The oxidation number is the conditional charge of an atom in a molecule, it receives the atom as a result of the complete acceptance of electrons, it is calculated from the assumption that all bonds are ionic in nature. How to determine the oxidation state?
Determination of oxidation state
There are charged particles, ions, positive charge which is equal to the number of electrons received from one atom. The negative charge of an ion is equal to the number of electrons accepted by one atom chemical element. For example, writing an element as Ca2+ means that the atoms of the elements have lost one, two or three elements. To find the composition of ionic compounds and molecular compounds, we need to know how to determine the oxidation state of elements. Oxidation states are negative, positive and zero. If we take into account the number of atoms, then algebra degree oxidation in the molecule is zero.
To determine the oxidation state of an element, you need to be guided by certain knowledge. For example, in metal compounds the oxidation state is positive. A highest degree oxidation corresponds to the group number of the periodic table where the element is located. Metals can have positive or negative oxidation states. This will depend on the factor by which atom the metal is connected. For example, if connected to a metal atom, then the degree will be negative, but if connected to a non-metal, then the degree will be positive.
The negative highest oxidation state of a metal can be determined by subtracting from the number eight the number of the group where the required element is located. As a rule, it is equal to the number of electrons located in the outer layer. The number of these electrons also corresponds to the group number.
How to calculate oxidation number
In most cases, the oxidation state of an atom of a particular element does not coincide with the number of bonds it forms, that is, it is not equal to the valency of that element. This can be clearly seen in the example of organic compounds.
Let me remind you that the valency of carbon in organic compounds is 4 (i.e. it forms 4 bonds), but the oxidation state of carbon, for example, in methanol CH 3 OH is -2, in CO 2 +4, in CH4 -4, in formic acid HCOOH + 2. Valence is measured by the number of covalent chemical bonds, including those formed by the donor-acceptor mechanism.
When determining the oxidation state of atoms in molecules, an electronegative atom, when one electron pair is displaced in its direction, acquires a charge of -1, but if there are two electron pairs, then there will be a charge of -2. The oxidation state is not affected by the bond between like atoms. For example:
- Connection C-C atoms equals their zero oxidation state.
- C-H bond – here, carbon, as the most electronegative atom, will have a charge of -1.
- Connection C-O charge carbon, being less electronegative, will be equal to +1.
Examples of determining the oxidation state
- In a molecule such as CH 3Cl there are three C-H bonds C). Thus, the oxidation state of the carbon atom in this compound will be equal to: -3+1=-2.
- Let's find the oxidation state of carbon atoms in the acetaldehyde molecule Cˉ³H3-C¹O-H. In this compound, the three C-H bonds will give a total charge on the C atom, which is equal to (Cº+3e→Cˉ³)-3. The double bond C=O (here oxygen will take electrons from the carbon atom, since oxygen is more electronegative) gives a charge on the C atom, it is equal to +2 (Cº-2e→C²), the bond C-H charge-1, which means the total charge on the C atom is: (2-1=1)+1.
- Now let's find the oxidation state in the ethanol molecule: Cˉ³H-Cˉ¹H2-OH. Here three C-H bonds will give a total charge on the C atom, it is equal to (Cº+3e→Cˉ³)-3. Two C-H bonds will give a charge on the C atom, which will be equal to -2, while the C→O bond will give a charge of +1, which means the total charge on the C atom is (-2+1=-1)-1.
Now you know how to determine the oxidation state of an element. If you have at least basic knowledge of chemistry, then this task will not be a problem for you.
Electronegativity, like other properties of atoms of chemical elements, changes with increasing serial number element periodically:
The graph above shows the periodicity of changes in the electronegativity of elements of the main subgroups depending on the atomic number of the element.
When moving down a subgroup of the periodic table, the electronegativity of chemical elements decreases, and when moving to the right along the period it increases.
Electronegativity reflects the non-metallicity of elements: the higher the electronegativity value, the more non-metallic properties the element has.
Oxidation state
How to calculate the oxidation state of an element in a compound?
1) The oxidation state of chemical elements in simple substances is always zero.
2) There are elements that exhibit a constant state of oxidation in complex substances:
3) There are chemical elements that exhibit a constant oxidation state in the vast majority of compounds. These elements include:
Element |
Oxidation state in almost all compounds |
Exceptions |
| hydrogen H | +1 | Hydrides of alkali and alkaline earth metals, for example: |
| oxygen O | -2 | Hydrogen and metal peroxides: Oxygen fluoride - |
4) The algebraic sum of the oxidation states of all atoms in a molecule is always zero. The algebraic sum of the oxidation states of all atoms in an ion is equal to the charge of the ion.
