What is the highest oxidation state of an element? Electronegativity
In chemical processes, the main role is played by atoms and molecules, the properties of which determine the outcome of chemical reactions. One of the important characteristics of an atom is the oxidation number, which simplifies the method of accounting for electron transfer in a particle. How to determine the oxidation state or formal charge of a particle and what rules do you need to know for this?
Any chemical reaction is caused by the interaction of atoms of different substances. The reaction process and its result depend on the characteristics of the smallest particles.
The term oxidation (oxidation) in chemistry means a reaction during which a group of atoms or one of them loses electrons or gains; in the case of acquisition, the reaction is called “reduction”.
The oxidation state is a quantity that is measured quantitatively and characterizes the redistributed electrons during a reaction. Those. During the process of oxidation, electrons in an atom decrease or increase, redistributing between other interacting particles, and the level of oxidation shows exactly how they are reorganized. This concept is closely related to the electronegativity of particles - their ability to attract and repel free ions.
Determining the level of oxidation depends on the characteristics and properties of a particular substance, so the calculation procedure cannot be unambiguously called easy or complex, but its results help conditionally record the processes of redox reactions. It should be understood that the resulting calculation result is the result of taking into account the transfer of electrons and has no physical meaning, and is not the true charge of the nucleus.
It is important to know! Inorganic chemistry often uses the term valence instead of the oxidation state of elements; this is not a mistake, but it should be borne in mind that the second concept is more universal.
The concepts and rules for calculating the movement of electrons are the basis for classifying chemical substances (nomenclature), describing their properties and drawing up communication formulas. But most often this concept is used to describe and work with redox reactions.
Rules for determining the degree of oxidation
How to find out the oxidation state? When working with redox reactions, it is important to know that the formal charge of a particle will always be equal to the value of the electron, expressed in a numerical value. This feature is due to the assumption that the electron pairs forming a bond are always completely shifted towards more negative particles. It should be understood that we are talking about ionic bonds, and in the case of a reaction, electrons will be divided equally between identical particles.
The oxidation number can have both positive and negative values. The thing is that during the reaction the atom must become neutral, and for this it is necessary to either add a certain number of electrons to the ion, if it is positive, or take them away if it is negative. To denote this concept, when writing a formula, an Arabic numeral with the corresponding sign is usually written above the element designation. For example, or etc.
You should know that the formal charge of metals will always be positive, and in most cases, you can use the periodic table to determine it. There are a number of features that must be taken into account in order to determine the indicators correctly.
Oxidation degree:
Having remembered these features, it will be quite simple to determine the oxidation number of elements, regardless of the complexity and number of atomic levels.
Useful video: determining the oxidation state
Mendeleev's periodic table contains almost all the necessary information for working with chemical elements. For example, schoolchildren use only it to describe chemical reactions. So, in order to determine the maximum positive and negative values of the oxidation number, you need to check the designation of the chemical element in the table:
- The maximum positive is the number of the group in which the element is located.
- The maximum negative oxidation state is the difference between the maximum positive boundary and the number 8.
Thus, it is enough to simply find out the extreme boundaries of the formal charge of a particular element. This action can be performed using calculations based on the periodic table.
It is important to know! One element can simultaneously have several different oxidation rates.
There are two main methods for determining the level of oxidation, examples of which are presented below. The first of them is a method that requires knowledge and ability to apply the laws of chemistry. How to arrange oxidation states using this method?
Rule for determining oxidation states
To do this you need:
- Determine whether a given substance is elemental and whether it is outside the bond. If so, then its oxidation number will be 0, regardless of the composition of the substance (individual atoms or multi-level atomic compounds).
- Determine whether the substance in question consists of ions. If so, then the degree of oxidation will be equal to their charge.
- If the substance in question is metal, then look at the indicators of other substances in the formula and calculate the metal readings using arithmetic operations.
- If the entire compound has one charge (essentially it is the sum of all particles of the elements represented), then it is enough to determine the indicators of simple substances, then subtract them from the total and get the metal data.
- If the relationship is neutral, then the total sum must be zero.
