Hybridizationcalled the hypothetical mixing process various types, but orbitals of a given atom that are close in energy with the appearance of the same number of new (hybrid 1) orbitals, identical in energy and shape.
Hybridization atomic orbitals occurs during the formation of covalent bonds.
Hybrid orbitals have the shape of a three-dimensional asymmetrical figure eight, strongly elongated to one side of the atomic nucleus: .
This form causes a stronger overlap of hybrid orbitals with the orbitals (pure or hybrid) of other atoms than in the case of pure atomic orbitals and leads to the formation of stronger covalent bonds. Therefore, the energy expended on the hybridization of atomic orbitals is more than compensated by the release of energy due to the formation of stronger covalent bonds involving hybrid orbitals. The name of hybrid orbitals and the type of hybridization are determined by the number and type of atomic orbitals participating in the hybridization, for example: sp-, sp 2 -, sp 3 -, sp 2 d- orsp 3 d 2 -hybridization.
The orientation of the hybrid orbitals, and therefore the geometry of the molecule, depend on the type of hybridization. In practice, the inverse problem is usually solved: first, the geometry of the molecule is established experimentally, after which the type and shape of the hybrid orbitals involved in its formation are described.
sp -Hybridization. Two hybrid sp- As a result of mutual repulsion, the orbitals are located relative to the atomic nucleus in such a way that the angle between them is 180° (Fig. 7).
Rice. 7. Mutual location in space of two sp- hybrid orbitals of one atom: A - surfaces covering regions of space where the probability of an electron being present is 90%; b - conditional image.
As a result of this arrangement of hybrid orbitals, molecules of composition AX 2, where A is the central atom, have linear structure, that is, the covalent bonds of all three atoms are located on the same straight line. For example, in a state sp- hybridization, the valence orbitals of the beryllium atom in the BeCl 2 molecule are located (Fig. 8). Linear configuration due to sp- The molecules BeH 2, Be(CH 3) 2, ZnCl 2, CO 2, HC≡N and a number of others also have hybridization of the valence orbitals of atoms.
Rice. 8. Triatomic linear molecule of beryllium chloride BeC1 2 (in the gaseous state): 1 - 3R- Cl atom orbital; 2 - two sp- hybrid orbitals of the Be atom.
s R 2 -Hybridization. Let us consider the hybridization of one s- and two R- orbitals. In this case, as a result of a linear combination of three orbitals, three hybrid orbitals arise sR 2 -orbitals. They are located in the same plane at an angle of 120° to each other (Fig. 9). sR 2 -Hybridization is characteristic of many compounds of boron, which, as shown above, in the excited state has three unpaired electrons: one s- and two R-electron. When overlapping sR 2 -orbitals of a boron atom with the orbitals of other atoms form three covalent bonds, equal in length and energy. Molecules in which the valence orbitals of the central atom are in the state sR 2 -hybridization, have a triangular configuration. The angles between covalent bonds are 120°. Able sR 2 -hybridization are the valence orbitals of boron atoms in the molecules BF 3, BC1 3, carbon and nitrogen atoms in the anions CO 3 2 -, NO 3 -.
Rice. 9. Mutual position in the space of three sR 2 -hybrid orbitals.
s R 3 -Hybridization. Substances in which the central atom contains four are very widespread. sR 3 -orbitals resulting from a linear combination of one s- and three R-orbitals. These orbitals are located at an angle of 109˚28′ to each other and are directed towards the vertices of the tetrahedron, in the center of which is the atomic nucleus (Fig. 10 a).
Formation of four equal covalent bonds due to overlap sR 3 -orbitals with orbitals of other atoms are typical for carbon atoms and other elements of group IVA; this determines the tetrahedral structure of the molecules (CH 4, CC1 4, SiH 4, SiF 4, GeH 4, GeBr 4, etc.).
Rice. 10. The influence of non-bonding electron pairs on the geometry of molecules:
a– methane (no non-bonding electron pairs);
b– ammonia (one non-bonding electron pair);
V– water (two non-bonding pairs).