5) The highest (maximum) oxidation state is equal to the group number. Exceptions that do not fall under this rule are elements of the secondary subgroup of group I, elements of the secondary subgroup of group VIII, as well as oxygen and fluorine.
Chemical elements whose group number does not coincide with their highest oxidation state (mandatory to remember)
6) The lowest oxidation state of metals is always zero, and the lowest oxidation state of non-metals is calculated by the formula:
lowest oxidation state of non-metal = group number − 8
Based on the rules presented above, you can establish the oxidation state of a chemical element in any substance.
Finding the oxidation states of elements in various compounds
Example 1
Determine the oxidation states of all elements in sulfuric acid.
Solution:
Let's write the formula of sulfuric acid:
The oxidation state of hydrogen in all complex substances is +1 (except metal hydrides).
The oxidation state of oxygen in all complex substances is -2 (except for peroxides and oxygen fluoride OF 2). Let us arrange the known oxidation states:
Let us denote the oxidation state of sulfur as x:
The sulfuric acid molecule, like the molecule of any substance, is generally electrically neutral, because the sum of the oxidation states of all atoms in a molecule is zero. Schematically this can be depicted as follows:
Those. we got the following equation:
Let's solve it:
Thus, the oxidation state of sulfur in sulfuric acid is +6.
Example 2
Determine the oxidation state of all elements in ammonium dichromate.
Solution:
Let's write the formula of ammonium dichromate:
As in the previous case, we can arrange the oxidation states of hydrogen and oxygen:
However, we see that the oxidation states of two chemical elements at once are unknown - nitrogen and chromium. Therefore, we cannot find oxidation states similarly to the previous example (one equation with two variables does not have a single solution).
Let us draw attention to the fact that this substance belongs to the class of salts and, accordingly, has an ionic structure. Then we can rightly say that the composition of ammonium dichromate includes NH 4 + cations (the charge of this cation can be seen in the solubility table). Consequently, since the formula unit of ammonium dichromate contains two positive singly charged NH 4 + cations, the charge of the dichromate ion is equal to -2, since the substance as a whole is electrically neutral. Those. the substance is formed by NH 4 + cations and Cr 2 O 7 2- anions.
We know the oxidation states of hydrogen and oxygen. Knowing that the sum of the oxidation states of the atoms of all elements in an ion is equal to the charge, and denoting the oxidation states of nitrogen and chromium as x And y accordingly, we can write:
Those. we get two independent equations:
Solving which, we find x And y:
Thus, in ammonium dichromate the oxidation states of nitrogen are -3, hydrogen +1, chromium +6, and oxygen -2.
You can read how to determine the oxidation states of elements in organic substances.
Valence
The valence of atoms is indicated by Roman numerals: I, II, III, etc.
The valence capabilities of an atom depend on the quantity:
1) unpaired electrons
2) lone electron pairs in the orbitals of valence levels
3) empty electron orbitals of the valence level
Valence possibilities of the hydrogen atom
Let us depict the electron graphic formula of the hydrogen atom:
It has been said that three factors can influence the valence possibilities - the presence of unpaired electrons, the presence of lone electron pairs in the outer level, and the presence of vacant (empty) orbitals in the outer level. We see one unpaired electron at the outer (and only) energy level. Based on this, hydrogen can definitely have a valence of I. However, in the first energy level there is only one sublevel - s, those. The hydrogen atom at the outer level has neither lone electron pairs nor empty orbitals.
Thus, the only valence that a hydrogen atom can exhibit is I.
Valence possibilities of the carbon atom
Let's consider the electronic structure of the carbon atom. In the ground state, the electronic configuration of its outer level is as follows:
Those. in the ground state at the outer energy level of the unexcited carbon atom there are 2 unpaired electrons. In this state it can exhibit a valence of II. However, the carbon atom very easily goes into an excited state when energy is imparted to it, and the electronic configuration of the outer layer in this case takes the form:
Despite the fact that a certain amount of energy is spent on the process of excitation of the carbon atom, the expenditure is more than compensated for by the formation of four covalent bonds. For this reason, valency IV is much more characteristic of the carbon atom. So, for example, the valency IV carbon in molecules has carbon dioxide, carbonic acid and absolutely all organic substances.
