Gillespie's theory of repulsion of localized electron pairs. Prediction of the geometric shape of molecules. Gillespie method. Molecular configuration according to the Gillespie method

To construct the MO energy diagram of a polyatomic molecule, it is necessary to know in advance its geometric structure. Usually it is established experimentally, but in many cases it can be very easily predicted. This possibility is provided by the method of repulsion of electron pairs of the valence shell, proposed in 1957 by R. Gillespie.

Gillespie's method is based on the idea that electrons, or more correctly, regions of increased electron density in the valence environment of the central atom, should be located in space so that their mutual repulsion is minimal. As we know, such regions correspond either to the covalent bonds formed by this atom, or to its unshared electron pairs. Therefore, we can say that in order to weaken the interelectronic repulsion, the bonds and lone pairs of the central atom should be located at the maximum possible distance from each other.

Configurations that provide the least mutual repulsion of two, three, four, five and six regions of increased electron density: 2-linear, 3-equilateral triangle, 4-tetrahedron, 5-trigonal bipyramid, 6-octahedron.

For example, in the BH3 boron hydride molecule, the least repulsion of the three regions of increased electron density corresponding to the B-H bonds is achieved provided that these bonds are directed towards the vertices of a regular triangle.

In the CH4 methane molecule, the smallest repulsion of the four regions of increased electron density (CH bonds) is achieved when they are arranged tetrahedral around the central carbon atom. Therefore, this molecule must have the shape of a tetrahedron with CH bonds directed to its vertices. The nitrogen atom in the ammonia molecule NH3 is surrounded by four regions of increased electron density. Three of them correspond to N-H bonds and one to a lone electron pair. As in the previous example, they should be directed towards the vertices of the tetrahedron. However, we "do not see" the vertex that is occupied by an unshared electron pair. Therefore, the ammonia molecule has the shape of a trigonal (i.e., triangular) pyramid with a nitrogen atom at its top and three hydrogen atoms at its base.

Another example is the carbon dioxide molecule CO2. It has two C=0 double bonds and, accordingly, two regions of increased electronic

density, each of which is formed by two electron pairs. Obviously, their least repulsion is achieved with a linear structure of the molecule.

Let us now formulate the procedure for determining the geometry of a molecule by the Gillespie method.

First, the number of regions of increased electron density (n) in the environment of the central atom is determined. To do this, count the number of bonds formed by him, both single and multiple, and the number of unshared electrons he has.

Then a figure or a polyhedron corresponding to the number n is chosen. The central atom is placed in the center of this figure or polyhedron.

Finally, those vertices that correspond to unshared electron pairs are mentally removed from the polyhedron or figure, and the observed shape of the molecule is obtained.

Let's look at a few more examples.

In the phosgene COCI2 molecule, the carbon atom forms a double bond with the oxygen atom and two single bonds with the chlorine atoms. Since there are no unshared electron pairs at the carbon atom in this molecule, the number n is equal to three and the molecule has the shape of a triangle.

In the nitrite anion NO2, the nitrogen atom forms bonds with two oxygen atoms: it spends two of its valence electrons on one, and one on the other. The nitrogen atom also has an unshared electron pair, which, together with two bonds, forms three regions of increased electron density, directed towards the vertices of the triangle. In one of them there is a lone pair, which is not taken into account when describing the mutual arrangement of atoms. Therefore, the NO2- anion has an angular shape with a bond angle between the N-O bonds close to 120 degrees. Around the central chlorine atom in the ClO3- chlorate anion there are three regions of electron density corresponding to three C1-0 bonds and one unshared electron pair. Therefore, this ion has a trigonal pyramidal structure similar to that of ammonia.

The phosphorus pentachdoride PCI5 molecule has the shape of a trigonal bipyramid, which is formed by five P-CI bonds.

The Gillespie method gives the best results in predicting the structure of compounds of non-transitional elements. However, even for them, forecasts made using it sometimes turn out to be erroneous, although the number of such erroneous forecasts is small. For example, the BaF2 molecule is predicted to have a linear structure, while the experimentally determined bond angle in this molecule is 100 degrees. On the contrary, the linear Li2O molecule is predicted by the Gillesia method as an angular one.

The geometry of polyatomic molecules determines their polarity. Molecules that have the shape of a regular geometric figure or a regular polyhedron are always non-polar. This is due to the fact that all shifts in the electron density on each bond towards the more electronegative element cancel each other out. So, for example, the PCI5 molecule has the shape of a regular polyhedron - a trigonal bipyramid and, therefore, is non-polar.

If the molecule has the shape of an irregular polyhedron or some angles are distorted in the correct one, then it turns out to be polar. For example, the P-Cl bonds in phosphorus trichloride form a trigonal pyramid with a phosphorus atom at the top (an irregular polyhedron), so the PCI3 molecule has a dipole moment.

Van der Waals forces

(The mechanism for the formation of in-in with an atomic or ionic bond is obvious: the formation of molecular orbitals, leading to a decrease in the potential energy of c-we as a result of the transition of electrons to lower energy levels, as well as the redistribution of electron density, which causes electrostatic attraction between ions.)

However, electrostatic attraction between neutral molecules is also possible, caused by van der Waals forces.

1) orientational interaction is carried out between polar molecules, which are oriented so as to approach opposite poles, resulting in attraction between them and a decrease in the potential energy of c-we when they approach-connect

2) inductive interaction. If a molecule does not have a permanent dipole moment, then it can arise in it, be induced (induced) under the influence of another, polar, molecule.

3) dispersion interaction. In any molecule, due to the fact that it is a cell with moving charges (nuclei, electrons), the so-called instantaneous microdipoles continuously arise, move and disappear. When molecules approach each other, their occurrence ceases to be completely random, independent; there is some consistency in their education.

Thus, the van der Waals ones are due to the correlation (consistency) of the motion of electrons in neighboring molecules due to the Coulomb interaction. They weaken very quickly with increasing distance between molecules. The relative contribution of each type of such forces depends mainly on 2 St-in molecules: polarity (the magnitude of the dipole moment) and polarizability (the ability to more or less easily change the relative spatial distribution of charges inside the molecule .)

46. ​​Zone structure of a solid body. Formation of energy bands in crystals of simple substances and in compounds with an ionic type of bond.

So far, we have considered the formation of a chemical bond in molecules formed from several atoms. However, many substances do not consist of molecules, but directly of atoms or ions. Chemical bonding in crystals of such substances is described using the band theory, which is a development of the MO method and considers the crystal as one very large molecule.

///Atomic crystals

Imagine how a crystal is formed from alkali metal atoms, each of which has only one valence s-orbital and one electron in this orbital. When two such atoms are combined, two molecular orbitals are formed: a bonding one, whose energy is less than the energy of the initial AO, and a loosening one, with a higher energy. If three atoms combine, then three MOs are formed, if four, then four MOs, and so on. In a crystal consisting of 1 mol of atoms, 6.022 * 10 orbitals should be formed.

It can be seen that in a molecule consisting of a small number of atoms, a rather large energy is required to transfer an electron to a free orbital. As the number of interacting atoms increases, the difference in MO energies becomes smaller, and with a very large number of atoms, it can be said that the orbitals form an almost continuous energy band. In accordance with the principle of least energy, electrons occupy the orbitals of the lower half of the band in pairs, leaving the upper half free. Electrons located in the filled part of the band, with the slightest excitation, can go over free orbitals with higher energy. Substances with an energy band partially filled with electrons conduct electric current well, i.e. have high electrical conductivity and are called metals.

Now consider the more complicated situation that arises when an atom has several valence orbitals, such as carbon (2s and 2p) or aluminum (3s and 3p). Depending on the properties of the connecting atoms, and on the type of the crystal lattice formed by them, the formed crystal MOs can merge into a single energy zone, or they can form several separate zones.

When a crystal is formed from N aluminum atoms, a single band arises, consisting of 4N orbitals, of which 3/2N are filled with electrons, and the rest are free. Therefore, aluminum is a metal with high electrical conductivity.

A diamond crystal has a different band structure. When N carbon atoms join, two bands are formed, each of which consists of 2N orbitals. Since the carbon atom has four valence electrons, it is easy to calculate that all levels of the band lying on the energy diagram in the region of lower energies are completely filled, and all levels of the upper band are free. The band occupied by electrons is called the valence band, and the free band is called the conduction band. Between the valence band and the conduction band there is a so-called band gap, in which there are no allowed energy states (i.e. MO) for electrons. Therefore, in order to acquire mobility, an electron must receive additional energy that exceeds the "width" of the band gap. In a diamond crystal, for example, this "width" is quite large and is approximately 5.5 eV, so diamond is a dielectric (insulator).

In order for a diamond to acquire electrical conductivity, it must either be irradiated with hard ultraviolet radiation or heated to a temperature of several thousand degrees. In addition, the transfer of electrons to the conduction band can occur under the action of a very high electric field. In this case, a phenomenon called dielectric breakdown occurs.

The germanium atom has the same electronic configuration as the carbon atom, and the crystal structure of germanium is similar to that of diamond. Therefore, two separate zones are also formed in crystalline germanium. The band gap of germanium (0.66 eV) is much smaller than that of diamond, and already at room temperature a small number of electrons (about one in 10 to the 21st power) are "thrown" from the valence band into the conduction band due to thermal motion. This is enough for germanium to have a noticeable electrical conductivity: it is 10 times higher than that of diamond, although 10 times less than that of typical metals. Germanium and a number of other substances with a not very wide (< 3 эВ) запрещенной зоной (Si, GaAs, PbS) называют полупроводниками.

A semiconductor is also gray tin - a crystalline modification that is stable at temperatures below 14 ° C. The structure of gray tin is also similar to that of diamond. However, the band gap in this substance is so small (about 0.1 eV) that even at room temperature a rather large number of electrons pass from the valence band to the conduction band. Therefore, the electrical conductivity of gray tin is only 15 times lower than that of metallic white tin, a modification that is stable at room and higher temperatures.

On the example of gray and white tin, we are faced with such an important circumstance that the concepts of "metal elements" and "substances with metallic properties" are not identical. Metal elements can form simple semiconductor substances (tin gray), and non-metal elements can form substances with a metallic type of chemical bond. For example, carbon and arsenic are non-metal elements, but the simple substances graphite and gray arsenic they form have metallic-type electrical conductivity.

////Ionic crystals

Now let's consider the structure of energy bands in crystals formed from atoms that differ greatly in electronegativity. Since the orbital energies of the initial atoms differ greatly, when their AOs overlap, two separate zones arise. The low energy zone (valence) mainly consists of orbitals of more electronegative atoms, and the high energy zone (conduction band) consists of orbitals of less electronegative atoms.

In crystalline NaCl, the valence band is completely occupied, while the conduction band is free. The band gap in sodium chloride is quite large, about 7 eV. Therefore, crystalline sodium chloride, like most other ionic substances, is a dielectric at room temperature.

Melts of ionic substances (for example, NaCl) conduct electricity. However, the electrical conductivity in these cases is due to the mobility of ions, not electrons.

///Structures of covalent, ionic and metallic crystals. Covalent, metallic and ionic radii of atoms

Crystal structures of substances with different types of chemical bonds are formed in accordance with different principles.