As an example, consider combining with an aluminum ion whose net charge is zero. The rules of chemistry confirm the fact that the Cl ion has an oxidation number of -1, and in this case there are three of them in the compound. This means that the Al ion must be +3 for the entire compound to be neutral.
This method is very good, since the correctness of the solution can always be checked by adding all the oxidation levels together.
The second method can be used without knowledge of chemical laws:
- Find data on particles for which there are no strict rules and the exact number of their electrons is unknown (this can be done by exclusion).
- Find out the indicators of all other particles and then find the desired particle from the total by subtraction.
Let's consider the second method using the example of the substance Na2SO4, in which the sulfur atom S is not determined, it is only known that it is different from zero.
To find what all oxidation states are equal to:
- Find known elements, keeping in mind traditional rules and exceptions.
- Na ion = +1, and each oxygen = -2.
- Multiply the number of particles of each substance by their electrons to obtain the oxidation states of all atoms except one.
- Na2SO4 contains 2 sodium and 4 oxygen; when multiplied, it turns out: 2 X +1 = 2 is the oxidation number of all sodium particles and 4 X -2 = -8 - oxygen.
- Add the results obtained 2+(-8) =-6 - this is the total charge of the compound without the sulfur particle.
- Represent the chemical notation as an equation: sum of known data + unknown number = total charge.
- Na2SO4 is represented as follows: -6 + S = 0, S = 0 + 6, S = 6.
Thus, to use the second method, it is enough to know the simple laws of arithmetic.
Oxidation table
To simplify the work and calculate oxidation indicators for each chemical substance, special tables are used where all the data is recorded.
It looks like this:
Useful video: learning to determine the oxidation state using formulas
Conclusion
Finding the oxidation number for a chemical is a simple task that requires only care and knowledge of the basic rules and exceptions. Knowing the exceptions and using special tables, this action will not take much time.
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PART AND
GENERAL CHEMISTRY
CHEMICAL BONDING AND STRUCTURE OF SUBSTANCE
Oxidation state
The oxidation state is the conditional charge on an atom in a molecule or crystal that would arise on it when all the polar bonds created by it were ionic in nature.
Unlike valence, oxidation states can be positive, negative, or zero. In simple ionic compounds, the oxidation state coincides with the charges of the ions. For example, in sodium chloride
NaCl (Na + Cl - ) Sodium has an oxidation state of +1, and Chlorine -1, in calcium oxide CaO (Ca +2 O -2). Calcium exhibits an oxidation state of +2, and Oxysene - -2. This rule applies to all basic oxides: the oxidation state of a metal element is equal to the charge of the metal ion (Sodium +1, Barium +2, Aluminum +3), and the oxidation state of Oxygen is -2. The oxidation state is indicated by Arabic numerals, which are placed above the symbol of the element, like valence, and the sign of the charge is indicated first, and then its numerical value:If the modulus of the oxidation state is equal to one, then the number “1” can be omitted and only the sign can be written:
Na + Cl - .Oxidation number and valence are related concepts. In many compounds, the absolute value of the oxidation state of elements coincides with their valency. However, there are many cases where the valence differs from the oxidation state.
In simple substances - non-metals, there is a covalent non-polar bond; the shared electron pair is displaced to one of the atoms, therefore the oxidation state of elements in simple substances is always zero. But the atoms are connected to each other, that is, they exhibit a certain valence, as, for example, in oxygen the valence of Oxygen is II, and in nitrogen the valence of Nitrogen is III:
In the hydrogen peroxide molecule, the valency of Oxygen is also II, and that of Hydrogen is I:
Definition of possible degrees oxidation of elements
The oxidation states that elements can exhibit in various compounds can in most cases be determined by the structure of the outer electronic level or by the element’s place in the Periodic Table.
Atoms of metallic elements can only donate electrons, so they exhibit positive oxidation states in compounds. Its absolute value in many cases (except d -elements) is equal to the number of electrons in the outer level, that is, the group number in the Periodic Table. Atoms d -elements can also donate electrons from a higher level, namely from unfilled d -orbitals. Therefore for d -elements, determining all possible oxidation states is much more difficult than for s- and p-elements. It is safe to say that the majority d -elements exhibit an oxidation state of +2 due to electrons in the outer electron level, and the maximum oxidation state in most cases is equal to the group number.