In addition to unpaired electrons and lone electron pairs, the presence of vacant ()valence level orbitals also affects the valence possibilities. The presence of such orbitals at the filled level leads to the fact that the atom can act as an electron pair acceptor, i.e. form additional covalent bonds through a donor-acceptor mechanism. For example, contrary to expectations, in the molecule carbon monoxide The CO bond is not double, but triple, as is clearly shown in the following illustration:
Valence possibilities of the nitrogen atom
Let us write the electronic graphic formula for the external energy level of the nitrogen atom:
As can be seen from the illustration above, the nitrogen atom in its normal state has 3 unpaired electrons, and therefore it is logical to assume that it is capable of exhibiting a valence of III. Indeed, a valence of three is observed in the molecules of ammonia (NH 3), nitrous acid (HNO 2), nitrogen trichloride (NCl 3), etc.
It was said above that the valence of an atom of a chemical element depends not only on the number of unpaired electrons, but also on the presence of lone electron pairs. This is due to the fact that a covalent chemical bond can be formed not only when two atoms provide each other with one electron, but also when one atom with a lone pair of electrons - donor () provides it to another atom with a vacant () orbital valence level (acceptor). Those. For the nitrogen atom, valence IV is also possible due to an additional covalent bond formed by the donor-acceptor mechanism. For example, four covalent bonds, one of which is formed by a donor-acceptor mechanism, are observed during the formation of an ammonium cation:
Despite the fact that one of the covalent bonds is formed according to the donor-acceptor mechanism, all N-H bonds in the ammonium cation are absolutely identical and do not differ from each other.
The nitrogen atom is not capable of exhibiting a valency equal to V. This is due to the fact that it is impossible for a nitrogen atom to transition to an excited state, in which two electrons are paired with the transition of one of them to a free orbital that is closest in energy level. The nitrogen atom has no d-sublevel, and the transition to the 3s orbital is energetically so expensive that the energy costs are not covered by the formation of new bonds. Many may wonder, what is the valency of nitrogen, for example, in molecules nitric acid HNO 3 or nitrogen oxide N 2 O 5? Oddly enough, the valency there is also IV, as can be seen from the following structural formulas:
The dotted line in the illustration shows the so-called delocalized π -connection. For this reason, terminal NO bonds can be called “one and a half bonds.” Similar one-and-a-half bonds are also present in the molecule of ozone O 3, benzene C 6 H 6, etc.
Valence possibilities of phosphorus
Let us depict the electronic graphic formula of the external energy level of the phosphorus atom:
As we see, the structure of the outer layer of the phosphorus atom in the ground state and the nitrogen atom is the same, and therefore it is logical to expect for the phosphorus atom, as well as for the nitrogen atom, possible valences equal to I, II, III and IV, as observed in practice.
However, unlike nitrogen, the phosphorus atom also has d-sublevel with 5 vacant orbitals.
In this regard, it is capable of transitioning to an excited state, steaming electrons 3 s-orbitals:
Thus, the valence V for the phosphorus atom, which is inaccessible to nitrogen, is possible. For example, the phosphorus atom has a valency of five in molecules of compounds such as phosphoric acid, phosphorus (V) halides, phosphorus (V) oxide, etc.
Valence possibilities of the oxygen atom
The electron graphic formula for the external energy level of an oxygen atom has the form:
We see two unpaired electrons at the 2nd level, and therefore valence II is possible for oxygen. It should be noted that this valence of the oxygen atom is observed in almost all compounds. Above, when considering the valency capabilities of the carbon atom, we discussed the formation of the carbon monoxide molecule. The bond in the CO molecule is triple, therefore, the oxygen there is trivalent (oxygen is an electron pair donor).
Due to the fact that the oxygen atom does not have an external d-sublevel, electron pairing s And p- orbitals is impossible, which is why the valence capabilities of the oxygen atom are limited compared to other elements of its subgroup, for example, sulfur.
Valence possibilities of the sulfur atom
External energy level of a sulfur atom in an unexcited state:
The sulfur atom, like the oxygen atom, normally has two unpaired electrons, so we can conclude that a valence of two is possible for sulfur. Indeed, sulfur has valency II, for example, in the hydrogen sulfide molecule H 2 S.
