An important characteristic of a molecule consisting of more than two atoms is its geometric configuration. It is determined by the mutual arrangement of atomic orbitals involved in the formation of chemical bonds.

To explain the geometric configuration of the molecule, the idea of ​​hybridization of the AO of the central atom is used. An excited beryllium atom has a configuration of 2s 1 2p 1, an excited boron atom has a configuration of 2s 1 2p 2, and an excited carbon atom has a configuration of 2s 1 2p 3. Therefore, we can assume that not the same, but different atomic orbitals can participate in the formation of chemical bonds. For example, in compounds such as BeCl 2, BCl 3, CCl 4 there must be bonds that are unequal in energy and direction. However, experimental data show that in molecules containing central atoms with different valence orbitals

(s, p, d), all connections are equivalent. To resolve this contradiction, Pauling and Slater proposed the concept of hybridization

Basic provisions of the concept of hybridization:

1. Hybrid orbitals are formed from different atomic orbitals that do not differ much in energy,

2. The number of hybrid orbitals is equal to the number of atomic orbitals involved in hybridization.

3. Hybrid orbitals are identical in electron cloud shape and energy.

4 Compared to atomic orbitals, they are more elongated in the direction of formation of chemical bonds and therefore provide better overlap of electron clouds.

It should be noted that orbital hybridization does not exist as a physical process. The hybridization method is a convenient model for a visual description of molecules.

Sp-hybridization

sp–Hybridization occurs, for example, during the formation of Be, Zn, Co and Hg (II) halides. In the valence state, all metal halides contain s- and p-unpaired electrons at the appropriate energy level. When a molecule is formed, one s- and one p-orbital form two hybrid sp-orbitals at an angle of 180 o (Fig. 5).

Fig.5 sp hybrid orbitals

Experimental data show that Be, Zn, Cd and Hg(II) halides are all linear and both bonds are of the same length.

sp 2 hybridization

As a result of the combination of one s-orbital and two p-orbitals, three hybrid sp 2 orbitals are formed, located in the same plane at an angle of 120 o to each other. This is, for example, the configuration of the BF 3 molecule (Fig. 6):

Fig.6 sp 2 hybrid orbitals

sp 3 hybridization

sp 3 -Hybridization is characteristic of carbon compounds. As a result of the combination of one s-orbital and three p-orbitals, four hybrid sp 3 orbitals are formed, directed towards the vertices of the tetrahedron with an angle between the orbitals of 109.5 degrees. Hybridization is manifested in the complete equivalence of the bonds of the carbon atom with other atoms in compounds, for example, in CH 4, CCl 4, C(CH 3) 4, etc. (Fig. 7).

Fig.7 sp 3 hybrid orbitals

The hybridization method explains the geometry of the ammonia molecule. As a result of the combination of one 2s and three 2p orbitals of nitrogen, four sp 3 hybrid orbitals are formed. The configuration of the molecule is a distorted tetrahedron, in which three hybrid orbitals participate in the formation of a chemical bond, but the fourth with a pair of electrons does not. The angles between N-H bonds are not equal to 90 o as in a pyramid, but they are not equal to 109.5 o, corresponding to a tetrahedron (Fig. 8):

Fig.8 sp 3 - hybridization in an ammonia molecule

When ammonia interacts with the hydrogen ion H + + ׃NH 3 = NH 4 + as a result of donor-acceptor interaction, an ammonium ion is formed, the configuration of which is a tetrahedron.

Hybridization also explains the difference in the angle between the O–H bonds in the corner water molecule. As a result of the combination of one 2s and three 2p orbitals of oxygen, four sp 3 hybrid orbitals are formed, of which only two are involved in the formation of a chemical bond, which leads to a distortion of the angle corresponding to the tetrahedron (rns.9):

Fig 9 sp 3 - hybridization in a water molecule

Hybridization can involve not only s- and p-orbitals, but also d- and f-orbitals.

