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The oxidation state is 3 in compounds. How to determine the degree of oxidation




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 electronegativity value, 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

Hydrogen

Helium / Helium

Lithium / Lithium

Beryllium / Beryllium

(-1), 0, (+1), (+2), (+3)

Carbon / Carbon

(-4), (-3), (-2), (-1), 0, (+2), (+4)

Nitrogen / Nitrogen

(-3), (-2), (-1), 0, (+1), (+2), (+3), (+4), (+5)

Oxygen / Oxygen

(-2), (-1), 0, (+1), (+2)

Fluorine / Fluorine

Sodium

Magnesium / Magnesium

Aluminum

Silicon / Silicon

(-4), 0, (+2), (+4)

Phosphorus / Phosphorus

(-3), 0, (+3), (+5)

Sulfur

(-2), 0, (+4), (+6)

Chlorine / Chlorine

(-1), 0, (+1), (+3), (+5), (+7), rarely (+2) and (+4)

Argon / Argon

Potassium / Potassium

Calcium / Calcium

Scandium / Scandium

Titanium / Titanium

(+2), (+3), (+4)

Vanadium / Vanadium

(+2), (+3), (+4), (+5)

Chromium / Chromium

(+2), (+3), (+6)

Manganese / Manganese

(+2), (+3), (+4), (+6), (+7)

Iron / Iron

(+2), (+3), rarely (+4) and (+6)

Cobalt / Cobalt

(+2), (+3), rarely (+4)

Nickel / Nickel

(+2), rarely (+1), (+3) and (+4)

Copper

+1, +2, rare (+3)

Gallium / Gallium

(+3), rare (+2)

Germanium / Germanium

(-4), (+2), (+4)

Arsenic / Arsenic

(-3), (+3), (+5), rarely (+2)

Selenium / Selenium

(-2), (+4), (+6), rarely (+2)

Bromine / Bromine

(-1), (+1), (+5), rarely (+3), (+4)

Krypton / Krypton

Rubidium / Rubidium

Strontium / Strontium

Yttrium / Yttrium

Zirconium / Zirconium

(+4), rarely (+2) and (+3)

Niobium / Niobium

(+3), (+5), rarely (+2) and (+4)

Molybdenum / Molybdenum

(+3), (+6), rarely (+2), (+3) and (+5)

Technetium / Technetium

Ruthenium / Ruthenium

(+3), (+4), (+8), rarely (+2), (+6) and (+7)

Rhodium

(+4), rarely (+2), (+3) and (+6)

Palladium / Palladium

(+2), (+4), rarely (+6)

Silver / Silver

(+1), rarely (+2) and (+3)

Cadmium / Cadmium

(+2), rare (+1)

Indium / Indium

(+3), rarely (+1) and (+2)

Tin / Tin

(+2), (+4)

Antimony / Antimony

(-3), (+3), (+5), rarely (+4)

Tellurium / Tellurium

(-2), (+4), (+6), rarely (+2)

(-1), (+1), (+5), (+7), rarely (+3), (+4)

Xenon / Xenon

Cesium / Cesium

Barium / Barium

Lanthanum / Lanthanum

Cerium / Cerium

(+3), (+4)

Praseodymium / Praseodymium

Neodymium / Neodymium

(+3), (+4)

Promethium / Promethium

Samaria / Samarium

(+3), rare (+2)

Europium / Europium

(+3), rare (+2)

Gadolinium / Gadolinium

Terbium / Terbium

(+3), (+4)

Dysprosium / Dysprosium

Holmium / Holmium

Erbium / Erbium

Thulium / Thulium

(+3), rare (+2)

Ytterbium / Ytterbium

(+3), rare (+2)

Lutetium / Lutetium

Hafnium / Hafnium

Tantalum / Tantalum

(+5), rarely (+3), (+4)

Tungsten / Tungsten

(+6), rare (+2), (+3), (+4) and (+5)

Rhenium / Rhenium

(+2), (+4), (+6), (+7), rare (-1), (+1), (+3), (+5)

Osmium / Osmium

(+3), (+4), (+6), (+8), rarely (+2)

Iridium / Iridium

(+3), (+4), (+6), rarely (+1) and (+2)

Platinum / Platinum

(+2), (+4), (+6), rarely (+1) and (+3)