In an atomic covalent crystal, the number of bonds formed by each of the atoms is usually equal to the number of its valence orbitals. For example, in a diamond crystal, each carbon atom is in a tetrahedral valence environment, as in saturated hydrocarbon molecules. In crystalline SiO2 quartz, each silicon atom forms four bonds with oxygen atoms, and each oxygen atom forms two bonds with silicon atoms.

In ionic substances, each anion tends to have as many cations as possible in its environment, and each cation seeks to have as many anions as possible. Therefore, in ionic crystals, the number of neighboring counterions always significantly exceeds the valency or oxidation state of the corresponding atom. For example, in a NaCl crystal, each Cl- anion is surrounded by six Na+ cations, and each sodium cation is surrounded by six chloride ions.

Finally, in metallic crystals, the valence electrons are so weakly bound to the atoms that the structure of the metal is often represented as a collection of cations surrounded by a "gas" of nearly free electrons. Therefore, in metals, atoms are usually arranged in such a way that, with a minimum volume of a crystal, the distances between atoms are greatest. In other words, it is more profitable for an atom in such a crystal to have many distant neighbors than a few close ones. For example, in the crystal structure of a-iron (a modification stable at temperatures below 769 ° C), each atom has eight nearest neighbors at a distance of 248 pm and six more distant ones, the distance to which is 287 pm. Under standard conditions, all alkali metals, barium, chromium, molybdenum, vanadium and a number of other metals have the same crystal structure.

Naturally, the distances between the same atoms in crystals with different bond types have different values. For example, in a gray tin covalent crystal, the Sn-Sn bond length is 280 pm, while in a white tin metal crystal, the shortest interatomic distance is 302 pm. Therefore, to predict the distances between atoms, atomic radii of various types are used - covalent, ionic and metallic. These radii are calculated values ​​determined from already known interatomic distances.

The covalent radius of an atom is taken to be half the length of a single bond between identical atoms. For example, the covalent radius of the hydrogen atom is considered to be half the H-H distance in the H2 molecule (37 pm), and the covalent radius of the carbon atom is half the C-C distance in the diamond crystal (77 pm). The C–H bond length calculated using these values ​​is 114 pm, which is in good agreement with the experimental value (109 pm in a CH4 molecule). Multiple bonds are shorter than single bonds, therefore, when calculating their lengths, either special corrections are introduced or special values ​​of covalent radii are used. The metal radius is also defined as half of the shortest internuclear distance in a metal crystal. The metallic radii of atoms are always larger than the covalent ones.

Ionic radii are found in a more complicated way. When an electron is attached to a neutral atom, the interelectronic repulsion in its valence shell increases, so the radius of the anion is greater than the covalent radius of the neutral atom. On the contrary, the size of the cation that has lost electrons is smaller than the size of the original atom. Therefore, it is believed that during the formation of an ionic crystal, large anions are packed close to each other, and the voids remaining between them are filled with cations. Accordingly, half of the shortest interanion distance is taken as the anion radius, and the difference between the shortest anion-cation distance and the anion radius is taken as the cation radius.

47. Complexes. Complex compounds - compounds with a number of characteristics:

1) In complexes, one can always distinguish the central atom and the atoms of the environment. The sum of their charges adds up the charge of the complex (positive 2+, negative 3-, or neutral).

2) Complex compounds are formed as a result of the combination of ordinary ions and molecules with each other, and polyatomic ions or molecules are part of the complex particles as a whole, with the preservation of all chemical bonds.

3) The central atom has more chemical bonds than its oxidation state or valence prescribes to it. (in 4- one iron ion2 is surrounded by 6 cyanide ions)

4) Complex particles in crystalline substances and solutions exist as a whole.

Complex compounds in nature: cryolite Na3AlF6 contains ion 3-, heme, chlorophyll, vitamin B-12, also complexes. Complexes are many enzymes. The complexes are used as catalysts, pigments, used to extract metals from ores, and to separate mixtures. (complexes do not include double salts !!)

Basic concepts:

Complexing agent - the central atom or ion in complex particles (usually a metal ion or atom, although there is also a non-metal (2-, -, -.))

Ligands - neutral ions or molecules associated with the complexing agent located around it (form its coordination environment). These are molecules or ions that can be donors of electron pairs (the atom that gives it is a donor atom). Ligands are monodentate (form one coordination bond) and polydentate (several, since they have several donor atoms)

Coordination number - the number of bonds that form a complexing agent with ligands (most often 6.4.2).

Formation of complexes in solutions and their stability.

Complex compounds can be obtained in various ways (for example, by reacting an anhydrous Mg(ClO4)2 salt with ammonia to give (ClO4)2). However, from a practical point of view, the most interesting is the formation of complexes in aqueous solutions containing complexing metal ions and ligands. The enthalpies of hydration of most cations are quite large, so water molecules are strongly bound to metal ions and can be considered as coordinated ligands. Such hydrated ions are called aqua complexes. Particles that form stronger bonds with metal ions than water molecules can displace them from the coordination environment of the complexing agent. The substitution process occurs in steps, each step is characterized by a corresponding equilibrium constant. For example:

1)2+ + NH3 = +H2O K=/=590

2) + + NH3= 2++ H2O K=/[ Ni(NH3)(H2O)52+ ]=170

3) + NH3= 2++ H2O K=/[ Ni(NH3)2(H2O)42+ ]=54

4) + NH3= 2++ H2O K=/[ Ni(NH3)3(H2O)32+ ]=16.6

5) + NH3= 2++ H2O K=/[ Ni(NH3)4(H2O)22+ ]=5.4

6) + NH3= 2++ H2O K=/[ Ni(NH3)5(H2O)2+ ]=1.12

The constants Ki are called stepwise complex formation constants (usually, the larger the number of the constant, the smaller its value).

The overall process constant)2+ +6 NH3=2++ H2O is denoted by the letter?(beta) and is called the stability constant of the complex. It is equal to the product of all step constants. But it can also be calculated according to the usual formula.

The stability constant can be written not only for the complex, but also for any stage of water replacement by other ligands.

Sometimes the tables give instability constants, which are the equilibrium constants of the reactions of destruction of complexes (replacement of ligands by water molecules); they are inverse with respect to the stability constants.

In order to determine the structure of a molecule by the Gillespie method, the following procedure is proposed.

1. Based on the formula of the molecule, the number of ligands n, with which the central atom forms a bond, is determined and the formula AX n E m is written indicating the value of n.

2. The total number of bonding and unshared electron pairs (n + m) is found by the formula:

(n + m) \u003d 1/2 (N c + N l - z) -,

where N c is the number of electrons of the central atom on its outer electron layer, N l is the number of electrons of the ligands involved in the formation of bonds with the central atom, is the number of -bonds in the molecule, z is the charge of the ion (in the case of determining the structure of the molecular anion).

3. The spatial arrangement of all electron pairs (bonding and unshared) is determined.

4. The number of unshared electron pairs m is found and the formula of the molecule AH n E m is specified (the value of m is indicated).

5. The geometry of the molecule is established.

Question 18) Effective charges of atoms in molecules. Dipole moment of bond, dipole moment of molecules. Dipole moment of a molecule and its structure by examples ….

EFFECTIVE CHARGE OF THE ATOM characterizes the difference between the number of electrons belonging to a given atom in a chemical. Comm., and the number of electrons free. atom. To estimate the effective charge of an atom, models are used in which the experimentally determined quantities are presented as f-tions of point non-polarizable charges localized on atoms; for example, the dipole moment of a diatomic molecule is considered as the product of the effective charge of the atom and the interatomic distance. Within the framework of such models, the effective charges of atoms can be calculated using optical data. or X-ray spectroscopy, NMR, etc. However, since the electron density in the chemical. conn. delocalized and there are no boundaries between atoms, it is impossible to describe decomp. connection characteristics one set of effective charges of atoms; the values ​​of this indicator determined by different experiments. methods may not match. The effective charges of atoms can also be determined on the basis of quantum chemistry. calculations.
The measure of bond polarity - its dipole moment () - is determined by the product

where q is the effective charge, l is the length of the dipole (the distance between two equal in magnitude and opposite in sign charges +q and –q).

The dipole moment is a vector quantity. The concepts of “bond dipole moment” and “dipole moment of a molecule” coincide only for diatomic molecules. The dipole moment of a complex molecule is equal to the vector sum of the dipole moments of all bonds. The dipole moment of a polyatomic molecule depends not only on the polarity of individual bonds in the molecule, but also on the geometric shape of the molecule.

For example, in a linear CO 2 molecule, each of the C–O bonds is polar, and the molecule as a whole is nonpolar
( (CO 2) \u003d 0), since the dipole moments of the bonds compensate each other (Fig. 8.1). In the corner H 2 O molecule, the bonds are located at an angle of 104.5 o and the vector sum of the dipole moments of the two bonds is expressed by the diagonal of the parallelogram (Fig. 8.1). If ¹ 0, then the molecule is polar.

What is important - the more symmetrical the molecule, the smaller its μ, for example, symmetrical molecules (CO 2; BCl 3; CCl 4; PCl 5; SF 6) are non-polar and have μ \u003d 0.

Question 19) Fundamentals of the molecular orbital method (MO LCAO). Explain paramagnetic properties ... and find the multiplicity of bonds in ... and ...

The main provisions of the MO method:

1) e are populated together in accordance with the Pauli principle: no more than 2 in 1 orbital e)

and Hund's rule: in the presence of an orbital with the same energy, these orbitals are first populated by 1 e

2) Each loosening e nullifies the action of the binding electron. If the gaps are greater or the same as the number of binders, then the molecule is not formed. Kr sv \u003d (number e (connection) - number e (open)) / 2

3) MMO allows you to determine the magnetic properties of a molecule: if a molecule has an unpaired electron, then it is a paramagnet, such a particle is pushed out of the magnetic field, a particle with an unpaired electron is called a radical, they are chemically active and poisonous. Molecules in which all e are paired are diamagnetic, they do not respond to a magnetic field.

Energy series MO

Question 20) The main provisions of the method of valence bonds in the description of chemical bonds in complex compounds. Look at the examples... and...

According to the MVS, during the formation of a CS, a d/a bond arises due to the lone electron pair of the ligands and free quantum cells of the complexing agent. In this case, the orbitals of the complexing agent undergo hybridization. In the case of sp, the hybrid is a regular triangle, sp3 is a tetrahedron, dsp2 is a square molecule, dsp3 is a trigonal bipyramid, and d2sp3 is an octahedron. The free e pairs possessed by the ligands fill the empty orbitals of the central ion. These orbitals are combined into hybrid combinations depending on the coordination number (c.h.).

Question 21) The main provisions of the theory of the crystal field in the description of the chemical bond in complex compounds. Look at the examples... and...

Basic provisions:

1. The bond between the c/o and ligands is considered as electrostatic.

2. Ligands are considered to be point ions or point dipoles, their electronic structure is ignored.

3. Ligands and c/o are considered to be rigidly fixed.

4. The electronic structure of the c/o is considered in detail.

Question 22) Equivalents in-in in exchange reactions (or in redox reactions). Equivalence factor, equivalent molar mass, equivalent molar volume. Give three examples. The law of equivalents.