With sp 3 d 2 hybridization, 6 equivalent clouds are formed. It is observed in such compounds as 4-, 4- (Fig. 10). In this case, the molecule has the configuration of an octahedron:

Rice. 10 d 2 sp 3 -hybridization in ion 4-

Ideas about hybridization make it possible to understand such structural features of molecules that cannot be explained in any other way. Hybridization of atomic orbitals (AO) leads to a displacement of the electron cloud in the direction of forming bonds with other atoms. As a result, the overlap areas of hybrid orbitals turn out to be larger than for pure orbitals and the bond strength increases.

Delocalized π bond

According to the MBC method, the electronic structure of a molecule looks like a set of different valence schemes (localized pair method). But, as it turned out, it is impossible to explain experimental data on the structure of many molecules and ions using only ideas about localized bonds. Research shows that only σ bonds are always localized. In the presence of π bonds, it may occur delocalization, in which the bonding electron pair simultaneously belongs to more than two atomic nuclei. For example, it was experimentally established that the BF 3 molecule has a flat triangular shape (Fig. 6). All three connections

B–F are equivalent, but the internuclear distance indicates that the bond is intermediate between single and double. These facts can be explained as follows. In the boron atom, as a result of the combination of one s-orbital and two p-orbitals, three hybrid sp 2 orbitals are formed, located in the same plane at an angle of 120 o to each other, but the free unhybridized p-orbital remains unused, and fluorine atoms have unshared electronic couples. Therefore, it is possible to form a π-bond through the donor-acceptor mechanism. The equivalence of all bonds indicates the delocalization of the π bond between three fluorine atoms.

The structural formula of the BF 3 molecule, taking into account the delocalization of the π-bond, can be depicted as follows (the non-localized bond is indicated by a dotted line):

Rice.11 Structure of the BF 3 molecule

A non-localized π-bond causes a non-integer multiplicity of the bond. In this case, it is equal to 1 1/3 since there is one σ-bond and 1/3 part of the π-bond between the boron atom and each fluorine atom.

In the same way, the equivalence of all bonds in the NO 3 – ion indicates the delocalization of the π-bond and negative charge to all oxygen atoms. In a flat triangular ion NO 3 - (sp 2 -hybridization of the nitrogen atom) delocalized

π bonds (depicted by dotted lines) are evenly distributed between all oxygen atoms (Fig. 12)

Rice. 12 Structural formula of the NO 3 ion - taking into account the delocalization of the π-bond

Similarly, delocalized π-bonds are uniformly distributed between all oxygen atoms in the anions: PO 4 3- (sp 3 -hybridization of the phosphorus atom → tetrahedron), SO 4 2- (sp 3 -hybridization of the sulfur atom → tetrahedron) (Fig. 13)

Fig.13 Structural formulas of SO 4 2- and PO 4 3- taking into account delocalization

A polyatomic molecule with the appearance of identical orbitals that are equivalent in their characteristics.

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Hybridization concept

Concept of hybridization of valence atomic orbitals was proposed by the American chemist Linus Pauling to answer the question why, if the central atom has different (s, p, d) valence orbitals, the bonds formed by it in polyatomic molecules with the same ligands turn out to be equivalent in their energy and spatial characteristics.

Ideas about hybridization occupy a central place in the method of valence bonds. Hybridization itself is not a real physical process, but only a convenient model that allows one to explain the electronic structure of molecules, in particular the hypothetical modifications of atomic orbitals during the formation of a covalent chemical bond, in particular, the alignment of the lengths of chemical bonds and bond angles in the molecule.

The concept of hybridization was successfully applied to the qualitative description of simple molecules, but was later extended to more complex ones. Unlike the theory of molecular orbitals, it is not strictly quantitative; for example, it is not able to predict the photoelectron spectra of even such simple molecules as water. Currently used mainly for methodological purposes and in synthetic organic chemistry.