Atoms of nonmetallic elements can exhibit both positive and negative oxidation states, depending on which atom of the element they form a bond with. If an element is more electronegative, then it exhibits a negative oxidation state, and if it is less electronegative, it exhibits a positive oxidation state.
The absolute value of the oxidation state of non-metallic elements can be determined by the structure of the outer electronic layer. An atom is capable of accepting so many electrons that eight electrons are located on its outer level: non-metallic elements of group VII accept one electron and exhibit an oxidation state of -1, group VI - two electrons and exhibit an oxidation state of -2, etc.
Non-metallic elements are capable of donating a different number of electrons: a maximum of as many as are located at the outer energy level. In other words, the maximum oxidation state of non-metallic elements is equal to the group number. Due to the circulation of electrons at the outer level of atoms, the number of unpaired electrons that an atom can give up in chemical reactions varies, so non-metallic elements are capable of exhibiting different intermediate values of oxidation state.
Possible oxidation states s- and p-elements
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PS Group |
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Highest oxidation state |
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Intermediate oxidation state |
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Lower oxidation state |
Determination of oxidation states in compounds
Any electrically neutral molecule, therefore the sum of the oxidation states of the atoms of all elements must be equal to zero. Let us determine the degree of oxidation in sulfur(I) V) oxide SO 2 tauphosphorus (V) sulfide P 2 S 5.
Sulfur(I V) oxide SO 2 formed by atoms of two elements. Of these, Oxygen has the largest electronegativity, so Oxygen atoms will have a negative oxidation state. For Oxygen it is equal to -2. In this case, Sulfur has a positive oxidation state. Sulfur can exhibit different oxidation states in different compounds, so in this case it must be calculated. In a molecule SO 2 two Oxygen atoms with an oxidation state of -2, so the total charge of the Oxygen atoms is -4. In order for the molecule to be electrically neutral, the Sulfur atom has to completely neutralize the charge of both Oxygen atoms, therefore the oxidation state of Sulfur is +4:
In the molecule there is phosphorus ( V) sulfide P 2 S 5 The more electronegative element is Sulfur, that is, it exhibits a negative oxidation state, and Phosphorus has a positive oxidation state. For Sulfur, the negative oxidation state is only 2. Together, the five atoms of Sulfur carry a negative charge of -10. Therefore two Phosphorus atoms have to neutralize this charge with a total charge of +10. Since there are two Phosphorus atoms in the molecule, each must have an oxidation state of +5:
It is more difficult to calculate the oxidation state in non-binary compounds - salts, bases and acids. But for this you should also use the principle of electrical neutrality, and also remember that in most compounds the oxidation state of Oxygen is -2, Hydrogen +1.
Let's look at this using potassium sulfate as an example. K2SO4. The oxidation state of Potassium in compounds can only be +1, and Oxygen -2:
Using the principle of electrical neutrality, we calculate the oxidation state of Sulfur:
2(+1) + 1 (x) + 4 (-2) = 0, whence x = +6.
When determining the oxidation states of elements in compounds, the following rules should be followed:
1. The oxidation state of an element in a simple substance is zero.
2. Fluorine is the most electronegative chemical element, therefore the oxidation state of Fluorine in all compounds is equal to -1.
3. Oxygen is the most electronegative element after Fluorine, therefore the oxidation state of Oxygen in all compounds except fluorides is negative: in most cases it is -2, and in peroxides - -1.
4. The oxidation state of Hydrogen in most compounds is +1, and in compounds with metal elements (hydrides) - -1.
5. The oxidation state of metals in compounds is always positive.
6. A more electronegative element always has a negative oxidation state.
7. The sum of the oxidation states of all atoms in a molecule is zero.
Valence is a complex concept. This term underwent a significant transformation simultaneously with the development of the theory of chemical bonding. Initially, valency was the ability of an atom to attach or replace a certain number of other atoms or atomic groups to form a chemical bond.
A quantitative measure of the valence of an element’s atom was the number of hydrogen or oxygen atoms (these elements were considered mono- and divalent, respectively) that the element attaches to form a hydride of the formula EH x or an oxide of the formula E n O m.
