In the case of homonuclear molecules, we carried out the combination of $AO$, using the rule according to which orbitals of the same energy interact most strongly. In heteronuclear molecules of the $AB$ type, the energy levels of the $A$ and $B$ atoms are not the same, so it is difficult to say unambiguously which orbitals will be combined. For the case of $LiH$ this is shown in Fig. 1.

Figure 1. Energy levels of $AO$, $Li$ and $H$

Polyatomic chemical particles (molecules, radicals, ions) with the general molecular formula $B_n$ are presented, containing one central atom $A$, two or more terminal atoms $B$ and, as a consequence, only $A-B \sigma$ bonds.

The geometric shape of the $AB_n$ particles is derived from the valence bond method, i.e., from the stereochemical arrangement of the axes of the valence hybrid orbitals of the central $A$ atom, and therefore the $\sigma $-bonds $A-B$.

Hybrid orbitals help to understand the spatial structure of molecules, for example, why the water molecule has an angular configuration, ammonia has a pyramidal configuration, and methane has a tetrahedral configuration.

Consideration of the relationship between hybridization and molecular shape

Beryllium hydride, $BeH_2$, has a linear structure. To construct its localized bonding molecular orbitals, two equivalent valence orbitals of the $Be$ atom will first be formed, directed toward the two hydrogen atoms, $H_a$ and $H_b$, respectively. This is accomplished by hybridizing, or mixing (making a linear combination of), the $2s-$ and $2p$ orbitals of $Be$, resulting in two equivalent "$sp$-hybrid" orbitals. One of these hybrid orbitals, $sp_a$, is directed toward the $H_a$ atom and overlaps strongly with the $1s_a$ orbital. Another hybrid orbital, $sp_b$, is directed toward the $H_b$ atom and overlaps strongly with the $1s_b$ orbital. With this line of reasoning, the two bonding molecular orbitals of $BeH_2$ are obtained by constructing two equivalent linear combinations, each of which is localized between two atoms:

These localized molecular orbitals are shown in Fig. 2. They contain four valence electrons, forming two localized bonding electron pairs, in agreement with the Lewis bond structure for $BeH_2$. Each of the linear $sp$-hybrid orbitals is half $p$ character and half $s$ character, and the two $sp$ orbitals allow the central $Be$ atom in $BeH_2$ to attach two hydrogen atoms to itself.

Now consider the $BH_3$ molecule (which is observed in mass spectrometric experiments and is a fragment of the $B_2H_6$ molecule). In this molecule, three hydrogen atoms are attached to the central boron atom. According to the theory of localized molecular orbitals, the bond in this molecule occurs as a result of hybridization of the $2s$ orbital and two $2p$ orbitals of the boron atom to form three equivalent $sp^2$ hybrid orbitals. Each hybrid orbital is one-third $s$-character and two-thirds $p$-character. Since any two $p$-orbitals lie in the same plane, and the $s$-orbital has no spatial direction, the three $sp^2$-hybrid orbitals lie in the same plane. These three $sp^2$ hybrid orbitals, overlapping with the three hydrogen $1s$ orbitals, form three equivalent localized bonding orbitals. Each of these bonding $(sp^2+1s)$ orbitals is occupied in the $BH_3$ molecule by a pair of electrons, as shown schematically in Fig. 4. Based on the concept of hybrid orbitals, it can be predicted that the $BH_3$ molecule should have a planar trigonal structure. The angle between the internuclear axes $H-B-H$, called the bond angle $H-B-H$, should be $120^\circ$.

Figure 2. Bonding pairs of electrons on localized bonds of the $BeH_2$ molecule formed with the participation of equivalent hybrid $sp$-orbitals of the $Be$ atom. Each $sp$-orbital of $Be$ forms a localized bonding molecular orbital with the $1s$-orbital of the hydrogen atom

Figure 3. Mutual overlap of orbitals. Hybrid orbitals: a - overlap of $s$-orbitals; b - overlap of $s-$ orbitals; c - overlap of $p-$ orbitals; r - $p$-hybrid orbital; d - $sp^2$-hybrid orbitals; e - $sp^3$-hybrid orbitals