This principle was reflected in the Gillespie-Nyholm theory of repulsion of electron pairs, the first and most important rule of which was formulated as follows:

“Electron pairs adopt an arrangement on the valence shell of an atom in which they are maximally distant from each other, that is, electron pairs behave as if they were mutually repelling.”

The second rule was that “all electron pairs included in the valence electron shell are considered to be located at the same distance from the nucleus”.

Types of hybridization

sp hybridization

Occurs when one s- and one p-orbital mix. Two equivalent sp-atomic orbitals are formed, located linearly at an angle of 180 degrees and directed in different directions from the nucleus of the central atom. The two remaining non-hybrid p-orbitals are located in mutually perpendicular planes and participate in the formation of π bonds or occupy lone pairs of electrons.

sp 2 -Hybridization

Occurs when one s- and two p-orbitals mix. Three hybrid orbitals are formed with axes located in the same plane and directed to the vertices of the triangle at an angle of 120 degrees. The non-hybrid p-atomic orbital is perpendicular to the plane and, as a rule, is involved in the formation of π bonds

sp 3 -Hybridization

Occurs when one s- and three p-orbitals are mixed, forming four sp 3 hybrid orbitals of equal shape and energy.

The axes of sp 3 hybrid orbitals are directed towards the vertices of the tetrahedron, while the nucleus of the central atom is located at the center of the circumscribed sphere of this tetrahedron. The angle between any two axes is approximately 109°28", which corresponds to the lowest electron repulsion energy. Also, sp 3 orbitals can form four σ bonds with other atoms or be filled with lone pairs of electrons. This state is typical for carbon atoms in saturated hydrocarbons and, accordingly, in alkyl radicals and their derivatives.

Hybridization and molecular geometry

The concept of hybridization of atomic orbitals underlies the Gillespie-Nyholm theory of repulsion of electron pairs. Each type of hybridization corresponds to a strictly defined spatial orientation of the hybrid orbitals of the central atom, which allows it to be used as the basis for stereochemical concepts in inorganic chemistry.

The table shows examples of correspondence between the most common types of hybridization and the geometric structure of molecules, under the assumption that all hybrid orbitals are involved in the formation of chemical bonds (there are no lone electron pairs).

Hybridization type Number
hybrid orbitals		 Geometry Structure Examples
sp 2 Linear

BeF 2 , CO 2 , NO 2 +
sp 2 3 Triangular

BF 3, NO 3 -, CO 3 2-
sp 3, d 3 s 4 Tetrahedral

CH 4, ClO 4 -, SO 4 2-, NH 4 +
dsp 2 4 Flat-square (2-) 2-
sp 3 d 5 Hexahedral

Most organic compounds have a molecular structure. Atoms in substances with a molecular type of structure always form only covalent bonds with each other, which is also observed in the case of organic compounds. Let us recall that covalent is a type of bond between atoms that is realized due to the fact that the atoms share part of their external electrons in order to acquire the electron configuration of a noble gas.

Based on the number of shared electron pairs, covalent bonds in organic substances can be divided into single, double and triple. These types of connections are indicated in the graphic formula by one, two or three lines, respectively:

The multiplicity of a bond leads to a decrease in its length, so a single C-C bond has a length of 0.154 nm, a double C=C bond has a length of 0.134 nm, and a triple C≡C bond has a length of 0.120 nm.

Types of bonds according to the method of overlapping orbitals

As is known, orbitals can have different shapes, for example, s-orbitals are spherical and p-dumbbell-shaped. For this reason, bonds can also differ in the way electron orbitals overlap:

ϭ bonds - are formed when orbitals overlap in such a way that the area of ​​their overlap is intersected by a line connecting the nuclei. Examples of ϭ-connections:

π-bonds - are formed when orbitals overlap, in two regions - above and below the line connecting the nuclei of atoms. Examples of π bonds:

How do you know when a molecule has π and ϭ bonds?