Lone electron pairs of hybrid orbital lei . In all the examples considered, the hybrid orbitals were “populated” by single electrons. However, there are often cases when a hybrid orbital is “occupied” by an electron pair. This affects the geometry of the molecules. Since a nonbonding electron pair is affected by the nucleus of only its atom, and a bonding electron pair is affected by two atomic nuclei, the nonbonding electron pair is closer to the atomic nucleus than the bonding one. As a result, the nonbonding electron pair repels the bonding electron pairs more than they repel each other. Graphically, for clarity, the large repulsive force acting between the non-bonding and bonding electron pairs can be depicted as larger in volume electron orbital non-bonding pair. A non-bonding electron pair is found, for example, on the nitrogen atom in the ammonia molecule (Fig. 10 b). As a result of interaction with bonding electron pairs, the H-N-H bond angles are reduced to 107.78° compared to 109.5° characteristic of a regular tetrahedron.
Bonding electron pairs experience even greater repulsion in a water molecule, where the oxygen atom has two non-bonding electron pairs. As a result, the H-O-H bond angle in a water molecule is 104.5° (Fig. 10 V).
If a non-bonding electron pair, as a result of the formation of a covalent bond through the donor-acceptor mechanism, turns into a bonding one, then the repulsive forces between this bond and other covalent bonds in the molecule are equalized; The angles between these bonds are also aligned. This occurs, for example, during the formation of ammonium cation:
Participation in hybridization d -orbitals. If the energy of atomic d- orbitals are not very different from energies s- And R- orbitals, then they can participate in hybridization. The most common type of hybridization involving d- orbitals is sR 3 d 2 - hybridization, as a result of which six hybrid orbitals of equal shape and energy are formed (Fig. 11 A), located at an angle of 90˚ to each other and directed towards the vertices of the octahedron, in the center of which is the atomic nucleus. Octahedron (Fig. 11 b) – is a regular octahedron: all edges in it are of equal length, all faces are regular triangles.
Rice. eleven. sR 3 d 2 - Hybridization
Less common sR 3 d- hybridization to form five hybrid orbitals (Fig. 12 A), directed to the vertices of the trigonal bipyramid (Fig. 12 b). A trigonal bipyramid is formed by connecting two isosceles pyramids with a common base - a regular triangle. Bold strokes in Fig. 12 b edges of equal length are shown. Geometrically and energetically sR 3 d- hybrid orbitals are unequal: three “equatorial” orbitals are directed towards the vertices regular triangle, and two “axial” ones - up and down perpendicular to the plane of this triangle (Fig. 12 V). The angles between the “equatorial” orbitals are equal to 120°, as in sR 2 - hybridization. The angle between the “axial” and any of the “equatorial” orbitals is 90°. Accordingly, covalent bonds that are formed with the participation of “equatorial” orbitals differ in length and energy from bonds in the formation of which “axial” orbitals participate. For example, in the PC1 5 molecule, the “axial” bonds are 214 pm long, and the “equatorial” bonds are 202 pm long.
Rice. 12. sR 3 d- Hybridization
Thus, considering covalent bonds as a result of overlapping atomic orbitals, it is possible to explain the geometry of the resulting molecules and ions, which depends on the number and type of atomic orbitals involved in the formation of bonds. The concept of hybridization of atomic orbitals, it is necessary to understand that hybridization is a conventional technique that allows you to clearly explain the geometry of a molecule through a combination of AOs.
A polyatomic molecule with the appearance of identical orbitals that are equivalent in their characteristics.
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Hybridization of electron orbitals
Cytology. Lecture 46. Orbital hybridization
Hybrid sp3 orbitals
Subtitles
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 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 is reflected in the Gillespie-Nyholm theory of repulsion of electron pairs, the first and most important rule 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 sides 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 unshared 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. They can form four σ bonds with other atoms or be filled with lone pairs of electrons.