Figure 4. Electron pairs shared across localized bonds in $BH_3$

Methane, $CH_4$, has four equivalent hydrogen atoms attached to the central carbon atom. To bond with four hydrogen atoms, carbon has to use all of its valence orbitals. By hybridizing one $2s-$ and three $2p$ orbitals, four equivalent $sp^3$ hybrid orbitals can be obtained. Each $sp^3$-hybrid orbital is one-quarter $s$-character and three-quarters $p$-character. All four $sp^3$ orbitals are directed toward the vertices of a regular tetrahedron, so $sp^3$ orbitals are sometimes called tetrahedral hybrids. As a result of the overlap of each $sp^3$-hybrid orbital with the $1s-$ orbital of the hydrogen atom, four localized bonding orbitals are formed. The best overlap between the $sp^3$ and $1s$ orbitals is obtained by placing four hydrogen atoms at the vertices of a regular tetrahedron, as shown in Fig. 5 (where a cube is depicted, the alternating vertices of which form the vertices of the mentioned tetrahedron). The methane molecule has eight valence electrons (four from the carbon atom and one from each of the four hydrogen atoms), which must be placed in four localized bonding orbitals. These eight electrons form four equivalent localized bonding electron pairs, shown schematically in Fig. 5.

The structure of the $CH_4$ molecule was determined by various experimental methods. All the data obtained lead to the conclusion about the tetrahedral structure of the $CH_4$ molecule (Fig. 6), in full agreement with the predictions of the theory of localized molecular orbitals. The $H-C-H$ bond angle is $109.5^\circ$, and the $C-H$ bond length is $1.093 A$.

Figure 5. Electron pairs shared at localized bonds in $CH_4$

Figure 6. Tetrahedral molecular structure of $CH_4$

The simplest organic compound is methane. Its molecule consists of five atoms - one carbon atom and four hydrogen atoms, evenly distributed in space around this central carbon atom. Here we are faced, first of all, with the most important postulate of organic chemistry - in all uncharged organic molecules, carbon is always tetravalent. Graphically, this is expressed in the fact that it must be connected to the chemical symbols of other elements or the same carbon by four dashes. In methane, all four hydrogen atoms are the same distance from the carbon atom and are as far apart in space as possible.

In a methane molecule, a carbon atom is located in the center of a regular tetrahedron, and four hydrogen atoms are at its vertices.

This is what a methane molecule looks like, taking into account the size of the atoms.

To build a model of a molecule, let's take a tetrahedron, that is, a regular tetrahedron made up of equilateral triangles, and place a carbon atom in its center. The hydrogen atoms will be located at the vertices of the tetrahedron. Let's connect all the hydrogens to the central carbon atom. The angle α between two such lines will be 109 degrees and 28 minutes.

So we built a methane model. But what are the actual sizes of molecules? In recent decades, with the help of physical research methods (we will discuss them later), it has been possible to accurately determine the interatomic distances in the molecules of organic compounds. In a methane molecule, the distance between the centers of a carbon atom and any hydrogen atom is 0.109 nm (1 nanometer, nm, is equal to 10 -9 m). To visualize what a molecule looks like in space, they use the Stewart-Brigleb models, in which atoms are depicted as balls of a certain radius.

Now let's ask ourselves this question: what forces bind the atoms in the molecule of an organic compound, why do the hydrogen atoms not break away from the carbon center?

A carbon atom consists of a positively charged nucleus (its charge is +6) and six electrons occupying different orbitals * around the nucleus, each of which corresponds to a certain energy level.

* (An orbital can be thought of as the region of space in which the probability of encountering an electron is greatest.)

Two electrons occupy the lowest orbital closest to the nucleus. They interact most strongly with “their” nucleus and do not take part in the formation of chemical bonds. The remaining four electrons are a different matter. It is believed that in the so-called unexcited carbon atom, i.e. in a separate atom that does not form any bonds with other atoms, these electrons are located as follows: two on the lower sublevel s and two at a higher sublevel R. Somewhat simplified and schematically, we can assume that the cloud that forms the electron located on s-sublevel, has the shape of a sphere. Clouds R-electrons look like three-dimensional eights, and these eights can be located in space along the axes x, y And z. In accordance with this, each atom has three R-orbitals: p x, p y And p z. So, each orbital in an atom has a certain shape and is located in space in a special way.