Gold / Gold

(+1), (+3), rarely (+2)

Mercury / Mercury

(+1), (+2)

Waist / Thallium

(+1), (+3), rarely (+2)

Lead / Lead

(+2), (+4)

Bismuth / Bismuth

(+3), rarely (+3), (+2), (+4) and (+5)

Polonium / Polonium

(+2), (+4), rarely (-2) and (+6)

Astatine / Astatine

Radon / Radon

Francium / Francium

Radium / Radium

Actinium / Actinium

Thorium / Thorium

Proactinium / Protactinium

Uranus / Uranium

(+3), (+4), (+6), rarely (+2) and (+5)

Examples of problem solving

EXAMPLE 1

Answer We will alternately determine the degree of phosphorus oxidation in each of the proposed transformation schemes, and then choose the correct answer.
  • 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

Exercise 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)

In chemistry, the terms "oxidation" and "reduction" mean reactions in which an atom or a group of atoms lose or, respectively, gain electrons. The oxidation state is a numerical value attributed to one or more atoms that characterizes the number of redistributed electrons and shows how these electrons are distributed between atoms during the reaction. Determining this quantity can be both a simple and quite complex procedure, depending on the atoms and the molecules consisting of them. Moreover, the atoms of some elements can have several oxidation states. Fortunately, there are simple unambiguous rules for determining the degree of oxidation, for the confident use of which it is enough to know the basics of chemistry and algebra.

Steps

Part 1

Determination of the degree of oxidation according to the laws of chemistry

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

    • For example, Al(s) and Cl 2 have an oxidation state of 0 because both are in a chemically uncombined elemental state.
    • Please note that the allotropic form of sulfur S 8, or octasulfur, despite its atypical structure, is also characterized by a zero oxidation state.

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

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

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

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

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

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

  4. Hydrogen has an oxidation state of +1, with a few exceptions. As with oxygen, there are also exceptions. As a rule, the oxidation state of hydrogen is +1 (unless it is in the elemental state H 2). However, in compounds called hydrides, the oxidation state of hydrogen is -1.

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

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

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

    • This is a good method of checking - if the sum of the oxidation states does not equal the total charge of the compound, then you are wrong somewhere.

    Part 2

    Determining the oxidation state without using the laws of chemistry

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

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

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

      • For Na 2 SO 4 , using our rules, we find that the charge (and hence the oxidation state) of the Na ion is +1, and for each of the oxygen atoms it is -2.
    3. In compounds, the sum of all oxidation states must equal the charge. For example, if the compound is a diatomic ion, the sum of the oxidation states of the atoms must be equal to the total ionic charge.
    4. It is very useful to be able to use the periodic table of Mendeleev and know where the metallic and non-metallic elements are located in it.
    5. The oxidation state of atoms in the elementary form is always zero. The oxidation state of a single ion is equal to its charge. Elements of group 1A of the periodic table, such as hydrogen, lithium, sodium, in elemental form have an oxidation state of +1; the oxidation state of group 2A metals, such as magnesium and calcium, in its elemental form is +2. Oxygen and hydrogen, depending on the type of chemical bond, can have 2 different oxidation states.

A chemical element in a compound, calculated from the assumption that all bonds are ionic.

The oxidation states can have a positive, negative or zero value, therefore the algebraic sum of the oxidation states of elements in a molecule, taking into account the number of their atoms, is 0, and in an ion - the charge of the ion.

1. The oxidation states of metals in compounds are always positive.

2. The highest oxidation state corresponds to the group number of the periodic system where this element is located (the exception is: Au+3(I group), Cu+2(II), from group VIII, the oxidation state +8 can only be in osmium Os and ruthenium Ru.

3. The oxidation states of non-metals depend on which atom it is connected to:

  • if with a metal atom, then the oxidation state is negative;
  • if with a non-metal atom, then the oxidation state can be both positive and negative. It depends on the electronegativity of the atoms of the elements.

4. The highest negative oxidation state of non-metals can be determined by subtracting from 8 the number of the group in which this element is located, i.e. the highest positive oxidation state is equal to the number of electrons on the outer layer, which corresponds to the group number.

5. The oxidation states of simple substances are 0, regardless of whether it is a metal or a non-metal.

Elements with constant oxidation states.