As we see, the sulfur atom appears at the external level d-sublevel with vacant orbitals. For this reason, the sulfur atom is able to expand its valence capabilities, unlike oxygen, due to the transition to excited states. Thus, when pairing a lone electron pair 3 p-sublevel the sulfur atom acquires electronic configuration external level of the following form:
In this state, the sulfur atom has 4 unpaired electrons, which tells us that sulfur atoms can exhibit a valence of IV. Indeed, sulfur has valency IV in molecules SO 2, SF 4, SOCl 2, etc.
When pairing the second lone electron pair located at 3 s-sublevel, the external energy level acquires the configuration:
In this state, the manifestation of valency VI becomes possible. Examples of compounds with VI-valent sulfur are SO 3, H 2 SO 4, SO 2 Cl 2, etc.
Similarly, we can consider the valence possibilities of other chemical elements.
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 system 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 - , Contact in hydrochloric acid covalent polar. 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 the 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- . |
The task of determining the oxidation state may turn out to be either a simple formality or complex puzzle. First of all, this will depend on the formula of the chemical compound, as well as the availability of basic knowledge of chemistry and mathematics.
Knowing the basic rules and algorithm of sequentially logical actions, which will be discussed in this article, when solving problems similar type, everyone can easily cope with this task. And after practicing and learning to determine the oxidation states of diverse chemical compounds, you can safely take on the task of balancing complex redox reactions by drawing up an electronic balance.
The concept of oxidation state
To learn how to determine the degree of oxidation, you first need to understand what this concept means?
- The oxidation number is used when writing in redox reactions when electrons are transferred from atom to atom.
- The oxidation state records the number of electrons transferred, indicating the conditional charge of the atom.
- The oxidation state and valency are often identical.
This designation is written on top of the chemical element, in its right corner, and is an integer with a “+” or “-” sign. A zero value of the oxidation state does not carry a sign.
Rules for determining the degree of oxidation
Let's consider the main canons for determining the oxidation state:
- Simple elementary substances, that is, those that consist of one type of atom, will always have a zero oxidation state. For example, Na0, H02, P04
- There are a number of atoms that always have one, constant, oxidation state. It is better to remember the values given in the table.
- As you can see, the only exception occurs with hydrogen in combination with metals, where it acquires an oxidation state of “-1” that is not characteristic of it.
- Oxygen also takes on the oxidation state "+2" in chemical compound with fluorine and “-1” in the compositions of peroxides, superoxides or ozonides, where the oxygen atoms are connected to each other.
- Metal ions have several oxidation states (and only positive ones), so it is determined by neighboring elements in the compound. For example, in FeCl3, chlorine has an oxidation state of “-1”, it has 3 atoms, so we multiply -1 by 3, we get “-3”. In order for the sum of the oxidation states of a compound to be “0”, iron must have an oxidation state of “+3”. In the formula FeCl2, iron will accordingly change its degree to “+2”.
- By mathematically summing the oxidation states of all atoms in the formula (taking into account the signs), a zero value should always be obtained. For example, in hydrochloric acid H+1Cl-1 (+1 and -1 = 0), and in sulfurous acid H2+1S+4O3-2 (+1 * 2 = +2 for hydrogen, +4 for sulfur and -2 * 3 = – 6 for oxygen; +6 and -6 add up to 0).
- The oxidation state of a monatomic ion will be equal to its charge. For example: Na+, Ca+2.
- The highest oxidation state, as a rule, correlates with the group number in D.I. Mendeleev’s periodic system.
Algorithm for determining the degree of oxidation
The order of finding the oxidation state is not complicated, but requires attention and certain actions.
Task: arrange the oxidation states in the compound KMnO4
- The first element, potassium, has a constant oxidation state of “+1”.
To check, you can look at periodic table, where potassium is in group 1 of elements. - Of the remaining two elements, oxygen tends to have an oxidation state of -2.
- We get the following formula: K+1MnxO4-2. It remains to determine the oxidation state of manganese.
So, x is the oxidation state of manganese unknown to us. Now it is important to pay attention to the number of atoms in the compound.
The number of potassium atoms is 1, manganese is 1, oxygen is 4.
Taking into account the electrical neutrality of the molecule, when the total (total) charge is zero,
1*(+1) + 1*(x) + 4(-2) = 0,
+1+1х+(-8) = 0,
-7+1x = 0,
(when transferring, we change the sign)
1x = +7, x = +7
Thus, the oxidation state of manganese in the compound is “+7”.
Task: arrange the oxidation states in the Fe2O3 compound.
