If its possible also Cyclohexene and Propyne?
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Let us think about the bonding between carbons in propene:
H2C=CH-CH3
There are two sigma bonds, and one pi bond involved. A pi bond indicates p orbitals on both bonded atoms. One p orbital on an atom indicates sp2 hybridization. The sigma bond of the double bond is formed from two sp2 orbitals, and the sigma bond of the adjacent single bond is formed from one sp2 and one sp3 orbital.
P orbitals are orthogonal (that is, perpendicular to the plane in which the other orbitals lie), so in an sp2 hybridized carbon, the p orbital is perpendicular to the plane of the sp2 orbitals, thus the sp2 orbitals are in a single plane, reduced to two dimensions. There are three of these sp2 orbitals, separated from each other by equal angles.
Going back to geometry, if you draw lines from the center of an equilateral triangle to it's corners, the angles formed by these lines are all 120 degrees. A triangle is a polygon with 3 sides, and the prefix for 3 is tri-, so we have "trigonal" (i.e. three sided) geometry at the sp2 carbons.
For sp3 hybridized carbons, there are no p orbitals to force the hybrid orbitals into a plane, so the sp3 orbitals are equally spaced in three dimensions, with four orbitals. If you take a triangle-based pyramid (four vertices, four triangular faces), and draw lines from the center to the vertices, the angles between these lines is 109.5 degrees. Since the pyramid is a three dimensional object, "gon" does not apply (a polygon is two dimensional, a polyhedron is three dimensional), so we have some sort of polyhedron. What does hedron mean? Hedron means face. How many faces does this triangle-based pyramid have? Four, which corresponds to the tetra- prefix. Thus, the sp3 hybridized carbon has "tetrahedral" (i.e. four faced) geometry.
So there you have the geometry for the carbons in propene. How about cyclohexene and propyne? Well, cyclohexene has a single double bond, so only two carbons have p orbitals occupied by the pi bond. That means two of the carbons have hybrid orbitals forced into a 2D plane, and the rest are sp3. Hopefully you recall the geometry at sp2 and sp3 carbons from a paragraph ago.
As far as propyne, the triple bond is composed of a sigma and two pi bonds, so two of the three carbons have two p orbitals, and the third has none. Since the two p orbitals have to be perpendicular to each other, that means the sp hybrid orbitals cannot be in the plane of either p orbital, so they are restricted to one dimension, that which is orthogonal to the two p orbitals. That means the sp hybrid orbitals are restricted to a line. You know the term that means being on a line already, so this geometry should be easy to name: linear.
Long story short, the geometry around carbons in a neutral organic compound depends on the hybridization of the carbon in question. sp3 is tetrahedral, sp2 is trigonal, sp is linear. Number of p orbitals determines hybridization.
H2C=CH-CH3
There are two sigma bonds, and one pi bond involved. A pi bond indicates p orbitals on both bonded atoms. One p orbital on an atom indicates sp2 hybridization. The sigma bond of the double bond is formed from two sp2 orbitals, and the sigma bond of the adjacent single bond is formed from one sp2 and one sp3 orbital.
P orbitals are orthogonal (that is, perpendicular to the plane in which the other orbitals lie), so in an sp2 hybridized carbon, the p orbital is perpendicular to the plane of the sp2 orbitals, thus the sp2 orbitals are in a single plane, reduced to two dimensions. There are three of these sp2 orbitals, separated from each other by equal angles.
Going back to geometry, if you draw lines from the center of an equilateral triangle to it's corners, the angles formed by these lines are all 120 degrees. A triangle is a polygon with 3 sides, and the prefix for 3 is tri-, so we have "trigonal" (i.e. three sided) geometry at the sp2 carbons.
For sp3 hybridized carbons, there are no p orbitals to force the hybrid orbitals into a plane, so the sp3 orbitals are equally spaced in three dimensions, with four orbitals. If you take a triangle-based pyramid (four vertices, four triangular faces), and draw lines from the center to the vertices, the angles between these lines is 109.5 degrees. Since the pyramid is a three dimensional object, "gon" does not apply (a polygon is two dimensional, a polyhedron is three dimensional), so we have some sort of polyhedron. What does hedron mean? Hedron means face. How many faces does this triangle-based pyramid have? Four, which corresponds to the tetra- prefix. Thus, the sp3 hybridized carbon has "tetrahedral" (i.e. four faced) geometry.
So there you have the geometry for the carbons in propene. How about cyclohexene and propyne? Well, cyclohexene has a single double bond, so only two carbons have p orbitals occupied by the pi bond. That means two of the carbons have hybrid orbitals forced into a 2D plane, and the rest are sp3. Hopefully you recall the geometry at sp2 and sp3 carbons from a paragraph ago.
As far as propyne, the triple bond is composed of a sigma and two pi bonds, so two of the three carbons have two p orbitals, and the third has none. Since the two p orbitals have to be perpendicular to each other, that means the sp hybrid orbitals cannot be in the plane of either p orbital, so they are restricted to one dimension, that which is orthogonal to the two p orbitals. That means the sp hybrid orbitals are restricted to a line. You know the term that means being on a line already, so this geometry should be easy to name: linear.
Long story short, the geometry around carbons in a neutral organic compound depends on the hybridization of the carbon in question. sp3 is tetrahedral, sp2 is trigonal, sp is linear. Number of p orbitals determines hybridization.