How to determine the charge of a chemical element in a compound. Highest oxidation state
Electronegativity, like other properties of atoms of chemical elements, changes periodically with an increase in the ordinal number of the element:
The graph above shows the periodicity of the change in the electronegativity of the elements of the main subgroups, depending on the ordinal number of the element.
When moving down the subgroup of the periodic table, the electronegativity of chemical elements decreases, when moving to the right along the period, it increases.
Electronegativity reflects the non-metallicity of elements: the higher the value of electronegativity, the more non-metallic properties are expressed in the element.
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 oxidation state in complex substances:
3) There are chemical elements that exhibit a constant oxidation state in the vast majority of compounds. These elements include:
Element |
The oxidation state in almost all compounds |
Exceptions |
| hydrogen H | +1 | Alkali and alkaline earth metal hydrides, 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 match their highest oxidation state (mandatory to memorize)
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 a non-metal = group number - 8
Based on the above rules, you can set the degree of oxidation 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 for sulfuric acid:
The oxidation state of hydrogen in all complex substances is +1 (except for metal hydrides).
The oxidation state of oxygen in all complex substances is -2 (except for peroxides and oxygen fluoride OF 2). Let's 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, nitrogen and chromium, are unknown. Therefore, we cannot find the oxidation states in the same way as in the previous example (one equation with two variables does not have a unique solution).
Let us pay attention to the fact that the indicated 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). Therefore, since there are two positive singly charged NH 4 + cations in the formula unit of ammonium dichromate, the charge of the dichromate ion is -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 the 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.
How to determine the oxidation states of elements in organic matter can be read.
Valence
The valency of atoms is indicated by Roman numerals: I, II, III, etc.
The valence possibilities of an atom depend on the quantity:
1) unpaired electrons
2) unshared electron pairs in the orbitals of valence levels
3) empty electron orbitals of the valence level
Valence possibilities of the hydrogen atom
Let's depict the electronic graphic formula of the hydrogen atom:
It was said that three factors can affect the valence possibilities - the presence of unpaired electrons, the presence of unshared electron pairs at the outer level, and the presence of vacant (empty) orbitals of the outer level. We see one unpaired electron in the outer (and only) energy level. Based on this, hydrogen can exactly have a valency equal to I. However, at the first energy level there is only one sublevel - s, those. the hydrogen atom at the outer level does not have either unshared electron pairs or empty orbitals.
Thus, the only valency that a hydrogen atom can exhibit is I.
Valence possibilities of a carbon atom
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, the outer energy level of an unexcited carbon atom contains 2 unpaired electrons. In this state, it can exhibit a valency equal to 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:
Although some energy is expended in 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, valence IV is much more characteristic of the carbon atom. So, for example, carbon has valency IV in the molecules of carbon dioxide, carbonic acid and absolutely all organic substances.
In addition to unpaired electrons and lone electron pairs, the presence of vacant () orbitals of the valence level also affects the valence possibilities. The presence of such orbitals in the filled level leads to the fact that the atom can act as an electron pair acceptor, i.e. form additional covalent bonds by the donor-acceptor mechanism. So, for example, contrary to expectations, in the carbon monoxide molecule CO, the bond is not double, but triple, which is clearly shown in the following illustration:
Valence possibilities of the nitrogen atom
Let's write down the electron-graphic formula of 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 can exhibit a valency equal to III. Indeed, a valency 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 unshared electron pairs. This is due to the fact that a covalent chemical bond can form not only when two atoms provide each other with one electron each, but also when one atom that has an unshared pair of electrons - donor () provides it to another atom with a vacant () orbital valence level (acceptor). Those. for the nitrogen atom, valency IV is also possible due to an additional covalent bond formed by the donor-acceptor mechanism. So, for example, four covalent bonds, one of which is formed by the donor-acceptor mechanism, is observed during the formation of the ammonium cation:
Despite the fact that one of the covalent bonds is formed by the donor-acceptor mechanism, all N-H bonds in the ammonium cation are absolutely identical and do not differ from each other.
A valency equal to V, the nitrogen atom is not able to show. This is due to the fact that the transition to an excited state is impossible for the nitrogen atom, in which the pairing of two electrons occurs with the transition of one of them to a free orbital, which is the 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 then is the valency of nitrogen, for example, in the molecules of nitric acid HNO 3 or nitric oxide N 2 O 5? Oddly enough, the valence 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, NO terminal bonds can be called "one and a half". Similar one-and-a-half bonds are also found in the ozone molecule O 3 , benzene C 6 H 6 , etc.
Valence possibilities of phosphorus
Let us depict the electron-graphic formula of the external energy level of the phosphorus atom:
As we can 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, which is observed in practice.
However, unlike nitrogen, the phosphorus atom also has d-sublevel with 5 vacant orbitals.
