EntranceValency and oxidation state of chemical elements. Electronegativity, oxidation state and valence of chemical elements
DEFINITION
Atom's ability to form chemical bonds called valency. A quantitative measure of valence is considered to be the number of different atoms in a molecule with which a given element forms bonds.
According to the exchange mechanism of the valence bond method, the valence chemical elements determined by the number of unpaired electrons contained in an atom. For s- and p-elements, these are electrons of the outer level; for d-elements, these are electrons of the outer and pre-external levels.
The values of the highest and lowest valencies of a chemical element can be determined using the Periodic Table D.I. Mendeleev. The highest valence of an element coincides with the number of the group in which it is located, and the lowest is the difference between the number 8 and the group number. For example, bromine is located in group VIIA, which means its highest valence is VII, and its lowest is I.
Paired (located two per atomic orbitals) electrons, when excited, can be separated in the presence of free cells of the same level (the separation of electrons into any level is impossible). Let's look at the example of elements of groups I and II. For example, the valence of elements of the main subgroup of group I is equal to one, since at the outer level the atoms of these elements have one electron:
3 Li 1s 2 2s 1
The valence of elements of the main subgroup of group II in the ground (unexcited) state is zero, since there are no unpaired electrons at the outer energy level:
4 Be 1s 2 2 s 2
When these atoms are excited, the paired s-electrons are separated into free cells of the p-sublevel of the same level and the valence becomes equal to two (II):
Oxidation state
To characterize the state of elements in compounds, the concept of oxidation state was introduced.
DEFINITION
The number of electrons displaced from an atom of a given element or to an atom of a given element in a compound is called oxidation state.
A positive oxidation state indicates the number of electrons that are displaced from a given atom, and a negative oxidation state indicates the number of electrons that are displaced toward a given atom.
From this definition it follows that in compounds with non-polar bonds the oxidation state of elements is zero. Examples of such compounds are molecules consisting of identical atoms (N 2, H 2, Cl 2).
The oxidation state of metals in the elemental 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 electric charge, since during the formation of these compounds there is an almost complete transfer of electrons from one atom to another: Na +1 I -1, Mg +2 Cl -1 2, Al +3 F -1 3, Zr +4 Br -1 4.
When determining the oxidation state of elements in compounds with polar covalent bonds, their electronegativity values are compared. Since during the formation of a chemical bond, electrons are displaced to the atoms of more electronegative elements, the latter have a negative oxidation state in compounds.
The concept of oxidation state for most compounds is conditional, since it does not reflect the real charge of the atom. However, this concept is very widely used in chemistry.
Most elements can exhibit varying degrees of oxidation in compounds. When determining their oxidation state, they use the rule according to which the sum of the oxidation states of elements in electrically neutral molecules is equal to zero, and in complex ions - the charge of these ions. As an example, let's calculate the degree of oxidation of nitrogen in compounds of the composition KNO 2 and HNO 3. The oxidation state of hydrogen and alkali metals in compounds is (+), and the oxidation state of oxygen is (-2). Accordingly, the oxidation degree of nitrogen is equal to:
KNO 2 1+ x + 2 × (-2) = 0, x=+3.
HNO 3 1+x+ x + 3 × (-2) = 0, x=+5.
Examples of problem solving
EXAMPLE 1
| Exercise | Valence IV is characteristic of: a) Ca; b) P; c) O; d)Si? |
| Solution | In order to give the correct answer to the question posed, we will consider each of the proposed options separately. a) Calcium is a metal. It is characterized by the only possible valency value, coinciding with the group number in the Periodic Table D.I. Mendeleev, in which it is located, i.e. The valence of calcium is II. The answer is incorrect. b) Phosphorus is a non-metal. Refers to a group of chemical elements with variable valence: the highest is determined by the group number in the Periodic Table D.I. Mendeleev, in which it is located, i.e. is equal to V, and the lowest is the difference between the number 8 and the group number, i.e. equal to III. The answer is incorrect. c) Oxygen is a non-metal. It is characterized by the only possible valency value equal to II. The answer is incorrect. d) Silicon is a non-metal. It is characterized by the only possible valency value, coinciding with the group number in the Periodic Table D.I. Mendeleev, in which it is located, i.e. The valence of silicon is IV. This is the correct answer. |
| Answer | Option (d) |
EXAMPLE 2
| Exercise | What is the valency of iron in the compound that is formed when it reacts with hydrochloric acid: a)I; b) II; c) III; d) VIII? |
| Solution | Let us write the equation for the interaction of iron with hydrochloric acid: Fe + HCl = FeCl 2 + H 2. As a result of the interaction, ferric chloride is formed and hydrogen is released. To determine the valency of iron using the chemical formula, we first count the number of chlorine atoms: We calculate the total number of chlorine valence units: We determine the number of iron atoms: it is equal to 1. Then the valency of iron in its chloride will be equal to: |
| Answer | The valency of iron in the compound formed during its interaction with hydrochloric acid is II. |
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 covalent polar 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:
Hydrogen +1. Exception: hydrides 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.
