Tetrahedral molecular shape table

Many elements of the periodic table, from titanium and tin to carbon and chlorine, exhibit tetrahedral electron group geometry and tetrahedral molecular shapes. In particular, silicon displays tetrahedral shapes in virtually all of its stable compounds. 

In any molecule in which there are no nonbonding pairs around the central atom, the molecular shape is the same as the molecular geometry. Thus, to use the examples from Table 6.2, all three two-substituent molecules have both a linear geometry and a linear shape. Both BH3 and H2CO have a triangular planar shape, CH4 has a tetrahedral shape, PF5 a triangular bipyramidal shape, and SF6 a square bipyramidal shape. 

The molecular shape indices (also called k, k, K-index) represent the overall molecular shape in three values based on counts of one-bond, two-bond, and three-bond fragments [58, 59]. Local topology (e.g., tetrahedral or planar coordination) as well as atom types are considered using parameter tables, including information such as valence radii and connectivity of the atoms. 

Three of the five basic molecular shapes are linear, trigonal planar and tetrahedral. Table 4.9 shows the arrangement of the electron pairs (charge centres) that results in minimum repulsion and the basic shapes of the molecules. The two other basic shapes adopted by molecules, trigonal bipyramidal and octahedral, are discussed in Chapter 14. 

In the electron-dot formula of water, H2O, there are also four electron groups, which have minimal repulsion when the electron-group geometry is tetrahedral. However, in H2O, two of the electron groups are lone pairs of electrons. Because the shape of H2O is determined by the two H atoms bonded to the central O atom, the H2O molecule has a bent shape with a bond angle of 109°. Table 10.3 gives the molecular shapes for molecules with two, three, and four bonded atoms. 

The VSEPR notation for the Cl2F+ ion is AX2E3. According to Table 11.1, molecules of this type exhibit an angular molecular geometry. Our next task is to select a hybridization scheme that is consistent with the predicted shape. It turns out that the only way we can end up with a tetrahedral array of electron groups is if the central chlorine atom is sp3 hybridized. In this scheme, two of the sp3 hybrid orbitals are filled, while the remaining two are half occupied. 

See Practice Problem Exercise 12.4) The four pairs of electrons around the sulfur require a tetrahedral arrangement. In this case two pairs are shared with hydrogen atoms, leaving two lone pairs. Thus the molecular structure is bent or V-shaped (case 5 in Table 12.4). 

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