Thus, the valence of the nitrogen atom in the ammonia molecule NH 3 is equal to three, and the sulfur atom in the H 2 S molecule is equal to two, since the valence of the hydrogen atom is equal to one.
In the compounds Na 2 O, BaO, Al 2 O 3, SiO 2, the valencies of sodium, barium and silicon are 1, 2, 3 and 4, respectively.
The concept of valency was introduced into chemistry before the structure of the atom became known, namely in 1853 by the English chemist Frankland. It has now been established that the valence of an element is closely related to the number of outer electrons of the atoms, since the electrons of the inner shells of the atoms do not participate in the formation of chemical bonds.
In the electronic theory of covalent bonds it is believed that valence of an atom is determined by the number of its unpaired electrons in the ground or excited state, participating in the formation of common electron pairs with electrons of other atoms.
For some elements, valence is a constant value. Thus, sodium or potassium in all compounds is monovalent, calcium, magnesium and zinc are divalent, aluminum is trivalent, etc. But most chemical elements exhibit variable valency, which depends on the nature of the partner element and the conditions of the process. Thus, iron can form two compounds with chlorine - FeCl 2 and FeCl 3, in which the valence of iron is 2 and 3, respectively.
Oxidation state- a concept that characterizes the state of an element in a chemical compound and its behavior in redox reactions; numerically, the oxidation state is equal to the formal charge that can be assigned to an element, based on the assumption that all the electrons in each of its bonds have transferred to a more electronegative atom.
Electronegativity- a measure of the ability of an atom to acquire a negative charge when forming a chemical bond or the ability of an atom in a molecule to attract valence electrons involved in the formation of a chemical bond. Electronegativity is not an absolute value and is calculated by various methods. Therefore, the electronegativity values given in different textbooks and reference books may differ.
Table 2 shows the electronegativity of some chemical elements on the Sanderson scale, and Table 3 shows the electronegativity of elements on the Pauling scale.
The electronegativity value is given below the symbol of the corresponding element. The higher the numerical value of an atom's electronegativity, the more electronegative the element is. The most electronegative is the fluorine atom, the least electronegative is the rubidium atom. In a molecule formed by atoms of two different chemical elements, the formal negative charge will be on the atom whose numerical value of electronegativity is higher. Thus, in a molecule of sulfur dioxide SO2, the electronegativity of the sulfur atom is 2.5, and the electronegativity of the oxygen atom is greater - 3.5. Therefore, the negative charge will be on the oxygen atom, and the positive charge will be on the sulfur atom.
In the ammonia molecule NH 3, the electronegativity value of the nitrogen atom is 3.0, and that of the hydrogen atom is 2.1. Therefore, the nitrogen atom will have a negative charge, and the hydrogen atom will have a positive charge.
You should clearly know the general trends in electronegativity changes. Since an atom of any chemical element tends to acquire a stable configuration of the outer electronic layer - an octet shell of an inert gas, the electronegativity of elements in a period increases, and in a group the electronegativity generally decreases with increasing atomic number of the element. Therefore, for example, sulfur is more electronegative compared to phosphorus and silicon, and carbon is more electronegative compared to silicon.
When drawing up formulas for compounds consisting of two non-metals, the more electronegative of them is always placed to the right: PCl 3, NO 2. There are some historical exceptions to this rule, for example NH 3, PH 3, etc.
The oxidation number is usually indicated by an Arabic numeral (with a sign in front of the number) located above the element symbol, for example:
To determine the degree of oxidation of atoms in chemical compounds, the following rules are followed:
- The oxidation state of elements in simple substances is zero.
- The algebraic sum of the oxidation states of atoms in a molecule is zero.
- Oxygen in compounds exhibits mainly an oxidation state of –2 (in oxygen fluoride OF 2 + 2, in metal peroxides such as M 2 O 2 –1).
- Hydrogen in compounds exhibits an oxidation state of + 1, with the exception of hydrides of active metals, for example, alkali or alkaline earth ones, in which the oxidation state of hydrogen is – 1.
- For monoatomic ions, the oxidation state is equal to the charge of the ion, for example: K + - +1, Ba 2+ - +2, Br – - –1, S 2– - –2, etc.