Equivalent - a real or conditional particle of substance X, which in a given acid-base reaction or exchange reaction is equivalent to one hydrogen ion H + (one OH ion - or a unit charge), and in this redox reaction is equivalent to one electron. In other words, the equivalent is the part of the molecule per electron in a given OVR or per proton (one hydroxyl, unit charge) in a given exchange reaction.

The equivalence factor feq(X) is a number showing what proportion of a real or conditional particle of substance X is equivalent to one hydrogen ion or one electron in a given reaction, i.e. the proportion that is the equivalent of a molecule, ion, atom, or formula unit of a substance. This value varies from zero to one.

1 mol equiv. contains 6.02.10 23 equivalents, and its mass in grams will be molar mass equivalent : M equiv = f equiv.M.

Molar volume equivalent is the volume occupied by 1 mol eq. in-va under normal conditions.

The law of equivalents: substances react in amounts proportional to their equivalents. If n (equiv 1) mole equivalents of one substance is taken, then the same number of mole equivalents of another substance n (equiv 2) will be required in this reaction, i.e. (The number of equivalents of substances that reacted and formed as a result of the reaction are equal)

neqv(A)= neqv(B)= neqv(C)= neqv(D)

Example: O2(0)----+4e----2O(-2)

Feqv =1/4; Meq = M* feq = 34*1/4 = 8g/mol; Veq = Vm*feq = 22.4*1/4=5.6 l/mol*eq

Question 23) The law of equivalents. Various forms of recording the law (reactions in-in in solutions, reactions in-in in a gaseous state). What is normal concentration and how is it related to molar concentration?

Question 22

1 mol equiv. contains 6.02.10 23 equivalents, and its mass in grams will be the molar mass of the equivalent: M eq = f eq.M. The number of mole equivalents of each of the participants in the process can be found as follows: ; , where m A and m B are the masses in A and B. And therefore, another record of the law of equivalents has: “the number of mole equivalents of participants in this process is a constant value”: n equiv.A = n equiv.B = n equiv.C = … = const.

If the participants in the process are in solution, then the number of mole equivalents of each of them can be found by multiplying the normal concentration of the substance by the volume of its solution. As a result, for this particular case, the law of equivalents takes the form:

For chemical calculations involving gases, along with molar masses, the value of 22.4 liters (the volume of 1 mol of gas under normal conditions) is actively used. Enter in the same way: .

Normal concentration (normality) n solution shows how many mole equivalents of dissolved matter are contained in 1 liter of solution.

n r.in mol equiv \u003d CH * V dis-ra

Relationship between Sn and Sm: at feq< 1 Сн >Cm

Question 24) Classification of redox reactions. Translate 2 examples of reactions of each type (do not use the equations from task No. 5).

OVR classification:

Group I - reactions of interatomic and intermolecular oxidation, reduction - these are reactions in which the exchange of e occurs between various atoms, molecules or ions.

2KBr+ Cl2 àBr2+ 2KCl

2) Group II - disproportionation reactions. In disproportionation reactions, molecules or ions of the same substance react with each other as a reducing agent and an oxidizing agent.

for example: (N -3 H 4) 2 Cr +6 2 O 7 N 0 2 + Cr +3 2 O 3 + 4H 2 0.

Сl2 + 2NaOH at NaCl + NaClO+H2O

3Cl2+ 6NaOH à 5 NaCl + NaClO3 + 3H2O

3) Group III - reactions of intramolecular oxidation, reduction, in which the oxidizing agent and reducing agent are in the same molecule: 2KClO3 à 3O2 + 2KCl

4) Group IV - counter-disproportionation reactions. Reactions that take place between the same element, in a positive and negative oxidation state, form an intermediate. Art. oxidation

2H2S + SO2 → 3S + 2H2O

(N 3- H 4) 2 N 3+ O 2 \u003d N 2 0 + H 2 O

Question 25) Typical reducing agents in OVR. What are the products of their oxidation? Give examples. OVR classification.

Reducing agent A substance whose molecules or ions donate electrons. Typical reducing agents:

Me atoms, NeMe ions, positively charged. ions NeMe, Me, which donate electrons. Group 1: Fr Cs K Na Li Ba Sr Ca Mg

Group 2: Br (-) S (2-) Cl (-) I (-) Se (2-)

Group 3: SO3 NO2 SnO2

Group 4: Sn Fe Cr Mn metal cations in low oxidation states

Group 5: H2 C N2H2 CO SO2

Ammonia and ammonium compounds.

HCl(conc only) KBr NaI CuS(2-)

Products of their oxidation: if the element is a reducing agent, its oxidation number rises.

Fe(+2) à Fe(+3)

Nitrite to Nitrate

sulfite to sulfate

NH3à to N2 or HNO3(or salts)

S(2-) à S(0) SO4 (acid or salts)

HHal à Br2 I2 Cl2

HNO2 à usually in NO

Classification 25 question

Examples:

2KMnO4+ 10KBr (reducing agent) + 8H2SO4 à 2MnSO4 + 5Br2+ 6K2SO4 +8H2O

KMnO4 + 16HCl à 5Cl2+2 MnCl2 + 2KCl +8H2O

Question 26) Typical oxidizers in OVR. What are their recovery products? OVR classification. Give examples.

An oxidizing agent is a substance whose molecules or ions accept electrons. Typical oxidizers:

1) in-va, the molecules of which contain atoms of elements in higher positive oxidation states, for example: KMn +7 O 4, KBi +5 O 3, K 2 Cr 2 +6 O 7, Pb +4 O 2; chromates, dichromates, manganates

2) metal cations of a higher charge (higher oxidation state), for example: Fe +3; Au+3; Sn +4 ;

3) halogens and oxygen (at elevated temperatures).

H2O2, H2SO4(c), HNO3

OVR classification: question 24

Recovery Products: If an element is an oxidizing agent, its oxidation state decreases; Among simple substances, oxidizing properties are characteristic of typical non-metals (F 2, Cl 2, Br 2, I 2, O 2, O 3). Halogens, acting as oxidizing agents, acquire an oxidation state of -1, and the oxidizing properties weaken from fluorine to iodine. Oxygen, recovering, acquires an oxidation state of –2 (H 2 O or OH–).

Concentrated sulfuric acid exhibits oxidizing properties due to sulfur in the highest oxidation state of +6. Concentrated H 2 SO 4 in reactions with metals can be reduced to SO 2 , S or H 2 S. The composition of the reduction products is determined by the metal activity, acid concentration and temperature. At normal temperature, concentrated sulfuric acid does not react with metals such as iron, chromium and aluminum (passivation phenomenon), and when heated, H 2 SO 4 does not react with gold and platinum. Inactive metals, standing in a series of standard electrode potentials to the right of hydrogen, reduce concentrated sulfuric acid to SO 2. Active metals (Ca, Mg, Zn, etc.) reduce concentrated sulfuric acid to free sulfur or hydrogen sulfide.

The KMnO 4 reduction products depend on the pH of the medium in which the reactions take place:

In acid à ​​salt Mn(+2) and salt K+

In alkaline à K2MnO4

In neutral à MnO2 + KOH

Dichromate, potassium chromate K 2 Cr 2 O 7 in an acidic environment (H 2 SO 4) is reduced to Cr 2 (SO 4) 3. In alkaline medium in Cr(OH)3 or complexes.

HNO3 (p) is mainly in NO, and concentrated in NO2.

KClO4 NaBrO3 KClO2 to anoxic salts KCl NaBr

Halogens in Cl-Br-I- (acids or salts)

Examples:

2KMnO 4 (oxidizer) + 5Na 2 SO 3 + 3H 2 SO 4 = 5Na 2 SO 4 + 2MnSO 4 + K 2 SO 4 + 3H 2 O;

K 2 Cr 2 O 7 + 6FeSO 4 + 7H 2 SO 4 = Cr 2 (SO 4) 3 + 3Fe 2 (SO 4) 3 + K 2 SO 4 + 7H 2 O.

4Ca + 5H 2 SO 4 (conc) = 4CaSO 4 + H 2 S + 4H 2 O.

S + 2H 2 SO 4 (conc) = 3SO 2 + 2H 2 O

FeO + 4HNO 3 (conc) = Fe (NO 3) 3 + NO 2 + 2H 2 O;

C + 4HNO 3 (conc) = CO 2 + 4NO 2 + 2H 2 O;

Question 27) General information about complex compounds: complexing agent, ligands, coordination number, inner and outer spheres. Classification of complex compounds. Give examples.

Higher-order compounds formed from simpler stable chemical forms (molecules, ions, atoms) are called complex compounds.

Central ion (atom) in set. connection called. complexing agent. (ion of d- or f-element, less often p- or s-element).

The ions or molecules immediately surrounding the c / o, called ligands, form together with c/o internal(coordination) sphere(highlighted).

Ions (molecules) outside the inner sphere form outer sphere set connections.

The total number of ligands in ext. sphere is called coordination number.

Classification of complex compounds. Give examples: see question 28.

Question 28) Classification of complex compounds: according to the type of coordinated ligands, according to the charge of the complex ion, according to the classes of compounds. Nomenclature of complex compounds. Give examples.

Classification of complex compounds:

1) according to the type of coordinated ligands: +) aqua complexes, in which the ligand is yavl. water molecule. For example: Cl3,

+) amino complexes (ammonia): NH 3 ligands. For example: Cl, (OH) 2 .

+) acid complexes, in which the ligands are anions of acid residues. For example: K, K 2 .

+) hydroxo complexes (OH). Example: Na3

+) mixed type [ Cr(H2O)5Cl]Cl2* H2O

2) by the charge of the complex ion: +) cationic - complexes have cationic c / o. For example: Cl2

+) anionic - complexes have anionic c / o. For example:

+) neutral - k / o have no charge. For example:

3) by classes of compounds:

+) acid complexes. For example: H 2 H[AuCl4].

+) base complexes. For example: OH (OH)2.

+) salt complexes. For example: K 3 Cl.

Nomenclature s.s:

First we name the anion, then the cation, but indicating the number and type of ligand, the name is read from right to left, the names of cations and anions are written separately, the valence of the complexing agent is indicated.

In the case of an anion, the root of the Latin name of the element is taken, “at” (stannate, aurat, plumbat) is added to it. For example: K 3 - potassium hexahydroxoaluminate (III); Cl is dichlorotetramminechromium (III) chloride.

Question 29) Hess' law, the conditions for its implementation. Enthalpies of formation, combustion, atomization (definition).

Hess' law: The change in enthalpy (int. Energy) of a chemical reaction does not depend on the path of the process, but depends only on the type and state of the starting materials and reaction products.

1. The only kind of work is extension work (no useful work).
2. Pressure and temperature do not change (p, T = const) - isobaric-isothermal process.
3. Volume and temperature do not change (V,T= const) - isochoric-isothermal process.
4. The chemical reaction must proceed to the end and irreversibly.
5. Heat is calculated per 1 mole of a substance, i.e. it is necessary to take into account the stoichiometric coefficients in the reaction equation.
6. Thermal effects are determined under normal conditions, T = 25ºС and p = 1 atm, and the most stable modification is also used.

enthalpy of formation H about f,298 (or H about arr, 298) is the change in enthalpy in the process of formation of a given substance (usually 1 mol), which is in the standard state, from simple substances, also in the standard state, and simple substances are present in the most thermodynamically stable states at a given temperature.