Alkali metals +1.
All metals of the second group +2. Exception: Hg +1, +2.
Aluminum +3.
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.
ValenceValenceis 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 structural formula water: H – O – N. Each dash is an electron pair, which means it shows valence, 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 valence 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.
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We learn to determine valency and oxidation state. Practice shows that many students have difficulty determining valency and oxidation state. The manual is aimed at mastering the fundamental chemical concepts of valency and oxidation state, developing the ability to give quantitative estimates and carry out calculations of valency and oxidation state using chemical formulas in inorganic and organic compounds, and also helps prepare students for passing the Unified State Exam. The manual is aimed at developing skills independent work With educational material, searching and using information, forming and developing creative potential, increasing interest in the discipline. Valency and oxidation state. Rules for determining the oxidation states of elements I. Valence Valency is the ability of atoms to attach to themselves a certain number of other atoms. Rules for determining valency 2) The metal atom comes first in the formula. 2) In the formulas of compounds, the non-metal atom exhibiting the lowest valency always comes in second place, and the name of such a compound ends in “id”. For example, Sao – calcium oxide, NaCl - sodium chloride, PbS – lead sulfide. Now you can write the formulas for any compounds of metals and non-metals. 3) The metal atom is placed first in the formula. II. Oxidation state Oxidation state- this is a conditional charge that an atom receives as a result of the complete donation (acceptance) of electrons, based on the condition that all bonds in the compound are ionic. Let's consider the structure of fluorine and sodium atoms: - 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 away valence electrons in order to complete the outer level? Do both atoms have an incomplete outer level? It is easier for a sodium atom to give up electrons, and for a fluorine atom to accept electrons before completing the outer level. F0 + 1ē → F-1 (the neutral atom accepts one negative electron and acquires the oxidation state “-1”, turning into negatively charged ion - anion ) Na0 – 1ē → Na+1 (the neutral atom gives up one negative electron and acquires the oxidation state “+1”, turning into positively charged ion - cation )
How to determine the oxidation state of an atom in PSCE? Determination rules oxidation state of an atom in PSCE: 1. Hydrogen usually exhibits oxidation number (CO) +1 (exception, compounds with metals (hydrides) – in hydrogen, CO is equal to (-1) Me+nHn-1) 2. Oxygen usually exhibits SO -2 (exceptions: O+2F2, H2O2-1 – hydrogen peroxide) 3. Metals only show + n positive CO 4. Fluorine always exhibits CO equal -1 (F-1) 5. For elements main subgroups: Higher CO (+) = group number N groups Lowest 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 equal to zero - Na0, P40, O20 II. IN complex substance the algebraic sum of the COs 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 [ S +6 O 4 -2 ]2- : (+6)*1+(-2)*4 = -2 Exercise 1 – determine the oxidation states of all atoms in the formula of sulfuric acid H2SO4? 1. Let’s put the known oxidation states of hydrogen and oxygen, and take CO of sulfur as “x” (+1)*1+(x)*1+(-2)*4=0 X=6 or (+6), therefore, sulfur has CO +6, i.e. S+6 Task 2 – determine the oxidation states of all atoms in the formula of phosphoric acid H3PO4? 1. Let’s put the known oxidation states of hydrogen and oxygen, and take the CO of phosphorus as “x” 2. Let’s compose and solve the equation according to rule (II): (+1)*3+(x)*1+(-2)*4=0 X=5 or (+5), therefore, phosphorus has CO +5, i.e. P+5 Task 3 – determine the oxidation states of all atoms in the formula of ammonium ion (NH4)+? 1. Let’s put the known oxidation state of hydrogen, and take CO2 of nitrogen as “x” 2. Let’s compose and solve the equation according to rule (II): (x)*1+(+1)*4=+1 X = -3, therefore, nitrogen has CO -3, i.e. N-3 Algorithm for compiling a formula by oxidation state Composing names of binary compounds Let’s compare the concepts of “valency” and “oxidation state”:
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Remember!