- Oxygen, as is known, has an oxidation state of “-2” and acts as an oxidizing agent. Taking into account the number of atoms (3), the total value for oxygen is “-6” (-2*3= -6), i.e. multiply the oxidation number by the number of atoms.
- To balance the formula and bring it to zero, 2 iron atoms will have an oxidation state of “+3” (2*+3=+6).
- The total is zero (-6 and +6 = 0).
Task: arrange the oxidation states in the Al(NO3)3 compound.
- There is only one aluminum atom and has a constant oxidation state of “+3”.
- There are 9 oxygen atoms in a molecule (3*3), the oxidation state of oxygen, as is known, is “-2”, which means that by multiplying these values, we get “-18”.
- It remains to equalize the negative and positive values, thus determining the degree of oxidation of nitrogen. -18 and +3, + 15 is missing. And given that there are 3 nitrogen atoms, it is easy to determine its oxidation state: divide 15 by 3 and get 5.
- The oxidation state of nitrogen is “+5”, and the formula will look like: Al+3(N+5O-23)3
- If it is difficult to determine the desired value in this way, you can compose and solve the equations:
1*(+3) + 3x + 9*(-2) = 0.
+3+3x-18=0
3x=15
x=5
So, the oxidation state is enough important concept in chemistry, symbolizing the state of atoms in a molecule.
Without knowledge of certain provisions or basics that allow you to correctly determine the degree of oxidation, it is impossible to cope with this task. Therefore, there is only one conclusion: thoroughly familiarize yourself with and study the rules for finding the oxidation state, clearly and concisely presented in the article, and boldly move on along the difficult path of chemical intricacies.
Topics of the Unified State Examination codifier: Electronegativity. Oxidation state and valence of chemical elements.
When atoms interact and form, electrons between them are in most cases unevenly distributed, since the properties of the atoms differ. More electronegative the atom attracts electron density more strongly to itself. An atom that has attracted electron density to itself acquires a partial negative charge δ — , its “partner” is a partial positive charge δ+ . If the difference in electronegativity of the atoms forming a bond does not exceed 1.7, we call the bond covalent polar . If the difference in electronegativities forming a chemical bond exceeds 1.7, then we call such a bond ionic .
Oxidation state is the auxiliary conditional charge of an element atom in a compound, calculated from the assumption that all compounds consist of ions (all polar bonds are ionic).
What does "conditional charge" mean? We simply agree that we will simplify things a little: we will consider any polar bonds to be completely ionic, and we will assume that the electron is completely leaving or coming from one atom to another, even if in fact this is not the case. And a conditionally electron leaves from a less electronegative atom to a more electronegative one.
For example, in the H-Cl bond we believe that hydrogen conditionally “gave up” an electron, and its charge became +1, and chlorine “accepted” an electron, and its charge became -1. In fact, there are no such total charges on these atoms.
Surely, you have a question - why invent something that doesn’t exist? This is not an insidious plan of chemists, everything is simple: this model is very convenient. Ideas about the oxidation state of elements are useful when compiling classifications chemical substances, description of their properties, compilation of formulas of compounds and nomenclature. Oxidation states are especially often used when working with redox reactions.
There are oxidation states higher, inferior And intermediate.
Higher the oxidation state is equal to the group number with a plus sign.
Lowest is defined as the group number minus 8.
AND intermediate An oxidation number is almost any whole number ranging from the lowest oxidation state to the highest.
For example, nitrogen is characterized by: the highest oxidation state is +5, the lowest 5 - 8 = -3, and intermediate oxidation states from -3 to +5. For example, in hydrazine N 2 H 4 the oxidation state of nitrogen is intermediate, -2.
Most often, the oxidation state of atoms in complex substances is indicated first by a sign, then by a number, for example +1, +2, -2 etc. When talking about the charge of an ion (assuming that the ion actually exists in a compound), then first indicate the number, then the sign. For example: Ca 2+ , CO 3 2- .
To find oxidation states, use the following rules :
- Oxidation state of atoms in simple substances equal to zero;
- IN neutral molecules the algebraic sum of oxidation states is zero, for ions this sum is equal to the charge of the ion;
- Oxidation state alkali metals (elements of group I of the main subgroup) in compounds is +1, oxidation state alkaline earth metals (elements of group II of the main subgroup) in compounds is +2; oxidation state aluminum in compounds it is equal to +3;
- Oxidation state hydrogen in compounds with metals (- NaH, CaH 2, etc.) is equal to -1 ; in compounds with non-metals () +1 ;
- Oxidation state oxygen equal to -2 . Exception make up peroxides– compounds containing the –O-O- group, where the oxidation state of oxygen is equal to -1 , and some other compounds ( superoxides, ozonides, oxygen fluorides OF 2 and etc.);
- Oxidation state fluoride in all complex substances is equal -1 .