Enthalpy of combustion - The standard enthalpy of combustion H o burn,298 is the enthalpy of combustion of a substance (usually 1 mol) in the standard state with the formation of CO 2 (g), H 2 O (l) and other substances, the composition of which must be specially indicated. All combustion products must also be in the standard state.

Enthalpy of atomization or ionization energy is the smallest energy required to remove an electron from a free atom in its lowest energy (ground) state to infinity.

Question 30) Hess' law. Consequences from the law of Hess. Under what conditions is this law enforced?

Hess' law: question 29

Consequences from the law of Hess:

Consequence 1. , where n i and nj

Consequence 2. , where n i and nj are the numbers of moles (coefficients in the equation).

1. The change in the enthalpy of reaction is equal to the sum of the enthalpies of formation of the reaction products minus the sum of the enthalpies of formation of the starting substances (summation is carried out taking into account stoichiometric coefficients).

2. The change in the enthalpy of the reaction is equal to the sum of the enthalpies of combustion of the starting materials minus the sum of the enthalpies of combustion of the reaction products (summation is carried out taking into account stoichiometric coefficients).

Execution condition question 29

Question 31) Standard thermodynamic characteristics. The concept of the standard state of individual liquid and crystalline substances, gases and solutions. Hess' law.

The branch of chemistry that studies the thermal effects of processes is thermochemistry.

Hess' law: question 29

An ideal solution is taken as the standard state of the solution, the activity of which dissolved is = 1, and the enthalpy = enthalpies in a really infinitely dilute solution.

Fugacity (volatility) is a quantity that is related to other thermodynamic characteristics of a real gas in the same way as pressure is related to it in the case of an ideal gas.

For the standard state of the gas in-va take the state of a hypothetically ideal gas, the volatility of which is equal to unity, and the enthalpy of a real gas at the same temperature and pressure tends to zero. Those. for std. comp. infinitely rarefied gas is accepted.

There was one problem - allotropic modifications. Which one should be taken as standard? They take the most stable form, except: white phosphorus is taken, and not the more stable red, because. he is more reactive; S (k. rhombus), not S (k. monoclinic); C (c. graphite), not C (c. diamond). If all participants in the process are found. to the standard. state, then the reaction is standard and is indicated by the upper right “zero”.

For liquid and crystalline individual substances, their state under normal pressure is taken as the standard state at each temperature.

Question 32) Enthalpy and Gibbs energy, their physical meaning, connection between them.

Enthalpy(heat content, Gibbs thermal function), thermodynamic potential characterizing the state of a thermodynamic system when entropy S and pressure p are chosen as the main independent variables. Designated H

The Gibbs energy is a thermodynamic function of the state of the system. Physical meaning: The Gibbs energy shows the spontaneity during a chemical reaction.

ΔG = Нт –TΔS

The physical meaning of enthalpy: the physical meaning of enthalpy: its change is the heat supplied to the system in an isobaric process (at constant pressure).

Let us analyze the equation G o T = H o T - T S o T. At low temperatures, T S o T is small. Therefore, the sign of G o T is determined mainly by the value of H o T (enthalpy factor). At high temperatures T S o T is a large value, the sign of D G o T is also determined by the entropy factor. Depending on the ratio of enthalpy (H o T) and entropy (T S o T) factors, there are four variants of processes.

1. If N o T< 0, S о Т >0, then G about T< 0 всегда (процесс может протекать самопроизвольно при любой температуре).

2. If H o T > 0, S o T< 0, то G о Т >0 always (the process does not proceed at any temperature).

3. If N o T< 0, S о Т < 0, то G о Т < 0 при Т < Н о / S о (процесс идет при низкой температуре за счет энтальпийного фактора).

4. If H o T > 0, S o T > 0, then G o T< 0 при Т >H o / S o (the process takes place at a high temperature due to the entropy factor).

Question 33) Gibbs energy as a thermodynamic state function. Definition and properties. Calculation of the standard Gibbs energy of the process from reference data. Criterion for spontaneous reactions.

G is the state function of the system, called the Gibbs energy. It is equal to the difference between the enthalpy and the product of entropy and temperature: G=H – T∙S The absolute value of G cannot be determined. ∆G=∆H – T∙∆S Calculate ∆G in the same way as ∆H, using standard Gibbs energies of formation. ΔG chemical reactions = ∑n i G i (products) − ∑n j G j (ref. in-in)

Holy functions:

1) unambiguous, finite, continuous function of the state of the system;

2) has the property of independence ΔG from the path of transition from the initial in-in to the products.

3) −A field ≥ G 2 − G 1 = ΔG.

The physical meaning of the Gibbs energy follows from the relationship: -Apol=U2+pV2-TS2-(U1-pV1-TS1) – the Gibbs energy in an equilibrium process, up to a sign, is equal to the useful work that the system can perform. In the case of non-equilibrium processes, the Gibbs energy will be (with the opposite sign) equal to the maximum possible useful work that the system can do.

: in systems that are at P, T = const, only processes accompanied by a decrease in the Gibbs energy can occur spontaneously
( G< 0). При достижении равновесия в системе G = 0.

Question 34) Criterion of spontaneous reactions, enthalpy and entropy factors of the process. Which reactions occur spontaneously in aqueous solutions?

Criterion of spontaneous flow of the process: question 34

ΔG = ΔH - TΔS. spontaneous flow of the process (ΔG< 0) возможно при:

If ∆H< 0 и ΔS >0, then always ΔG< 0 и реакция возможна при любой температуре.

If ∆H > 0 and ∆S< 0, то всегда ΔG >0, and a reaction with the absorption of heat and a decrease in entropy is impossible under any circumstances.

In other cases (ΔH< 0, ΔS < 0 и ΔH >0, ΔS > 0), the sign of ΔG depends on the relation between ΔH and TΔS. The reaction is possible if it is accompanied by a decrease in the isobaric potential; at room temperature, when the T value is small, the TΔS value is also small, and usually the enthalpy change is larger than TΔS. Therefore, most reactions occurring at room temperature are exothermic. The higher the temperature, the greater the TΔS, and even endothermic reactions become feasible.

The more negative ΔG, the simpler, under milder conditions, the process proceeds.

Enthalpy and entropy factors of the process:

If ΔH<0 отражает стремление к объединению частиц в более крупные агрегаты, то ΔS>0 reflects the desire for a random arrangement of particles, for their disaggregation. The transition of the system to a state with a minimum energy when ΔS=0, if ΔH=0, then the system spontaneously passes into the most disordered state. Each of these opposite tendencies, quantified by ΔН and ΔS, depends on the nature of the substance and the conditions of the process (t-ra, pressure, ratio between the reagents, etc.).

Product ТΔS (kJ/mol) yavl. entropy factor of the process, ΔН - enthalpy factor. In a state of equilibrium: ΔH = TΔS. This equation is yavl. equilibrium condition characterizes such a state of a given system when the rates of opposite processes occurring in it become equal. From this equation: the calculation of the change in entropy in an equilibrium process is possible from directly measured quantities. The ΔH of the phase transition can be determined experimentally using a calorimeter.

Reactions occurring spontaneously in aqueous solutions: For which the change in Gibbs energy is less than zero.

Question 35) chemical balance. True (stable) and apparent (kinetic) equilibrium, their signs. Give examples.

The equilibrium state is called the ionic state of the system, which does not change in time, and this invariance is not due to the occurrence of any external process. Equilibrium remains unchanged until external conditions change. Distinguish between true (stable) and apparent (kinetic) equilibrium.

True equilibrium remains unchanged not due to the absence of processes, but due to their flow simultaneously in two opposite directions at the same speed. True equilibrium has the following characteristics:

1. If there is no external influence, then the system is unchanged in time.

2. Any small external influence causes a change in the equilibrium of the system. If the external influence is removed, the system returns to its original state.

3. The state of the system will be the same regardless of which side it approaches equilibrium from.

The apparent equilibrium is also unchanged in time in the absence of external influence, however, the second and third features are not characteristic of it. An example of a system in apparent equilibrium is a supersaturated solution: it is enough for a mote to enter such a solution or shaking, and the release of excess dissolved matter from the solution begins.

When external conditions change, the equilibrium changes in accordance with the new conditions, or, as they say, "shifts."

Question 36) chemical balance. The Le Chatelier-Brown principle and the shift of equilibrium. Consider the example of the reaction ......

Chemical equilibrium question 35

The shift in equilibrium obeys a pattern called Le Chatelier's principle: “if a system that is in true equilibrium is acted upon from the outside, changing any of the parameters that determine the state of equilibrium, then the direction of the process that weakens the effect of the impact will increase in the system and the position of the system will shift in the same direction.”

Equilibrium shift:

1) An increase in the temperature of the equilibrium system enhances the course of the endothermic process, cooling - vice versa.

2) a change in pressure significantly affects only the equilibria of gas systems. An increase in pressure for them leads to a shift in equilibrium towards a smaller volume, a drop in pressure leads to a shift in the direction of a larger volume.

3) increase in the concentration of ref. c-c leads to a shift of equilibrium to the right (towards products).

Question 37) Chemical equilibrium constant. The ratio of the values ​​of K p and K c for gaseous equilibria. Communication and equilibrium constants.

The quantitative characteristic of chemical equilibrium is the constant of chemical equilibrium

2 SO2 (g) + O 2 (g) 2 SO 3 (g)

At the moment of equilibrium, the concentration does not change, and the partial pressure of gases also does not change.

Ks =

The partial pressure of a gas is the pressure that the gas would produce if it occupied the entire volume.

Relationship between Kc and Kp for homogeneous processes:

P1 = n1*R*T/V = C1*RT

Kp \u003d Kc (RT) ^ Δn

Δn is the difference between the coefficients in the formulas of substances in the right and left parts in the chemical equation

The chemical equilibrium constant is related to the change in the Gibbs energy by the equation:

G T o = - RTlnK .

The lower the Gibbs energy, the higher the equilibrium constant

K= 1 ΔG=0 chemical equilibrium

K<1 ΔG>0 (equilibrium shifts to the left) K>1 ΔG<0 (равновесие сместится вправо)

Question 38) Equilibrium of dissociation of associated (weak) electrolytes by example.... Degree of dissociation, dissociation constant. Ostwald's dilution law.

Degree of dissociation a is the ratio of the number of molecules that have decayed into ions N¢ to the total number of dissolved molecules N:

The degree of dissociation is expressed as a percentage or fractions of a unit. If a = 0, then there is no dissociation and the substance is not an electrolyte. If a = 1, then the electrolyte completely decomposes into ions.

Weak electrolytes in aqueous solutions include acids: carbonic, sulphurous, hydrogen sulfide, sulfuric (in the second stage), orthophosphoric, all carboxylic acids; bases: hydroxides of magnesium, beryllium, aluminum, ammonium, all hydroxides of d-elements.

The dissociation constant - the equilibrium constant in the process of dissociation is the same.

A m B n(k) = mA n+ + nA m-

K diss =

The dissociation constant characterizes the ability of the electrolyte to dissociate into ions. The larger the dissociation constant, the more ions in a weak electrolyte solution. For example, in a solution of nitrous acid HNO 2 there are more H + ions than in a solution of hydrocyanic acid HCN, since K (HNO 2) \u003d 4.6 10 - 4, and K (HCN) \u003d 4.9 10 - 10.