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 lowest energy electron repulsion. 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 organic 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 | 4 | Tetrahedral |
|
CH 4, ClO 4 -, SO 4 2-, NH 4 + |
| dsp 2 | 4 | Flat-square |
Problem 261.
What types of carbon AO hybridization correspond to the formation of CH molecules 4, C 2 H 6, C 2 H 4, C 2 H 2?
Solution:
a) In CH molecules 4 and C 2 H 6 The valence electron layer of a carbon atom contains four electron pairs:
Therefore, the electron clouds of the carbon atom in the CH 4 and C 2 H 6 molecules will be maximally distant from each other during sp3 hybridization, when their axes are directed towards the vertices of the tetrahedron. In this case, in the CH4 molecule, all the vertices of the tetrahedron will be occupied by hydrogen atoms, so that the CH4 molecule has a tetrahedral configuration with a carbon atom in the center of the tetrahedron. In the C 2 H 6 molecule, hydrogen atoms occupy three vertices of the tetrahedron, and the common electron cloud of another carbon atom is directed towards the fourth vertex, i.e. two carbon atoms are connected to each other. This can be represented by diagrams:
b) In the C 2 H 4 molecule there is a valence electron layer of the carbon atom, as in the CH 4 and C 2 H 6 molecules. contains four electron pairs:
When C 2 H 4 is formed, three covalent bonds are formed according to the usual mechanism, i.e. are - connections, and one - - connection. When a C 2 H 4 molecule is formed, each carbon atom has two hydrogen atoms - bonds and two bonds to each other, one - and one - bonds. Hybrid clouds that match this type hybridization, are located in the carbon atom so that the interaction between electrons is minimal, i.e. as far apart as possible. This arrangement of carbon atoms (two double bonds between carbon atoms) is characteristic of sp 2 hybridization of carbon AO. During sp 2 hybridization, the electron clouds in carbon atoms are oriented in directions lying in the same plane and making angles of 120 0 with each other, i.e. in directions to the vertices of a regular triangle. In the ethylene molecule, the formation of - bonds involves three sp 2 -hybrid orbitals of each carbon atom, two between two hydrogen atoms and one with the second carbon atom, and - the bond is formed due to the p-electron clouds of each carbon atom. Structural formula molecules C 2 H 4 will look like:
c) In the C 2 H 2 molecule, the valence electron layer of the carbon atom contains four pairs of electrons:
The structural formula of C 2 N 2 is:
Each carbon atom is connected by one electron pair to a hydrogen atom and three electron pairs to another carbon atom. Thus, in an acetylene molecule, carbon atoms are connected to each other by one -bond and two -bonds. Each carbon atom is connected to hydrogen by an - bond. The formation of - bonds involves two sp-hybrid AOs, which are located relative to each other so that the interaction between them is minimal, i.e. as far apart as possible. Therefore, during sp-hybridization, the electron clouds between carbon atoms are oriented in opposite directions relative to each other, i.e. angle between C-C connections is 180 0. Therefore, the C 2 H 2 molecule has a linear structure:
Problem 262.
Indicate the type of hybridization of silicon AO in SiH 4 and SiF 4 molecules. Are these molecules polar?
Solution:
In SiH 4 and SiF 4 molecules, the valence electron layer contains four pairs of electrons:
Therefore, in both cases, the electron clouds of the silicon atom will be maximally distant from each other during sp 3 hybridization, when their axes are directed towards the vertices of the tetrahedron. Moreover, in the SiH 4 molecule all the vertices of the tetrahedron are occupied by hydrogen atoms, and in the SiF 4 molecule - by fluorine atoms, so that these molecules have a tetrahedral configuration with a silicon atom in the center of the tetrahedron:
In tetrahedral molecules SiH 4 and SiF 4, the dipole moments of the Si-H and Si-F bonds mutually cancel each other, so that the total dipole moments of both molecules will be equal to zero. These molecules are non-polar, despite the polarity of the Si-H and Si-F bonds.
Problem 263.
