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The ethylene molecule has sigma bonds. There are bonds in the ethylene molecule

The ethylene molecule has sigma bonds. There are bonds in the ethylene molecule

Consists of one sigma and one pi-bond, triple - of one sigma- and two orthogonal pi-bonds.

The concept of sigma and pi bonds was developed by Linus Pauling in the 30s of the last century.

L. Pauling's concept of sigma and pi bonds became an integral part of the theory of valence bonds. At present, animated images hybridizations of atomic orbitals have been developed.

However, L. Pauling himself was not satisfied with the description of sigma and pi bonds. At a symposium on theoretical organic chemistry, dedicated to the memory of F. A. Kekule (London, September 1958), he abandoned the σ, π-description, proposed and substantiated the theory of a bent chemical bond . The new theory clearly took into account the physical meaning of the covalent chemical bond.

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Pi-bonds and hybridized sp2 orbitals

Structure carbon atom. Sigma - and pi-bonds. Hybridization. Part 1

Chemistry. Covalent chemical bond in organic compounds. Foxford Online Learning Center

Subtitles

In the last video, we talked about the sigma bond. Let me draw 2 nuclei and orbitals. Here's the sp3 hybrid orbital of this atom, most of it here. And here, too, sp3-hybrid orbital. Here is a small part, here is a large part. A sigma bond is formed where the orbitals overlap. How can another type of connection be formed here? This will require some explanation. This is the sigma bond. It is formed when 2 orbitals overlap on the axis connecting the nuclei of atoms. Another type of bond can be formed by two p-orbitals. I will draw the nuclei of 2 atoms and one p-orbital each. Here are the cores. Now I will draw the orbitals. The P orbital is like a dumbbell. I'll draw them a little closer to each other. Here is a p-orbital in the shape of a dumbbell. This is one of the p-orbitals of the atom. I'll draw more of her. Here is one of the p-orbitals. Like this. And this atom also has a p-orbital parallel to the previous one. Let's say it's like this. Like this. Should have corrected it. And these orbitals overlap. That's it. 2 p-orbitals are parallel to each other. Here hybrid sp3-orbitals are directed at each other. And these are parallel. So p-orbitals are parallel to each other. They overlap here, up and down. This is a P-bond. I will sign. This is a 1 P bond. It is written with a single Greek small letter "P". Well, or so: "P-connection." And this - P bond is formed due to the overlap of p-orbitals. Sigma bonds are regular single bonds, and P bonds are added to them to form double and triple bonds. For a better understanding, consider the ethylene molecule. Its molecule is arranged like this. 2 carbons linked by a double bond, plus 2 hydrogens each. To better understand bond formation, we need to draw orbitals around carbon atoms. So that's it... First I'll draw the sp2 hybrid orbitals. I'll explain what's going on. In the case of methane, 1 carbon atom is bonded to 4 hydrogen atoms, thus forming a three-dimensional tetrahedral structure, like this. This atom is pointed at us. This atom lies in the plane of the page. This atom lies behind the plane of the page, And this one sticks up. This is methane. The carbon atom forms sp3 hybrid orbitals, each of which forms a single sigma bond with one hydrogen atom. Now let's write the electronic configuration of the carbon atom in the methane molecule. Let's start with 1s2. Next should be 2s2 and 2p2, but in fact everything is more interesting. See. There are 2 electrons on the 1s orbital, and instead of 2s and 2p orbitals with 4 electrons in total, they will have sp3 hybrid orbitals: here is one, here is the second, here is the third sp3 hybrid orbital and the fourth. An isolated carbon atom has a 2s orbital and 3 2p orbitals along the x-axis, along the y-axis, and along the z-axis. In the last video, we saw that they mix to form bonds in the methane molecule and the electrons are distributed like this. There are 2 carbon atoms in the ethylene molecule, and at the end it is clear that this is an alkene with a double bond. In this situation, the electronic configuration of carbon looks different. Here's the 1s orbital, and it's still full. It has 2 electrons. And for the electrons of the second shell, I will take a different color. So what's on the second shell? There are no s- and p-orbitals here, because these 4 electrons must be made unpaired in order to form bonds. Each carbon atom forms 4 bonds with 4 electrons. 