In order to interact with other atoms and form chemical bonds with them, the carbon atom must first of all transform into a special one, excited state. In this case, one electron jumps from s-orbitals on p-orbital. As a result, one electron occupies a spherical s-orbital, and the three remaining electrons form three figure-eight orbitals. However, this position is energetically unfavorable for the atom. The lower energy of the atom corresponds to four identical orbitals, symmetrically located in space. Therefore, mixing, averaging, or, as they say, occurs hybridization available orbitals, and the result is four new identical orbitals.

These hybrid orbitals are also similar to figure eights, but the figure eights are lopsided: the electron density is almost entirely shifted to one side. Such hybridized orbitals are denoted sp 3(by the number of electrons from different non-hybrid orbitals involved in their formation: one with s-orbitals and three - s R-orbitals).

How is the methane molecule structured? To each of the four hybrid orbitals directed from the carbon atom in different directions (or rather, to the corners of an imaginary tetrahedron that can be built around it), hydrogen atoms H are suitable. The hydrogen atom is a nucleus with a charge of +1 (for the light isotope of ordinary hydrogen - simply a proton), and one electron occupying a spherical orbital around the proton. Clouds of "carbon" and "hydrogen" electrons overlap, and this means the formation of a chemical bond. The more the electron clouds of different atoms overlap, the stronger the bond. Now it becomes clear why hybridized orbitals are more advantageous - after all, such a one-sided, protruding figure eight can overlap much more strongly with the cloud of a hydrogen electron than non-hybrid orbitals that are less extended in space. Note that these arguments are somewhat conditional: a pure, so to speak, single and unexcited carbon atom does not really exist. Therefore, there is no point in discussing how all these orbital transformations, called hybridization, actually occur. However, for the convenience of describing chemical bonds using formulas and numbers, such conventions turn out to be useful. We will see this again and again.

How to get methane?

One of the simplest ways is to act on aluminum carbide with water:

However, aluminum carbide is too expensive a starting material to obtain such a common, cheap product as methane, and there is no need to obtain it from other compounds - after all, natural gas consists of 85-98% methane.

Methane is one of the main “building blocks” from which organic compounds can be built. What are these compounds and how can they be obtained from methane?

In general, methane is a relatively inert substance, and the range of chemical reactions that can be carried out with it is small.

Let's take a mixture of two gases - methane and chlorine and place it in a glass vessel. If this vessel is kept in the dark, then no reaction is observed. But let's try to illuminate the bottle with sunlight...

A quantum of light interacts with a chlorine molecule, as a result the molecule splits into two parts - two chlorine atoms:

The resulting atoms are much more active than molecules; they immediately attack methane molecules and capture hydrogen atoms. In this case, molecules of hydrogen chloride HCl and very unstable, very active particles, the so-called methyl radicals ⋅CH 3, are formed:

The result is a chlorine atom already known to us (its future fate is not difficult to predict: it attacks new methane molecules, and everything repeats itself) and chloromethane, or methyl chloride, a methane derivative in which one of the hydrogen atoms is replaced by chlorine.

The reaction we described belongs to the category of so-called chain reactions, in which each stage, as in a chain, is connected with the previous and subsequent ones. Active particles - the product of one stage (here these are chlorine atoms and methyl radicals ⋅CH 3) - are used in the next stage as starting substances. The discovery of chain reactions was one of the major events in the history of chemical science, and academician N. N. Semenov and English scientist S. N. Hiishelwood were awarded the Nobel Prize for their contribution to the study of such reactions and the creation of their theory.

If we introduce such quantities of reagents into the reaction that there are two molecules of methane per chlorine molecule, then we will basically get methyl chloride CH 3 Cl. If you take chlorine in excess, then the substitution reaction will go further and you will get, in addition to methyl chloride, methylene chloride CH 2 Cl 2, chloroform CHCl 3 and, finally, the product of complete replacement of hydrogen with chlorine, carbon tetrachloride CCl 4:

But let's not forget about our task: to build various complex molecules from simple bricks - methane molecules. For this we need methyl chloride. If you act on this compound with sodium metal, then from every two CH 3 Cl molecules one ethane molecule is formed, in which there is a carbon-carbon bond:

What is ethane? This is methane in which one of the hydrogens is replaced by a methyl ⋅CH 3 radical. And this radical itself, as we already know, is obtained by removing one hydrogen atom from methane.

