Pyramidal carbon methane

The same kind of sp3 hybridization that describes the bonds to carbon in the tetrahedral methane molecule also describes bonds to nitrogen in the trigonal pyramidal ammonia molecule, to oxygen in the bent water molecule, and to all other atoms that VSEPR theory predicts to have a tetrahedral arrangement of four charge clouds. 

For example, hh a disubstituted compound CH Rj (Fig. 8) (1) if the molecule is planar, then two isomers are possible. This planar configuration can be either square or rectangular in each case, there are two isomers only. (2) If the molecule is pyramidal, then two isomers are also possible. There are only two isomers, whether the base is square or rectangular. (3) If the molecule is tetrahedral, then only one form is possible. The carbon atom is at the center of the tetrahedron. In actuality, only one disubstituted isomer is known. Therefore, only the tetrahedral model for a disubstituted methane agrees with the evidence of the isomer number. 

For tetrasubstituted compounds of the type CRjR2R3R4 (Fig. 9) (1) if the molecule is planar, then three isomers are possible. (2) If the molecule is p3uramidal, then six isomers are possible. Each of the forms in Fig. 9, top, drawn as a pyramid, is not superimposable on its mirror image. Thus, three pairs of enantiomers are possible (one of which is shown in Fig. 9, middle). (3) If the molecule is tetrahedral, two isomers are possible, related to one another as object to mirror image. In actuality, only two tetrasubsti-tuted isomers of methane are known (pair of enantiomers). This is strong evidence for the tetrahedral model for the carbon atom. Similar reasoning leads to the same conclusion for trisubstituted methanes. 

How many isomers of formula CH2YZ would be expected from each of the following structures for methane (a) Carbon at the center of a rectangle (b) Carbon at the center of a square (c) carbon at the apex of a square base pyramid (d) carbon in the center of a tetrahedron. 

The three molecules of interest are methane (4A), ammonia (6A), and water (7A), shown first in the Lewis electron dot representations. Using the VSEPR model, these three molecules are drawn again using the wedge-dashed line notation. Methane (CH4, 4B) has no unshared electrons on carbon but there are electrons in the C-H covalent bonds. Assume that repulsion of the electrons in the bonds leads to a tetrahedral arrangement to minimize electronic repulsion. Ammonia (H3N, 6B) has a tetrahedral array around nitrogen if the electron pair is taken into account. If only the atoms are viewed, however, 6B has the pyramidal shape shown. Water (HOH, 7B) has two electron pairs that occupy the corners of a tetrahedral shape, as shown. 

Here H is the physical width of the pore, which is defined as the distance from the plane passing through carbon atoms of the outermost layer of one wall to the corresponding plane of the opposite wall. This formula was suggested by Everett and Fowl [16] and Kaneko et al. [17] for 1-Site model. For the 5-Site models, the accessible volume is calculated based on the pyramid configuration of methane becanse it is energetically favorable. 

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