Tetrahedral hybridisation

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. 

A carbon atom combining with four other atoms clearly does not use the one 2s and the three 2p atomic orbitals that would now be available, for this would lead to the formation of three directed bonds, mutually at right angles (with the three 2p orbitals), and one different, non-directed bond (with the spherical 2s orbital). Whereas in fact, the four C—H bonds in, for example, methane are known to be identical and symmetrically (tetrahedrally) disposed at an angle of 109° 28 to each other. This may be accounted for on the basis of redeploying the 2s and the three 2p atomic orbitals so as to yield four new (identical) orbitals, which are capable of forming stronger bonds (cf. p. 5). These new orbitals are known as sp3 hybrid atomic orbitals, and the process by which they are obtained as hybridisation ... 

In steric terms there is a relief of crowding on going from the initial halide, with a tetrahedral disposition of four substituents about the sp3 hybridised carbon atom, to the carbocation, with a planar disposition of only three substituents (cf. five for the SN2 T.S.) about the now sp2 hybridised carbon atom. The three substituents are as far apart from each other as they can get in the planar carbocation, and the relative relief of crowding (halide - carbocation) will increase as the substituents increase in size (H- Me- Me3C). The SN1 reaction rate would thus be expected to increase markedly (on both electronic and steric grounds) as the series of halides is traversed. It has not, however, proved possible to confirm this experimentally by setting up conditions such that the four halides of Fig. 4.1 (p. 82) all react via the SN1 pathway. 

Crystal data for all four amides are uniformly similar. While amides 31b and 31f (IXnI = 65.6° and 65.3°, respectively) are more pyramidal than the urea and carbamate (IxnI = 57.1° and 56.3°, respectively) all four possess sp3-hybridised nitrogens. Average angles at nitrogen for the two amides are smaller than that required by pure tetrahedral geometry as exemplified by the tetrahedral nitrogen in l-aza-2-adamantanone 1. 

Depending on the ligand field strength and the number of ligand systems that can be accommodated, hybridisations other than octahedral are possible. The tetracyanocuprate ion in line (f) of Table 5.1 has a tetrahedral configuration arising from sp3 hybridisation, as... 

So far we have discussed conformations of a molecule obtained by rotation along sp -sp3 bond i.e., between two tetrahedral carbon atoms. But there are many compounds in which one carbon is in a state of sp2 hybridisation. Examples are substituted alkenes where one carbon atom is tetrahedral and the other trigonal, for example propene ... 

The carbon atom now has four unpaired electrons and can form four bonds, but we still have not explained why these bonds adopt a tetrahedral arrangement. To do this, we introduce the concept of hybridisation. 

We have already explained. In terms of hybridisation, how a carbon atom can form four sp hybrid orbitals (see p. 47). We can apply this concept to explain the bonding in alkanes. Ethane is taken as an example of a typical alkane. The four sp hybrid orbitals on each carbon atom will overlap end-on with four other orbitals three hydrogen Is orbitals and one sp hybrid orbital on the other carbon atom. Four cr bonds will be formed and they will adopt a tetrahedral arrangement. This is illustrated for ethane in the diagram. 

A nucleophile attacks the electrophilic carbon atom of the polar carbonyl group from a direction approximately perpendicular to the plane of sp hybridised orbitals of carbonyl carbon (Fig. 12.2). The hybridisation of carbon changes from sjf to s f in this process, and a tetrahedral alkoxlde intermediate is produced. This... 

Possible explanations for this unusual behaviour have been offered by first-principles band structure calculations and synchrotron X-ray structural studies. The former revealed that, because of hybridisation between Ba and C orbitals, the rigid-band model is not appropriate for the description of the electronic properties and the calculated N(ev) for K3Ba3C60 and Rb3Ba3C60 are almost identical [69]. The structural analysis revealed positional disorder of the Ba2+ and K+ ions in the distorted tetrahedral sites of the bcc structure and the existence of short Ba-C and K-C contacts, consistent with strong hybridisation between the K, Ba and the C60 states [70]. It is important to notice that in K3Ba3C60 there is a perfect matching between the size of K+ and Ba2+ ions, while in both the Rb+ and Cs+ analogues, there is a considerable mismatch, which leads to fundamental structural... 

Because amines are tetrahedral so they are chiral if they have three different substituents. However, it is not possible to separate the enantiomers of a chiral amine because amines can easily undergo pyramidal inversion. This process interconverts the enantiomers. The inversion involves a change of hybridisation where the nitrogen becomes sp2 hybridised rather than sp3 hybridised. Because of this, the molecule becomes planar and the lone pair of electrons occupy a p orbital. Once the hybridisation reverts back to sp3, the molecule can either revert back to its original shape or invert. 

The four sp3 orbitals for these three atoms (i.e. N, O and Cl) form a tetrahedral arrangement having one or more of the hybridised orbitals occupied by a lone pair of electrons. For an isolated atom, nitrogen forms a pyramidal shape where the bond angles are slightly less than 109.5 (c. 107 )fig.(a). This compression of the bond angles is because of the orbital containing the lone... 

Alkanes These are the open chain organic compounds having the general formula Cn H2n + 2. In them all the bonds are o-bonds and so they are also called saturated hydrocarbons. All the carbon atoms in any alkane are sp3 hybridised and so their shape in tetrahedral. Since C-C and C-H o-bonds are strong. So the alkanes are unreactive to most of the chemical reagents. 

---