If we now replace one of the hydrogens (any atom) in ethane with methyl, we get a new substance - propane CH 3 -CH 2 -CH 3. We know how this can be done practically: first, in methane and ethane, replace one hydrogen with chlorine and then act on a mixture of methyl and ethyl chloride with sodium (this reaction is called the Wurtz reaction in honor of the French chemist who discovered it):

Let's go further. Let's replace one of the hydrogen atoms in propane with chlorine. It turns out that now it doesn’t matter which atom is replaced! By replacing hydrogen at the extreme carbon atom (there are two such atoms) or at the middle one, we get two different compounds: normal propyl chloride ( n-propyl chloride) and isopropyl chloride:

Let us now replace the chlorine atoms in each of these compounds with methyl groups. We will get two different butanes - normal (i.e. not branched) butane ( n-butane) and iso-butane:

Let’s add one more “brick” to the resulting molecules. Let's start with n-butane. Here you can replace one of the outermost hydrogen atoms with methyl. We get normal pentane. You can replace one of the middle hydrogens. Let's come to iso-pentane. Apparently from n-Butane you won't get anything new anymore. Let's turn to iso-butane. If we replace one of the extreme hydrogens in it (in CH 3 groups), then we will arrive at the already mentioned iso-pentane, and replacing the middle single hydrogen atom, we get neopentane:

You can continue this procedure ad infinitum. All these connections are called hydrocarbons(more precisely - saturated, saturated hydrocarbons, or alkanes), because they consist of only two elements - carbon and hydrogen. In any alkane the number of hydrogen atoms is 2 n+ 2, where n- number of carbon atoms. Therefore, the formula of a saturated hydrocarbon can be written in general form as follows: C n N 2n+2 .

In building our structures, we must say, we stopped just in time. The fact is that the number of possible isomers increases catastrophically quickly with an increase in the number of carbon atoms in the alkane molecule. So, for decane, the hydrocarbon C 10 H 22, 75 different isomers are possible, the number of isomers for the hydrocarbon C 20 H 42 (eicosane) is 366,319. The number of possible isomers for tetracontane, the hydrocarbon C 40 H 82, is even difficult to imagine: 62 491 178 805 831.

Now it becomes clear why such a huge number of organic compounds are already known today - several million - and why in this respect organic chemistry has far surpassed inorganic chemistry. But so far we have only talked about the simplest representatives of organic substances - saturated hydrocarbons.

We removed a number of isomeric hydrocarbons from methane using the Wurtz reaction. However, in practice no one does this. The fact is that the simplest hydrocarbons, along with methane, are contained in natural gas, the composition of which is different for different deposits. For example, the gas from the North Stavropol field contains 85% methane, about 5% ethane, 2.5% propane and 1.4% pentane and heavier hydrocarbons. The gas from the Gazlinskoye field consists of 98% methane, and only 1.6% ethane. There are many hydrocarbons in oil, but more on that in the following chapters.

Lower hydrocarbons - methane, ethane, propane and butane - are colorless, odorless gases or with a faint smell of gasoline. Hydrocarbons from pentane to pentadecane C 15 H 32 are liquids and, finally, higher hydrocarbons at ordinary temperatures are solids.

As the number of carbon atoms increases, the boiling and melting point of the compound increases.

Saturated hydrocarbons have another name - paraffins, reflecting their chemical inertness (in Latin parum affinis- low affinity). And yet they are quite widely used in the chemical industry to produce a wide variety of substances. The main directions of industrial use of methane are shown in the diagram.

Before we finish talking about methane and saturated hydrocarbons, let’s answer one question: how is the bond in paraffins between two carbon atoms, for example, in ethane? Everything is simple here - around each carbon atom there are, as in methane, four hybridized sp 3-orbitals, three of them bond with hydrogen atoms, and one overlaps with exactly the same orbital of another carbon atom. The C-C bond length is 0.154 nm.

>> Chemistry: Valence states of the carbon atom

You already know that electron orbitals are characterized by different energy values, different geometric shapes and directions in space. Thus, the 1s orbital has a lower energy. This is followed by the 2s orbital, which has higher energy. Both of these orbitals are spherical in shape. Naturally, the 2s orbital is larger than the 1s orbital: the higher energy is a consequence of the larger average distance between the electrons and the nucleus. Three dumbbell-shaped 2s orbitals with equal energy are directed along the coordinate axes. Therefore, the axis of each 2p orbital is perpendicular to the axes of the other two 2p orbitals.