With the covalent type of bond, there is always a ϭ-bond between any two atoms, and a π-bond only in the case of multiple (double, triple) bonds. Wherein:

  • Single bond is always a ϭ-bond
  • A double bond always consists of one ϭ-bond and one π-bond
  • A triple bond is always formed by one ϭ and two π bonds.

Let us indicate these types of bonds in the propinic acid molecule:

Hybridization of carbon atom orbitals

Orbital hybridization is a process in which orbitals that initially have different shapes and energies are mixed, instead forming the same number of hybrid orbitals that are equal in shape and energy.

So, for example, when mixing one s- and three p- four orbitals are formed sp 3-hybrid orbitals:

In the case of carbon atoms, hybridization always takes part s- orbital, and the number p-orbitals that can take part in hybridization vary from one to three p- orbitals.

How to determine the type of hybridization of a carbon atom in an organic molecule?

Depending on how many other atoms a carbon atom is bonded to, it is either in a state sp 3, or able sp 2, or able sp- hybridization:

Let's practice determining the type of hybridization of carbon atoms using the example of the following organic molecule:

The first carbon atom is bonded to two other atoms (1H and 1C), which means it is in a state sp-hybridization.

  • The second carbon atom is bonded to two atoms - sp-hybridization
  • The third carbon atom is bonded to four other atoms (two C and two H) – sp 3-hybridization
  • The fourth carbon atom is bonded to three other atoms (2O and 1C) – sp 2-hybridization.

Radical. Functional group

The term radical most often means a hydrocarbon radical, which is the remainder of a hydrocarbon molecule without one hydrogen atom.

The name of a hydrocarbon radical is formed based on the name of the corresponding hydrocarbon by replacing the suffix –an to suffix –il .

Functional group - a structural fragment of an organic molecule (a certain group of atoms), which is responsible for its specific chemical properties.

Depending on which of the functional groups in the molecule of a substance is the eldest, the compound is classified into one class or another.

R – designation of hydrocarbon substituent (radical).

Radicals can contain multiple bonds, which can also be considered functional groups, since multiple bonds contribute to the chemical properties of a substance.

If a molecule of an organic substance contains two or more functional groups, such compounds are called polyfunctional.


b – torsion angle between planes passing through triplets of atoms 1, 2, 3 and 2, 3, 4.

The linear form is characteristic of diatomic molecules. To predict the spatial structure of a polyatomic molecule, one must know not only the bond length, but also the values ​​of the bond and torsion angles.

If the molecule consists of three or more atoms, i.e. it contains two or more chemical bonds, then angles are formed between their directions (up to 180 0) which are called bond angles(a).

Torsion angle (b)– dihedral angle between two planes passing through any selected triplets of atoms.

Examples of molecular geometry are presented in Fig. 4.11.

Let us consider the effect of hybridization on the geometric shape of molecules.

If we take into account that the p-orbitals are directed to each other at an angle of 90 0, it would be necessary to propose that the bonds in molecules, for example, H 2 O, NH 3, should also be directed to each other at right angles. However, it is not. Moreover, the expected inequality of bonds formed due to orbitals of different shapes is often not justified experimentally. It has been experimentally established that if an atom forms several bonds of the same type due to electrons of different energy sublevels, then these bonds turn out to be energetically equivalent.

Rice. 4.11. Molecular geometry:

(a) – linear; (b) – triangular; (c) – tetrahedral;

(d) – trigonal-bipyramidal; (e) – octahedral;

(e) – pentagonal-bipyramidal

The quantum mechanical theory of atomic structure cannot explain this fact, and a hypothesis arose to bring the theory into line with experiment orbital hybridization .

According to this hypothesis, different orbitals of one atom involved in the formation of s-bonds are aligned in shape and energy. From several different orbitals, the same number of hybrid orbitals are formed, having the same shape and the same energy. Hybrid orbitals are distributed evenly in the space around the nucleus.

