Hybridization and Bonding

In summary, the hybridization of carbon can vary depending on the bonding partners, such as sp2 in ethene and sp3 in CO2. The trigonal bipyramidal structure of PCl5 is the most optimal due to the arrangement of bonding orbitals. Axial and equatorial bonds refer to the orientation of bonds in relation to the central atom. There is no one formula to calculate hybridization as it depends on the specific molecule and bonding partners.
  • #1
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Question 1:

When hybridization takes place, how do all the hybridized orbitals have the same energy?

Here is an extract from my book:

sp2 Hybridisation in C2H4: In the formation
of ethene molecule, one of the sp2 hybrid
orbitals of carbon atom overlaps axially with
sp2 hybridised orbital of another carbon atom
to form C–C sigma bond. While the other two
sp2 hybrid orbitals of each carbon atom are
used for making sp2–s sigma bond with two
hydrogen atoms. The unhybridised orbital
(2px or 2py) of one carbon atom overlaps
sidewise with the similar orbital of the other
carbon atom to form weak π bond, which
consists of two equal electron clouds
distributed above and below the plane of
carbon and hydrogen atoms.

The last time I checked, carbon was sp3 hybridized, and what is the "unhybridized orbital"? Why does the hybridization of carbon change with the formation of multiple bonds?

Question 2:

PCl5 has a trigonal bipyramidal structure, I do not understand how this is possible. The electronic configuration of the valence shell of P is 3s2 3p3. In its exited state, (I think) one electron from the 3s orbital will jump to the 4s orbital. So, totally, there are 2 s orbitals and 3 p orbitals taking part in bond formation with the 5 Cl atoms. The p orbitals will be bonded mutually perpendicularly to each other, where as the 2 s orbitals will be bonded in a position in which there is minimum repulsion from the other bonded pairs of electrons. But, the trigonal bipyramidal structure does not satisfy this condition. Why?


Question 3:

What is the difference between axial and equatorial bonds?

Question 4:

Can someone give me a formula to calculate the hybridization which will work for all the compounds? (I know 3 formulae, but each one has limitations)

Thanks in advance.
 
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  • #2
1. Carbon isn't always sp3 hybridized. It's dependent on what it's bonded to. What about in CO2, where it shares a double bond with each oxygen atom? In that case, it's sp1 hybridized, as there are only 2 bonding regions.

A carbon with 4 bonding regions is sp3, 3 is sp2, 2 is sp1.

2. Have you looked at the molecule? Trigonal bipyramidal is the optimal structure. If you move any chlorines, you'd be decreasing bond angles.

http://en.wikipedia.org/wiki/File:Phosphorus-pentachloride-3D-balls.png

3. Take glucose, for example-

http://rmni.iqfr.csic.es/guide/eNMR/sugar/gif/gluco.GIF

In the alpha confirmation, the bond between the rightmost carbon and the OH group is perpendicular to the bonds in the middle of the molecule. That's an axial bond. On the right, you'll see that the carbon-OH bond is parallel to the central bonds, which is equatorial.

I cannot answer 4.
 
  • #3
Carbon "isn't" hybridized. It is a model to describe the bonding. Usually, you can describe bonding with many different models. E.g. it is equally possible to describe bonding in ethylene (C2H4) using sp3 hybrid orbitals on C.
 
  • #4
aroc91 said:
2. Have you looked at the molecule? Trigonal bipyramidal is the optimal structure. If you move any chlorines, you'd be decreasing bond angles.

The structure of CH4 is not square planar but tetrahedral, because px, py and pz are mutually perpendicular to each other, and there is no specific direction for s orbital, so it is formed where there is minimum bond pair repulsion. I was trying to use a similar logic for PCl5.
 
  • #5


I am happy to provide a response to these questions on hybridization and bonding.

Question 1: Hybridization occurs when atomic orbitals combine to form new hybrid orbitals with different shapes and energies. In the case of sp2 hybridization in C2H4, the three sp2 hybrid orbitals have the same energy because they are formed by mixing one s orbital and two p orbitals, which have the same energy. This results in three equivalent sp2 hybrid orbitals with the same energy level. The unhybridized orbitals (2px or 2py) are not involved in bonding and therefore do not affect the energy levels of the hybrid orbitals.

Question 2: The electronic configuration of PCl5 is actually 3s2 3p3 3d0. In its excited state, one electron from the 3s orbital does not jump to the 4s orbital as there is no 4s orbital in the valence shell of P. Instead, one electron from the 3p orbital jumps to the 3d orbital, resulting in five valence electrons for bonding with the five Cl atoms. The trigonal bipyramidal structure is possible because the five bonding orbitals (three from the 3p and two from the 3d) are oriented in a way that minimizes repulsion between the bonding pairs of electrons.

Question 3: Axial bonds refer to the bonds that are perpendicular to the plane of a molecule, while equatorial bonds refer to the bonds that are in the same plane as the molecule. In molecules with a trigonal bipyramidal geometry, the axial bonds are oriented at 90 degrees from the equatorial bonds. This difference in orientation can affect the stability and reactivity of a molecule.

Question 4: There is no single formula that can accurately predict the hybridization of all compounds. The three common formulas (VSEPR theory, Valence Bond theory, and Molecular Orbital theory) have their limitations and are most accurate for simple molecules. For more complex molecules, a combination of these theories and experimental data is often used to determine the hybridization.
 

1. What is hybridization and why is it important in chemistry?

Hybridization is a process in which atomic orbitals mix to form new hybrid orbitals, resulting in the formation of new covalent bonds. It is important in chemistry because it helps explain the shapes and properties of molecules, as well as their bonding behavior.

2. How does hybridization affect the bond angles in a molecule?

Hybridization determines the geometry of a molecule, which in turn affects the bond angles. For example, molecules with sp3 hybridization have a tetrahedral geometry and bond angles of approximately 109.5°, while molecules with sp2 hybridization have a trigonal planar geometry and bond angles of approximately 120°.

3. What is the difference between sigma and pi bonds?

Sigma bonds are formed by the end-to-end overlap of atomic orbitals, while pi bonds are formed by the side-to-side overlap of atomic orbitals. Sigma bonds are typically stronger and more stable than pi bonds.

4. How does hybridization affect the bond strength and stability of a molecule?

Hybridization can affect the bond strength and stability of a molecule by changing the type of bond formed. For example, a molecule with sp3 hybridization will have stronger and more stable sigma bonds compared to a molecule with sp2 hybridization, which will have both sigma and weaker pi bonds.

5. Can hybridization occur in all types of chemical bonds?

Yes, hybridization can occur in all types of chemical bonds, including covalent, ionic, and metallic bonds. However, it is most commonly observed in covalent bonds, where the sharing of electrons between atoms is necessary for hybridization to occur.

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