To place correctly oxidation states, you need to keep four rules 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 that 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 the element is located (for example, phosphorus is in group V, the highest s.d. of phosphorus is +5). Important exceptions: F, O.
4) The search for oxidation states of other elements is based on a simple rule:
In a neutral molecule, the sum of the oxidation states of all elements is zero, and in an ion - the charge of the ion.
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 create the simplest equation: x + 3 (+1) = 0. The solution is obvious: x = -3. Answer: N -3 H 3 +1.
Example 2. Indicate the oxidation states of all atoms in the H 2 SO 4 molecule.
Solution. The oxidation states of hydrogen and oxygen are already known: H(+1) and O(-2). We create an equation to determine the oxidation state of sulfur: 2 (+1) + x + 4 (-2) = 0. Solving this equation, we find: x = +6. Answer: H +1 2 S +6 O -2 4.
Example 3. Calculate the oxidation states of all elements in the Al(NO 3) 3 molecule.
Solution. The algorithm remains unchanged. The composition of the “molecule” of aluminum nitrate includes one Al atom (+3), 9 oxygen atoms (-2) and 3 nitrogen atoms, the oxidation state of which we have to calculate. The corresponding equation is: 1 (+3) + 3x + 9 (-2) = 0. Answer: Al +3 (N +5 O -2 3) 3.
Example 4. Determine the oxidation states of all atoms in the (AsO 4) 3- ion.
Solution. In this case, the sum of oxidation states will no longer be equal to zero, but to the charge of the ion, i.e., -3. Equation: x + 4 (-2) = -3. Answer: As(+5), O(-2).
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 a mathematical point of view, the answer will be negative. A linear equation with two variables cannot have a unique solution. But we are solving more than just an equation!
Example 5. Determine the oxidation states of all elements in (NH 4) 2 SO 4.
Solution. The oxidation states of hydrogen and oxygen are known, but sulfur and nitrogen are not. A classic example of a problem with two unknowns! We will consider ammonium sulfate not as a single “molecule”, but as a combination of two ions: NH 4 + and SO 4 2-. The charges of ions are known to us; each of them contains only one atom with an unknown oxidation state. Using the experience gained in solving previous problems, we can easily find the oxidation states of nitrogen and sulfur. Answer: (N -3 H 4 +1) 2 S +6 O 4 -2.
Conclusion: if a molecule contains several atoms with unknown oxidation states, try to “split” the molecule into several parts.
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 a neighboring carbon atom. Along the C-H bond, the electron density shifts towards the carbon atom (since the electronegativity of C exceeds the EO of hydrogen). If this displacement were complete, the carbon atom would acquire a charge of -3.
The C atom in the -CH 2 OH group is bonded to two hydrogen atoms (a shift in electron density towards C), one oxygen atom (a shift in electron density towards O) and one carbon atom (it can be assumed that the shift in electron density in this case not happening). The oxidation state of carbon is -2 +1 +0 = -1.
Answer: C -3 H +1 3 C -1 H +1 2 O -2 H +1.
Do not confuse the concepts of “valency” and “oxidation state”!
Oxidation number is often confused with valence. Don't make this mistake. I will list the main differences:
- the oxidation state has a sign (+ or -), the valency does not;
- the oxidation state can be zero even in a complex substance; valence equal to zero means, as a rule, that an atom of a given element is not connected to other atoms (we will not discuss any kind of inclusion compounds and other “exotics” here);
- oxidation state 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; the valence of 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 equal to -1.
A short test on the topic "Oxidation state"
Take a few minutes to check your understanding of this topic. You need to answer five simple questions. Good luck!
One of the basic concepts in chemistry, widely used in drawing up equations of redox reactions, is oxidation state atoms.
For practical purposes (when composing equations for redox reactions), it is convenient to represent the charges on atoms in molecules with polar bonds as integers equal to the charges that would arise on the atoms if the valence electrons were completely transferred to more electronegative atoms, i.e. e. if the bonds were completely ionic. These charge values are called oxidation states. The oxidation state of any element in a simple substance is always 0.
In the molecules of complex substances, some elements always have a constant oxidation state. Most elements are characterized by variable oxidation states, differing in both sign and magnitude, depending on the composition of the molecule.
Often the oxidation state is equal to the valency and differs from it only in sign. But there are compounds in which the oxidation state of an element is not equal to its valence. As already noted, in simple substances the oxidation state of an element is always zero, regardless of its valence. The table compares the valencies and oxidation states of some elements in various compounds.