Orbitals of various shapes can take part in hybridization. Let us consider only the hybridization of s- and p-atomic orbitals. The orbital that appears when the s- and p-orbitals are “aligned” is an unequal-armed “figure eight” (Fig. 4.12). It is more extended in one direction from the core than in the other. Since the degree of overlap of the valence orbitals is higher in this case, the chemical bond formed by the hybrid orbital should be stronger than that formed by ordinary s- and p-orbitals.

Rice. 4.12. Sp hybrid orbital shape

Depending on the number of interacting orbitals in an atom, a different number of hybrid orbitals are formed as a result of hybridization. Consequently, the shapes of the molecules will be different. Let's look at a number of simple examples.

During the formation of beryllium halide molecules, for example, BeCl 2, one s- and one p-electrons, which appear when the atom is excited, take part in the formation of chemical bonds from the central atom:


Such excitation is justified if the energy released during the subsequent formation of a chemical bond compensates for the energy expended on excitation of the atom (jump of the s-electron to the p-orbital).

The presence of unpaired electrons would provide two bonds between the beryllium atom and two chlorine atoms (having unpaired 3p electrons), but these bonds would not be equal.

With the expenditure of little energy, instead of the original s- and p-orbitals of the beryllium atom, two equivalent sp-orbitals are formed. Hybrid sp orbitals are elongated (Fig. 4.13, a) in directions opposite to each other (bond angle 180 O). Both bonds formed are energetically equivalent.

Since the energy released during the formation of Be–Cl bonds is greater than the sum of the energy expended on the excitation of a beryllium atom and the hybridization of its 2s and 2p orbitals, the formation of a BeCl 2 molecule is energetically favorable.

The case under consideration is called sp hybridization . Molecules formed with the participation of sp-hybrid orbitals are linear. The simplest example of this kind is a linear acetylene molecule C 2 H 2 formed due to sp-hybrid orbitals of carbon atoms (the remaining p-orbitals of carbon atoms do not participate in hybridization in this case, but form p-bonds).

When one s- and two p-orbitals hybridize, three equivalent sp 2 hybrid orbitals are formed. An example is the formation of a boron trichloride molecule. When excited, three unpaired electrons appear in a boron atom:

The orbitals in which these electrons are located are averaged in shape and energy, forming three sp 2 hybrid orbitals located at an angle of 120 O to each other (Fig. 4.13, b). This angle is optimal: it ensures maximum mutual separation and minimum repulsive energy of the three hybrid orbitals, therefore ensuring the minimum energy of the system.

Thus, molecules formed by sp 2 hybrid orbitals represent a regular triangle, in the center of which there is a central atom (in our case, boron), and at the vertices are the remaining atoms (chlorine). All three bonds in the molecule are equivalent.

Examples of molecules with sp 2 -hybrid orbitals of the carbon atom are organic substances: ethylene C 2 H 4, benzene C 6 H 6, etc. (in these cases, three orbitals of the carbon atom are hybridized, and the fourth is involved in the formation of a p-bond).

If four orbitals take part in the formation of hybrid orbitals (for example, in the CH 4 methane molecule), sp 3 hybridization occurs. An excited carbon atom has 4 unpaired electrons (one s and three p electrons):

Rice. 4.13. Diagram of the arrangement of orbitals for sp- (a), sp 2 - (b) and

sp 3 - hybridization (c)

If all four orbitals take part in hybridization, then the four formed sp 3 -hybrid orbitals, due to mutual repulsion, are oriented to each other at an angle of 109 O 28 / (Fig. 4.9, c). In this case, the carbon atom occupies a place in the center of a regular tetrahedron, and its partners are located at the vertices (in the case of methane, hydrogen atoms).