Element

Characteristic oxidation state

Exceptions

Metal hydrides: LIH-1

oxidation state called the conditional charge of the particle under the assumption that the bond is completely broken (has an ionic character).

H- Cl = H + + Cl - ,

The bond in hydrochloric acid is covalent polar. The electron pair is more biased towards the atom Cl - , because it is more electronegative whole element.

How to determine the degree of oxidation?

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

The oxidation state is indicated above the element: Br 2 0 , Na 0 , O +2 F 2 -1 ,K + Cl - etc.

It can be negative and positive.

The oxidation state of a simple substance (unbound, free state) is zero.

The oxidation state of oxygen in most compounds is -2 (the exception is peroxides H 2 O 2, where it is -1 and compounds with fluorine - O +2 F 2 -1 , O 2 +1 F 2 -1 ).

- Oxidation state a simple monatomic ion is equal to its charge: Na + , Ca +2 .

Hydrogen in its compounds has an oxidation state of +1 (exceptions are hydrides - Na + H - and type connections C +4 H 4 -1 ).

In metal-non-metal bonds, the atom that has the highest electronegativity has a negative oxidation state (electronegativity data are given on the Pauling scale): H + F - , Cu + Br - , Ca +2 (NO 3 ) - etc.

Rules for determining the degree of oxidation in chemical compounds.

Let's take a connection KMnO 4 , it is necessary to determine the oxidation state of the manganese atom.

Reasoning:

  1. Potassium is an alkali metal in group I of the periodic table, and therefore has only a positive oxidation state of +1.
  2. Oxygen is known to have an oxidation state of -2 in most of its compounds. This substance is not a peroxide, which means it is no exception.
  3. Makes an equation:

K+MnXO 4 -2

Let X- unknown to us the degree of oxidation of manganese.

The number of potassium atoms is 1, manganese - 1, oxygen - 4.

It is proved that the molecule as a whole is electrically neutral, so its total charge must be equal to zero.

1*(+1) + 1*(X) + 4(-2) = 0,

X = +7,

Hence, the oxidation state of manganese in potassium permanganate = +7.

Let's take another example of an oxide Fe2O3.

It is necessary to determine the oxidation state of the iron atom.

Reasoning:

  1. Iron is a metal, oxygen is a non-metal, which means that it is oxygen that will be an oxidizing agent and have a negative charge. We know that oxygen has an oxidation state of -2.
  2. We consider the number of atoms: iron - 2 atoms, oxygen - 3.
  3. We make an equation where X- the oxidation state of the iron atom:

2*(X) + 3*(-2) = 0,

Conclusion: the oxidation state of iron in this oxide is +3.

Examples. Determine the oxidation states of all atoms in the molecule.

1. K2Cr2O7.

Oxidation state K+1, oxygen O -2.

Given indexes: O=(-2)×7=(-14), K=(+1)×2=(+2).

Because the algebraic sum of the oxidation states of elements in a molecule, taking into account the number of their atoms, is 0, then the number of positive oxidation states is equal to the number of negative ones. Oxidation states K+O=(-14)+(+2)=(-12).

It follows from this that the number of positive powers of the chromium atom is 12, but there are 2 atoms in the molecule, which means that there are (+12):2=(+6) per atom. Answer: K 2 + Cr 2 +6 O 7 -2.

2.(AsO 4) 3-.

In this case, the sum of the oxidation states will no longer be equal to zero, but to the charge of the ion, i.e. - 3. Let's make an equation: x+4×(- 2)= - 3 .

Answer: (As +5 O 4 -2) 3-.

To place correctly oxidation states There are four rules to keep in mind.

1) In a simple substance, the oxidation state of any element is 0. Examples: Na 0, H 0 2, P 0 4.

2) You should remember the elements for which are characteristic constant oxidation states. All of them are listed in the table.


3) The highest oxidation state of an element, as a rule, coincides with the number of the group in which this element is located (for example, phosphorus is in group V, the highest SD of phosphorus is +5). Important exceptions: F, O.

4) The search for the oxidation states of the remaining elements is based on a simple rule:

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

A few simple examples for determining oxidation states

Example 1. It is necessary to find the oxidation states of elements in ammonia (NH 3).