1,2,3,4. But now the s-orbital hybridizes not with 3 p-orbitals, but with 2 of them. Here is the 2sp2 orbital. S-orbital mixes with 2 p-orbitals. 1s and 2p. And one p-orbital remains the same. And this remaining p-orbital is responsible for the formation of the P-bond. The presence of a P-bond leads to a new phenomenon. The phenomenon of lack of rotation around the axis of communication. Now you will understand. I will draw both carbon atoms in a volume. Now you will understand everything. I'll take a different color for this. Here is a carbon atom. Here is its core. I'll mark it with the letter C, it's carbon. First comes the 1s orbital, this little sphere. Then there are hybrid 2sp2 orbitals. They lie in the same plane, forming a triangle, well, or "pacific". I will show it in scale. This orbital points here. This one is directed there. They have a second, small part, but I won't draw it, because it's easier. They are similar to p-orbitals, but one of the parts is much larger than the second. And the last one is here. It looks a bit like a Mercedes badge if you draw a circle here. This is the left carbon atom. It has 2 hydrogen atoms. Here is 1 atom. There he is, right here. With one electron per 1s orbital. Here is the second hydrogen atom. This atom will be here. And now the right carbon atom. Now we draw it. I'll draw the carbon atoms close together. This is the carbon atom. Here is its 1s orbital. It has the same electronic configuration. 1s orbital around and the same hybrid orbitals. Of all the orbitals of the second shell, I drew these 3. I have not drawn the P-orbital yet. But I will. I'll draw the connections first. The first will be this bond formed by the sp2-hybrid orbital. I'll paint with the same color. This bond is formed by the sp2-hybrid orbital. And this is a sigma bond. The orbitals overlap on the bond axis. Everything is simple here. And there are 2 hydrogen atoms: one bond here, the second bond here. This orbital is slightly larger because it is closer. And this hydrogen atom is here. And that's also sigma bonds, if you notice. The S orbital overlaps with sp2, the overlap lies on the axis connecting the nuclei of both atoms. One sigma bond, the other. Here's another hydrogen atom, also linked by a sigma bond. All bonds in the figure are sigma bonds. In vain I sign them. I will mark them with small Greek letters "sigma". And here too. So this link, this link, this link, this link, this link is a sigma link. And what about the remaining p-orbital of these atoms? They do not lie in the plane of the Mercedes sign, they stick up and down. I'll take a new color for these orbitals. For example, purple. Here is the p-orbital. It is necessary to draw it more, very large. In general, the p-orbital is not that big, but I draw it like this. And this p-orbital is located, for example, along the z-axis, and the rest of the orbitals lie in the xy plane. The z-axis is up and down. The lower parts should also overlap. I'll draw more of them. Like this and like this. These are p orbitals and they overlap. This is how this connection is formed. This is the second component of the double bond. And here it is necessary to explain something. It's a P-bond, and that too. It's all the same P-bond. j The second part of the double bond. What's next? By itself, it is weak, but in combination with a sigma bond, it brings atoms closer together than an ordinary sigma bond. Therefore, a double bond is shorter than a single sigma bond. Now the fun begins. If there was one sigma bond, both groups of atoms could rotate around the axis of the bond. For rotation around the bond axis, a single bond is suitable. But these orbitals are parallel to each other and overlap, and this P-bond does not allow rotation. If one of these groups of atoms rotates, the other rotates with it. The P-bond is part of a double bond, and double bonds are rigid. And these 2 hydrogen atoms cannot rotate separately from the other 2. Their location relative to each other is constant. That's what's happening. I hope you now understand the difference between sigma and p bonds. For a better understanding, let's take the example of acetylene. It is similar to ethylene, but it has a triple bond. One hydrogen atom on each side. Obviously, these bonds are sigma bonds formed by sp orbitals. The 2s orbital hybridizes with one of the p orbitals, the resulting sp hybrid orbitals form sigma bonds, here they are. The remaining 2 bonds are P-bonds. Imagine another p-orbital pointing at us, and here another one, their second halves are directed away from us, and they overlap, and here one hydrogen atom. Maybe I should make a video about it. I hope I didn't confuse you too much.