The carbon atoms that make up organic compounds will always be tetravalent, have the electronic configuration 1s 2 2s 2 2p 2 and can be in three valence states.

Let us consider the first valence state of the carbon atom using the example of the methane molecule CH4.

When a methane molecule CH4 is formed, the carbon atom goes from the ground state to an excited state and has four unpaired electrons: one and three p-electrons, which participate in the formation of four a-bonds with four hydrogen atoms. In this case, it should be expected that three C-H bonds formed due to the pairing of three p-electrons of carbon atoms with three “electrons of three hydrogen atoms (s-p) should differ from the fourth (s-s) bond in strength, length, direction. Calculation of the electron density in methane crystals shows that all bonds in its molecule are equivalent and directed towards the vertex of the tetrahedron. This is explained by the fact that during the formation of a methane molecule, covalent bonds arise due to the interaction of not “pure”, but so-called hybrid, i.e., orbitals averaged in shape and size (and therefore in energy).

Hybridization of orbitals is the process of aligning them in shape and energy.

The number of hybrid orbitals is equal to the number of original orbitals. Compared to them, hybrid orbitals are more elongated in space, which ensures their more complete overlap with the orbitals of neighboring atoms.

In the methane molecule and in other alkanes, as well as in all organic molecules at the site of a single bond, the carbon atoms will be in a state of sp 3 hybridization, i.e., the orbitals of one s- and three p-electrons have undergone hybridization at the carbon atom and four have been formed identical hybrid orbitals.

As a result of the overlap of four hybrid sp 3 orbitals of the carbon atom and s orbitals of four hydrogen atoms, a tetrahedral methane molecule with four identical a-bonds at an angle of 109°28 is formed. If in a methane molecule one hydrogen atom is replaced by a CH3 group, then an ethane molecule is obtained CH3-CH3.

A carbon atom that has three hydrogen atoms and one carbon atom is called primary.

In the ethane molecule there is a single (sometimes called ordinary, ordinary) non-polar carbon-carbon bond with a length of 0.154 nm.

In the propane molecule CH3-CH2-CH3, at the central carbon atom there are two hydrogen atoms and two carbon atoms. Such an atom is called secondary.

If a carbon atom is bonded to three carbon atoms, then it is said to be a tertiary atom:

CH3 - CH - CH3
CH3

Carbon with four carbon atoms is called quaternary:

CH3
CH3 - C - CH3
CH3

Let us consider the second valence state of the carbon atom using the example of the ethylene molecule C2H4. As you remember, there is a double bond between the carbon atoms, which is reflected in the structural formula by two identical lines:

The bonds reflected by these dashes, although covalent, are different in the way they overlap - one of them is a, the other is - P.

In the ethylene molecule, each carbon atom is connected not to four, but to three other atoms (with one carbon atom and two hydrogen atoms), therefore only three electron orbitals enter into hybridization: one b and two p, i.e. sp 2 -hybridization. These three orbitals are located in the same plane at an angle of 120° relative to each other. The orbitals of each carbon atom overlap with the s-orbitals of two hydrogen atoms and with one of the same sp2-rm6-ride orbitals of the neighboring carbon atom and form three a-bonds at the same angle of 120°. Consequently, the ethylene molecule will have a planar structure. Two p-orbitals of carbon atoms that do not participate in hybridization will overlap in two regions perpendicular to the plane of the molecule ("lateral overlap") and form P-connection.

However, the “lateral” overlap of p-orbitals occurs to a lesser extent than p-orbitals along the bond line, and, in addition, it is formed at a greater distance from the nuclei of the bonding atoms. Therefore, the I-connection will be less strong than P-connection. And yet under the influence P-bonds bring carbon atoms even closer to each other: in molecules of methane CH4 and ethane C2H6, the distance between the nuclei of atoms (bond length) is 0.154 nm, and in molecules of ethylene C2H4 - 0.134 nm.