Valence -is the ability of an atom to form a certain number of bonds with other atoms.
Rules for determining valency
1. In molecules of simple substances: H2, F2, Cl2, Br2, I2 is equal to one.
2. In molecules of simple substances: O2, S8 is equal to two.
3. In the molecules of simple substances: N2, P4 and CO - carbon monoxide (II) - is equal to three.
4. In the molecules of simple substances that carbon forms (diamond, graphite), as well as in the organic compounds that it forms, the valency of carbon is four.
5. Included complex substances Hydrogen is monovalent, oxygen is mainly divalent. To determine the valency of atoms of other elements in the composition of complex substances, you need to know the structure of these substances.
Oxidation state is the conditional charge of the atoms of a chemical element in a compound, calculated on the basis of the assumption that all compounds (with ionic and covalent polar bonds) consist only of ions.
The highest oxidation state of an element is equal to the group number.
Exceptions:
fluorine the highest oxidation state is zero in a simple substance F20
oxygen highest oxidation state +2 in oxygen fluoride O+2F2
The lowest oxidation state of an element is eight minus the group number(by the number of electrons that an atom of an element can accept to complete the eight electron level)
Rulesdetermination of oxidation state (hereinafter denoted: st. ok.)
General rule: The sum of all oxidation states of elements in a molecule, taking into account the number of atoms, is zero(The molecule is electrically neutral.) , in an ion - equal to the charge of the ion.
I. The oxidation state of simple substances is zero: Sa0 , O20 ,Cl20
II. Art. OK. in binarycconnections:
Less electronegative element is put first. (Exceptions: C-4H4+ methane and N-3H3+ammonia)
It must be remembered that
Art. OK. metal is always positive
Art. OK. metals of groups I, II, III of the main subgroups is constant and equal to the group number
For the remaining art. OK. calculated by general rule.
More electronegative element is placed in second place, its art. OK. is equal to eight minus the group number (according to the number of electrons it accepts to complete the eight electron level).
Exceptions: peroxides, for example, Н2+1О2-1, Ba+2O2-1, etc.; metal carbides of groups I and II Ag2+1C2-1, Ca+2C2-1, etc. (In the school course, the compound FeS2 is found - pyrite. This is iron disulfide. The oxidation state of sulfur in it is (-1) Fe+2S2-1). This happens because in these compounds there are bonds between the same atoms -O-O-, -S-S-, a triple bond in carbides between carbon atoms. The oxidation state and valency of the elements in these compounds do not coincide: carbon has a valence of IV, oxygen and sulfur have a valence of II.
III. Oxidation state in Me bases+ n(HE)nequal to the number of hydroxo groups.
1. in the hydroxo group st. OK. oxygen -2, hydrogen +1, charge of hydroxo group 1-
2. art. OK. metal is equal to the number of hydroxyl groups
IV. Oxidation state in acids:
1st Art. OK. hydrogen +1, oxygen -2
2. art. OK. the central atom is calculated according to the general rule by solving the simple equation
For example, H3+1PxO4-2
3∙(+1) + x + 4∙(-2) = 0
3 + x – 8 = 0
x = +5 (don't forget the + sign)
You can remember that for acids with the highest oxidation state of the central element corresponding to the group number, the name will end with -naya:
Н2СО3 coal Н2С+4О3
Н2SiО3 silicon (excl.) Н2Si+4О3
НNO3 nitrogen НN+5О3
H3PO4 phosphorus H3P+5O4
Н2SO4 sulfuric Н2S+6О4
HClO4 chlorine HCl+7O4
НMnО4 manganese НMn+7О4
It remains to remember:
НNO2 nitrogenous НN+3О2
Н2SO3 sulfurous Н2S+4О3
HClO3 chloric HCl+5O3
HClO2 chloride HCl+3O2
HClHychlorous HCl+1O
V. Oxidation state in salts
at the central atom is the same as in the acid residue. It is enough to remember or define Art. OK. element in acid.
