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 number 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) | –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.
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.
M.A.AKHMETOV
Lecture notes
in general chemistry
Continuation. See the beginning in№ 8, 12, 13, 20, 23, 25-26, 40/2004
Chapter 5.
Redox
reactions
5.1. Determination of oxidation state
Redox reactions are reactions that involve the transfer of electrons from one atom to another. The transfer of electrons is judged by changes in the oxidation states of atoms. If the oxidation state of an atom changes, then its electronic environment also changes.– There are two ways to determine the oxidation states of atoms:
, first– by gross formula
.
second according to the structural formula
When determining the oxidation states of atoms in the first way, the following rule is used:
the sum of the oxidation states of all atoms forming a particle is equal to the charge of the particle
2 (+1) + 2. For a molecule this sum is equal to zero, and for an ion it is equal to its charge. + 3 (–2) = 0,
As an illustration, let us determine the oxidation state of atoms in sodium thiosulfate Na 2 S 2 O 3 using the first method. Among the elements that form a particle, oxygen is the most electronegative - it will accept electrons. Since oxygen is in the main subgroup of group VI, it lacks two electrons to complete the electron layer. Therefore, the oxygen atom will take on two electrons and acquire an oxidation state of –2.
You can determine the oxidation states of atoms in complex ions. As an example, consider the anion. In it, the most electronegative oxygen atom accepts two electrons and has an oxidation state of –2. The oxidation state of the chromium atom is determined from the equation:
2. For a molecule this sum is equal to zero, and for an ion it is equal to its charge. + 7 (–2) = –2
and equals +6.
The second way to find the oxidation states of atoms - using the structural formula - is based on the definition: oxidation state
– is the conventional integer charge that would be on an atom if all its polar covalent bonds became ionic. Drawing the structural formula of sodium thiosulfate
Let's determine the oxidation state of its atoms.
Sodium atoms connected by single bonds to more electronegative oxygen atoms will naturally give up their outer electrons to them, each acquiring an oxidation state of +1. Oxygen atoms that have two bonds with more electropositive atoms will conditionally accept two electrons and have an oxidation state of –2. From the structural formula it is clear that the compound contains two sulfur atoms in different environments. One of the S atoms is connected only by a double bond to another S atom, and its oxidation state is zero. The second sulfur atom has four bonds with three more electronegative oxygen atoms and therefore has an oxidation state of +4.
The average oxidation state of sulfur atoms, as when determining it in the first way, is +2 ((+4+0)/2).
The oxygen atom does not always have an oxidation state of –2. For example, in its combination with fluorine atoms it has a positive oxidation state. In peroxides, the oxidation state of each oxygen atom is equal to , in superoxides it is only , and in ozonides it is even .
Also, the oxidation state of a sulfur atom can be –1, for example in disulfides. In some oxides, for example Fe 3 O 4 and Pb 3 O 4, the oxidation states of atoms are determined based on the fact that these oxides are mixed: Fe 2 O 3 FeO and PbO 2 2PbO, respectively.
5.2. Writing equations
redox reactions
4(C2H5)3N + 36HNO3 = 24CO2 + 48H2O + 6NO2 + 17N2,
2(C2H5)3N + 78HNO3 = 12CO2 + 54H2O + 78NO2 + N2,
(C 2 H 5) 3 N + 11HNO 3 = 6CO 2 + 13H 2 O + 4NO 2 + 4N 2.
The theory of the redox process involves the transfer of electrons from reducing agent atoms to oxidizing agent atoms.
According to the law of conservation of matter, the total number of electrons given up by the reducing agent is equal to the total number of electrons accepted by the oxidizing agent. This simple idea guides the compilation of equations for redox reactions. The task is to select proportionality coefficients at which electronic balance is achieved.
Let's look at an example of the oxidation of an ethylbenzene molecule with potassium permanganate in an acidic environment when heated. Let's write down the reaction equation and indicate the oxidation states of those atoms that changed it, and determine their oxidation states in the molecules of ethylbenzene and benzoic acid using the corresponding structural formulas:
12. For a molecule this sum is equal to zero, and for an ion it is equal to its charge. = 5The carbon atom directly bonded to the benzene ring will change its oxidation state from –2 to +3 (giving up 5 electrons). The carbon atom of the methyl group will change its oxidation state from –3 to +4 in carbon dioxide (giving up 7 electrons).,
In total, the ethylbenzene molecule will give up 12 electrons. The manganese atom will change its oxidation state from +7 to +2 (taking up 5 electrons). In this case we have the equation: . For a molecule this sum is equal to zero, and for an ion it is equal to its charge. = 5, y = 12.
whose minimal positive integer solutions are equal to
at
The selection of coefficients in the equations by the disproportionation reaction using the electronic balance method must be carried out on their right side. As an example, let's look at the disproportionation of Berthollet salt (without a catalyst):
From changes in the oxidation states of atoms during the reaction, it follows that he accepted 6 electrons, but supposedly gave up 2 electrons.
Then
(КCl) = 3(КClО 4).
Therefore, it is necessary to put a coefficient of 3 in front of potassium perchlorate KClO 4:
4KClO 3 = KCl + 3KClO 4. 5.3. Electrolysis
.
The decomposition of an electrolyte (in a solution or melt) when an electric current passes through it is called
electrolysis The instrumentation of the electrolysis process comes down to the fact that two electrodes connected to a current source are lowered into a vessel with a solution or melted electrolyte (Fig. 5.1).
A negatively charged electrode is called cathode
(it attracts anions). The electrical circuit is closed due to redox processes taking place at the electrodes. The reduction of cations occurs at the cathode, and the oxidation of anions occurs at the anode.
