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Valence-electron contribution

In the BeF2 molecule, there are two electron-pair bonds. These electron pairs are located in the two sp hybrid orbitals. In each orbital, one electron is a valence electron contributed by beryllium the other electron comes from the fluorine atom. [Pg.186]

In a molecule that has lone pairs or a single nonbonding electron on the central atom, the valence electrons contribute to the electron arrangement about the central atom but only bonded atoms are considered in the identification of the shape. Lone pairs distort the shape of a molecule so as to reduce lone pair-bonding pair repulsions. [Pg.226]

A quantitative consideration on the origin of the EFG should be based on reliable results from molecular orbital or DPT calculations, as pointed out in detail in Chap. 5. For a qualitative discussion, however, it will suffice to use the easy-to-handle one-electron approximation of the crystal field model. In this framework, it is easy to realize that in nickel(II) complexes of Oh and symmetry and in tetragonally distorted octahedral nickel(II) complexes, no valence electron contribution to the EFG should be expected (cf. Fig. 7.7 and Table 4.2). A temperature-dependent valence electron contribution is to be expected in distorted tetrahedral nickel(n) complexes for tetragonal distortion, e.g., Fzz = (4/7)e(r )3 for com-... [Pg.244]

Fig. 7.7 Schematic diagrams for common electron configurations of Ni " complexes in the one-electron approximation. The resulting valence electron contributions V z are obtained from Table 4.2... Fig. 7.7 Schematic diagrams for common electron configurations of Ni " complexes in the one-electron approximation. The resulting valence electron contributions V z are obtained from Table 4.2...
According to Wade s rules, one non-bonding electron pair is allocated to each metal atom in polyionic metal clusters of the p-block elements. Thus the planar five-membered M5 anions are 16-electron species comprised of two valence electrons contributed by each metal atom and the 6- charge. Since there are eight electron pairs for bonding five cluster atoms, Wade s rules predict an arachno structure ( + 3 electron pairs for n cluster atoms), i.e. a pentagonal bipyramid with two vacant sites. [Pg.167]

Count the total number of valence electrons contributed by the electrically neutral atoms. If the species being considered is an ion, add one electron to the total for each negative charge subtract one for each positive charge. [Pg.3]

The configuration wave function f > is usually represented by a determinant for the "active space where only the v valence electrons contribute to one column each n columns for the f electrons, v-n-1 columns for the d electrons and one column for the (s-p) conduction electrons for each RE atom n - nf. [Pg.251]

Table 1 summarizes physical properties of TiC and UC. The M-C bond length of UC is 2.481 A, while that of TiC is 2.164 A. The difference of 0.317 A between their bond lengths is comparable to that of 0.222 A between the atomic radii of the corresponding metal elements. Both compounds have high melting points, but there is a quite difference in micro-hardness and resistivity between TiC and UC. Since the valence electi ons of Ti atom are attiibuted to the 3d-electrons and those of U atom to the 5f- and 6d-electrons, it is valuable to consider how these valence electrons contribute to the physical properties of their compounds. [Pg.124]

Thus six BH electron pair, one BB electron pair and four BHB bridge bonds together account for twenty-two valence electrons contributed by four boron and ten hydrogen atoms. Thus skeletal electrons are 22. [Pg.144]

Rather than let each valence electron contribute equally to a, in L,.,., one can take the contribution of a, for the / , orbitals that project from the oxygen... [Pg.367]

Detailed calculations show that the value of Z ff for the outermost electrons in Si and Cl atoms is 4.29-b and 6.I2-F, respectively, (a) What value do you estimate for Z g experienced by the outermost election in both Si and Cl by assuming core elections contribute 1.00 and valence electrons contribute 0.00 to the screening constant (b) What values do you estimate for Z using Slater s rules (c) Which approach gives a more accurate estimate of Z (d) Which method of approximation more accurately accounts for the steady increase in Z g that occurs upon moving left to right across a period (e) Predict Z g for a valence election in P, phosphorus, based on the calculations for Si and Cl. [Pg.280]

Consider the isoelectronic ions F and Na. (a) Which ion is smaller (b) Using Equation 7.1 and assuming that core electrons contribute 1.00 and valence electrons contribute 0.00 to the screening constant, S, calculate for the 2p electrons in both ions, (c) Repeat this calculation using Slater s rules to estimate the screening constant, S. (d) For isoelectronic ions,... [Pg.281]

The valence electron contribution is evident in most cases, but there are exceptions such as the frequently encountered cations TT + and Pb +. In the case of transition elements their valence electron contribution can only be calculated from the observed structural features of the anionic tetrahedron complex. This may in case be verified by magnetic measurements. [Pg.197]

Two i/) -hybridized carbon atoms form a sigma (cr) bond between them by overlap of one sp orbital from each carbon. The remaining carbon sp orbitals form cr bonds to four hydrogens through overlap with the hydrogen It orbitals. These five cr bonds account for 10 of the 12 valence electrons contributed by the two carbons and four hydrogens, and comprise the 0 >bond framework of the molecule. [Pg.37]

The simplest paramagnetic molecular state with A = 0 is The rotational quantum number N is equal to R from the above discussion, and from here forward we will use N for the discussion of nuclear rotation, in accordance with standard notation for E states. The lone valence electron contributes 1 p.B of magnetic moment, making E molecules suitable for trapping. The rotational ground state of E is a pure N = 0) state, which eliminates helium-induced Zeeman relaxation from first-order effects, in accordance with Equation 13.6. In order to change the spin projection Ms, there must be a mechanism by which the spin can be coupled to the rotational wavefunction, which can be altered by the electrostatic interaction. [Pg.493]


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See also in sourсe #XX -- [ Pg.2 , Pg.130 ]




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