VI. The oxidation state of an element in a complex ion is equal to the charge of the ion.
For example, NH4+Cl-: we write the ion NxH4+1
x + 4∙(+1) = +1
Art. OK. nitrogen -3
For example, define Art. OK. elements in potassium hexacyanoferrate(III) K3
Potassium has +1: K3+1, hence the charge of the ion is 3-
Iron has +3 (indicated in the name) 3-, hence (CN)66-
One group (CN)-
More electronegative nitrogen: it has -3, hence (CxN-3)-
Art. OK. carbon +2
VII. Degree oxidation carbon in organic compounds is varied and is calculated based on the fact that Art. OK. hydrogen is +1, oxygen -2
For example, C3H6
3∙x + 6∙1 = 0
Art. OK. carbon -2 (with the valency of carbon being IV)
Exercise.Determine the oxidation state and valence of phosphorus in hypophosphorous acid H3PO2.
Let's calculate the oxidation state of phosphorus.
Let's denote it by x. Let's substitute the oxidation state of hydrogen +1, and oxygen -2, multiplying by the corresponding number of atoms: (+1) ∙ 3 + x + (-2) ∙ 2 = 0, hence x = +1.
Let us determine the valency of phosphorus in this acid.
It is known that it is a monoprotic acid, so only one hydrogen atom is bonded to the oxygen atom. Considering that hydrogen in compounds is monovalent and oxygen is divalent, we obtain a structural formula from which it is clear that phosphorus in this compound has a valency of five.
Graphical method for determining oxidation state
in organic matter
In organic substances, the oxidation states of elements can be determined algebraic method, and it turns out average value of oxidation state. This method is most applicable if all carbon atoms organic matter at the end of the reaction they acquired the same degree of oxidation (combustion reaction or complete oxidation).
Consider this case:
Example 1. Carbonization of deoxyribose with concentrated sulfuric acid with further oxidation:
С5Н10О4 + H2SO4 ® CO2 + H2O + SO2
Let's find the oxidation state of carbon x in deoxyribose: 5x + 10 – 8 = 0; x = - 2/5
In the electronic balance we take into account all 5 carbon atoms:
I.Valence (repetition)
Valency is the ability of atoms to attach to themselves a certain number of other atoms.
Rules for determining valency
elements in connections
1. Valence hydrogen mistaken for I(unit). Then, in accordance with the formula of water H 2 O, two hydrogen atoms are attached to one oxygen atom.
2. Oxygen in its compounds always exhibits valence II. Therefore, the carbon in the compound CO 2 (carbon dioxide) has a valence of IV.
3. Higher valence equal to group number .
4. Lowest valence is equal to the difference between the number 8 (the number of groups in the table) and the number of the group in which this element is located, i.e. 8 - N groups .
5. For metals located in “A” subgroups, the valence is equal to the group number.
6. Nonmetals generally exhibit two valences: higher and lower.
For example: sulfur has the highest valency VI and the lowest (8 – 6) equal to II; phosphorus exhibits valences V and III.
7. Valence can be constant or variable.
The valency of elements must be known in order to compose chemical formulas of compounds.
Remember!
Features of compilation chemical formulas connections.
1) The lowest valence is shown by the element that is located to the right and above in D.I. Mendeleev’s table, and the highest valence is shown by the element located to the left and below.
For example, in combination with oxygen, sulfur exhibits the highest valency VI, and oxygen the lowest valency II. Thus, the formula for sulfur oxide will be SO 3.
In the compound of silicon with carbon, the first exhibits the highest valency IV, and the second - the lowest IV. So the formula– SiC. This is silicon carbide, the basis of refractory and abrasive materials.
2) The metal atom comes first in the formula.
2) In the formulas of compounds, the non-metal atom exhibiting the lowest valency always comes in second place, and the name of such a compound ends in “id”.