- In compounds with a covalent polar bond, the oxidation state of the more electronegative atom has a minus sign, and the less electronegative atom has a plus sign.
- In organic compounds, the oxidation state of hydrogen is +1.
Let us illustrate the above rules with several examples.
Example 1. Determine the degree of oxidation of elements in the oxides of potassium K 2 O, selenium SeO 3 and iron Fe 3 O 4.
Potassium oxide K 2 O. The algebraic sum of the oxidation states of atoms in a molecule is zero. The oxidation state of oxygen in oxides is –2. Let us denote the oxidation state of potassium in its oxide as n, then 2n + (–2) = 0 or 2n = 2, hence n = +1, i.e., the oxidation state of potassium is +1.
Selenium oxide SeO 3. The SeO 3 molecule is electrically neutral. The total negative charge of the three oxygen atoms is –2 × 3 = –6. Therefore, to reduce this negative charge to zero, the oxidation state of selenium must be +6.
Fe3O4 molecule electrically neutral. The total negative charge of the four oxygen atoms is –2 × 4 = –8. To equalize this negative charge, the total positive charge on the three iron atoms must be +8. Therefore, one iron atom must have a charge of 8/3 = +8/3.
It should be emphasized that the oxidation state of an element in a compound can be a fractional number. Such fractional oxidation states are not meaningful when explaining bonding in a chemical compound, but can be used to construct equations for redox reactions.
Example 2. Determine the degree of oxidation of elements in the compounds NaClO 3, K 2 Cr 2 O 7.
The NaClO 3 molecule is electrically neutral. The oxidation state of sodium is +1, the oxidation state of oxygen is –2. Let us denote the oxidation state of chlorine as n, then +1 + n + 3 × (–2) = 0, or +1 + n – 6 = 0, or n – 5 = 0, hence n = +5. Thus, the oxidation state of chlorine is +5.
The K 2 Cr 2 O 7 molecule is electrically neutral. The oxidation state of potassium is +1, the oxidation state of oxygen is –2. Let us denote the oxidation state of chromium as n, then 2 × 1 + 2n + 7 × (–2) = 0, or +2 + 2n – 14 = 0, or 2n – 12 = 0, 2n = 12, hence n = +6. Thus, the oxidation state of chromium is +6.
Example 3. Let us determine the degree of oxidation of sulfur in the sulfate ion SO 4 2–. The SO 4 2– ion has a charge of –2. The oxidation state of oxygen is –2. Let us denote the oxidation state of sulfur as n, then n + 4 × (–2) = –2, or n – 8 = –2, or n = –2 – (–8), hence n = +6. Thus, the oxidation state of sulfur is +6.
It should be remembered that the oxidation state is sometimes not equal to the valence of a given element.
For example, the oxidation states of the nitrogen atom in the ammonia molecule NH 3 or in the hydrazine molecule N 2 H 4 are –3 and –2, respectively, while the valency of nitrogen in these compounds is three.
The maximum positive oxidation state for elements of the main subgroups, as a rule, is equal to the group number (exceptions: oxygen, fluorine and some other elements).
The maximum negative oxidation state is 8 - the group number.
