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2pz orbital

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]

Sr) and the nonbonding orbital formed by the Oxygen 2s and 2pz orbitals eombining to form the "lone pair" orbital direeted along the z-axis away from the two Hydrogen atoms). [Pg.173]

The three integrals shown above can be seen to be equal and to be of the exchange-integral form by expressing the integrals in terms of integrals over cartesian functions and recognizing identities due to the equivalence of the 2px, 2py, and 2pz orbitals. For example. [Pg.285]

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]

FIGURE 1 3 Boundary surfaces of the 2p orbitals The wave function changes sign at the nucleus The two halves of each orbital are indicated by different colors The yz plane is a nodal surface for the Ip orbital The probability of finding a electron in the yz plane is zero Anal ogously the xz plane is a nodal surface for the 2py orbital and the xy plane is a nodal surface for the 2pz orbital You may examine different presentations of a 2p orbital on Learning By Modeling... [Pg.9]

Normally, you would expects all 2p orbitals in a given first row atom to be identical, regardless of their occupancy. This is only true when you perform calculations using Extended Hiickel. The orbitals derived from SCE calculations depend sensitively on their occupation. Eor example, the 2px, 2py, and 2pz orbitals are not degenerate for a CNDO calculation of atomic oxygen. This is especially important when you look at d orbital splittings in transition metals. To see a clear delineation between t2u and eg levels you must use EHT, rather than other semiempirical methods. [Pg.148]

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]

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]

Consistent with the involvement of the boron 2pz-orbital in this transition is the almost total reduction of the long-wavelength absorption of (PhCH2)3B above 250 nm by its complexation with ammonia. Strong chemical support for a multiply bonded boron in the excited state (48b) can be deduced from photolysis results with tribenzylborane (49).48 While photolysis of... [Pg.369]

Note that by the investigators design, the bridgehead protons ( ) are not acidic, since any resulting carbanion would be orthogonal to the boron 2pz-orbital and thus incapable of 7r-bonding stabilization. [Pg.371]

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]

In writing these configurations for diatomic molecules of second-row elements, we have omitted the electrons from the Is orbitals because they are not part of the valence shells of the atoms. When considering the oxygen molecule, for which the a orbital arising from the combinations of the 2pz orbitals lies lower in energy than the 7r orbitals, we find that the electron configuration is... [Pg.79]

The bonding in the XeF2 molecule can be explained quite simply in terms of a 3-center, 4 electron bond that spans all three atoms in the molecule. The bonding in this molecular orbital description involves the filled 5pz orbital of Xe and the half-filled 2pz orbitals of the two F-atoms. The linear combination of these three atomic orbitals affords one bonding, one non-bonding and one anti-bonding orbital, as depicted below ... [Pg.570]

The cyclopentyl cation (39) undergoes a rapid degenerate rearrangement which can be frozen out at cryogenic temperatures as shown by solid state CPMAS 13C NMR spectra.57 MP2/6-31G(d,p) calculations show that cyclopentyl cation has a twisted conformation 4058 in which the axial hydrogens are bend toward the carbocation center. This is due to the pronounced geometrical distortion caused by the hyper-conjugative interaction of the /i-cr-C-H-bond with the formally vacant 2pz-orbital at the C+ carbon of this secondary carbocation. [Pg.142]

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]

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]

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]

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]

We may, however, examine the 2x2 Kekule-only matrices for these two cases. For the pure 2pz orbitals we have, in hartrees, the secular equation... [Pg.201]

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

See other pages where 2pz orbital is mentioned: [Pg.148]    [Pg.173]    [Pg.184]    [Pg.223]    [Pg.223]    [Pg.317]    [Pg.6]    [Pg.242]    [Pg.357]    [Pg.367]    [Pg.368]    [Pg.369]    [Pg.379]    [Pg.382]    [Pg.3]    [Pg.27]    [Pg.74]    [Pg.78]    [Pg.154]    [Pg.258]    [Pg.260]    [Pg.179]    [Pg.35]    [Pg.36]    [Pg.43]    [Pg.80]    [Pg.86]    [Pg.40]    [Pg.161]    [Pg.216]   
See also in sourсe #XX -- [ Pg.61 ]




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