The nitrogen atom has five electrons in its outer energy level:


The modern quantum chemical theory of chemical bonding suggests that during the formation of the ammonia molecule NH 3, the orbitals of the nitrogen atom undergo sp 3 hybridization. In this case, we talk about hybridization not of electrons, but of orbitals, so it can be observed both in the case of orbitals containing one electron each, and in the case of orbitals occupied by two electrons, or completely free of them. The three hydrogen atoms of ammonia occupy the three vertices of the resulting tetrahedron of sp 3 hybrid orbitals. The fourth vertex of the tetrahedron is occupied by a hybrid electron cloud that is not involved in the formation of a chemical bond. Since not all vertices of the tetrahedron are identical, the bond angle in the ammonia molecule is less than the tetrahedral one and equal to 107 0, i.e. the molecule is a slightly distorted tetrahedron (trigonal pyramid, see Fig. 4.10, b).

In a water molecule, the H–O–H bond angle is also close to tetrahedral (104.5 0). This is explained by the fact that the orbitals of the oxygen atom undergo sp 3 hybridization, and two sp 3 hybrid orbitals overlap with the s orbitals of two hydrogen atoms, and two are occupied by lone electron pairs.

The energy characteristics of bonds in hydrogen halides show that even in this case, the orbitals of the halogen atoms are subject to sp 3 hybridization, and the bond is formed by the s orbital of the hydrogen atom and the sp 3 hybrid orbital of the halogen atom. It seems that there is no need to apply the theory of hybridization to diatomic molecules, but the bond in hydrogen halides is stronger than the calculation gives for the bond formed by a “pure” p-orbital.

Examples of the influence of lone electron pairs of the central atom on the geometry of the molecule are also considered in Fig. 4.14. and in table. 4.3.

Thus, the correspondence of the theory of hybridization with experimental studies (for example, data on the energy of chemical bonds) confirms the importance of the concept of hybridization itself. Hybridization determines the chemical and crystal chemical structure of substances, and, consequently, their chemical properties.

Rice. 4.14. The influence of lone electron pairs (E) of the central atom on the spatial configuration of molecules:

(a) – tetrahedron; (b) – trigonal pyramid; (c) – angular shape;

(d) – trigonal bipyramid; (e) – distorted tetrahedron; (e) – T-shape; (g) – linear form; (h) – octahedron; (i) – tetragonal pyramid; (k) – square

Table 4.3

Number of electron pairs of the central atom

and spatial configuration of ABn molecules

Number of electron pairs of atom A Arrangement of electron pairs Number of binding pairs Number of lone pairs Geometric shape and composition of the molecule* Examples
Linear Linear AB 2 BeH2, BeCl2
Triangular Flat triangle AB 3 Angular triangle AB 2 E BF 3 SnCl 2
Tetrahedral Tetrahedron AB 4 Trigonal pyramid AB 3 E Angular AB 2 E 2 CCl 4, CH 4 H 3 N, NF 3 H 2 O, OF 2
Trigonal-bipyra-midal Trigonal bipyramid AB 5 Irregular tetrahedron AB 4 E T-shaped AB 3 E 2 Linear AB 2 E 3 PCl 5 SF 4 ClF 3 XeF 2 , IF
Octahedral Octahedron AB 6 Square pyramid AB 5 E Flat square AB 4 E 2 SF 6, SiF IF 5, SbF XeF 4, ICl
Pentagonal-bipi-ramidal Pentagonal bipyramid AB 7 Irregular octahedron AB 6 E IF 7 XeF 6

*E – lone electron pair.

Hybridization of JSC- this is the alignment of valence AOs in shape and energy during the formation of a chemical bond.

1. Only those AOs whose energies are sufficiently close (for example, 2s- and 2p-atomic orbitals) can participate in hybridization.

2. Vacant (free) AOs, orbitals with unpaired electrons and lone electron pairs can participate in hybridization.

3. As a result of hybridization, new hybrid orbitals appear, which are oriented in space in such a way that after they overlap with the orbitals of other atoms, electron pairs are as far apart as possible. This state of the molecule corresponds to the minimum energy due to the maximum repulsion of like-charged electrons.