In SO 2 and SO 3 molecules, the sulfur atom is in a state of sp 2 hybridization. Are these molecules polar? What is their spatial structure?
Solution:
During sp 2 hybridization, hybrid clouds are located in the sulfur atom in directions lying in the same plane and making angles of 120 0 with each other, i.e. directed towards the vertices of a regular triangle.
a) In the SO 2 molecule, two sp 2 -hybrid AOs form a bond with two oxygen atoms, the third sp 2 -hybrid orbital will be occupied by a free electron pair. This electron pair will shift the electron plane and the SO 2 molecule will take the shape of an irregular triangle, i.e. angle OSO will not be equal to 120 0. Therefore, the SO 2 molecule will have an angular shape with sp 2 hybridization of the atomic orbitals, the structure:
In the SO 2 molecule, mutual compensation of dipole moments S-O connections not happening; the dipole moment of such a molecule will have a value greater than zero, i.e. the molecule is polar.
b) In the corner SO 3 molecule, all three sp2-hybrid AOs form a bond with three oxygen atoms. The SO3 molecule will have the shape of a flat triangle with sp2 hybridization of the sulfur atom:
In a triangular SO 3 molecule, the dipole moments of the S-O bonds cancel each other out, so that the total dipole moment will be zero, the molecule is polar.
Problem 264.
When SiF4 interacts with HF, a strong acid H 2 SiF 6 is formed, which dissociates into H + and SiF 6 2- ions. Can In a similar way does the reaction occur between CF 4 and HF? Indicate the type of hybridization of silicon AO in the SiF 6 2- ion.
Solution:
a) When excited, the silicon atom goes from the 1s 2 2s 2 2p 6 3s 2 3p 3 state to the 1s 2 2s 2 2p 6 3s 1 3p 4 3d 0 state, and the electronic structure of the valence orbitals corresponds to the scheme:
Four unpaired electrons of an excited silicon atom can participate in the formation of four covalent bonds according to the usual mechanism with fluorine atoms (1s 2 2s 2 2p 5), each having one unpaired electron, to form a SiF 4 molecule.
When SiF 4 interacts with HF, the acid H 2 SiF 6 is formed. This is possible because the SiF 4 molecule has free 3d orbitals, and the F- (1s 2 2s 2 2p 6) ion has free pairs of electrons. The connection is carried out according to the donor-acceptor mechanism due to a pair of electrons from each of the two F - (HF ↔ H + + F -) ions and free 3d orbitals of the SiF 4 molecule. In this case, the SiF 6 2- ion is formed, which with the H + ions forms an acid molecule H 2 SiF 6.
b) Carbon (1s 2 2s 2 2p 2) can form, like silicon, a CF 4 compound, but the valence capabilities of the carbon atom will be exhausted (there are no unpaired electrons, free pairs of electrons and free valence orbitals at the valence level). The structure diagram of the valence orbitals of an excited carbon atom has the form:
When CF 4 is formed, all valence orbitals of carbon are occupied, so an ion cannot be formed.
In the SiF 4 molecule, the valence electron layer of the silicon atom contains four pairs of electrons:
The same is observed for the CF 4 molecule. therefore, in both cases, the electron clouds of silicon and carbon atoms will be as far apart as possible from each other during sp3 hybridization. When their axes are directed to the vertices of the tetrahedron:
The method of hybridization of atomic orbitals is based on the assumption that during the formation of a molecule, instead of the original atomic and -electron clouds, equivalent “mixed” or hybrid electron clouds are formed that are elongated towards neighboring atoms, due to which their more complete overlap with the electron clouds of these atoms is achieved . Such deformation of electron clouds requires energy. But a more complete overlap of the valence electron clouds leads to the formation of a stronger chemical bond and, therefore, to an additional gain of energy. If this energy gain is sufficient to more than compensate for the energy expended on the deformation of the initial atomic electron clouds, such hybridization ultimately leads to a decrease in the potential energy of the resulting molecule and, consequently, to an increase in its stability.
