The s Orbitals

Recall that the / quantum number for the s orbitals is 0 therefore, the W quantum number must be 0. Thus, for each value of n, there is only one s orbital. 

So how do s orbitals differ as the value of n changes One way to address this question is to look at the radial probability function, also called the radial probability density, which is defined as the probability that we will find the electron at a specific distance from the nucleus. 

For the Is orbital, we see that the probability rises rapidly as we move away from the nucleus, maximizing at about 0.5 A. Thus, when the electron occupies the Is orbital, it is most likely to be found this distance from the nucleus. Notice also that in the Is orbital the probability of finding the electron at a distance greater than about 3 A from the nucleus is essentially zero. 

How many maxima would you expect to find in the radial probability function for the 4s orbital of the hydrogen atom How many nodes would you expect in this function  

A FIGURE 6.18 Radial probability distributions for the Is, 2s, and 3s orbitals of hydrogen. These graphs of the radial probability function plot probability of finding the electron as a function of distance from the nucleus. As n increases, the most likely distance at which to find the electron (the highest peak) moves farther from the nucleus. 

Recall that the I quantum number for the s orbitals is 0 therefore, the mi quantum number must be 0. Thus, for each value of n, there is only one s orbital. So how do s orbitals differ as the value of n changes For example, how does the electron-density distribution of the hydrogen atom change when the electron is excited from the Is orbital to the 2s orbital To address this question, we will look at the radial probability density, which is the probability that the electron is at a specific distance from the nucleus. 

Comparing the radial probability distributions for the Is, 2s, and 3s orbitals reveals three trends  

For an ns orbital, the number of peaks is equal to n, with the outermost peak being larger than inner ones. 

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We wish to construct linear combinations of the atomic orbitals such that the overall wavefunction meets the Bloch requirement. Suppose the s orbitals in our lattice are labelled X , where the wth orbital is located at position x = na. An acceptable linear combination of these orbitals that satisfies the Bloch requirements is ... 

FIGURE 2 8 sp Hybridization (a) Electron configuration of carbon in its most stable state (b) Mixing the s orbital with the three p orbitals generates four sp hybrid orbitals The four sp hybrid orbitals are of equal energy therefore the four valence electrons are distributed evenly among them The axes of the four sp orbitals are directed toward the corners of a tetrahedron... 

A UHF wave function may also be a necessary description when the effects of spin polarization are required. As discussed in Differences Between INDO and UNDO, a Restricted Hartree-Fock description will not properly describe a situation such as the methyl radical. The unpaired electron in this molecule occupies a p-orbital with a node in the plane of the molecule. When an RHF description is used (all the s orbitals have paired electrons), then no spin density exists anywhere in the s system. With a UHF description, however, the spin-up electron in the p-orbital interacts differently with spin-up and spin-down electrons in the s system and the s-orbitals become spatially separate for spin-up and spin-down electrons with resultant spin density in the s system. 

This point is illustrated in Figure 8.13 which shows the X-ray photoelectron spectrum of a 2 1 mixture of CO and CO2 gases obtained with MgXa (1253.7 eV) source radiation. The ionization energy for removal of an electron from the s orbital on a carbon atom, referred to as the C s ionization energy, is 295.8 eV in CO and 297.8 eV in CO2, these being quite comfortably resolved. The O s ionization energy is 541.1 eV in CO and 539.8 eV in CO2, which are also resolved. 

Orbitals 7 and 9 (the latter is the LUMO) of formaldehyde exhibit this same character. Orbital 7 is a bonding 7t orbital, and orbital 9 is a Tt . However, the n orbital formed of the pj orbitals from the carbon and the oxygen (which also lie in the YZ plane) is not the HOMO. Instead, an orbital formed from Pj, orbitals from the carbon and the oxygen and from the s orbitals on the hydrogens is the highest occupied orbital. The contributions from the carbon and oxygen are situated along the double bond while the HOMO in ethylene was perpendicular to this bond. 