Let us consider the third valence state of the carbon atom using the example of the acetylene molecule C2H2, in which a triple bond CH=CH is realized: one a-bond and two p-bonds. The acetylene molecule has a linear structure, since each carbon atom in it is connected by a-bonds to only two other atoms - a carbon atom and a hydrogen atom, while BP hybridization occurs, in which only two orbitals participate - one s and one p . Two hybrid orbitals are oriented relative to each other at an angle of 180° and form two P-bond with the s-orbital of the hydrogen atom and one more P- connections located in mutually perpendicular planes.

The appearance of a third bond causes further convergence of carbon atoms - the distance between them (the length of the C=-C bond) in the acetylene molecule is 0.120 nm.

1. What types of hybridization of electron orbitals of a carbon atom do you know?

2. The order of connection of atoms in molecules is reflected by structural formulas. Determine the type of hybridization of each carbon atom in the 1,2 butadiene molecule if its structural formula is

3. How many orbitals of the second energy level of the carbon atom are not involved in nuclear hybridization; in ya2-hybridization; in yr3 hybridization?

4. What are the angles between the axes of the carbon atom for:

a) sp 2 hybrid orbitals;

b) sp-hybrid orbitals;

c) sp-hybrid and non-hybrid p-orbitals;

d) non-hybrid p-orbitals;

e) sp 3 hybrid orbitals?

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Page 1

The theory of valency played a crucial role in the development of the theory of chemistry in general and organic chemistry in particular. Based on the theory of valency, Kekule assumed that the carbon atom is tetravalent, and in 1858 he tried, based on this assumption, to imagine the structure of the simplest organic molecules and radicals. In the same 1858, the Scottish chemist Archibald Scott Cooper (1831-1892) proposed depicting the forces connecting atoms (or bonds, as they are commonly called) in the form of dashes. After the first organic molecule was “built,” it became abundantly clear why organic molecules tend to be much larger and more complex than inorganic molecules.

According to Kekule's ideas, carbon atoms can be connected to each other using one or more of their four valence bonds, forming long chains - straight or branched. Apparently, no other atoms possess this remarkable ability to the extent that carbon possesses it.

So, imagining that each carbon atom has four valence bonds, and each hydrogen atom has one such bond, we can depict the three simplest hydrocarbons (compounds whose molecules are formed only by carbon and hydrogen atoms), methane CH4, ethane C2H6 and propane C3H8, in the following way:

By increasing the number of carbon atoms, this sequence can be continued, almost indefinitely. By adding oxygen (two valence bonds) or nitrogen (three valence bonds) to the hydrocarbon chain, we can imagine the structural formulas of the molecules of ethyl alcohol (C2H6O) and methylamine (CH5N):

Assuming the possibility of two bonds (double bond) or three bonds (triple bond) between neighboring atoms, we can depict the structural formulas of compounds such as ethylene (C2H4), acetylene (C2H2), methyl cyanide (C2H3N), acetone (C3H6O) and acetic acid (C2H4O2):

The usefulness of the structural formulas was so obvious that many organic chemists adopted them immediately. They declared completely obsolete all attempts to depict organic molecules as structures built from radicals. As a result, it was found necessary to show its atomic structure when writing the formula of a compound.

Russian chemist Alexander Mikhailovich Butlerov (1823-1886) used this new system of structural formulas in his theory of the structure of organic compounds. In the 60s of the last century, he showed how, using structural formulas, one can clearly explain the reasons for the existence of isomers (see Chapter 5). For example, ethyl alcohol and dimethyl ether have the same empirical formula C2H6O, but the structural formulas of these compounds differ significantly:

It is therefore not surprising that a change in the arrangement of atoms results in two sets of very different properties. In ethyl alcohol, one of the six hydrogen atoms is attached to an oxygen atom, while in dimethyl ether, all six hydrogen atoms are attached to carbon atoms. The oxygen atom holds the hydrogen atom more weakly than the carbon atom, so that sodium metal added to ethyl alcohol replaces the hydrogen (one-sixth of the total). Sodium added to dimethyl ether does not displace hydrogen at all. Thus, when composing structural formulas, one can be guided by chemical reactions, and structural formulas, in turn, can help to understand the essence of the reactions.

