Be—F bond

The beryllium-fluorine bond is also highly ionic in character. However, there are two such Be-F bonds and the electrical properties of the entire molecule depend upon how these two bonds are oriented relative to each other. We must find the geometrical sum of these two bond dipoles. 

Consider, for example, BeF2, which is a symmetrical, linear molecule. Each Be—F bond is polar (because fluorine has a greater electronegativity than beryllium). Due to the linear shape of this molecule, however, the polarities of the two bonds are directly opposite each other. The two bonding polarities exactly counteract each other, so that BeF2 is a non-polar molecule. The shape of a molecule, combined with the polarity of its individual bonds, therefore, determine polarity. 

For the more tight clusters SiO2 C70 and CS2 C7o, the local minimum of the PES corresponds to symmetric structure D h, the M(SiO) and R CS) distances are, respectively, 0.013 and 0.045 A shorter and their vstr frequencies experience a blue shift by 80 cm-1 (Si-O) and 145-245 cm-1 (C-S). As in the above clusters, the frequency of the deformation vibration nu e ) is red shifted by 78 (SiCA) and 145 (CS2) cm-1. Like the C-O and Be-F bonds, the Si-O bonds compressed in the cage become more polar. In the CS2 C7o cluster, a noticeable charge is transferred from the cage to the sulfur atoms, each of which acquires about 0.01 e (Table 11). 

The VSEPR model correctly predicts that Bep2 is linear with two identical Be—F bonds. How can we use valence-bond theory to describe the bonding The electron configuration of F (ls 2s 2p ) indicates an unpaired electron in a 2p orbital. This electron can be paired with an unpaired Be electron to form a polar covalent bond. Which orbitals on the Be atom, however, overlap with those on the F atoms to form the Be — F bonds ... 

The Be atom now has two unpaired electrons and can therefore form two polar covalent bonds with F atoms. The two bonds would not be identical, however, because a Be 2s orbital would be used to form one of the bonds and a 2p orbital would be used to form the other. Therefore, although the promotion of an electron allows two Be — F bonds to form, we still have not explained the structure of BeF2. 

F 2p atomic orbital Overlap region— F 2p atomic orbital A Figure 9.16 Formation of two equivalent Be—F bonds in BeF2. 

What is the orientation of the two unhybridized p orbitals on Be with respect to the two Be—F bonds ... 

One can analyze the vibrational modes of structural units made of longer and longer chains, more and more complex, as, for example, chains or leaflets of Si04 tetrahedra. The list of predicted modes can then be compared to the experimental observation (see Fig. 14 and Ref. 15). This method allowed the determination of the deformations of the Si04 tetrahedra and the comparison of the force constants of the Si—O, Ge—O, and Be—F bonds. 

Fig. 17.10 Solid lines are the CD functions for the formation of the Be - F bond in F2Be FCl NCH black) and F2B FC1 green). The dotted black line corresponds to the formation of the beryllium bond in the binary complex when atomic positions are frozen to the ones of the ternary complex. In all cases, the z-axis is the one joining the beryllium and the fluorine atom (FCl) and the Be atom lies in the raigin black dot in the z axis). The vertical line identifies the isodensity boundary of the ternary complex...

The CD curves associated to the formation of the Be - F bond in these three systems are shown in Fig. 17.13 as black, blue and red solid lines, respectively for N-base = NCH, NHCH2 and NH3. Remarkably enough, all curves are positive in the Be-F (FCl) region, demonstrating the presence of a significant Be F charge transfer. In the Be-F region, the black curve always assumes smaller values. 

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