Consider, as an example of hybridization, the formation of a beryllium fluoride molecule. Each fluorine atom that is part of this molecule has one unpaired electron,
which is involved in the formation of covalent bonds. A beryllium atom in an unexcited state has no unpaired electrons:
Therefore, to participate in the formation of chemical bonds, the beryllium atom must go into excited state :
The resulting excited atom has two unpaired electrons: the electron cloud of one of them corresponds to the state, the other -. When these electron clouds overlap with the p-electron clouds of two fluorine atoms, covalent bonds can form (Fig. 38).
However, as already mentioned, with the expenditure of some energy, instead of the original s- and p-orbitals of the beryllium atom, two equivalent hybrid orbitals (-orbitals) can be formed. The shape and arrangement of these orbitals are shown in Fig. 39, from which it is clear that the hybrid -orbitals are elongated in opposite directions.
The overlap of hybrid -electron clouds of the beryllium atom with p-electron clouds of fluorine atoms is shown in Fig. 40.
Rice. 38. Scheme of overlap of -electron clouds of fluorine atoms with and -electron clouds of a beryllium atom (for each bond separately). The areas of overlap of electron clouds are shaded.
Rice. 39. Shape ( schematic illustration) and the relative position of the hybrid -electron clouds of the beryllium atom (for each hybrid orbital separately).
Rice. 40. Scheme of the formation of chemical bonds in a molecule. To simplify the figure, the hybrid electron clouds of the beryllium atom are not depicted in full.
Due to the elongated shape of hybrid orbitals, more complete overlap of interacting electron clouds is achieved, which means stronger chemical bonds are formed. The energy released during the formation of these bonds is greater than the total energy expenditure for the excitation of the beryllium atom and the hybridization of its atomic orbitals. Therefore, the process of molecule formation is energetically favorable.
The considered case of hybridization of one s- and one p-orbital, leading to the formation of two -orbitals, is called -hybridization. As Fig. The 39, -orbitals are oriented in opposite directions, which leads to the linear structure of the molecule. Indeed, the molecule is linear, and both bonds in this molecule are equivalent in all respects.
Other cases of hybridization of atomic orbitals are also possible, but the number of resulting hybrid orbitals is always equal to the total number of initial atomic orbitals involved in hybridization. Thus, when one s- and two p-orbitals are hybridized (-hybridization - read “es-pe-two”), three equal -orbitals are formed. In this case, hybrid electron clouds are located in directions lying in the same plane and oriented at angles of 120° to each other (Fig. 41). Obviously, this type of hybridization corresponds to the formation of a flat triangular molecule.
An example of a molecule in which β-hybridization occurs is the boron fluoride molecule. Here, instead of the original one s- and two p-orbitals of the excited boron atom
Three equal -orbitals are formed. Therefore, the molecule is built in the shape of a regular triangle, with a boron atom in the center and fluorine atoms at the vertices. All three bonds in the molecule are equivalent.
If one s- and three p-orbitals (- hybridization) are involved in hybridization, then the result is the formation of four hybrid -orbitals, elongated in the directions towards the vertices of the tetrahedron, i.e., oriented at angles to each other (Fig. 42). Such hybridization occurs, for example, in an excited carbon atom during the formation of a methane molecule.
Rice. 41. Mutual arrangement of hybrid electron clouds.
Rice. 42. Mutual arrangement of hybrid electron clouds.
Therefore, the methane molecule has the shape of a tetrahedron, and all four bonds in this molecule are equivalent.
Let's return to considering the structure of the water molecule. During its formation, -hybridization of oxygen atomic orbitals occurs. That is why the HOH bond angle in a molecule is close not to, but to the tetrahedral angle. The slight difference between this angle and 109.5° can be understood if we take into account the unequal state of the electron clouds surrounding the oxygen atom in the water molecule. In fact, in the methane molecule (I)
all eight electrons occupying hybrid -orbitals in the carbon atom are involved in the formation of covalent bonds. This causes a symmetrical distribution of electron clouds with respect to the nucleus of the carbon atom. Meanwhile, in the molecule, only four of the eight electrons occupying the hybrid -orbitals of the oxygen atom form bonds, and two electron pairs remain unshared, i.e., they belong only to the oxygen atom. This leads to some asymmetry in the distribution of electron clouds surrounding the oxygen atom and, as a consequence, to a deviation of the angle between bonds from .