Butlerov paid especially much attention to one type of isomerism, called tautomerism (dynamic isomerism), in which some substances always act as mixtures of two compounds. If one of these compounds is isolated in its pure form, it will immediately partially transform into another compound. Butlerov showed that tautomerism is caused by the spontaneous transition of a hydrogen atom from an oxygen atom to a neighboring carbon atom (and vice versa).

In order to fully prove the validity of the system of structural formulas, it was necessary to determine the structural formula of benzene, a hydrocarbon containing six carbon atoms and six hydrogen atoms. It was not possible to do this right away. It seemed that there was no such structural formula that, while meeting the requirements of valency, would at the same time explain the greater stability of the compound. The first versions of the structural formulas of benzene were very similar to the formulas of some hydrocarbons - very unstable compounds and not similar in chemical properties to benzene.

Continuation. See the beginning in № 15, 16/2004

Lesson 5. Hybridization
carbon atomic orbitals

A covalent chemical bond is formed using shared bonding electron pairs like:

Form a chemical bond, i.e. Only unpaired electrons can create a common electron pair with a “foreign” electron from another atom. When writing electronic formulas, unpaired electrons are located one at a time in an orbital cell.
Atomic orbital is a function that describes the density of the electron cloud at each point in space around the atomic nucleus. An electron cloud is a region of space in which an electron can be detected with a high probability.
To harmonize the electronic structure of the carbon atom and the valence of this element, concepts about the excitation of the carbon atom are used. In the normal (unexcited) state, the carbon atom has two unpaired 2 R 2 electrons. In an excited state (when energy is absorbed) one of 2 s 2 electrons can go to free R-orbital. Then four unpaired electrons appear in the carbon atom:

Let us recall that in the electronic formula of an atom (for example, for carbon 6 C – 1 s 2 2s 2 2p 2) large numbers in front of the letters - 1, 2 - indicate the number of the energy level. Letters s And R indicate the shape of the electron cloud (orbital), and the numbers to the right above the letters indicate the number of electrons in a given orbital. All s-spherical orbitals:

At the second energy level except 2 s-there are three orbitals 2 R-orbitals. These 2 R-orbitals have an ellipsoidal shape, similar to dumbbells, and are oriented in space at an angle of 90° to each other. 2 R-Orbitals denote 2 p x, 2p y and 2 p z in accordance with the axes along which these orbitals are located.

When chemical bonds are formed, the electron orbitals acquire the same shape. Thus, in saturated hydrocarbons one s-orbital and three R-orbitals of the carbon atom to form four identical (hybrid) sp 3-orbitals:

This - sp 3 -hybridization.
Hybridization– alignment (mixing) of atomic orbitals ( s And R) with the formation of new atomic orbitals called hybrid orbitals.

Hybrid orbitals have an asymmetric shape, elongated towards the attached atom. Electron clouds repel each other and are located in space as far as possible from each other. In this case, the axes of four sp 3-hybrid orbitals turn out to be directed towards the vertices of the tetrahedron (regular triangular pyramid).
Accordingly, the angles between these orbitals are tetrahedral, equal to 109°28".
The vertices of electron orbitals can overlap with the orbitals of other atoms. If electron clouds overlap along a line connecting the centers of atoms, then such a covalent bond is called sigma()-connection. For example, in the ethane molecule C 2 H 6, a chemical bond is formed between two carbon atoms by overlapping two hybrid orbitals. This is a connection. In addition, each of the carbon atoms with its three sp 3-orbitals overlap with s-orbitals of three hydrogen atoms, forming three -bonds.

In total, three valence states with different types of hybridization are possible for a carbon atom. Except sp 3-hybridization exists sp 2 - and sp-hybridization.
sp 2 -Hybridization- mixing one s- and two R-orbitals. As a result, three hybrids are formed sp 2 -orbitals. These sp 2-orbitals are located in the same plane (with axes X, at) and are directed to the vertices of the triangle with an angle between the orbitals of 120°. Unhybridized
R-the orbital is perpendicular to the plane of the three hybrid sp 2-orbitals (oriented along the axis z). Upper half R-orbitals are above the plane, the lower half is below the plane.
Type sp 2-carbon hybridization occurs in compounds with a double bond: C=C, C=O, C=N. Moreover, only one of the bonds between two atoms (for example, C=C) can be an - bond. (The other bonding orbitals of the atom are directed in opposite directions.) The second bond is formed as a result of overlapping non-hybrid R-orbitals on both sides of the line connecting the atomic nuclei.