For weak I-I electrolytes (HCN, HNO 2, CH 3 COOH), the value of the dissociation constant K d is related to the degree of dissociation and electrolyte concentration c Ostwald equation:

For practical calculations, provided that<<1 используется приближенное уравнение

Question 39) Principles of constructing the scale of standard thermodynamic functions of ion formation in aqueous solutions. How to determine the standard enthalpy of formation…. in aqueous solution.

An infinitely dilute solution is a solution in which for each molecule of the p-th matter there is an infinitely large number of moles of the p-la. Its characteristics:

1) all electrolytes in it are completely dissociated.

2) interaction between ions is completely absent.

Any property of an individual ion, for example thermodynamic, cannot be objectively determined. In such situations, they resort to the construction of a scale of relative values, in which the value of the property under consideration for one of the systems is postulated, and the value of the property of other systems is counted from the accepted value. In particular, for aqueous solutions of electrolytes, the scale of thermodynamic func. arr. ions is built on the basis of the following assumptions:

ΔH 0 (H + ∞ solution) = 0

S 0 (H + ∞ solution) = 0

ΔG 0 (H + ∞ solution) = 0

Based on this, we get: )

We found ΔNobr0Cl- (solution, st.c), then we find ΔNobr0NaCl (solution, st.c), and along it - ΔNobr0Na+ (solution, st.c), etc.

We obtain the scale of enthalpies of formation of ions.

And how did you find ΔNobr0НCl (solution, st.c)? ½ H2(g) + ½ Cl2(g) + ∞H2O(g)=HCl (½ H2(g) + ½ Cl2(g))= ΔHobr0HCl(g) ----->HCl(g)--- -(+ ∞Н2О(l)= ΔНsolv0НCl(g))---->HCl Obtained: ΔНоbr0НCl(solution, st.c)= ΔНоbr0НCl(g)+ ΔНsolv0НCl(g)

How did you find the last one? ΔNobr0НCl (solution, HCl*nH2O)= ΔNobr0НCl(g)+ ΔНsolv0НCl(g) to form a solution of the composition HCl*nH2O.
In a similar way, the standard Gibbs energies of ion formation and the standard entropies of ions were found. Although entropies in-in can only be positive, ion entropy can be negative because they are relative values.

Question 40) pH and pOH scales. Calculation of the pH of solutions of non-associated electrolytes using examples ....

Equilibrium of the water dissociation process

H 2 O H + + OH -

is described by the constant K w , which is called the “ionic product of water”. The ionic product of water is:

K w \u003d [H + ] [OH - ].

For practical purposes, it is convenient to use not the concentration of hydrogen ions, but its hydrogen index - the negative decimal logarithm - pH. The pH value is:

pH - an indicator of the acidity or alkalinity of the environment

pH = - lg.

H + - activity of hydrogen ions

1) In a neutral environment pH = - lg10 -7 = 7

2) In an acidic pH environment< 7

3) In an alkaline environment, pH > 7

However, for practical purposes, when calculating the pH of dilute solutions, the equation is usually used

pH + pOH = 14,

where pOH = - lg [O H-].

4) However, in a strongly alkaline environment, the pH may be slightly higher than 14, and in a very acidic environment, it may take negative values.

Question 41) Equilibrium of water dissociation. Ionic product of water. pH and pOH scales.

Pure water, although poor (compared to electrolyte solutions), can conduct electricity. This is due to the ability of a water molecule to disintegrate (dissociate) into two ions, which are the conductors of electric current in pure water:

H 2 O ↔ H + + OH -

The dissociation is reversible, that is, the H + and OH - ions can again form a water molecule. Eventually comes dynamic an equilibrium in which the number of decayed molecules is equal to the number formed from H + and OH - ions.

Ionic product of water:

1) Kd =

2) Kd H2O can be found from thermodynamic. functions:

ΔG 0 diss = -RT*lnK diss (use tabular values)

3)α<<1 в растворе const

V h 2 o \u003d 1l m h 2 o \u003d 1000 g

equals =1000(g/l)/18(g/mol)=55.6 mol/l. = const

Kdiss* = const = * = Kw

* = Kw is the ionic product of water

Kw \u003d Kdiss * \u003d 1.8 * 10 -16 * 55.56 \u003d 10 -14

* = Kw = 10 -14

pH and pOH scale question 40

Question 42) Equilibrium of dissociation of complex compounds. Stability constant and instability constant. Reactions of formation of complex compounds. Give examples of obtaining hydroxocomplex, aminocomplex and acidocomplex.

Reactions for the formation of complex compounds: complex compounds are formed and exist in solutions with a relatively large excess of ligands. Usually it is taken several times more than the amount required in accordance with the stoichiometric ratio. As a result, the dissociation of the complex compound is suppressed and it is stabilized.

Different bond strengths in the inner and outer spheres of the complex compound lead to a difference in the nature of the dissociation of these parts of the molecule. In the outer sphere in aqueous solutions, all complex compounds are strong electrolytes, while dissociation in the inner sphere occurs to a small extent.

K 2 2K + + [ Zn(CN) 4 ] 2- ; 2- Zn 2+ + 4CN - .

The equilibrium constant for the last process (dissociation of a complex ion) is called the instability constant: K equal = K nest = .

The equilibrium constant of the reverse process: Zn 2+ + 4CN - \u003d 2- naz. stability constant: K equal. = To set = .

The more K mouth. (less K nest.), the stronger the complex compound, the weaker it dissociates. It is clear that the work To mouth. and K nest. equal to one.

Examples:

1) Hydroxo complexes- complex compounds containing hydroxide ions OH - as ligands. Hydroxo complexes are formed in the reactions protolysis from aqua complexes:

3+ + H 2 O 2+ + H 3 O +

or when dissolved amphoteric hydroxides in aqueous solutions of alkali metal hydroxides:

Zn (OH) 2 + 2 OH - \u003d 2 -

2) This group of complex compounds can be divided into two parts: complexes with oxygen-containing ligands and complexes with anoxic(predominantly halide or pseudohalide) ligands. For example, acid complexes with oxygen-containing ligands include the dithiosulfate argentate(I) ion, which is obtained by the exchange reaction:

Ag + + 2 SO 3 S 2 - \u003d 3 -

and hexanitrocobaltate (III) ion, which precipitates as small yellow crystals of potassium salt when mixing solutions containing cobalt (II) chloride, potassium nitrite and acetic acid:

CoCl 2 + 7 KNO 2 + 2 CH 3 COOH =
\u003d K 3 ¯ + NO + 2 KCl + 2 CH 3 COOK + H 2 O

3) Ammonia complexes are usually obtained by reacting metal salts or hydroxides with ammonia in water or non-aqueous solutions, or processing the same salts in the crystalline state gaseous ammonia:

AgCl (t) + 2 NH 3. H 2 O \u003d Cl + 2 H 2 O

Cu (OH) 2 (t) + 4 NH 3. H 2 O \u003d (OH) 2 + 4 H 2 O

NiSO4 + 6NH3. H 2 O \u003d SO 4 + 6 H 2 O

CoCl 2 + 6 NH 3 (g) = Cl 2

In cases where the ammonia complex is unstable in an aqueous solution, it can be obtained in liquid ammonia:

AlCl 3 (s) + 6 NH 3 (l) \u003d Cl 3 (s)

Question 43) Buffer solutions and their properties. Calculation of the pH of the buffer solution of the composition ....

Buffer solutions are solutions that have a constant pH value, not dependent on dilution and small additions of strong acids and alkalis.

1) They can be: mixtures of a weak acid and its salt

CH3COOH + CH3COONa pH<7

2) Weak base and its salt

NH4OH+NH4Cl pH<7

Example: HCN + KCN

Kdiss HCN = * /

kdiss * /

In the presence of a strong electrolyte, KCN diss. HCN is suppressed according to the Le Chatelier Brownie principle. Then the concentration of HCN can be equated to the concentration of acid outcome.

Equals ̴̴ C HCN ref = C acid

Equal ̴ C KCN ref = C salt

Kdiss * (Acids/Salts)

Kdiss * (With base/With salt)

Acids / C salts \u003d const when diluting the solution

H+ + CN- à HCN in acid medium

OH- + H+ à H2O in alkaline medium

Virtually no effect on pH

As seen:

1) when diluted with water, it leads to the same decrease With to-you and With salt, and the ratio With to-you / With salt will not change and the pH will remain the same.

2) add a few drops of HCl to the buffer solution, while part of the salt will turn into to-that; as a result With to-you increases slightly and With salt will decrease, and the ratio With to-you / With salt

3) A similar thing will happen when a few drops of NaOH solution are poured into the buffer mixture: With salt increases With to-you will decrease a little, and the ratio With to-you / With salt and, accordingly, the pH of the buffer solution will change little.

Question 44) Equilibrium of dissolution and dissociation of a sparingly soluble electrolyte. Solubility product. The relationship between PR and solubility (for example ... ..).

AmBn(k)↔ AmBn↔ mA(n+) + nB(m-)- sat. sol.
intermediate solution
AmBn(k) ↔ mA(n+) + nB(m-)

K equals = m * n /

PR = solubility product (PR).

Since = const, then

K equals * = m * n = PR

PR = m * n

Thus, solubility product (SP) is the equilibrium constant of dissolution and dissociation of a sparingly soluble electrolyte. It is numerically equal to the product of concentrations (activities) of ions in powers of stoichiometric coefficients in a saturated aqueous solution of a given sparingly soluble electrolyte. Let be solubility (concentration of a saturated solution at a given temperature) electrolyte is R mol/l. Then:

PR \u003d (mp) m (np) n \u003d m m * n n * p m + n

From here we find the relationship between PR and solubility: .

If a PC > PR- precipitate will fall PC< ПР − the precipitate will dissolve, PC = PR- equilibrium will be established. (

Question 45) Conditions for precipitation and dissolution of sparingly soluble electrolytes. Connection of PR with solubility by example ....

If a PC > PR- precipitate will fall PC< ПР − the precipitate will dissolve, PC = PR- equilibrium will be established. ( PC = n. m . PC is the product of concentrations).

Question 46) The product of solubility as the equilibrium constant of dissolution and dissociation of a poorly soluble compound. Connection of PR with solubility by example ....

Question 44-45

Question 47) Complete (irreversible) hydrolysis. Mutual enhancement of hydrolysis (co-hydrolysis). Give examples.

Hydrolysis- exchange reaction of the interaction of a solute (for example, salt) with water. Hydrolysis occurs when salt ions are able to form slightly dissociated electrolytes with H + and OH - water ions.

The hydrolysis of some salts formed by weak bases and weak acids proceeds irreversibly.

Irreversibly hydrolyzed, for example, aluminum sulfide:

Al 2 S 3 + 6H 2 O 2Al (OH) 3 + 3H 2 S.

Hydrolysis proceeds irreversibly if a salt formed by a heavy metal and a salt formed by a weak volatile acid are simultaneously introduced into the solution, for example,

2AI CI3+3Na2 S+ H 2 O \u003d Al 2 S 3 + 6NaCI

CH3COONH4 = CH3COO + NH4

SbCl 3 + H 2 O SbOCl + 2HCl.