Oxidation state of an atom (element) in a compound is the conditional charge calculated under the assumption that the compound consists only of ions. When determining the oxidation state, it is conventionally assumed that the valence electrons in a compound are transferred to more electronegative atoms, and therefore the compounds consist of positively and negatively charged ions. In reality, in most cases, there is not a complete donation of electrons, but only a displacement of an electron pair from one atom to another. Then we can give another definition: The oxidation state is the electric charge that would arise on an atom if the electron pairs with which it is connected to other atoms in the compound were transferred to more electronegative atoms, and the electron pairs connecting identical atoms were divided between them.
When calculating oxidation states, a number of simple rules are used:
1 . The oxidation state of elements in simple substances, both monatomic and molecular, is zero (Fe 0, O 2 0).
2 . The oxidation state of an element in the form of a monoatomic ion is equal to the charge of this ion (Na +1, Ca +2, S –2).
3 . In compounds with a covalent polar bond, a negative charge refers to the more electronegative atom, and a positive charge to the less electronegative atom, and the oxidation states of the elements take the following values:
The oxidation state of fluorine in compounds is always -1;
The oxidation state of oxygen in compounds is -2 (); with the exception of peroxides, where it is formally equal to -1 (), oxygen fluoride, where it is equal to +2 (), as well as superoxides and ozonides, in which the oxidation state of oxygen is -1/2;
The oxidation state of hydrogen in compounds is +1 (), with the exception of metal hydrides, where it is -1 ( 
);
For alkali and alkaline earth elements, the oxidation states are +1 and +2, respectively.
Most elements can exhibit variable oxidation states.
4 . The algebraic sum of oxidation states in a neutral molecule is equal to zero, in a complex ion it is equal to the charge of the ion.
For elements with a variable oxidation state, its value is easy to calculate, knowing the formula of the compound and using rule No. 4. For example, it is necessary to determine the degree of oxidation of phosphorus in phosphoric acid H 3 PO 4. Since oxygen has CO = –2, and hydrogen has CO = +1, then for phosphorus to have a zero sum, the oxidation state must be +5:
For example, in NH 4 Cl, the sum of the oxidation states of all hydrogen atoms is 4×(+1), and the oxidation state of chlorine is -1, therefore, the oxidation state of nitrogen must be equal to -3. In the SO 4 2– sulfate ion, the sum of the oxidation states of the four oxygen atoms is -8, so sulfur must have an oxidation state of +6 for the total charge of the ion to be -2.
The concept of oxidation state for most compounds is conditional, because does not reflect the real effective charge of an atom, but this concept is very widely used in chemistry.
The maximum, and for non-metals, the minimum oxidation state has a periodic dependence on the serial number in D.I. PSHE. Mendeleev, which is due to the electronic structure of the atom.
| Element | Oxidation state values and examples of compounds |
| F | –1 (HF, KF) |
| O | –2 (H 2 O, CaO, CO 2); –1 (H 2 O 2); +2 (OF 2) |
| N | –3 (NH 3); –2(N 2 H 4); –1 (NH 2 OH); +1 (N 2 O); +2 (NO); +3 (N 2 O 3, HNO 2); +4 (NO 2); +5 (N 2 O 5, HNO 3) |
| Cl | –1 (HCl, NaCl); +1 (NaClO); +3 (NaClO 2); +5 (NaClO 3); +7 (Cl 2 O 7, NaClO 4) |
| Br | –1 (KBr); +1 (BrF); +3 (BrF 3); +5 (KBrO 3) |
| I | –1 (HI); +1 (ICl); +3 (ICl 3); +5 (I 2 O 5); +7 (IO 3 F, K 5 IO 6) |
| C | –4 (CH 4); +2 (CO); +4 (CO 2 , CCl 4) |
| Si | –4 (Ca 2 Si); +2 (SiO); +4 (SiO 2, H 2 SiO 3, SiF 4) |
| H | –1 (LiH); +1 (H 2 O, HCl) |
| S | –2 (H 2 S, FeS); +2 (Na 2 S 2 O 3); +3 (Na 2 S 2 O 4); +4 (SO 2, Na 2 SO 3, SF 4); +6 (SO 3, H 2 SO 4, SF 6) |
| Se, Te | –2 (H 2 Se, H 2 Te); +2 (SeCl 2, TeCl 2); +4 (SeO 2, TeO 2); +6 (H 2 SeO 4, H 2 TeO 4) |
| P | –3 (PH 3); +1 (H 3 PO 2); +3 (H 3 PO 3); +5 (P 2 O 5 , H 3 PO 4) |
| As, Sb | –3 (GaAs, Zn 3 Sb 2); +3 (AsCl 3, Sb 2 O 3); +5 (H 3 AsO 4, SbCl 5) |
| Li, Na, K | +1 (NaCl) |
| Be, Mg, Ca | +2 (MgO, CaCO 3) |
| Al | +3 (Al 2 O 3, AlCl 3) |
| Cr | +2 (CrCl 2); +3 (Cr 2 O 3, Cr 2 (SO 4) 3); +4 (CrO 2); +6 (K 2 CrO 4, K 2 Cr 2 O 7) |
| Mn | +2 (MnSO 4); +3 (Mn 2 (SO 4) 3); +4 (MnO 2); +6 (K 2 MnO 4); +7 (KMnO 4) |
| Fe | +2 (FeO, FeSO 4); +3 (Fe 2 O 3, FeCl 3); +4 (Na 2 FeO 3) |
| Cu | +1 (Cu 2 O); +2 (CuO, CuSO 4, Cu 2 (OH) 2 CO 3) |
| Ag | +1 (AgNO 3) |
| Au | +1 (AuCl); +3 (AuCl 3, KAuCl 4) |
| Zn | +2 (ZnO, ZnSO 4) |
| Hg | +1 (Hg 2 Cl 2); +2 (HgO, HgCl 2) |
| Sn | +2 (SnO); +4 (SnO 2, SnCl 4) |
| Pb | +2 (PbO, PbSO 4); +4 (PbO 2) |
In chemical reactions, the rule of preserving the algebraic sum of the oxidation states of all atoms must be fulfilled. In the complete equation of a chemical reaction, the oxidation and reduction processes must exactly compensate each other. Although the degree of oxidation, as noted above, is a rather formal concept, it is used in chemistry for the following purposes: firstly, to compile equations of redox reactions, secondly, to predict the redox properties of elements in a compound.