For example, Sao – calcium oxide, NaCl - sodium chloride, PbS – lead sulfide.
Now you can write the formulas for any compounds of metals and non-metals.
3) The metal atom is placed first in the formula.
II. Oxidation state (new material)
Oxidation state- this is a conditional charge that an atom receives as a result of the complete donation (acceptance) of electrons, based on the condition that all bonds in the compound are ionic.
Let's 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 away valence electrons in order to complete the outer level?
Do both atoms have an incomplete outer level?
It is easier for a sodium atom to give up electrons, and for a fluorine atom to accept electrons before completing the outer 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 gives up 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 PSHE D.I. Mendeleev?
Determination rules oxidation state of an atom in PSHE D.I. Mendeleev:
1. Hydrogen usually exhibits oxidation number (CO) +1 (exception, compounds with metals (hydrides) - in hydrogen, CO is equal to (-1) Me + n H n -1)
2. Oxygen usually exhibits SO -2 (exceptions: O +2 F 2, H 2 O 2 -1 - hydrogen peroxide)
3. Metals only show + n positive CO
4. Fluorine always exhibits CO equal -1 (F -1)
5. For elements main subgroups:
Higher CO (+) = group number N groups
Lowest 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 equal to zero - Na 0 , P 4 0 , O 2 0
II. IN complex substance the algebraic sum of the COs 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 the known oxidation states of hydrogen and oxygen, and take CO of sulfur as “x”
H +1 S x O 4 -2
(+1)*1+(x)*1+(-2)*4=0
X = 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 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. Let’s compose and solve the equation according to rule (II):
(+1)*3+(x)*1+(-2)*4=0
X = 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 ammonium ion (NH 4) +?
1. Let’s put the known oxidation state of hydrogen, and take CO2 of nitrogen as “x”
(N x H 4 +1) +
2. Let’s compose and solve the equation according to rule (II):
(x)*1+(+1)*4=+1
X = -3, therefore, nitrogen has C O -3, i.e. N-3
The concept is widely used in chemistry electronegativity (EO) — the property of atoms of a given element to attract electrons from atoms of other elements in compounds is called electronegativity. The electronegativity of lithium is conventionally taken as unity, the EO of other elements is calculated accordingly. There is a scale of values of EO elements.
The numerical values of EO elements have approximate values: this is a dimensionless quantity. The higher the EO of an element, the more clearly its non-metallic properties appear. According to EO, the elements can be written as follows:
F > O > Cl > Br > S > P > C > H > Si > Al > Mg > Ca > Na > K > Cs
Fluorine has the greatest EO value. Comparing the EO values of elements from francium (0.86) to fluorine (4.1), it is easy to notice that EO obeys the Periodic Law. In the Periodic Table of Elements, EO in a period increases with the element number (from left to right), and in the main subgroups it decreases (from top to bottom). In periods, as the charges of the atomic nuclei increase, the number of electrons on the outer layer increases, the radius of the atoms decreases, therefore the ease of electron loss decreases, the EO increases, and therefore the non-metallic properties increase.
The difference in electronegativity of the elements in a compound (ΔX) will allow us to judge the type of chemical bond.
If the value Δ X = 0 – covalent nonpolar bond.
With a difference in electronegativity up to 2.0 the bond is called polar covalent, For example: H-F connection in a hydrogen fluoride molecule HF: Δ X = (3.98 – 2.20) = 1.78
Connections with electronegativity differences greater than 2.0 are considered ionic. For example: Na-Cl bond in NaCl compound: Δ X = (3.16 – 0.93) = 2.23.
Electronegativity depends on the distance between the nucleus and the valence electrons, and on how close the valence shell is to completion. The smaller the radius of an atom and the more valence electrons, the higher its EO.
Fluorine is most electronegative element. Firstly, it has 7 electrons in its valence shell (only 1 electron is missing from the octet) and, secondly, this valence shell is located close to the nucleus.
The atoms of alkali and alkaline earth metals are the least electronegative. They have large radii and their outer electron shells are far from complete. It is much easier for them to give up their valence electrons to another atom (then the outer shell will become complete) than to “gain” electrons.