What do orbitals look like There are four different kinds of orbitals, denoted s, p, d, and f] each with a different shape. Of the four, we ll be concerned primarily with s and p orbitals because these are the most common in organic and biological chemistry. The s orbitals are spherical, with the nucleus at their center p orbitals are dumbbell-shaped and four of the five d orbitals are doverleaf-shaped, as shown in Figure 1.3. The fifth d orbital is shaped like an elongated dumbbell with a doughnut around its middle. 

Figure 1.3 Representations of s, p, and d orbitals. The s orbitals are spherical, the p orbitals are dumbbell-shaped, and four of the five d orbitals are cloverleafshaped. Different lobes of p orbitals are often drawn for convenience as teardrops, but their true shape is more like that of a doorknob, as indicated.

The asymmetry of sp3 orbitals arises because, as noted previously, the two lobes of a p orbital have different algebraic signs, + and -. Thus, when a p orbital hybridizes with an s orbital, the positive p lobe adds to the s orbital but the negative p lobe subtracts from the s orbital. The resultant hybrid orbital is therefore unsymmetrical about the nucleus and is strongly oriented in one direction. 

These 20 cases do not represent anomalies to the order of orbital filling which is invariably governed by the n + ( rule but are anomalous in the sense that the s orbital is not completely filled before the corresponding d orbital begins to fill. 

The plot with the most electron density closest to the origin (0, 0—the nucleus) arises from the s-orbital. Curve (b) corresponds to the 3s-orbital curve (a) corresponds to the 3p-orbital. 

There are 2.56 d orbitals available for bond formation. To form 5.78 bonds these would hybridize with the s orbital and 2.22 of the less stable p orbitals. In copper, with one electron more than nickel, there is available an additional 0.39 electron after the hole in the atomic d orbitals is filled. This might take part in bond formation, with use of additional Ap orbital. However, the increase in interatomic distance from nickel to copper suggests that it forms part of an unshared pair with part of the bonding electrons, thus decreasing the effective number of bonds. 

It may be mentioned that the possibility of bivalence of tin in grey tin and the mercury alloy, suggested by the bipositive oxidation state of the element in many of its compounds, is ruled out because it leads to too small a value of R 1)—smaller than that for quadrivalent tin, whereas a larger value would be expected as the result of the appropriation of much of the s orbital by the unshared pair. 

As to the first, we note the interaction of the s orbital of atom A with the s orbital of B, the with the and the, pair of A with the p y pair of B. In principle, of course, we could have considered the possibility of an interaction between, say, the s orbital on A with a p orbital on B as shown in Fig. 6-2. The sketch shows that net overlap between these orbitals is zero and so no bonding or antibonding molecular orbitals are formed in this way. Now the labels s and p here... 

Thus, l,6-methano[10]annulene (77) and its oxygen and nitrogen analogs 78 and 79 have been prepared and are stable compounds that undergo aromatic substitution and are diatropic. For example, the perimeter protons of 77 are found at 6.9-7.3 5, while the bridge protons are at —0.5 5. The crystal structure of 77 shows that the perimeter is nonplanar, but the bond distances are in the range 1.37-1.42A. It has therefore been amply demonstrated that a closed loop of 10 electrons is an aromatic system, although some molecules that could conceivably have such a system are too distorted from planarity to be aromatic. A small distortion from planarity (as in 77) does not prevent aromaticity, at least in part because the s orbitals so distort themselves as to maximize the favorable (parallel) overlap of p... 

The atomic basis consists in a double-zeta set expanded with polarization functions (DZP) and augmented by diffuse functions (DZPR). Exponents and contraction coefficient are from McLean and Chandler 1980 [18] diffuse functions, centered on the heavy atoms with exponents of 0.023 for the s orbitals and 0.021 for the p orbitals are from Dunning and Hay 1977 [34]. Extension of the DZP basis set with two sets of diffuse s (0.0437, 0.0184) and p (0.0399, 0.0168) functions (DZPRR) has also been tested. 

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