Mutual enhancement of hydrolysis. Assume that equilibria have been established in different vessels:

CO 3 2– + H 2 O HCO 3 – + OH –

Al 3+ + H 2 O AlOH 2+ + H +

Both salts are slightly hydrolyzed, but if the solutions are mixed, then the binding of H + and OH - ions occurs. In accordance with the principle of Le - Chatelier, both equilibria are shifted to the right, and hydrolysis proceeds completely:

2 AlCl 3 + 3 Na 2 CO 3 + 3 H 2 O \u003d 2 Al (OH) 3 + 3 CO 2 + 6 NaCl

This is called mutual hydrolysis enhancement.

2FeCl 3 + 3Na 2 CO 3 + 3H 2 O \u003d 2Fe (OH) 3 + 6NaCl + 3CO 2
2Fe 3+ + 3CO 3 2- + 3H 2 O \u003d 2Fe (OH) 3 + 3CO 2
Al 2 (SO 4) 3 + 3Na 2 CO 3 + 3H 2 O \u003d 2Al (OH) 3 + 3Na 2 SO 4 + 3CO 2
2Al 3+ + 3CO 3 2- + 3H 2 O \u003d 2Al (OH) 3 + 3CO 2
Cr 2 (SO 4) 3 + 3Na 2 S + 6H 2 O \u003d 2Cr (OH) 3 + 3Na 2 SO 4 + 3H 2 S
2Cr 3+ + 3S 2- + 6H 2 O \u003d 2Cr (OH) 3 + 3H 2 S

2CuCl 2 + 2Na 2 CO 3 + H 2 O → (CuOH) 2 CO 3 + CO 2 + 4NaCl

Question 48) Complete (irreversible) hydrolysis. Give two examples. Joint hydrolysis of two salts with the formation of a) metal hydroxide (+3), b) basic metal carbonate (+2).

a) 2NaCl 3 (cr) + 3Na 2 CO 3 (cr) + 3H 2 O (l) \u003d 2Al (OH) 3 (tv) + 3CO 2 (g) + 6NaCl (p - p)

b) 2CuCl 2 + 2Na 2 CO 3 + H 2 O = (CuOH) 2 CO 3 ↓ + CO 2 + 4NaCl

Question 49) Hydrolysis of salts simultaneously by cation and anion (reversible hydrolysis). Calculation of the hydrolysis constant, degree of hydrolysis and pH of solutions of such salts using the example of ...

Salts formed by a strong base and a weak acid, for example, CH 3 COONa, Na 2 CO 3, Na 2 S, KCN, are hydrolyzed by the anion:

CH 3 COOHa + HOH CH 3 COOH + NaOH (pH > 7).

AB + H2O = AOH + HB

Kravn \u003d [AON] * / *

Kequi[H2O] = K hydr = const

Khydr = * /

Kt + + A - + H 2 O \u003d KtOH + HA

The hydrolysis constant has the form:

Let the total concentration of the salt hydrolyzed both cation and anion simultaneously be equal to with mol/l, the degree of hydrolysis is h. Then:

From here: . The value of the hydrolysis constant does not depend on the concentrations of hydrolyzed salts, or, in other words, the degree of hydrolysis of a salt undergoing hydrolysis by cation and anion simultaneously will be the same at any concentration of salt in solution.

Question 51) Hydrolysis of salts by anion. Methods for suppressing hydrolysis. Calculation of the hydrolysis constant, degree of hydrolysis and pH of solutions of salts hydrolyzed by the anion using the example of ....

Salt is formed by a strong base and a weak acid (NaCN K2SO3 Na3PO4)

NaCN = Na+ + CN-

CN- + HOH = HCN + OH-

Ways to suppress hydrolysis:

1) solution cooling;

2) adding acid to the solution to suppress cation hydrolysis, adding alkali to the solution to suppress anion hydrolysis.

General view of anion hydrolysis:

Then the hydrolysis constant:

The product gives us the constant of the ionic product of water - K W, and the fraction is the acid dissociation constant. Thus we get:

Because \u003d and \u003d Csoli

Because αhydr<1 мала

kg==

We find the pH of the solution

The degree of hydrolysis () is equal to the ratio of the number of hydrolyzed molecules to the total number of dissolved molecules. In the case of anion hydrolysis, it is not large. The weaker the acid (or base), the greater the degree of hydrolysis.

A simple and convenient method for predicting the geometry of molecules is the repulsion model of localized electron pairs or the R. D. Gillespie method, which is based on the VS method. The initial data for this method are: the number of other atoms associated with the central atom; valence possibilities of all bound atoms; the number of electrons in the outer layer of the central atom.

The main provisions of the Gillespie method are as follows.

Each electron pair, both forming a bond and unshared, occupies a certain place in space (localized electron pair). Due to repulsion, electron pairs are located in such a way as to be as far apart as possible from each other, and unshared electron pairs occupy a larger volume than shared ones. Double and triple bonds are considered as single bonds, although they occupy a larger volume.

The procedure for working on the Gillespie method is approximately as follows. Let us designate an atom of the structure as A, any other atom connected with it - by the letter B, etc.; lone electron pair - E, the total number of chemical bond partners of the central atom - P, and the number of unshared electron pairs - t. Then the molecule considered in the simplest case with respect to the central atom will have the form AB w E ;// . Usually, the most polyvalent atom is chosen as the central atom. Complex, bulky molecules within the framework of the Gillespie method are considered in parts. As a result of summation P and t according to the method proposed above, the initial model of the geometry of a molecule or ion is determined, and then the actual geometry of the particle.

The spatial configuration of molecules depending on the number of electron pairs is given in Table. 3.1.

Table 3.1

Molecular configuration according to the Gillespie method

The end of the table. 3.1

Number of electron pairs	Location electronic	molecules	Geometry molecules
	tetrahedral-		Tetrahedron
			Trigonal
			pyramid
		av 2 e 2
	trigonal		Trigonal
	bipyramidal-		bipyramid
		av 4 e,	Disphenoid
			T-shaped
			Linear
	Octahedral-
			Square
			pyramid
	Pentagonal	Av 7	Pentagonal
	bipyrams-		bipyramid
		ab 6 e,	One-cap
			octahedron

Let us demonstrate the capabilities of the Gillespie method using the example of several molecules.

Ammonia (NH 3): the central atom is nitrogen, t =(5 - 3*1)/2 == 1; hence the type of the molecule is AB 3 E t , the initial model is a tetrahedron, the molecule is a trigonal pyramid, the angle between the H - N - H bonds is less than tetrahedral (109 ° 28 ") due to the lone pair of electrons occupying a larger volume, and is about 107 .3°.

Water (H 2 0): the central atom is oxygen, t = (6 --2 - 1) / 2 = 2; this implies the type of molecule - AB 9 E 9, the initial model of which is a tetrahedron, the molecule is angular, the bond angle between the chemical bonds H - O - H is even smaller due to the presence of two lone pairs of electrons on the oxygen atom and is equal to 104.5 °.

Tin chloride (SnCl 9): the central atom is tin, t== (4-2 -1) / 2 = 1; the type of the molecule is AB 2 E G the initial model is a regular triangle, the molecule is angular, the bond angle between the chemical bonds Cl - Sn - Cl is 120°.

Carbon monoxide (1U) (C0 9): the central atom is carbon, t =(4 - 2 2) / 2 = 0; the type of the molecule is AB 2, the molecule is linear, the angle between the bonds O \u003d C \u003d O is 180 °.

It should be noted that the Gillespie method has significant limitations. Its main disadvantages:

• not applicable to most connections d- and 5-elements;
• inapplicability to compounds with a significant ionicity of the chemical bond. Thus, the Li 2 0 molecule is linear, but, as belonging to the AB 2 E 2 type, it must be angular;
• the impossibility of predicting the "inertness" (lack of direction, stereoactivity) of an unshared electron pair. Thus, the Pblg^SbBr^" and TeBr 6 2_ ions belong to the AB 6 Ej type, but in reality they turn out to be regular octahedral structures. low electronegativity.

Based on electrostatic concepts Gillespie proposed a more general theory of the spatial structure of molecules. Basic provisions:

• 1. The geometry of a molecule or ion is determined only by the number of electron pairs at the valence level of the central atom.
• 2. Electron pairs occupy such an arrangement on the valence shell of an atom when they are as far apart as possible, i.e., electron pairs behave as if they repel each other.
• 3. The region of space occupied by a non-bonding (lone) pair of electrons has large dimensions than the area occupied by the bonding electron pair.
• 4. The size of the region of space occupied by a bonding pair of electrons decreases with an increase in the electronegativity of the ligand and with a decrease in the electronegativity of the central atom.
• 5. Two electron pairs of a double bond occupy a larger region of space than one electron pair of a single bond.

The designations that are used to describe the geometric configuration of molecules: A is a polyvalent atom; X - atoms associated with atom A;

n is the number of X atoms; E - lone pair of electrons; m is the number of unshared electron pairs.

Then the formula of the molecule according to Gillespie is written as follows: AX n E m .

The geometry of the molecule depends on the sum (n + m). The number n, which determines the number of X atoms directly attached to the A atom, is the same as its coordination number. Each electron pair is taken as a point charge. The central atom A is placed in the center of a sphere of a certain radius, which for the same type of attached X atoms is equal to the length of the A-X bond. Point electron pairs are located on the surface of the sphere.

Applying the rule of maximum removal of electron pairs on the sphere from each other, it is possible to derive the geometry of the simplest molecules and ions, gradually increasing the sum of shared and unshared pairs (Fig. 4 and Table 1). valence hybridization polarity covalent

It makes no sense to consider the AX molecule, since it will always be linear, regardless of the number of unshared electron pairs in atom A.

The AX 2 molecule is also always linear, since the maximum repulsion of two electron pairs will place them at the ends of the diameter of the conditional sphere.

Three bonding electron pairs, as far as possible from each other, form a regular triangle (AX 3 molecule). In this case, the angle X-A-X is 120 o. Molecules BF 3 , AlF 3 have such a structure. If one of the binding electron pairs is replaced by a lone pair of electrons, then the molecule will be described by the formula AX 2 E and have an angular structure, and, according to Gillespie's third rule, the X-A-X angle will become less than 120 o. An example of such a geometry is the SnF 2 molecule.

Rice. 4.

Four binding pairs of electrons will form a tetrahedron in space. According to Gillespie's theory, this type of molecule is AX 4 . The angle X-A-X will be 109 about 28?. Typical representatives of this type of molecules are CH 4 , CCl 4 , SnF 4 molecules. By successively reducing the number of bonding pairs of electrons and increasing the number of unshared electron pairs, for molecules of the AX 3 E type we obtain a trigonal-pyramidal structure (ammonia molecule NH 3), and for molecules of the AX 2 E 2 type - angular (water molecule H 2 O).

The coordination number "five" is realized in molecules of the AX 5 type. Examples of such molecules are phosphorus pentafluoride or pentachloride (PF 5 , PCl 5). Five halogen atoms in space occupy the vertices of a trigonal bipyramid. Three atoms are located in the equatorial plane, forming an isosceles triangle, and two - respectively above and below this plane. The distance A-X from the center of the molecule to one of the tops of the pyramid, called axial, is greater than the similar equatorial one.