Electronegativity can be expressed quantitatively and the elements can be ranked in increasing order. Most often used electronegativity scale proposed by the American chemist L. Pauling.
Oxidation state
Complex substances consisting of two chemical elements are called binary(from Latin bi - two), or two-element (NaCl, HCl). In the case of an ionic bond in a NaCl molecule, the sodium atom transfers its outer electron to the chlorine atom and becomes an ion with a charge of +1, and the chlorine atom accepts an electron and becomes an ion with a charge of -1. Schematically, the process of converting atoms into ions can be depicted as follows:
During a chemical interaction in an HCl molecule, the shared electron pair is shifted towards a more electronegative atom. For example, 
, i.e., the electron will not completely transfer from the hydrogen atom to the chlorine atom, but partially, thereby determining the partial charge of the atoms δ:
H +0.18 Cl -0.18 . If we imagine that in the HCl molecule, as well as in the NaCl chloride, the electron has completely transferred from the hydrogen atom to the chlorine atom, then they would receive charges +1 and -1:
Such conditional charges are called oxidation state. When defining this concept, it is conventionally assumed that in covalent polar compounds the bonding electrons are completely transferred to a more electronegative atom, and therefore the compounds consist only of positively and negatively charged atoms.
The oxidation state is the conditional charge of the atoms of a chemical element in a compound, calculated on the basis of the assumption that all compounds (both ionic and covalently polar) consist only of ions. The oxidation number can have a negative, positive or zero value, which is usually placed above the element symbol at the top, for example:
Those atoms that have accepted electrons from other atoms or to which common electron pairs are displaced have a negative oxidation state value. i.e. atoms of more electronegative elements. A positive oxidation state is given to those atoms that donate their electrons to other atoms or from which shared electron pairs are drawn, i.e. atoms of less electronegative elements. Atoms in molecules of simple substances and atoms in a free state have a zero oxidation state, for example:
In compounds, the total oxidation state is always zero.
Valence
The valency of an atom of a chemical element is determined primarily by the number of unpaired electrons participating in the formation of a chemical bond.
Valence possibilities atoms are defined:
The number of unpaired electrons (one-electron orbitals);
The presence of free orbitals;
The presence of lone pairs of electrons.
IN organic chemistry the concept of “valence” replaces the concept of “oxidation state”, which is usually used to work with in inorganic chemistry. However, this is not the same thing. Valence has no sign and cannot be zero, while the oxidation state is necessarily characterized by a sign and can have a value equal to zero.
Basically, valency refers to the ability of atoms to form a certain number of covalent bonds. If an atom has n unpaired electrons and m lone electron pairs, then this atom can form n + m covalent bonds with other atoms, i.e. its valence will be equal to n + m. When estimating the maximum valency, one should proceed from the electronic configuration of the “excited” state. For example, the maximum valency of a beryllium, boron and nitrogen atom is 4.
Constant valences:
- H, Na, Li, K, Rb, Cs - Oxidation state I
- O, Be, Mg, Ca, Sr, Ba, Ra, Zn, Cd - Oxidation state II
- B, Al, Ga, In - Oxidation state III
Valence Variables:
- Cu - I and II
- Fe, Co, Ni - II and III
- C, Sn, Pb - II and IV
- P- III and V
- Cr- II, III and VI
- S- II, IV and VI
- Mn- II, III, IV, VI and VII
- N- II, III, IV and V
- Cl- I, IV, VIAndVII
Using valencies, you can create a formula for a compound.
A chemical formula is a conventional recording of the composition of a substance using chemical symbols and indices.
For example: H 2 O is the formula of water, where H and O are the chemical signs of the elements, 2 is an index that shows the number of atoms of a given element that make up the water molecule.
When naming substances with variable valence, its valence must be indicated, which is placed in brackets. For example, P 2 0 5 - phosphorus oxide (V)
I. Oxidation state free atoms and atoms in molecules simple substances equal to zero—Na 0 , R 4 0 , ABOUT 2 0
II. IN complex substance the algebraic sum of the CO of all atoms, taking into account their indices, is equal to zero = 0. and in complex ion its charge.
For example:
Let's look at several compounds as an example and find out the valence chlorine:
Reference material for taking the test:
Mendeleev table
Solubility table