In this regard, it is able to pass into an excited state, steaming electrons 3 s-orbitals:
Thus, the valency V for the phosphorus atom, which is inaccessible to nitrogen, is possible. So, for example, a phosphorus atom has a valence of five in the molecules of such compounds as phosphoric acid, phosphorus (V) halides, phosphorus (V) oxide, etc.
Valence possibilities of the oxygen atom
The electron-graphic formula of the external energy level of the oxygen atom has the form:
We see two unpaired electrons at the 2nd level, and therefore valency II is possible for oxygen. It should be noted that this valency of the oxygen atom is observed in almost all compounds. Above, when considering the valence possibilities of the carbon atom, we discussed the formation of the carbon monoxide molecule. The bond in the CO molecule is triple, therefore, oxygen is trivalent there (oxygen is an electron pair donor).
Due to the fact that the oxygen atom does not have an external level d-sublevels, depairing of electrons 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
The external energy level of the sulfur atom in the unexcited state:
The sulfur atom, like the oxygen atom, has two unpaired electrons in its normal state, so we can conclude that a valency of two is possible for sulfur. Indeed, sulfur has valency II, for example, in the hydrogen sulfide molecule H 2 S.
As we can see, the sulfur atom at the outer level has 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. So, when unpairing a lone electron pair 3 p-sublevel the sulfur atom acquires electronic configuration outer level like this:
In this state, the sulfur atom has 4 unpaired electrons, which tells us about the possibility of sulfur atoms showing a valency equal to IV. Indeed, sulfur has valency IV in the molecules SO 2, SF 4, SOCl 2, etc.
When unpairing the second lone electron pair located on 3 s- sublevel, the external energy level acquires the following configuration:
In such a state, the manifestation of valence VI already becomes possible. An example 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.
To characterize the redox ability of particles, such a concept as the degree of oxidation is important. The OXIDATION STATE is the charge that an atom in a molecule or ion could have if all its bonds with other atoms were broken, and the common electron pairs left with more electronegative elements.
Unlike the real-life charges of ions, the oxidation state shows only the conditional charge of an atom in a molecule. It can be negative, positive or zero. For example, the oxidation state of atoms in simple substances is "0" (, 
,
,
). In chemical compounds, atoms can have a constant oxidation state or a variable. For metals of the main subgroups I, II and III of groups of the Periodic system in chemical compounds, the oxidation state is usually constant and equal to Me +1, Me +2 and Me +3 (Li +, Ca +2, Al +3), respectively. The fluorine atom always has -1. Chlorine in compounds with metals always has -1. In the vast majority of compounds, oxygen has an oxidation state of -2 (except for peroxides, where its oxidation state is -1), and hydrogen +1 (except for metal hydrides, where its oxidation state is -1).
The algebraic sum of the oxidation states of all atoms in a neutral molecule is equal to zero, and in an ion it is equal to the charge of the ion. This relationship makes it possible to calculate the oxidation states of atoms in complex compounds.
In the sulfuric acid molecule H 2 SO 4, the hydrogen atom has an oxidation state of +1, and the oxygen atom is -2. Since there are two hydrogen atoms and four oxygen atoms, we have two "+" and eight "-". Six "+" are missing to neutrality. It is this number that is the oxidation state of sulfur - 
. The potassium dichromate K 2 Cr 2 O 7 molecule consists of two potassium atoms, two chromium atoms and seven oxygen atoms. Potassium has an oxidation state of +1, oxygen has -2. So we have two "+" and fourteen "-". The remaining twelve "+" fall on two chromium atoms, each of which has an oxidation state of +6 ( 
).
Typical oxidizing and reducing agents
From the definition of reduction and oxidation processes, it follows that, in principle, simple and complex substances containing atoms that are not in the lowest oxidation state and therefore can lower their oxidation state can act as oxidizing agents. Similarly, simple and complex substances containing atoms that are not in the highest oxidation state and therefore can increase their oxidation state can act as reducing agents.
The strongest oxidizing agents are:
1) simple substances formed by atoms having a large electronegativity, i.e. typical non-metals located in the main subgroups of the sixth and seventh groups of the periodic system: F, O, Cl, S (respectively F 2 , O 2 , Cl 2 , S);
2) substances containing elements in higher and intermediate
positive oxidation states, including in the form of ions, both simple, elemental (Fe 3+) and oxygen-containing, oxoanions (permanganate ion - MnO 4 -);
3) peroxide compounds.
Specific substances used in practice as oxidizers are oxygen and ozone, chlorine, bromine, permanganates, dichromates, oxyacids of chlorine and their salts (for example, 
,
,
), Nitric acid ( 
), concentrated sulfuric acid ( 
), manganese dioxide ( 
), hydrogen peroxide and metal peroxides ( 
,
).