The bond angle between bonds lying in the equatorial plane is 120°, and the bond angle between bonds lying in the axial plane is 180°. For molecules that are derivatives of a trigonal bipyramid, there are two alternative arrangement possibilities for unshared electron pairs. With an axial arrangement, it experiences repulsion from three nearby atoms, and in an equatorial location, from two. Therefore, the first unshared pairs of electrons always occupy the equatorial position as the most energetically favorable. An example is the sulfur tetrafluoride SF 4 molecule, which has the form of a swing or a disphenoid. In molecules of the AX 3 E 2 type, such as ClF 3 or ICl 3 , the second unshared electron pair is also located in the equatorial plane. Therefore, all four atoms are in the same plane, resembling the letter T in shape. Due to the fact that the lone pair of electrons occupies a region in space more size, there is a distortion of the corresponding bond angles in the direction of their decrease. The third lone pair of electrons, also occupying a position in the equatorial plane, turns the T-shaped molecule into a linear one. A representative of molecules of the AX 2 E 3 type is the xenon difluoride molecule XeF 2 .

The most favorable arrangement of six X atoms around the central A atom is octahedral. Molecules of the AX 6 type, such as SF 6 , have the shape of an octahedron. The first lone pair of electrons will occupy any of the vertices of the octahedron, turning it into a square pyramid. An example of an AX 5 E type molecule is IF 5 . For the second unshared electron pair, there are two possibilities for location: next to the first (cis-position) and opposite it (trans-position). The maximum repulsion of electron pairs forces them to take a trans position. As a result, molecules of the AX 4 E 2 type have the shape of a square, for example, XeF 4 .

Table 1.

Number of electron pairs	Coordination			Molecule Type	Molecule shape
	Linear				Linear
	Tetrahedron				Tetrahedron Trigonal bipyramid
	Trigonal bipyramid				Trigonal bipyramid Disphenoid T-shaped Linear
					Square bipyramid flat square

The geometry of a molecule is determined by the repulsion of electron pairs in the valence shell of the central atom.

The lone pair occupies the largest place in space.

A multiple bond is similar to a lone pair, but larger.

The size of the bonding electron pair among the elements of one subgroup decreases with increasing electronegativity of the atom.

Classification of electron pairs: ... there are bonding and lone

Classification of atoms in a molecule: a) Center M - an atom with the maximum valency (in the classical sense), with the maximum coordination number. b) Ligands X, each of which is associated with the center (we set the coordination number allocated to one ligand equal to one. This is a monodentate ligand. There are also polydentate ligands that capture several coordination sites near the center at once). The valency of even a monodentate ligand can be more than unity.

Incompleteness (flaw) of the LEP theory: Formulated only for molecules in spin-paired states. Its application to compounds of d-elements is ambiguous...

The geometric consequences of the theory of transmission lines based on the model of electron repulsion are determined using the simplest electrostatic approach.

The model underlying the theory of power transmission lines: repulsive identical electric charges (localized pairs) float on the surface of a sphere, in the center of which a central atom is placed. In this case, for a different number of pairs, the following orbital orientations are possible:

Two floating charges will be located at the poles of the sphere, and both charges and the center of the sphere lie on the same line,

Three floating charges will be located at the vertices of an equilateral triangle on the equatorial plane of the sphere, and the center of the sphere falls exactly in the center of the triangle,

Four are at the vertices of a tetrahedron inscribed in a sphere, and the center of the sphere is at the center of the tetrahedron,

Five floating charges can lead to two almost equivalent situations, namely:

a) they will be located at the vertices of a trigonal bipyramid, and the central atom will be located in its center;

b) they will be located at the vertices of a square pyramid inscribed in a sphere, and the center of the sphere coincides with its center of gravity.

Six floating charges will be located at the vertices of an octahedron inscribed in a sphere, and the center of the entire system will be in the center of the octahedron.

Each pair occupies a separate coordination place in space. Pairs are divided into: 1) - binding, 2) - unshared.

All electron pairs of a multiple bond occupy one common coordination site. In this sense, they are like one pair of larger volume.

There are also more complex spatial configurations, but they are already important for exotic, special, cases.

That's all the basic rules.

The spatial distribution of electron pairs in the shell of the central atom in all cases is in complete agreement with this simple model. This conclusion has been verified theoretically in detail.

The number of electron pairs in the shell of an atom is integer, it is limited, it is determined by the valence state of the atom.

Let us turn to examples illustrating the predictions of the transmission line theory.

Consider the compounds of different groups of the Periodic System.

1) Molecules of compounds of elements of the 2nd group have the simplest structure generated by the valence configuration of the central atom of the form MX 2 .

The center itself has 2 valence electrons, and 2 ligands provide 1 electron for?-bonding (“shared ownership” with the center). All these electrons, according to the rules of the power transmission line, should be assigned to the center. These are 2 valence pairs X. Their maximum repulsion generates a linear structure. These are the molecules BeH 2 , BeCl 2 , HgCl 2 , ... in the gas phase. All three atoms are located along the same axis.

2) For molecules of compounds of elements of the 3rd group, the configuration of the central atom contains 3 valence electrons, and from each ligand, 1 more valence electron enters the?-bonding. The shell of the central atom should be assigned only 6 valence electrons - 3 pairs. Their repulsion generates a molecular structure like a flat triangle. The general formula for such a system is MX 3 (complete structure). These are the molecules BF 3 (gas phase), AlCl 3 (gas phase).

3) For molecules of compounds of elements of the 4th group, the central atom contains 4 valence electrons. In the spin-paired state of the electron shell of the molecule, there can be 4 or 2 ligands. Depending on their number, 2 variants of the structure are possible.

3.1. In the tetravalent state of the center with four ligands in the field of the center there are only 8 binding electrons. The general formula of such a system is MX 4 (full structure). Tetrahedral structures of molecules such as СCl 4 , SiFl 4 , GeCl 4 correspond to four pairs.

3.2. In the divalent state of the center, the total number of electrons with monovalent ligands is 6, of which 2 pairs serve 2 α-bonds with X ligands, and 1 pair is idle - pair E. Such a structure can be described by the formula MX 2 E (fragment structure). To accommodate all pairs, 3 coordination places are required. This is a fragment of a triangular structure. 1 slot for a blank, and 2 slots for ligands. The basis of the structure is a flat triangle. The idle pair repels bonding pairs. The geometry of the molecule is angular, but the valence angle is less than 120 0 due to the larger size of the idler pair (the valence dummy needs more space in space, and it pushes the bonding "labor" pairs apart - it's ridiculous, but such is the law of nature!). Such are the molecules of SnCl 2 , PbCl 2 , GeCl 2 in the gas phase. Their structures are fragments of a “triangle”. That's what they call it - fragmentation.

4) There are several possibilities in the molecules of compounds of elements of the 5th group. The central atom has 5 valence electrons. If it is 5-valent, then 5 more come from 5 ligands. The resulting 5 pairs occupy 5 coordination places. Such a structure can be described by the formula MX 5 (complete structure). The geometry of the molecules of these compounds corresponds to a trigonal bipyramid. These are PF 5 , AsCl 5 , SbBr 5 .

Here, a qualitative geometric difference between the ligands arises. We will call two of them axial and three equatorial. We will introduce the corresponding indices for the binding pairs X a and X e , serving?-bonds with ligands (see the figure at the end of the text). Equatorial positions are preferred for accommodating both lone pairs and multiple bonds, and bulkier substituents.

If the central atom-element of the 5th group is only trivalent, then 1 electron enters the center shell from each ligand, in total, 8 electrons are assigned to the center shell - 4 pairs. Three bonding pairs (X) and one unshared E. Such a structure can be described by the formula MX 3 E (fragment structure). All pairs are oriented towards the vertices of the tetrahedron. One place is occupied by a “dummy” pair, pushing back connections and distorting the “correct” structure. Ligands are located only in three positions out of five. The basis of the geometry of the nuclear core is a trigonal pyramid, and from the point of view of the theory of power lines, it is a fragmentation tetrahedron. These are NH 3 , PCl 3 , PF 3 , AsCl 3 , SbBr 3 (and also, with the center from the 6th group, the isoelectronic ion H 3 O +).

(The coordinating place of the pacifier - the concept of the molecular structure is ignored - does not seem to count, although it is it that determines the structure).

7) Molecules of compounds of elements of the 6th group. The central atom has 6 valence electrons. The hexavalent center receives 6 more from 6 ligands. These 6 pairs require 6 focal points. Formula of the valence shell of the central atom MX 6 (full structure). The geometry of such molecules is octahedral. These are SF 6 , SeCl 6 ... .

If the central atom from the 6th group is only 4-valent, then the shell formula of the center is MX 4 E (fragmentation structure). It corresponds to the geometry of the bisphenoid. The center should be assigned 10 electrons - 5 pairs, and the main structure for them is a trigonal bipyramid with its 5 coordination sites. Ligands are located in 4 coordination sites, and, as always, they are crowded by the fifth pair - “dummy”. It occupies a more “comfortable” equatorial place. There are two more places left on the equator, but the valence angle between them is less than 120 0 - they will be replaced by an idle pair. The two remaining atoms occupy axial positions, and since they are also pushed away from the axis by the empty space, the bond angle between them is less than 180 0 . The bisphenoid resembles a yoke with two arms hanging across it. According to the theory of power lines, this is a fragment of a trigonal bipyramid. These are the structures of SF 4 , SeCl 4 , TeCl 4 , ...

9) Finally, a divalent atom of the 6th group is also possible. Two ligands will add two electrons, and there will be 8 electrons in the field of the center. Forming 4 pairs, they generate 4 coordination sites at the vertices of the tetrahedron. Center shell formula MX 2 E 2 (fragmentation structure). 2 pairs occupy 2 vertices of the tetrahedron, and 2 ligands are located in the 2 remaining vertices, and the bond angle is again reduced compared to the tetrahedral one. It is less than 104 0 28". These structures are angular. These are the molecules H 2 O, H 2 S, SCl 2, SeCl 2, TeCl 2, ...

10) Structures with 7 valence pairs near the center are diverse. Variants of the basic geometry vary significantly depending on the central atom or ion. There is no longer a single simple polyhedron suitable for constructing a complete structure based on a center from the 7th group. The only seemingly simple case of a complete structure is that the IF 7 molecule has a pentagonal bipyramid structure. But XeF 6 also has 7 pairs in the valence shell of the center, but already a different structure - a single-capped octahedron (see Fig.)

Fragment structures are clear and quite standard. Let's turn to them.

If 5 monovalent ligands fall into the field of an element of the 7th group, 6 valence pairs are obtained. 1 of them is single. Center configuration formula MX 5 E. The basis of the geometric structure is a fragment of an octahedron. In its “flawed” axial vertex, there is 1 idle pair, pushing the other 5 connecting pairs away from ideal positions. This results in the structure of a square pyramid, or rather a “square umbrella”, since the central atom lies below the plane of the four ligands.

These are ClFl 5 ; BrCl 5 ; BrF 5 ; ...

