To obtain tlie whole orbital we must multiply R(r) by the appropriate angular part. For example, we would use the following expressions for the Is, 2s and 2pz orbitals [Pg.75]

Table 15.4. Comparison of some one- and two-electron matrix elements for pure and SCVB 2pz orbitals. All energies are in hartrees. |

The question arises as to the directions of these two orbitals. From Eq. (11) it was anticipated that there would be considerable occupancy of the nitrogen 2pz orbital hence, the value of 2a/Bo = 0.361 is associated with C22, the fractional occupancy of the p, orbital. The form of the ji tensor requires that the other p orbital be in the x direction or cl = 0.078. This occupancy of the px orbital was not anticipated in Eq. (11). [Pg.278]

Two fluorine atoms join together to increase their number of valence electrons to eight. When their half - filled 2pz orbitals overlap, a bond is formed. As a result, each fluorine atom completes its octet and together they form the stable fluorine [Pg.8]

The intramolecular rearrangement of allylic boranes (Eq. 5) clearly involves a multiple boron-carbon bond in the transition state (45), as the boron 2pz-orbital interacts with the 7r-bonding MO. [Pg.368]

You might recall from Section 1.9 that a carbon-carbon triple bond results from the interaction of two sp-hybridized carbon atoms. The two sp hybrid orbitals of carbon lie at an angle of 180° to each other along an axis perpendicular to the axes of the two unhybridized 2py and 2pz orbitals. When two sp-hybridized carbons approach each other, one sp-sp a bond and two p-p -rr bonds are [Pg.261]

In this case since carbon has only two unpaired electrons, it seems likely that it will only form only two covalent bonds, but it is known that carbon can form four covalent bonds. To form four bonds, one electron is promoted from the 2s orbital to the 2pz orbital. Then the one 2s orbital and three 2p orbitals mix together to form four new sp3 hybrid orbitals as shown in Figure 5. So in this case of hybridization, three p and one s orbital combine to give four identical sp3 orbitals. [Pg.25]

For a function to transform according to a specific irreducible representation means that the function, when operated upon by a point-group symmetry operator, yields a linear combination of the functions that transform according to that irreducible representation. For example, a 2pz orbital (z is the C3 axis of NH3) on the nitrogen atom [Pg.590]

The picture from the results of Table 11.8 is not significantly different. Structure 3 from before is now 1 (with the same coefficient, of course), but we have a mixture of the same atomic configurations. The new structures 2 and 3 show standard two-electron bonds involving 2s and 2pz orbitals on opposite atoms. This feature is not so clear from the standard tableaux functions. [Pg.152]

EPR. The frozen solution EPR spectrum of 75 shown in Fig. 13 exhibits rhombic symmetry typical of a Ni(III) bis(dithiolene) complex (126), with g-tensor values of gx = 2.13, gy = 2.04, g, = 1.99. The nickel dimer can thus be viewed as a classical bis(dithiolene) moiety with a 3B3g ground state where the odd electron orbital composed primarily of a metal dyZ orbital and sulfur 2pz orbitals. The Ni(II) in the pz, as expected, is EPR silent (19, 122). [Pg.519]

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