Solution. We already know (see 2) that Art. OK. hydrogen is +1. It remains to find this characteristic for nitrogen. Let x be the desired oxidation state. We compose the simplest equation: x + 3 (+1) \u003d 0. The solution is obvious: x \u003d -3. Answer: N -3 H 3 +1.


Example 2. Specify the oxidation states of all atoms in the H 2 SO 4 molecule.

Solution. The oxidation states of hydrogen and oxygen are already known: H(+1) and O(-2). We compose an equation for determining the degree of oxidation of sulfur: 2 (+1) + x + 4 (-2) \u003d 0. Solving this equation, we find: x \u003d +6. Answer: H +1 2 S +6 O -2 4 .


Example 3. Calculate the oxidation states of all elements in the Al(NO 3) 3 molecule.

Solution. The algorithm remains unchanged. The composition of the "molecule" of aluminum nitrate includes one atom of Al (+3), 9 oxygen atoms (-2) and 3 nitrogen atoms, the oxidation state of which we have to calculate. Corresponding equation: 1 (+3) + 3x + 9 (-2) = 0. Answer: Al +3 (N +5 O -2 3) 3.


Example 4. Determine the oxidation states of all atoms in the (AsO 4) 3- ion.

Solution. In this case, the sum of the oxidation states will no longer be equal to zero, but to the charge of the ion, i.e., -3. Equation: x + 4 (-2) = -3. Answer: As(+5), O(-2).

What to do if the oxidation states of two elements are unknown

Is it possible to determine the oxidation states of several elements at once using a similar equation? If we consider this problem from the point of view of mathematics, the answer will be negative. A linear equation with two variables cannot have a unique solution. But we are not just solving an equation!

Example 5. Determine the oxidation states of all elements in (NH 4) 2 SO 4.

Solution. The oxidation states of hydrogen and oxygen are known, but sulfur and nitrogen are not. A classic example of a problem with two unknowns! We will consider ammonium sulfate not as a single "molecule", but as a combination of two ions: NH 4 + and SO 4 2-. We know the charges of ions, each of them contains only one atom with an unknown degree of oxidation. Using the experience gained in solving previous problems, we can easily find the oxidation states of nitrogen and sulfur. Answer: (N -3 H 4 +1) 2 S +6 O 4 -2.

Conclusion: if the molecule contains several atoms with unknown oxidation states, try to "split" the molecule into several parts.

How to arrange oxidation states in organic compounds

Example 6. Indicate the oxidation states of all elements in CH 3 CH 2 OH.

Solution. Finding oxidation states in organic compounds has its own specifics. In particular, it is necessary to separately find the oxidation states for each carbon atom. You can reason as follows. Consider, for example, the carbon atom in the methyl group. This C atom is connected to 3 hydrogen atoms and an adjacent carbon atom. On the C-H bond, the electron density shifts towards the carbon atom (because the electronegativity of C exceeds the EO of hydrogen). If this displacement were complete, the carbon atom would acquire a charge of -3.

The C atom in the -CH 2 OH group is bonded to two hydrogen atoms (electron density shift towards C), one oxygen atom (electron density shift towards O) and one carbon atom (we can assume that the shifts in electron density in this case not happening). The oxidation state of carbon is -2 +1 +0 = -1.

Answer: C -3 H +1 3 C -1 H +1 2 O -2 H +1.

Do not confuse the concepts of "valence" and "oxidation state"!

Oxidation state is often confused with valence. Don't make that mistake. I will list the main differences:

  • the oxidation state has a sign (+ or -), valence - no;
  • the degree of oxidation can be equal to zero even in a complex substance, the equality of valency to zero means, as a rule, that the atom of this element is not connected to other atoms (we will not discuss any kind of inclusion compounds and other "exotics" here);
  • the degree of oxidation is a formal concept that acquires real meaning only in compounds with ionic bonds, the concept of "valence", on the contrary, is most conveniently applied in relation to covalent compounds.

The oxidation state (more precisely, its modulus) is often numerically equal to the valence, but even more often these values ​​do NOT coincide. For example, the oxidation state of carbon in CO 2 is +4; valency C is also equal to IV. But in methanol (CH 3 OH), the valency of carbon remains the same, and the oxidation state of C is -1.

A small test on the topic "The degree of oxidation"

Take a few minutes to check how you have understood this topic. You need to answer five simple questions. Good luck!