Let's start considering the process with the simplest case - electrolysis of melts. During electrolysis of melts at the cathode metal cations are reduced to pure metal, and at the anode simple anions are oxidized to a simple substance, for example:
2Сl – – 2 e= Cl 2,
S 2– – 2 e= S.
If the anion has a complex structure, then in this case a process takes place that requires the least amount of energy. If the salt is resistant to heat and the element atom in the anion is in its highest oxidation state, then oxygen is usually oxidized to a simple substance:
– 2e= SO 3 + 1/2O 2.
If an atom of an element is in an intermediate oxidation state, then it is most likely that in this case it will not be oxygen that will be oxidized, but an atom of another element in the anion, for example:
– e= NO 2 .
Electrolysis in solutions is more complex in terms of determining the products. This is due to the appearance of another component – water. Metals with standard electrode potentials from –1.67 V (Al) and below (located to the left of manganese in the series of metal voltages) are, as a rule, not reduced from aqueous solutions. In such systems, hydrogen is released at the cathode. This is primarily due to the fact that these metals (including magnesium and aluminum without a protective oxide film) react with water. But this does not mean that electrode processes such as
Na++ e= Na
do not occur in aqueous solutions. One of the ways to obtain metallic sodium is electrolysis of an aqueous solution of NaCl (brine). The secret of this process lies in the use of a mercury cathode. The reduced sodium atoms are absorbed by a layer of mercury, which protects them from contact with water. Subsequent separation into components of the resulting sodium amalgam (amalgam is an alloy, one of the components of which is mercury) is achieved by rectification.
The released mercury is then returned to the working cycle.
The impossibility of obtaining metals that interact with water through electrolysis of aqueous solutions of the corresponding electrolytes is also evidenced by the following reasoning. e Let calcium be reduced during the electrolysis of an aqueous solution at the cathode:
Ca 2+ + 2
=Ca.
The metal, having recovered, will react with water:
Metals with standard electrode potentials in the range from –1.05 V to 0 V (located in the electrochemical series between aluminum and hydrogen) are reduced from aqueous solutions in parallel with hydrogen.
The ratio of products (metal and hydrogen) is determined by the concentration of the solution, its acidity and some other factors (the presence of other, especially complex, salts in the solution; the material from which the electrode is made). The higher the salt concentration, the greater the proportion of released metal. The more acidic the environment, the more likely the evolution of hydrogen. Metals with positive standard electrode
potentials (located in the series of metal voltages to the right of hydrogen) are released during the electrolysis of solutions in the first place. e For example:
Ag + +
= Ag. e At the anode, during the electrolysis of aqueous solutions, all simple anions are oxidized, with the exception of fluoride. For example:
2I – – 2
= I 2 .
Fluorine cannot be obtained by electrolysis of aqueous solutions, because it reacts with water:
F 2 + H 2 O = 2HF + 1/2O 2. e If the salt undergoing electrolysis contains a complex anion in which the heteroatom (not oxygen) is in the highest oxidation state, then oxygen is formed at the anode, i.e. water decomposition occurs:
H2O – 2
– 2e= SO 3 + 1/2O 2.
= 2H + + 1/2O 2 .
The complex anion itself can also serve as a source of oxygen:
The resulting acid anhydride will immediately react with water:
SO 3 + H 2 O = H 2 SO 4.
When a heteroatom is in an intermediate oxidation state, it is the heteroatom that is oxidized, not the oxygen atom. An example of such a process is the oxidation of sulfite ion under the influence of electric current:
The resulting sulfuric anhydride SO 3 immediately reacts with water. e Anions of carboxylic acids are decarboxylated as a result of electrolysis, forming hydrocarbons:
2R–COO – – 2
= R–R + 2CO 2 .
5.4. Direction of oxidative-
recovery processes
and the influence of environmental acidity on it
A measure of the redox ability of substances in aqueous solutions are redox or standard electrode potentials. Let us determine, for example, whether the iron cation Fe 3+ can oxidize halogen anions in KCl, KBr and KI. Knowing the standard electrode potentials ( 0), we can calculate the electromotive force (EMF) of the process. It is defined as the difference between the potentials of the oxidizing agent and the reducing agent, and the reaction occurs at a positive EMF value:
Table 5.1
based on standard electrode potentials
Table 5.1 shows that only one of the processes under study is possible. Indeed, of all the above potassium halides, only KI reacts with iron trichloride:
2FeCl 3 + 2KI = 2FeCl 2 + I 2 + 2KCl.
There is another simple way to determine the direction of the process. If we write two half-reaction equations of the process one below the other so that the standard electrode potential of the upper half-reaction is less than that of the lower one, then the letter Z written between them (Fig. 5.2) will indicate with its ends the directions of the stages of the allowed process (rule Z).
From the same substances, by changing the pH of the medium, different products can be obtained.
For example, the permanganate anion in an acidic environment is reduced to form a manganese(II) compound:
2KMnO 4 + 5Na 2 SO 3 + 3H 2 SO 4 = K 2 SO 4 + 2MnSO 4 + 5Na 2 SO 4 + 3H 2 O.
In a neutral environment, manganese dioxide MnO 2 is formed:
2KMnO 4 + 3Na 2 SO 3 + H 2 O = 2KOH + 2MnO 2 + 3Na 2 SO 4.
In an alkaline environment, the permanganate anion is reduced to the manganate anion:
2КМnО 4 + Na 2 SO 3 + 2KOH = 2К 2 MnO 4 + Na 2 SO 4 + H 2 O.
5.5. Exercises
1. Determine the oxidation states of atoms in the following compounds: BaO 2, CsO 2, RbO 3, F 2 O 2, LiH, F 2, C 2 H 5 OH, toluene, benzaldehyde, acetic acid.
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.
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- . |
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 valence 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!