Carried out by the overlapping of s-atomic orbitals along the line of connection of atoms, pi-bonds arise when the p-atomic orbitals overlap on both sides of the line of connection of atoms. It is believed that the pi bond is realized in multiple bonds - a double bond consists of one sigma and one pi bond, a triple bond consists of one sigma and two orthogonal pi bonds.

The concept of sigma and pi bonds was developed by Linus Pauling in the 30s of the last century. One s- and three p- valence electrons of the carbon atom undergo hybridization and become four equivalent sp 3 hybridized electrons, through which four equivalent chemical bonds are formed in the methane molecule. All bonds in the methane molecule are equidistant from each other, forming a tetrahedral configuration.

In the case of double bond formation, sigma bonds are formed by sp 2 hybridized orbitals. The total number of such bonds on a carbon atom is three and they are located in the same plane. The angle between the bonds is 120°. The pi-bond is located perpendicular to the specified plane (Fig. 1).

In the case of triple bond formation, sigma bonds are formed by sp-hybridized orbitals. The total number of such bonds on a carbon atom is two and they are at an angle of 180° to each other. Two pi-bonds of a triple bond are mutually perpendicular (Fig. 2).

In the case of the formation of an aromatic system, for example, benzene C 6 H 6, each of the six carbon atoms is in the state of sp 2 - hybridization and forms three sigma bonds with bond angles of 120 °. The fourth p-electron of each carbon atom is oriented perpendicular to the plane of the benzene ring (Fig. 3.). In general, a single bond arises, extending to all carbon atoms of the benzene ring. Two regions of pi bonds of high electron density are formed on both sides of the plane of sigma bonds. With such a bond, all carbon atoms in the benzene molecule become equivalent and, therefore, such a system is more stable than a system with three localized double bonds. A non-localized pi bond in the benzene molecule causes an increase in the bond order between carbon atoms and a decrease in the internuclear distance, that is, the chemical bond length d cc in the benzene molecule is 1.39 Å, while d C-C = 1.543 Å, and d C=C = 1.353 Å.

L. Pauling's concept of sigma and pi bonds became an integral part of the theory of valence bonds. Animated images of the hybridization of atomic orbitals have now been developed.

However, L. Pauling himself was not satisfied with the description of sigma and pi bonds. At a symposium on theoretical organic chemistry dedicated to the memory of F. A. Kekule (London, September 1958), he abandoned the σ, π description, proposed and substantiated the theory of a bent chemical bond. The new theory clearly took into account the physical meaning of the covalent chemical bond, namely the Coulomb electron correlation.

Notes

The electronic structure of the ethylene molecule

The carbon atoms in an alkene molecule are linked by a double bond. These atoms are in the state of sp 2 hybridization. The double bond between them is formed from two pairs of shared electrons, i.e. it is a four-electron bond. It is a combination of covalent σ-bonds and π-bonds. The σ bond is formed due to the axial overlap of the sp2 hybrid orbitals, and the π bond is due to the lateral overlap of the unhybridized p orbitals of two carbon atoms (Fig. 1).

Rice. 1. The structure of the ethylene molecule.

Five σ-bonds of two sp 2 -hybridized carbon atoms lie in the same plane at an angle of 120 o and make up the σ-skeleton of the molecule. Above and below this plane, the electron density of the π-bond is symmetrically located, which can also be depicted as a plane perpendicular to the σ-skeleton.

When a π-bond is formed, carbon atoms approach each other, because the internuclear space in a double bond is more saturated with electrons than in a σ-bond. This contracts the atomic nuclei and therefore the length of the double bond (0.133 nm) is less than the single bond (0.154 nm).

Examples of problem solving

EXAMPLE 1

Exercise

As a result of the addition of iodine to ethylene, 98.7 g of the iodo derivative were obtained. Calculate the mass and amount of the ethylene substance taken for the reaction.

Solution

We write the reaction equation for the addition of iodine to ethylene:

H 2 C \u003d CH 2 + I 2 → IH 2 C - CH 2 I.

As a result of the reaction, an iodo derivative, diiodoethane, was formed. We calculate its amount of substance (the molar mass is - 282 g / mol):

n(C 2 H 4 I 2) \u003d m (C 2 H 4 I 2) / M (C 2 H 4 I 2);

n (C 2 H 4 I 2) \u003d 98.7 / 282 \u003d 0.35 mol.