The most powerful reducing agents are:
1) simple substances whose atoms have low electronegativity ("active metals");
2) metal cations in low oxidation states (Fe 2+);
3) simple elemental anions, for example, sulfide ion S 2- ;
4) oxygen-containing anions (oxoanions) corresponding to the lowest positive oxidation states of the element (nitrite 
, sulfite 
).
Specific substances used in practice as reducing agents are, for example, alkali and alkaline earth metals, sulfides, sulfites, hydrogen halides (except HF), organic substances - alcohols, aldehydes, formaldehyde, glucose, oxalic acid, as well as hydrogen, carbon, monoxide carbon ( 
) and aluminum at high temperatures.
In principle, if a substance contains an element in an intermediate oxidation state, then these substances can exhibit both oxidizing and reducing properties. It all depends on
"partner" in the reaction: with a sufficiently strong oxidizing agent, it can react as a reducing agent, and with a sufficiently strong reducing agent, as an oxidizing agent. So, for example, the nitrite ion NO 2 - in an acidic environment acts as an oxidizing agent with respect to the ion I -:
2
+
2
+ 4HCl→ 
+
2
+ 4KCl + 2H 2 O
and as a reducing agent in relation to the permanganate ion MnO 4 -
5
+
2
+ 3H 2 SO 4 → 2 
+
5
+ K 2 SO 4 + 3H 2 O
The video course "Get an A" includes all the topics necessary for a successful passing the exam in mathematics for 60-65 points. Completely all tasks 1-13 profile exam mathematics. Also suitable for passing the Basic USE in mathematics. If you want to pass the exam with 90-100 points, you need to solve part 1 in 30 minutes and without mistakes!
Preparation course for the exam for grades 10-11, as well as for teachers. Everything you need to solve part 1 of the exam in mathematics (the first 12 problems) and problem 13 (trigonometry). And this is more than 70 points on the Unified State Examination, and neither a hundred-point student nor a humanist can do without them.
All necessary theory. Quick Ways solutions, traps and USE secrets. All relevant tasks of part 1 from the Bank of FIPI tasks have been analyzed. The course fully complies with the requirements of the USE-2018.
The course contains 5 large topics, 2.5 hours each. Each topic is given from scratch, simply and clearly.
Hundreds of exam tasks. Text problems and probability theory. Simple and easy to remember problem solving algorithms. Geometry. Theory, reference material, analysis of all types of USE tasks. Stereometry. Cunning tricks for solving, useful cheat sheets, development of spatial imagination. Trigonometry from scratch - to task 13. Understanding instead of cramming. Visual explanation of complex concepts. Algebra. Roots, powers and logarithms, function and derivative. Base for solving complex problems of the 2nd part of the exam.
Part I
1. The oxidation state (s. o.) is conditional charge of the atoms of a chemical element in a complex substance, calculated on the basis of the assumption that it consists of simple ions.
Should know!
1) In connections with. O. hydrogen = +1, except for hydrides.
2) In compounds with. O. oxygen = -2, except for peroxides 
and fluorides
3) The oxidation state of metals is always positive.
For metals of the main subgroups of the first three groups With. O. constant:
Group IA metals - p. O. = +1,
Group IIA metals - p. O. = +2,
Group IIIA metals - p. O. = +3.
4) For free atoms and simple substances p. O. = 0.
5) Total s. O. all elements in the compound = 0.
2. Method of formation of names two-element (binary) compounds.
4. Complete the table "Names and formulas of binary compounds."
5. Determine the degree of oxidation of the highlighted element of the complex compound.
Part II
1. Determine the oxidation states of chemical elements in compounds according to their formulas. Write down the names of these substances.
2. Separate substances FeO, Fe2O3, CaCl2, AlBr3, CuO, K2O, BaCl2, SO3into two groups. Write down the names of substances, indicating the degree of oxidation.
3. Establish a correspondence between the name and oxidation state of an atom of a chemical element and the formula of the compound.
4. Make formulas of substances by name.
5. How many molecules are contained in 48 g of sulfur oxide (IV)?
6. Using the Internet and other sources of information, prepare a report on the use of any binary connection according to the following plan:
1) formula;
2) name;
3) properties;
4) application.
H2O water, hydrogen oxide.
Water under normal conditions is a liquid, colorless, odorless, in a thick layer - blue. The boiling point is about 100⁰С. It is a good solvent. A water molecule consists of two hydrogen atoms and one oxygen atom, this is its qualitative and quantitative composition. This complex substance, it is characterized by the following Chemical properties: interaction with alkali metals, alkaline earth metals. Exchange reactions with water are called hydrolysis. These reactions have great importance in chemistry.
7. The oxidation state of manganese in the K2MnO4 compound is:
3) +6
8. Chromium has the lowest oxidation state in a compound whose formula is:
1) Cr2O3
9. Chlorine exhibits the maximum oxidation state in a compound whose formula is:
3) Сl2O7
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 - , Communication in hydrochloric acid 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. Oxidation state 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:
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:
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-. |