When an ammonia molecule is formed, the atomic orbitals of the central atom (nitrogen) also occur. That is why the bond angle is close to tetrahedral. The slight difference in this angle from 109.5° is explained, as in the water molecule, by the asymmetry in the distribution of electron clouds around the nucleus of the nitrogen atom: out of four electron pairs, three participate in the formation of N - H bonds, and one remains unshared.
As shown in Fig. 39, 41 and 42, the hybrid electron clouds are shifted relative to the atomic nucleus.
Therefore, the center of electric charge of a lone electron pair located in a hybrid orbital does not coincide with the position atomic nucleus, i.e., with the center of the positive charge present in the atom. This shift in the charge of a lone electron pair leads to the appearance of a dipole moment, which makes a significant contribution to the total dipole moment of the molecule. It follows from this that the polarity of a molecule depends not only on the polarity of individual bonds and their mutual arrangement (see § 40), but also on the presence of lone electron pairs in hybrid orbitals and on the spatial arrangement of these orbitals.
For elements of the third and subsequent periods, -orbitals can also participate in the formation of hybrid electron clouds. The case of -hybridization is especially important, when one, three and two -orbitals participate in the formation of hybrid orbitals. In this case, six equivalent hybrid orbitals are formed, elongated in the directions towards the vertices of the octahedron. The octahedral structure of the molecule, ions and many others is explained by the hybridization of the atomic orbitals of the central atom.
| Single (σ) | Double (σ+π) | Triple (σ + π + π) |
| C–C C–H C–O H–Cl | C=O C=C O=O | С≡С С≡N N≡N |
Hybridization
If an atom is connected to other atoms by the IDENTICAL BONDS, but orbitals are involved in their formation different types, then the HYBRIDIZATION method is used.
Example:The CH 4 molecule has the shape of a regular tetrahedron, in which all 4 bonds have the same length, strength, and are at the same angles to each other.
However, a tetravalent carbon atom has electrons in three p orbitals and one s orbital. They are different in energy, shape and located in space differently.
To explain, the concept of HYBRIDIZATION is used:
From four atomic orbitals, 4 new ones are formed,
hybrid orbitals, which in space are located at the MAXIMUM DISTANCE FROM EACH OTHER. This is a regular tetrahedron, the angles between the bonds are 109° 29´.
Since one s and three p-shells are involved in the formation of four bonds, this type of hybridization is designated sp 3
Depending on the number and type of orbitals that take part in hybridization, the following types of hybridization are distinguished:
1) sp-hybridization. One s orbital and one p orbital are involved. The molecule has a linear structure, the bond angle is 180 0.
2) sp 2 hybridization. One s orbital and two p orbitals are involved. The molecule is located in a plane (the ends of the hybrid orbitals are directed towards the vertices equilateral triangle), bond angle – 120 0.
3) sp 3 hybridization. One s orbital and three p orbitals are involved. The molecule has a tetrahedral shape, the bond angle is 109.28 0.
How to determine the type of hybridization?
1. Hybridization involves sigma bonds and LONE ION PAIRS.
2. Total number participating orbitals sigma bonds + electron pairs = number of hybrid orbitals and determines the type of hybridization.
Exercise: determine the type of hybridization of the carbon atom in the phosgene molecule.
O=C – Cl
1) carbon forms 2 single bonds (these are sigma bonds) and one double bond (sigma + pi). All 4 electrons of carbon are involved in the formation of these bonds.
2) thus, THREE SIGMA connections will take part in hybridization. This sp 2 - hybridization, the molecule has the shape flat triangle. The pi bond is located perpendicular to the plane of this triangle.
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