Factor 4—Orbitals

The three factors we have learned so far will not explain the difference in acidity between the two highlighted protons in the compound below  

In order to determine which proton is more acidic, we remove each proton and compare the resulting conjugate bases  

In both cases, the negative charge is on a carbon, so factor 1 does not help, hi both cases, the charge is not stabilized by resonance, so factor 2 does not help, hi both cases, there are no inductive effects to consider, so factor 3 does not help. The answer here comes from looking at the type of orbital that is accommodating the charge. 

Let s quickly review the shape of hybridized orbitals, sp, sp, and sp orbitals all have roughly the same shape, but they are different in size  

Notice that the sp orbital is smaller and tighter than the other orbitals. It is closer to the nucleus of the atom, which is located at the point where the front lobe (white) 

If we pull off the protons and look at the conjugate bases to compare them, we see this  

Notice that the sp orbital is smaller and tighter than the other orbitals. It is closer to the nucleus of the atom, which is located at the point where the front lobe (white) meets the back lobe (gray). Therefore, a lone pair of electrons residing in an sp orbital will be held closer to the positively charged nucleus and wall be stabilized by being close to the nucleus. 

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SkillBuilder 3.8 Assessing Relative Stability. Factor 4 Orbital... 

Rosen [1931] presented the next significant modification to the wavefunction. Wang s basic treatment was further complicated by the inclusion of a polarization factor orbitals deformed along the molecular axis replaced the original spherically symmetric ones. These orbitals have the form... 

A second detennining factor in the Femii contact mechanism is the requirement tliat the wavefrmction of the bonding orbital has a significant density at each nucleus, in order for the nuclear and the electron magnets to interact. One consequence of this is that K correlates with nuclear volume and therefore rises sharply for heavier nuclei. Thus the constants m the XFI series with X = Si, Ge, Sn and are... 

In equation (bl. 15.24), r is the vector coimecting the electron spin with the nuclear spin, r is the length of this vector and g and are the g-factor and the Boln- magneton of the nucleus, respectively. The dipolar coupling is purely anisotropic, arising from the spin density of the impaired electron in an orbital of non-... 

This lineshape analysis also implies tliat electron-transfer rates should be vibrational-state dependent, which has been observed experimentally [44]- Spin-orbit relaxation has also been identified as an important factor in controlling tire identity of botli electron and vibrational-state distributions in radiationless ET reactions. 

With 4) containing a normalization factor and all permutations over the atomic orbital wave functions i (1 = 1,2,... 2n). Likewise, if all electron pairs were exchanged in a cyclic manner, the product wave function, 4>b, has the fonn ... 

This greater reactivity of the silanes may be due to several factors, for example, the easier approach of an oxygen molecule (which may attach initially to the silane by use of the vacant silicon d orbitals) and the formation of strong Si—O bonds (stronger than C—O). 

Two factors affect the stability of this orbital. The first is the stabilizing influence of the positively charged nuclei at the center of the AOs. This factor requires that the center of the AO be as close as possible to the nucleus. The other factor is the stabilizing overlap between the two constituent AOs, which requires that they approach each other as closely as possible. The best compromise is probably to shift the center of each AO slightly away from its own nucleus towards the other atom, as shown in figure 7-23a. However, these slightly shifted positions are only correct for this particular MO. Others may require a slight shift in the opposite direction. 

To avoid having the wave function zero everywhere (an unacceptable solution ), the spin orbitals must be fundamentally difl erent from one another. For example, they cannot be related by a constant factor. You can write each spin orbital as a product of a space function W hich depen ds on ly on the x, y, and z. coordin ates of th e electron—and a spin fun ction. The space function is usually called themolecnlarorbitah While an in finite number of space functions are possible, only two spin funclions are possible alpha and beta. 

Even with the minimal basis set of atomic orbitals used m most sem i-empirical calculatitm s. the n urn ber of molecii lar orbitals resulting from an SCFcalciilation exceeds the num ber of occupied molecular orbitals by a factor of about two. The n um ber of virtual orbitals in an ah initio calculation depends on the basis set used in this calculation. 

Olh cr I cacLiori s arc con trolled kin ctically, and the most stable product is not the major one observed. In these cases, you must look at the reactant side of the reaction coordinate to discover factors determ in in g th e ou tcorn e. Kloptn an an d Salem developed an analysis of reactivity in terms of two factois an electrostatic in leraclion approxim ated by atom ic ch arges an d a Kron tier orbital interaction, Fleming s book provides an excellent introduction to Ih ese ideas. 

Conversely, these factors dictate that molecular orbital calculations on metals yield less reliable results than with organic corn poiin ds. 

VVc can now see why the normalisation factor of the Slater determinantal wavefunction is I v/N . If each determinant contains N terms then the product of two Slater determinants, ldeU rminant][determinant], contains (N ) terms. However, if the spin orbitals form an oi lhonormal set then oidy products of identical terms from the determinant will be nonzero when integrated over all space. We Ccm illustrate this with the three-electron example, k ljiiiidering just the first two terms in the expansion we obtain the following ... 

Plot this orbital with appropriate scale factors to deteiiiiine the behavior of tE in rectangular coordinates. Describe its behavior in spherical polar coordinates. 

The remainder of the input file gives the basis set. The line, 1 0, specifies the atom center 1 (the only atom in this case) and is terminated by 0. The next line contains a shell type, S for the Is orbital, tells the system that there is 1 primitive Gaussian, and gives the scale factor as 1.0 (unsealed). The next line gives Y = 0.282942 for the Gaussian function and a contiaction coefficient. This is the value of Y, the Gaussian exponential parameter that we found in Computer Project 6-1, Part B. [The precise value for y comes from the closed solution for this problem S/Oir (McWeeny, 1979).] There is only one function, so the contiaction coefficient is 1.0. The line of asterisks tells the system that the input is complete. 

So, for any atom, the orbitals can be labeled by both 1 and m quantum numbers, which play the role that point group labels did for non-linear molecules and X did for linear molecules. Because (i) the kinetic energy operator in the electronic Hamiltonian explicitly contains L2/2mer2, (ii) the Hamiltonian does not contain additional Lz, Lx, or Ly factors. 

Recall that each of these results is subject to multiplication by a factor of (-l) p to account for possible ordering differences in the spin-orbitals in > and >. 

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