By John G. Verkade
Realizing molecular orbitals (MOs) is a prerequisite to appreciating many actual and chemical homes of topic. This commonly revised moment variation of A Pictorial method of Molecular Bonding offers the author's cutting edge method of MOs, producing them pictorially for a large choice of molecular geometries. an incredible enhancement to the second one variation is the computer- and Macintosh-compatible Nodegame software program, that's coordinated with the textual content and aids in pictorially instructing molecular orbital concept utilizing generator orbitals.
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2. Generator and' Molecular Orbitals The electronic structure and energy of a molecule is determined by the occupied MOs. What do MOs look like? Like AOs, they possess lobes and nodes. The shapes of the nodes of AOs are a consequence of the spherical symmetry of the potential energy of an electron around one nucleus. Because molecules have more than one nucleus, the potential energy of an electron will not have spherical symmetry and hence the patterns of its MO nodes and lobes will be less than spherically symmetrical.
Bonds which are weak, average, and strong have dissociation energies of about 2, 5, and 8 eV, respectively. Weak bonds tend to be easily stretched and compressed compared to strong bonds and their oscillations also tend to be larger than those of strong bonds. These characteristics are reflected in Figure 3-2 in which the potential well is seen to be wider than in Figure 3-1 b. Molecular vibrations tend to be quite small 53 Generator and Molecular Orbitals RO R2 U(R) f-----~-,------,----_,_--=_--- R Figure 3-2.
L_V x x (a) (b) (c) Figure 2-10. Drawings depicting a vector r (a), a unit vector e (b), and the projection of r onto e (c). Pe (r) = (ex' r) = ~x, x Pe (r) = (e y • r) = 1JY, Pe (r) = (e z ' r) = (z Y or Pe/r) = (ej · r). z (2-19) Here ex = (1,0,0), ey = (0, 1,0), and ez = (0,0, 1) are the unit vectors in the x, Y, and z directions. Equations 2-19 then permit us to rewrite Equations 2-17 as Equation 2-20: kr (2pj) = KPe(r)eJ where} = x, y, z. (2-20) We already know, of course, that (2px) , (2py), and (2pz) possess identical shapes and differ only in the directions along which they are oriented.