Listed above are situations when we consider the oxidation state constant . All other chemical elements have an oxidation state — variable, and depends on the order and type of atoms in the compound.
Examples:
Exercise: determine the oxidation states of the elements in the potassium dichromate molecule: K 2 Cr 2 O 7 .
Solution: The oxidation state of potassium is +1, the oxidation state of chromium is denoted as X, the oxidation state of oxygen is -2. The sum of all oxidation states of all atoms in a molecule is equal to 0. We get the equation: +1*2+2*x-2*7=0. Solving it, we get the oxidation state of chromium +6.
In binary compounds, the more electronegative element has a negative oxidation state, and the less electronegative element has a positive oxidation state.
note that The concept of oxidation state is very arbitrary! The oxidation number does not indicate the real charge of an atom and has no real physical meaning. This is a simplified model that works effectively when we need, for example, to equalize the coefficients in the equation chemical reaction, or for algorithmizing the classification of substances.
Oxidation number is not valence! The oxidation state and valency do not coincide in many cases. For example, the valence of hydrogen in simple matter H2 is equal to I, and the oxidation state, according to rule 1, is equal to 0.
This basic rules, which will help you determine the oxidation state of atoms in compounds in most cases.
In some situations, you may have difficulty determining the oxidation state of an atom. Let's look at some of these situations and look at how to resolve them:
- In double (salt-like) oxides, the degree of an atom is usually two oxidation states. For example, in iron scale Fe 3 O 4, iron has two oxidation states: +2 and +3. Which one should I indicate? Both. To simplify, we can imagine this compound as a salt: Fe(FeO 2) 2. In this case, the acidic residue forms an atom with an oxidation state of +3. Or the double oxide can be represented as follows: FeO*Fe 2 O 3.
- In peroxo compounds, the oxidation state of oxygen atoms connected by covalent nonpolar bonds, as a rule, changes. For example, in hydrogen peroxide H 2 O 2 and alkali metal peroxides, the oxidation state of oxygen is -1, because one of the bonds is covalent nonpolar (H-O-O-H). Another example is peroxomonosulfuric acid (Caro acid) H 2 SO 5 (see figure) contains two oxygen atoms with an oxidation state of -1, the remaining atoms with an oxidation state of -2, so the following entry will be more understandable: H 2 SO 3 (O2). Chromium peroxo compounds are also known - for example, chromium (VI) peroxide CrO(O 2) 2 or CrO 5, and many others.
- Another example of compounds with ambiguous oxidation states is superoxides (NaO 2) and salt-like ozonides KO 3. In this case, it is more appropriate to talk about the molecular ion O 2 with a charge of -1 and O 3 with a charge of -1. The structure of such particles is described by some models that are taught in the Russian curriculum in the first years of chemical universities: MO LCAO, the method of superimposing valence schemes, etc.
- In organic compounds, the concept of oxidation state is not very convenient to use, because There are a large number of covalent nonpolar bonds between carbon atoms. However, if you draw structural formula molecules, then the oxidation state of each atom can also be determined by the type and number of atoms with which the atom is directly bonded. For example, the oxidation state of primary carbon atoms in hydrocarbons is -3, for secondary atoms -2, for tertiary atoms -1, and for quaternary atoms - 0.
Let's practice determining the oxidation state of atoms in organic compounds. To do this, it is necessary to draw the complete structural formula of the atom, and select the carbon atom with its closest environment - the atoms with which it is directly connected.
- To simplify calculations, you can use the solubility table - it shows the charges of the most common ions. In most Russian chemistry exams (USE, GIA, DVI), the use of solubility tables is permitted. This is a ready-made cheat sheet, which in many cases can significantly save time.
- When calculating the oxidation state of elements in complex substances, we first indicate the oxidation states of elements that we know for sure (elements with a constant oxidation state), and denote the oxidation state of elements with a variable oxidation state as x. The sum of all charges of all particles is zero in a molecule or equal to the charge of an ion in an ion. From this data it is easy to create and solve an equation.