If 3 monovalent ligands fall into the field of an element of the 7th group, we arrive at 5 valence pairs. Two of them are single. Center configuration formula MX 3 E 2 . The basis of the geometric structure is a fragment of a trigonal bipyramid (or a double fragment of a bisphenoid). In its “flawed” peaks there are equatorial idle pairs, crowding 3 connecting ones. Two axial pairs are deviated from the common axis with the center in the same direction. It turns out the structure of the letter T with bond angles less than 90 0 . These are ClFl 3 ; BrCl 3 ; BrF 3 ...

11) Are no exception to the theory of power transmission lines of the connection of inert gases. It makes sense to consider fragmentation structures such as molecules XeF 2 , XeF 4 , ... their geometry is determined by the same rules.

Five pairs fall into the shell of the Xe atom in the XeF 2 molecule. The basis of the structure is a fragment of a trigonal bipyramid. 3 lone pairs occupy equatorial positions, located at three vertices in the plane of an equilateral triangle. Bonding pairs of XeF bonds are less bulky and occupy axial positions on the same axis as the center. The nuclear core is linear, but the true basis of geometry is the molecule - the trigonal bipyramid.

The shell of the Xe atom in the XeF 4 molecule contains 6 pairs. The basis of the structure is an octahedron. 2 lone pairs are located on the same axis with the center. Less bulky bonding pairs of XeF bonds occupy 4 positions in a common plane with the center. The molecule is square, but its power transmission line base is an octahedron.

The shell of the Xe atom in the XeF 6 molecule contains 7 pairs. The basis of this structure in the theory of transmission lines is the so-called one-capped octahedron. This is a less symmetrical structure, but, nevertheless, it retains a third-order symmetry axis passing through the axis of the orbital of the lone pair, as well as through the centers of two planes in which there are triangles formed by two triplets of atoms of the ligands F" and F".

12) There are connections with a large number of electronic pairs ...

Examples of these are given in Ronald Gillespie's excellent classic The Geometry of Molecules, as well as in the more recent monograph by R. Gillespie and I. Hargittai, The Valence Shell Electron Pair Repulsion Model and the Structure of Molecules.

Summary: Structures corresponding to a certain number of electron pairs in the valence shell of the central atom can be complete and fragmentary.

The geometry of the complete structure is a regular polyhedron. At all vertices of the polyhedron, the complete structure has ligands, their number is equal to the formal valence of the central atom, i.e. the group number of the central atom.

The geometry of the fragment structure is a fragment of a regular polyhedron. Lone pairs are located at one or more vertices of a theoretical polyhedron. The remaining vertices are occupied by ligands, their number is less than the group number of the central atom. The bond angles are smaller than in a regular polyhedron. Fragment structures are formed at a reduced valency of the central atom.

The following examples are for exercises.

Compounds with multiple bonds.

Others are treated similarly, including some structures that once seemed incredible. In some of them, the ligands have multiple bonds.

Molecules can be considered as examples.

SF3N; SF4O; SO2; IF5O; IO 3 - ; IF 2 O 2 - ; XeO 2 ; XeF 2 O 2 ;...

1) SF 3 N molecule.

The center has 6 own valence electrons, from the ligands 3 + 3 more. Only 6 pairs. The SN triple bond is served by 3 pairs that occupy 1 coordination site. The remaining 3 pairs are bonding and serve 3 single SF bonds occupying 3 coordination positions. There are 4 focal points in total. The basis of the structure is a tetrahedron. The triple bond occupies a larger volume and, repelling, pushes the SF bonds together, the F atoms are brought together, and the bond angles FSF are less than the tetrahedral value 109 0 28". The bond angle FSN, on the contrary, is greater than 109 0 28".

2) SF 4 O molecule.

The center has 6 valence electrons, from the ligands another 4 + 2. Only 6 pairs. Two of them belong to the double bond, occupying one common coordination site in space. The remaining 4 pairs are binding and occupy 4 coordination sites. There are 5 focal points in total. The basis of the structure is a trigonal bipyramid. The double bond =O has a larger volume and is located on the equator, repelling other bonds. All FSF bond angles are less than ideal. At the equator - less than 120 0, and the axial atoms are deviated from the axis in the direction of the other two equatorial atoms, and the bond angle between them is less than 180 0

3) SO 2 molecule.

The center has 6 valence electrons, from the ligands another 2 + 2. Only 5 pairs. There are 4 pairs of 2 double bonds in 2 coordination sites. One coordination place is occupied by a single pair. There are 3 focal points in total. The geometry of the molecule is angular. The rolled angle OSO is greater than the ideal value of 120 0 due to the mutual repulsion of the electrons of the S=O double bonds.

4) IF 5 O molecule.

The center has 7 valence electrons, from the ligands another 5 + 2. There are 7 pairs in total, 1 bond is double and 2 of its pairs occupy 1 coordination site. Repelling, it pushes out the rest of the ligands (binding pairs) occupying 5 coordination sites. There are 0 idle pairs. The basis of the structure is a slightly deformed octaer, in which 4 fluorine atoms are close to a single one, lying on a common axis with the center and the O atom. All bond angles FIF are less than 90 0 , and the bond angle FIO is greater than 90 0 .

5) Molecule IO 3 -.

The center has 7 valence electrons, 2 + 2 + 2 more from the ligands, and in the anion state there is one more excess electron. Only 7 pairs. 6 pairs are allocated for servicing 3 double bonds in three coordination sites. One pair remains single. As a result, 4 coordination sites are used to accommodate 3 double bonds and 1 lone pair. The basis of the structure is a fragmentation tetrahedron. The geometry of the nuclear core is a trigonal pyramid. Multiple bonds repel the lone pair from each other, and the bond angle OIO is greater than the ideal tetrahedral angle 109 0 28".

6) Molecule IF 2 O 2 -.

The center has 7 valence electrons, ligands have 2 + 4 more and plus 1 excess electron in the anion state. Only 7 pairs. 4 pairs are supplied to service 2 double bonds at 2 coordination sites, 1 unshared pair remains. As a result, to accommodate 4 ligands and the 1st lone pair, 5 coordination sites are needed. At the equator there are 2 I=O bonds and 1 lone pair. The two I-F bonds are tilted away from the F-I-F axis in the direction of the single lone pair, as repulsion from the two I=O double bonds prevails due to their larger volume.

7) XeO 2 molecule.

The center has 8 valence electrons, from the ligands another 2 + 2. Only 6 pairs. 4 pairs are supplied to service 2 double bonds at 2 coordination sites, 2 unshared remain. As a result, 4 coordination sites are required to accommodate the ligands and the 1st lone pair. The geometry of the molecule is a fragmentation tetrahedron with the O-Xe-O bond angle increased compared to the tetrahedral one, since the volume of multiple bonds is greater than lone pairs.

8) XeF 2 O 2 molecule.

The center has 8 valence electrons, from the ligands another 2 + 4. Only 7 pairs. To serve 4 coordination sites occupied by ligands, 2+4=6 pairs are allocated, leaving 1 idle pair. As a result, only 5 coordination sites are needed to accommodate the ligands and the 1st lone pair. The basis of the structure is a bisphenoid (a fragment of a trigonal bipyramid). At the equator - O atoms and 1 idle pair. F atoms can occupy only axial positions, but at the same time they are deviated from the axis of the F-Xe-F bonds in the direction of the idler pair, repelling more from the bulkier Xe=O bonds.

So it is possible to discuss the structure of many molecules, and molecular ions. For molecular ions, additional electrons should be added (for anions) or subtracted (for cations) electrons, forming the necessary “stereochemical electronic LEP-configuration” of the central atom. (Caution! The term is non-standard)

Summary: The stated rules and techniques can be given a simple

mathematical form.

Let the central atom M in the molecule be the atoms of the sp-element and contain a total of G electrons per AO of the valence level, and ligands (L";L”;L"”) contained in quantities (l";l”;l"”), form bonds of different multiplicity, for example (1,2,3). We write the formula of the molecule as M(L") l "(L”) l” (L"”) l " ” .

The number B of all bonding pairs of the X center with the ligands is equal to the total number of bonds formed by the ligands (and the center): B =(1l"+2l"+3l"").

The number of all electrons assigned to the valence shell of the central atom is equal to the sum of all G electrons of the valence level of the center with the number B. Half of the sum is the desired number of all pairs: A=(G+B)/2=(G+1l"+2l” +3l"")/2.

The number of unshared pairs E of the center M, which, like ligands, also occupy coordination sites near the center, is equal to the difference:

AB=(G+1l"+2l”+3l"”)/2(1l"+2l”+3l"”)=(G1l"2l”3l"”)/2.

The number k of coordination sites formed in the field of the center is equal to the number k of polyhedral vertices - the stencil of the molecular structure. It consists of the number of ligands and the number of lone pairs, i.e.

k =l"+l"+l""+AB.

So we get a structure of the form MX n E m in the form MX l " +l”+l " ” E A-B .

Bond angle distortion is caused by:

first of all, multiple bonds,

secondly, lone pairs,

then bonding pairs in descending order of their volume (increasing electronegativity).

bulky complex substituents distort the structure like lone pairs.

Hybridization of the AO of the central atom.

In the theory of transmission lines, all electrons in the valence shell are assigned to the central atom. From here it is sometimes easy to determine the number and nature of the atomic orbitals that are allocated near the center for their placement.

In the valence state of the central atom, these AOs are hybridized - mixed with unit coefficients.

Sometimes it is possible to determine the type of hybridization and the orientation of the?-bonds of the molecule generated by it.

Structures of the theory of transmission lines and hybridization of the AO of the central atom.

Considering the valence configurations of the central atom, all the electrons of the valence shell assigned to it should be placed on the AO of its valence and nearby vacant sublevels. We get:

MX 2 within two AOs (s+p). The sp hybrid forms 2 equivalent?-AO,

oriented in opposite directions along a common axis.

It turns out LINE.

MX 3 and MX 2 E within three AOs (s+2p). Hybrid sp 2 , forms 3 equivalent?-AO, oriented to the vertices of an equilateral triangle.

It turns out a TRIANGLE.

MX4; MX 3E; MX 2 E 2 this requires four AOs (s+3p). Hybrid sp3,

forms 4 equivalent? -AO. It turns out TETRAHEDRON.

MX5; MX4E; MX 3 E 2 ; MX 2 E 3 this requires five AOs (s+3p+d). One

of the five AOs, this is the d-AO of the nearest higher level. Hybrid dsp 3,

forms two types?-AO, and therefore three ligands are equatorial and two -

axial. It turns out TRIGONAL BIPYRAMID.

MX6; MX5E; MX 4 E 2 within five AOs (s+3p+2d). Two AOs out of six are two d-

AO of the nearby higher level. Hybrid d 2 sp 3, forms six

equivalent? -AO. It turns out an octahedron.

ATTENTION! The connection between the theory of transmission lines with the theory of hybridization and the theory of directed valences is very important, and sometimes it can be established from the theory of transmission lines. One gets the impression of a logically complete conclusion like a hybrid of the central atom, but this is not always achievable! Nevertheless, the theory of transmission lines is built for connections of non-transitional elements. And already on a hybrid of the dsp 2 type, it "stumbles" - this option, natural for complexes, is not foreseen in it ... Do not overestimate any of the approaches. There are no comprehensive qualitative concepts in modern quantum chemistry.

The main structures of the molecules of non-transition elements,

predicted by the theory of transmission lines