4. The type of hybridization (the number of AOs undergoing hybridization) is determined by the number of atoms “attacking” a given atom and the number of lone electron pairs in a given atom.

Example. BF 3. At the moment of bond formation, a rearrangement of the AO of atom B occurs, transforming into an excited state: B 1s 2 2s 2 2p 1 ® B* 1s 2 2s 1 2p 2 .


Hybrid joint-stock companies are located at an angle of 120°. The molecule has a regular shape triangle(flat, triangular):

3. sp 3 -hybridization. This type of hybridization is typical for atoms of the 4th group ( e.g. carbon, silicon, germanium) in molecules of the EH 4 type, as well as for the C atom in diamond, alkane molecules, for the N atom in the NH 3, NH 4 + molecule, the O atom in the H 2 O molecule, etc.

Example 1. CH 4. At the moment of bond formation, a restructuring of the AO of the C atom occurs, which goes into an excited state: C 1s 2 2s 2 2p 2 ® C* 1s 2 2s 1 2p 3 .

Hybrid joint-stock companies are located at an angle of 109 about 28".

Example 2. NH 3 and NH 4 +.

Electronic structure of the N atom: 1s 2 2s 2 2p 3. 3 AOs containing unpaired electrons and 1 AO containing a lone electron pair undergo hybridization. Due to the stronger repulsion of the lone electron pair from the electron pairs of s-bonds, the bond angle in the ammonia molecule is 107.3 o (closer to tetrahedral rather than direct).

The molecule has the shape of a trigonal pyramid:

The concept of sp 3 hybridization makes it possible to explain the possibility of the formation of ammonium ion and the equivalence of bonds in it.

Example 3. H 2 O.

Electronic structure of the O atom 1s 2 2s 2 2p 4. 2 AOs containing unpaired electrons and 2 AOs containing lone electron pairs undergo hybridization. The bond angle in a water molecule is 104.5 o (also closer to tetrahedral rather than straight).

The molecule has an angular shape:

The concept of sp 3 hybridization makes it possible to explain the possibility of the formation of an oxonium (hydronium) ion and the formation of 4 hydrogen bonds by each molecule in the ice structure.

4. sp 3 d-hybridization.This type of hybridization is typical for atoms of elements of group 5 (starting with P) in molecules of the EC 5 type.

Example. PCl 5. Electronic structure of the P atom in the ground and excited states: P 1s 2 2s 2 2p 6 3s 2 3p 3 ® P* 1s 2 2s 2 2p 6 3s 1 3p 3 3d 1 . Molecule shape - hexahedron (more precisely - trigonal bipyramid):

5. sp 3 d 2 hybridization.This type of hybridization is typical for atoms of elements of group 6 (starting with S) in molecules of the EC 6 type.

Example. SF 6. Electronic structure of the S atom in the ground and excited states: S 1s 2 2s 2 2p 6 3s 2 3p 4 ® P* 1s 2 2s 2 2p 6 3s 1 3p 3 3d 2 .

Molecule shape - octahedron :

6. sp 3 d 3 hybridization.This type of hybridization is typical for atoms of group 7 elements (starting with Cl) in molecules of the EC 7 type.

Example. IF 7. Electronic structure of the F atom in the ground and excited states: I 5s 2 3p 5 ® I* 5s 1 3p 3 3d 3 . Molecule shape - decahedron (more precisely - pentagonal bipyramid):

7. sp 3 d 4 hybridization.This type of hybridization is typical for atoms of group 8 elements (except He and Ne) in molecules of the EC 8 type.

Example. XeF 8. Electronic structure of the Xe atom in the ground and excited states: Xe 5s 2 3p 6 ® Xe* 5s 1 3p 3 3d 4.

Molecule shape - dodecahedron:

There may be other types of AO hybridization.