a rule, is two oxidation states. For example, in iron oxide Fe 3 O 4 iron has two oxidation states: +2 and +3. Which one to indicate? Both. To simplify, this compound can be represented as a salt: Fe (FeO 2) 2. In this case, the acid residue forms an atom with an oxidation state of +3. Or a double oxide can be represented as follows: FeO * Fe 2 O 3.
- In peroxo compounds, the degree of oxidation of oxygen atoms connected by covalent nonpolar bonds, as a rule, changes. For example, in hydrogen peroxide H 2 O 2, and alkali metal peroxides, the oxidation state of oxygen is -1, because one of the bonds is covalent non-polar (H-O-O-H). Another example is peroxomonosulfuric acid (Caro acid) H 2 SO 5 (see figure) contains two oxygen atoms with an oxidation state of -1, the remaining atoms with an oxidation state of -2, so the following entry will be more understandable: H 2 SO 3 (O2). Chromium peroxo compounds are also known - for example, chromium (VI) peroxide CrO (O 2) 2 or CrO 5, and many others.
- Another example of compounds with ambiguous oxidation states are superoxides (NaO 2) and salt-like ozonides KO 3 . In this case, it is more appropriate to talk about the molecular ion O 2 with a charge of -1 and O 3 with a charge of -1. The structure of such particles is described by some models, which in the Russian curriculum pass the first courses of chemical universities: MO LCAO, the method of superposition of valence schemes, etc.
- In organic compounds, the concept of oxidation state is not very convenient to use, because there are a large number of covalent non-polar bonds between carbon atoms. However, if you draw the structural formula of a molecule, then the oxidation state of each atom can also be determined by the type and number of atoms with which this atom is directly bonded. For example, for primary carbon atoms in hydrocarbons, the oxidation state is -3, for secondary -2, for tertiary atoms -1, for quaternary - 0.
Let's practice determining the oxidation state of atoms in organic compounds. To do this, you need to draw the full structural formula of the atom, and select the carbon atom with its immediate environment - the atoms with which it is directly connected.
- To simplify the calculations, you can use the solubility table - the charges of the most common ions are indicated there. In most Russian chemistry exams (USE, GIA, DVI), the use of a solubility table is allowed. This is a ready-made cheat sheet, which in many cases can save a lot of time.
- When calculating the oxidation state of elements in complex substances, we first indicate the oxidation states of the elements that we know for sure (elements with a constant oxidation state), and the oxidation state of elements with a variable oxidation state is denoted as x. The sum of all charges of all particles is equal to zero in a molecule or equal to the charge of an ion in an ion. It is easy to form and solve an equation from these data.