Many elements are characterized by several values of oxidation states, and by calculating its oxidation state, one can predict redox properties: an element in the highest negative oxidation state can only donate electrons (oxidize) and be a reducing agent, in the highest positive oxidation state it can only accept electrons (reduce). ) and be an oxidizing agent, in intermediate oxidation states - both oxidize and reduce.
Oxidation-reduction is a single, interconnected process. Oxidation corresponds to an increase in the oxidation state of the element, and recovery - its reduction.
Many textbooks adhere to the interpretation of oxidation as the loss of electrons, and reduction as their gain. This approach, proposed by the Russian scientist Pisarzhevsky (1916), is applicable to electrochemical processes on electrodes and relates to the discharge (charging) of ions and molecules.
However, the explanation of changes in oxidation states as processes of removal and addition of electrons is generally incorrect. It can be applied to some simple ions like
Cl - - ®Cl 0 .
To change the oxidation state of atoms in complex ions like
CrO 4 2 - ®Cr +3
a decrease in the positive oxidation state of chromium from +6 to +3 corresponds to a smaller real increase in the positive charge (on Cr in CrO 4 2 - real charge "+0.2 electron charge, and on Cr +3 - from +2 to +1.5 in different connections).
The transfer of charge from the reducing agent to the oxidizing agent, equal to the change in the oxidation state, occurs with the participation of other particles, for example, H + ions:
CrO 4 2 - + 8H + + 3 ®Cr +3 + 4H 2 O.
The entry presented is called half-reactions .
Related information.
The chemical element in a compound, calculated from the assumption that all bonds are ionic.
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 equal to 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 number of the group of the periodic table where the element is located (exceptions are: Au +3(I group), Cu +2(II), from group VIII the oxidation state +8 can only be found 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 either positive or 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 the element is located, i.e. the highest positive oxidation state is equal to the number of electrons in 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 a particle under the assumption that the bond is completely broken (has an ionic character). H- Cl = H + + Cl - , The bond in hydrochloric acid is polar covalent. The electron pair is more shifted towards the atom Cl - , because it is a more electronegative element. How to determine the oxidation state?Electronegativity is the ability of atoms to attract electrons from other elements. The oxidation number 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 for most compounds is -2 (the exception is peroxides H 2 O 2, where it is equal to -1 and compounds with fluorine - O +2 F 2 -1 , O 2 +1 F 2 -1 ). - Oxidation state of 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-nonmetal bonds, the negative oxidation state is that atom that has greater electronegativity (data on electronegativity are given in 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 the connection KMnO 4 , it is necessary to determine the oxidation state of the manganese atom. Reasoning:
K+Mn X O 4 -2 Let X- unknown to us oxidation state of manganese. The number of potassium atoms is 1, manganese - 1, oxygen - 4. It has been proven that the molecule as a whole is electrically neutral, so its total charge must be zero. 1*(+1) + 1*(X) + 4(-2) = 0, X = +7, This means that 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:
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 equal to 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 chromium atom has 12 positive powers, but there are 2 atoms in the molecule, which means there are (+12) per atom: 2 = (+6). Answer: K 2 + Cr 2 +6 O 7 -2. 2.(AsO 4) 3- . In this case, the sum of oxidation states will no longer be equal to zero, but to the charge of the ion, i.e. - 3. Let's make an equation: x+4×(- 2)= - 3 . Answer: (As +5 O 4 -2) 3- . |
In chemistry, the terms “oxidation” and “reduction” refer to reactions in which an atom or group of atoms loses or gains electrons, respectively. The oxidation state is a numerical value assigned to one or more atoms that characterizes the number of redistributed electrons and shows how these electrons are distributed between atoms during a reaction. Determining this value can be either a simple or quite complex procedure, depending on the atoms and the molecules consisting of them. Moreover, the atoms of some elements may have several oxidation states. Fortunately, there are simple, unambiguous rules for determining the oxidation state; to use them confidently, a knowledge of the basics of chemistry and algebra is sufficient.