Training tasks
1. In which compound the oxidation state of phosphorus is +5?
1) HPO 3
2) H3PO3
3) Li 3 P
4) AlP
2. In which compound does the oxidation state of phosphorus equal to –3?
1) HPO 3
2) H3PO3
3) Li 3 PO 4
4) AlP
3. In which compound is the oxidation state of nitrogen equal to +4?
1) HNO2
2) N 2 O 4
3) N 2 O
4) HNO3
4. In which compound is the oxidation state of nitrogen equal to –2?
1) NH 3
2) N 2 H 4
3) N 2 O 5
4) HNO2
5. In which compound the oxidation state of sulfur is +2?
1) Na 2 SO 3
2)SO2
3) SCl 2
4) H2SO4
6. In which compound the oxidation state of sulfur is +6?
1) Na 2 SO 3
2) SO 3
3) SCl 2
4) H 2 SO 3
7. In substances whose formulas are CrBr 2, K 2 Cr 2 O 7, Na 2 CrO 4, the oxidation state of chromium is respectively equal to
1) +2, +3, +6
2) +3, +6, +6
3) +2, +6, +5
4) +2, +6, +6
8. The minimum negative oxidation state of a chemical element is usually equal to
1) period number
3) the number of electrons missing to complete the outer electron layer
9. The maximum positive oxidation state of chemical elements located in the main subgroups, as a rule, is equal to
1) period number
2) the serial number of the chemical element
3) group number
4) the total number of electrons in the element
10. Phosphorus exhibits the maximum positive oxidation state in the compound
1) HPO 3
2) H3PO3
3) Na3P
4) Ca 3 P 2
11. Phosphorus exhibits minimal oxidation state in the compound
1) HPO 3
2) H3PO3
3) Na 3 PO 4
4) Ca 3 P 2
12. The nitrogen atoms in ammonium nitrite, located in the cation and anion, exhibit oxidation states, respectively
1) –3, +3
2) –3, +5
3) +3, –3
4) +3, +5
13. The valency and oxidation state of oxygen in hydrogen peroxide are respectively equal
1) II, –2
2) II, –1
3) I, +4
4) III, –2
14. The valence and degree of oxidation of sulfur in pyrite FeS2 are respectively equal
1) IV, +5
2) II, –1
3) II, +6
4) III, +4
15. The valency and oxidation state of the nitrogen atom in ammonium bromide are respectively equal to
1) IV, –3
2) III, +3
3) IV, –2
4) III, +4
16. The carbon atom exhibits a negative oxidation state when combined with
1) oxygen
2) sodium
3) fluorine
4) chlorine
17. exhibits a constant state of oxidation in its compounds
1) strontium
2) iron
3) sulfur
4) chlorine
18. The oxidation state +3 in their compounds can exhibit
1) chlorine and fluorine
2) phosphorus and chlorine
3) carbon and sulfur
4) oxygen and hydrogen
19. The oxidation state +4 in their compounds can exhibit
1) carbon and hydrogen
2) carbon and phosphorus
3) carbon and calcium
4) nitrogen and sulfur
20. The oxidation state equal to the group number in its compounds exhibits
1) chlorine
2) iron
3) oxygen
4) fluorine
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 the elements included in them is equal to the electric charge, since during the formation of these compounds there is an almost complete transition 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.
Highest oxidation state
For elements that exhibit different oxidation states in their compounds, there are concepts of highest (maximum positive) and lowest (minimum negative) oxidation states. The highest oxidation state of a chemical element usually numerically coincides with the group number in D.I. Mendeleev’s Periodic Table. Exceptions are fluorine (oxidation state is -1, and the element is located in group VIIA), oxygen (oxidation state is +2, and the element is located in group VIA), helium, neon, argon (oxidation state is 0, and the elements are located in VIII group), as well as elements of the cobalt and nickel subgroup (oxidation state is +2, and the elements are located in group VIII), for which the highest oxidation state is expressed by a number whose value is lower than the number of the group to which they belong. Elements of the copper subgroup, on the contrary, have a highest oxidation state greater than one, although they belong to group I (the maximum positive oxidation state of copper and silver is +2, gold +3).
Examples of problem solving
EXAMPLE 1
- In hydrogen sulfide, the oxidation state of sulfur is (-2), and in a simple substance - sulfur - 0:
Change in the oxidation state of sulfur: -2 → 0, i.e. sixth answer.
- In a simple substance - sulfur - the oxidation state of sulfur is 0, and in SO 3 - (+6):
Change in the oxidation state of sulfur: 0 → +6, i.e. fourth answer option.
- In sulfurous acid, the oxidation state of sulfur is (+4), and in a simple substance - sulfur - 0:
1×2 +x+ 3×(-2) =0;
Change in the oxidation state of sulfur: +4 → 0, i.e. third answer option.