According to the reaction equation n(C 2 H 4 I 2): n(C 2 H 4) = 1:1, i.e. n (C 2 H 4 I 2) \u003d n (C 2 H 4) \u003d 0.35 mol. Then the mass of ethylene will be equal to (molar mass - 28 g / mol):

m(C 2 H 4) = n (C 2 H 4) ×M (C 2 H 4);

m(C 2 H 4) \u003d 0.35 × 28 \u003d 9.8 g.

Answer

The mass of ethylene is 9.8 g, the amount of ethylene substance is 0.35 mol.

EXAMPLE 2

Exercise

Calculate the volume of ethylene, reduced to normal conditions, that can be obtained from technical ethyl alcohol C 2 H 5 OH weighing 300 g. Note that industrial alcohol contains impurities, the mass fraction of which is 8%.

Solution

We write the reaction equation for the production of ethylene from ethyl alcohol:

C 2 H 5 OH (H 2 SO 4) → C 2 H 4 + H 2 O.

Find the mass of pure (without impurities) ethyl alcohol. To do this, we first calculate mass fraction:

ω pure (C 2 H 5 OH) \u003d ω impure (C 2 H 5 OH) - ω impurity;

ω pure (C 2 H 5 OH) = 100% - 8% = 92%.

m pure (C 2 H 5 OH) \u003d m impure (C 2 H 5 OH) ×ω pure (C 2 H 5 OH) / 100%;

m pure (C 2 H 5 OH) = 300 × 92 / 100% = 276 g.

Let's determine the amount of ethyl alcohol substance (molar mass - 46 g / mol):

n(C 2 H 5 OH) \u003d m (C 2 H 5 OH) / M (C 2 H 5 OH);

n(C 2 H 5 OH) = 276/46 = 3.83 mol.

According to the reaction equation n(C 2 H 5 OH): n(C 2 H 4) = 1:1, i.e. n (C 2 H 5 OH) \u003d n (C 2 H 4) \u003d 3.83 mol. Then the volume of ethylene will be equal to:

V(C 2 H 4) = n(C 2 H 4) × V m ;

V (C 2 H 4) \u003d 3.83 × 22.4 \u003d 85.792 liters.

Answer

The volume of ethylene is 85.792 liters.

what is the difference between sigma and pi bonds in ethylene and acetylene molecules → and got the best answer

Answer from Yatiana Ivanova[guru]
Sigma bonds can be formed by hybrid and/or non-hybrid orbitals, but are always directed along the axis connecting the centers of atoms. At the carbon atom, sigma bonds form only hybrid orbitals. Pi bonds can only be formed by non-hybrid p-orbitals located perpendicular to the plane of the molecule (and the line connecting the centers of atoms). The pi bond is characterized by lateral overlap of p-orbitals above and below the plane of the molecule. The area of overlap of electron clouds (orbitals) in the sigma bond is larger than in the pi bond, therefore the sigma bond is stronger. Features of bonds in ethylene and acetylene molecules. - In both molecules, carbon atoms are bonded to hydrogen atoms by sigma bonds (hybrid orbitals of carbon atoms overlap with non-hybrid s-orbitals of hydrogen atoms). - Both molecules have multiple bonds between carbon atoms, i.e. there is both a sigma bond and pi -bonds.- In the C2H4 ethylene molecule, the carbon atom is in the sp2-hybridization state (3 hybrid sp2-orbitals and 1 non-hybrid p-orbital); in the C2H2 acetylene molecule, the carbon atom is in a state of sp-hybridization (2 hybrid sp-orbitals and 2 non-hybrid p-orbitals).- Therefore, in the ethylene molecule, there is a double bond between carbon atoms (1 sigma- and 1 pi-bond), acetylene molecule - a triple bond (1 sigma and 2 pi bonds located in mutually perpendicular planes). Length C-C connections in an acetylene molecule it is less than in an ethylene molecule, and the bond strength is greater. Both ethylene and acetylene are characterized by addition reactions along a pi bond, which is less strong and, therefore, more reactive than a sigma bond. In the acetylene molecule, the addition occurs in 2 stages: first through one pi bond, then through the other.

Answer from Ksenia[guru]
Sigma bonds are single, simple, strong bonds.
Pi ties are not strong ties.
And in what substances does not matter.