Tetrahedral hybrid orbitals

Know the meaning of molecular formula, structural formula, structural (or constitutional) isomers, continuous and branched chain, formal charge, resonance, contributing structures, sigma (a) bond, sp3-hybrid orbitals, tetrahedral carbon. 

The element before carbon in Period 2, boron, has one electron less than carbon, and forms many covalent compounds of type BX3 where X is a monovalent atom or group. In these, the boron uses three sp hybrid orbitals to form three trigonal planar bonds, like carbon in ethene, but the unhybridised 2p orbital is vacant, i.e. it contains no electrons. In the nitrogen atom (one more electron than carbon) one orbital must contain two electrons—the lone pair hence sp hybridisation will give four tetrahedral orbitals, one containing this lone pair. Oxygen similarly hybridised will have two orbitals occupied by lone pairs, and fluorine, three. Hence the hydrides of the elements from carbon to fluorine have the structures... 

Here, the bonding between carbon atoms is briefly reviewed fuller accounts can be found in many standard chemistry textbooks, e.g., [1]. The carbon atom [ground state electronic configuration (ls )(2s 2px2py)] can form sp sp and sp hybrid bonds as a result of promotion and hybridisation. There are four equivalent 2sp hybrid orbitals that are tetrahedrally oriented about the carbon atom and can form four equivalent tetrahedral a bonds by overlap with orbitals of other atoms. An example is the molecule ethane, CjH, where a Csp -Csp (or C-C) a bond is formed between two C atoms by overlap of sp orbitals, and three Csp -Hls a bonds are formed on each C atom. Fig. 1, Al. 

The concept of hybridization explains how carbon forms four equivalent tetrahedral bonds but not why it does so. The shape of the hybrid orbital suggests the answer. When an 5 orbital hybridizes rvith three p orbitals, the resultant sp3 hybrid orbitals are unsyimmetrical about the nucleus. One of the two... 

FIGURE 3.14 Each C H bond in methane is formed by the pairing of an electron in a hydrogen U-orbital and an electron in one of the four sp hybrid orbitals of carbon. Therefore, valence-bond theory predicts four equivalent cr-bonds in a tetrahedral arrangement, which is consistent with experimental results. 

So far, we have not considered whether terminal atoms, such as the Cl atoms in PC15, are hybridized. Because they are bonded to only one other atom, we cannot use bond angles to predict a hybridization scheme. However, spectroscopic data and calculation suggest that both s- and p-orbitals of terminal atoms take part in bond formation, and so it is reasonable to suppose that their orbitals are hybridized. The simplest model is to suppose that the three lone pairs and the bonding pair are arranged tetrahedrally and therefore that the chlorine atoms bond to the phosphorus atom by using sp hybrid orbitals. 

Four equivalent ip -hybridized orbitals achieve maximum distance from one another when they arrange in a tetrahedral structure ... 

An inner atom with a steric number of 4 has tetrahedral electron group geometry and can be described using S p hybrid orbitals. 

Both inner atoms have steric numbers of 4 and tetrahedral electron group geometry, so both can be described using s p hybrid orbitals. All four hydrogen atoms occupy outer positions, and these form bonds to the inner atoms through 1 s-s p overlap. The oxygen atom has two lone pairs, one in each of the two hybrid orbitals not used to form O—H bonds. 

Figure 3.17 Geometry of hybrid orbitals, (a) digonal sp hybrids oppositely directed along the same axis (b) trigonal sp2 hybrids pointing along three axes in a plane inclined at 120° (c) tetrahedral sp3 hybrids pointing towards the comers of a regular tetrahedron. (Reproduced with permission from R. McWeeny, Coulson s Valence, 1979, Oxford University Press, Oxford.)...

From your prior study of chemistry, the hybrid orbital types sp and sp3 used to describe these cases are probably familiar. It is not unusual to hear someone say that CH4 is tetrahedral because the carbon atom is sp3 hybridized. However, CH4 is tetrahedral because that structure represents the configuration of lowest energy, and our way of describing a set of orbitals that matches that geometry is by combining 95... 

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