EXAMPLE 2
| Exercise | Nitrogen exhibits valency III and oxidation state (-3) in the compound: a) N 2 H 4 ; b) NH 3; c) NH 4 Cl; d) N 2 O 5 |
| Solution | In order to give the correct answer to the question posed, we will alternately determine the valency and oxidation state of nitrogen in the proposed compounds. a) the valence of hydrogen is always equal to I. The total number of units of valence of hydrogen is equal to 4 (1 × 4 = 4). Let us divide the obtained value by the number of nitrogen atoms in the molecule: 4/2 = 2, therefore, the valency of nitrogen is II. This answer option is incorrect. b) the valence of hydrogen is always equal to I. The total number of units of hydrogen valence is equal to 3 (1 × 3 = 3). Let us divide the obtained value by the number of nitrogen atoms in the molecule: 3/1 = 2, therefore, the valency of nitrogen is III. The oxidation degree of nitrogen in ammonia is (-3): This is the correct answer. |
| Answer | Option (b) |
Oxidation state. Determination of the oxidation state of an element's atom using the chemical formula of the compound. Drawing up the formula of a compound based on the known oxidation states of elemental atoms
The oxidation state of an element is the conditional charge of an atom in a substance, calculated on the assumption that it consists of ions. To determine the oxidation state of elements, you need to remember certain rules:
1. The oxidation state can be positive, negative or zero. It is indicated by an Arabic numeral with a plus or minus sign above the element symbol.
2. When determining oxidation states, we proceed from the electronegativity of the substance: the sum of the oxidation states of all atoms in the compound is zero.
3. If a compound is formed by atoms of one element (in a simple substance), then the oxidation state of these atoms is zero.
4. The atoms of some chemical elements are usually assigned steel oxidation states. For example, the oxidation state of fluorine in compounds is always -1; lithium, sodium, potassium, rubidium and cesium +1; magnesium, calcium, strontium, barium and zinc +2, aluminum +3.
5. The oxidation state of hydrogen in most compounds is +1, and only in compounds with some metals it is equal to -1 (KH, BaH2).
6. The oxidation state of oxygen in most compounds is -2, and only in some compounds is it assigned an oxidation state of -1 (H2O2, Na2O2 or +2 (OF2).
7. Atoms of many chemical elements have variable oxidation states.
8. The oxidation state of the metal atom in compounds is positive and is numerically equal to its valency.
9. The maximum positive oxidation state of an element is usually equal to the number of the group in the periodic table in which the element is found.
10. The minimum oxidation state for metals is zero. For nonmetals, in most cases below the negative oxidation state is equal to the difference between the group number and the number eight.
11. The oxidation state of an atom forms a simple ion (consists of one atom) and is equal to the charge of this ion.
Using the above rules, we will determine the oxidation states of chemical elements in the composition of H2SO4. This is a complex substance consisting of three chemical elements - hydrogen H, sulfur S and oxygen O. Let us note the oxidation states of those elements for which they are constant. In our case, these are hydrogen H and oxygen O.
Let us determine the unknown oxidation state of sulfur. Let the oxidation state of sulfur in this compound be x.
Let’s create equations by multiplying for each element its index by the oxidation state and equating the extracted amount to zero: 2 (+1) + x + 4 (-2) = 0
2 + X – 8 = 0
x = +8 – 2 = +6
Therefore, the oxidation number of sulfur is plus six.
In the following example, we will find out how to create a formula for a compound with known oxidation states of elemental atoms. Let's create the formula for ferrum (III) oxide. The word “oxide” means that to the right of the iron symbol you need to write the oxygen symbol: FeO.
Let us note the oxidation states of chemical elements above their symbols. The oxidation state of iron is indicated in the name in brackets (III), therefore equal to +3, the oxidation state of oxygen in oxides is -2.
Let's find the least common multiple of the numbers 3 and 2, this is 6. Divide the number 6 by 3, we get the number 2 - this is the index for iron. Divide the number 6 by 2, we get the number 3 - this is the index for oxygen.
In the following example, we will find out how to create a formula for a compound with known oxidation states of element atoms and ion charges. Let's create the formula for calcium orthophosphate. The word “orthophosphate” means that to the right of the Calcium symbol you must write the acidic residue of orthophosphate acid: CaPO4.
Let's note the oxidation state of calcium (rule number four) and the charge of the acid residue (according to the solubility table).
Let's find the least common multiple of numbers 2 and 3, this is 6. Divide the number 6 by 2, we get the number 3 - this is the index for calcium. Divide the number 6 by 3, we get the number 2 - this is the index for the acid residue.