Covalent bond formed by lateral overlap R-orbitals of neighboring carbon atoms is called pi()-connection.

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Due to less orbital overlap, the -bond is less strong than the -bond.
sp-Hybridization– this is mixing (alignment in shape and energy) of one s- and one
R-orbitals to form two hybrid sp-orbitals. sp-The orbitals are located on the same line (at an angle of 180°) and directed in opposite directions from the nucleus of the carbon atom. Two
R-orbitals remain unhybridized. They are placed mutually perpendicular
directions of connections. On the image sp-orbitals are shown along the axis y, and the unhybridized two
R-orbitals – along the axes X And z.

A carbon-carbon triple bond CC consists of an -bond formed by overlapping
sp-hybrid orbitals, and two -bonds.
The relationship between such parameters of the carbon atom as the number of attached groups, the type of hybridization and the types of chemical bonds formed is shown in Table 4.

Table 4

Covalent carbon bonds

Number of groups related with carbon	Type hybridization	Types participating chemical bonds	Examples of compound formulas
4	sp 3	Four - connections
3	sp 2	Three - connections and one - connection
2	sp	Two - connections and two - connections	H–CC–H

Exercises.

1. Which electrons of atoms (for example, carbon or nitrogen) are called unpaired?

2. What does the concept of “shared electron pairs” mean in compounds with a covalent bond (for example, CH 4 or H 2 S )?

3. What electronic states of atoms (for example, C or N ) are called basic, and which are excited?

4. What do the numbers and letters mean in the electronic formula of an atom (for example, C or N )?

5. What is an atomic orbital? How many orbitals are there in the second energy level of the C atom? and how do they differ?

6. How are hybrid orbitals different from the original orbitals from which they were formed?

7. What types of hybridization are known for the carbon atom and what do they consist of?

8. Draw a picture of the spatial arrangement of orbitals for one of the electronic states of the carbon atom.

9. What chemical bonds are called and what? Specify-And-connections in connections:

10. For the carbon atoms of the compounds below, indicate: a) type of hybridization; b) types of its chemical bonds; c) bond angles.

Answers to exercises for topic 1

Lesson 5

1. Electrons that are located one at a time in an orbital are called unpaired electrons. For example, in the electron diffraction formula of an excited carbon atom there are four unpaired electrons, and the nitrogen atom has three:

2. Two electrons involved in the formation of one chemical bond are called shared electron pair. Typically, before a chemical bond is formed, one of the electrons in this pair belonged to one atom, and the other electron belonged to another atom:

3. Electronic state of an atom in which the order of filling electron orbitals is observed: 1 s 2 , 2s 2 , 2p 2 , 3s 2 , 3p 2 , 4s 2 , 3d 2 , 4p 2, etc., are called underlying condition. IN excited state one of the valence electrons of the atom occupies a free orbital with a higher energy; such a transition is accompanied by the separation of paired electrons. Schematically it is written like this:

While in the ground state there were only two unpaired valence electrons, in the excited state there are four such electrons.

5. An atomic orbital is a function that describes the density of the electron cloud at each point in space around the nucleus of a given atom. At the second energy level of the carbon atom there are four orbitals - 2 s, 2p x, 2p y, 2p z. These orbitals differ:
a) the shape of the electron cloud ( s– ball, R– dumbbell);
b) R-orbitals have different orientations in space - along mutually perpendicular axes x, y And z, they are designated p x, p y, p z.

6. Hybrid orbitals differ from the original (non-hybrid) orbitals in shape and energy. For example, s-orbital – the shape of a sphere, R– symmetrical figure eight, sp-hybrid orbital – asymmetric figure eight.
Energy differences: E(s) < E(sp) < E(R). Thus, sp-orbital – an orbital averaged in shape and energy, obtained by mixing the original s- And p-orbitals.

7. For a carbon atom, three types of hybridization are known: sp 3 , sp 2 and sp (see text of lesson 5).

9. -bond - a covalent bond formed by head-on overlapping of orbitals along a line connecting the centers of atoms.
-bond – a covalent bond formed by lateral overlap R-orbitals on both sides of the line connecting the centers of the atoms.
-Bonds are shown by the second and third lines between connected atoms.