Steps
Part 1
Determination of oxidation state according to the laws of chemistry- For example, Al(s) and Cl 2 have an oxidation state of 0 because both are in a chemically unbound elemental state.
- Please note that the allotropic form of sulfur S8, or octasulfur, despite its atypical structure, is also characterized by a zero oxidation state.
-
Determine whether 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 Cl ion has a charge of -1, and thus its oxidation state is -1.
-
Please note that metal ions can have several oxidation states. The atoms of many metallic elements can be ionized to varying degrees. For example, the charge of ions of a metal such as iron (Fe) is +2 or +3. The charge of metal ions (and their oxidation state) can be determined by the charges of ions of other elements with which the 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 there are 3 such ions in the compound, for the substance in question to be overall neutral, the Al ion must have a charge of +3. Thus, in this case, the oxidation state of aluminum is +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 a few exceptions to this rule:
- If oxygen is in its elemental state (O2), its oxidation state is 0, as is the case for other elemental substances.
- If oxygen is included peroxide, its oxidation state is -1. Peroxides are a group of compounds containing a simple oxygen-oxygen bond (that is, the peroxide anion O 2 -2). For example, in the composition of the H 2 O 2 (hydrogen peroxide) molecule, oxygen has a charge and oxidation state of -1.
- When combined with fluorine, oxygen has an oxidation state of +2, read the rule for fluorine below.
-
Hydrogen has an oxidation state of +1, with some exceptions. As with oxygen, there are exceptions here too. Typically, the oxidation state of hydrogen is +1 (unless it is in the elemental state H2). However, in compounds called hydrides, the oxidation state of hydrogen is -1.
- For example, in H2O the oxidation state of hydrogen is +1 because the oxygen atom has a -2 charge 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 overall electrical neutrality, the charge of the hydrogen atom (and thus its oxidation state) must be equal to -1.
-
Fluorine Always has an oxidation state of -1. As already noted, the oxidation state of some elements (metal ions, oxygen atoms in peroxides, etc.) 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 foreign electrons. Thus, their charge remains unchanged.
-
The sum of oxidation states in a compound is equal to its charge. The oxidation states of all atoms in a chemical compound must add up to the charge of that 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 way to check - if the sum of the oxidation states does not equal the total charge of the compound, then you have made a mistake somewhere.
Part 2
Determination of oxidation state without using the laws of chemistry-
Find atoms that do not have strict rules regarding oxidation numbers. For some elements there are no firmly established rules for finding the oxidation state. 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 specified), you can determine the oxidation number 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 a given atom.
- For example, in the compound Na 2 SO 4 the charge of the sulfur atom (S) is unknown - we only know that it is not zero, since sulfur is not in an elemental state. This compound serves as a good example to illustrate the algebraic method of determining oxidation state.
-
Find the oxidation states of the remaining 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 rules in the case of O, H atoms, and so on.
- For Na 2 SO 4, using our rules, we find that the charge (and therefore the oxidation state) of the Na ion is +1, and for each of the oxygen atoms it is -2.
- 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 equal the total ionic charge.
- It is very useful to be able to use the periodic table and know where metallic and non-metallic elements are located in it.
- The oxidation state of atoms in elemental 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 their elemental form have an oxidation state of +1; Group 2A metals such as magnesium and calcium have an oxidation state of +2 in their elemental form. Oxygen and hydrogen, depending on the type of chemical bond, can have 2 different oxidation states.
Determine whether 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.
