Planar molecules bonding

Some final remarks on vector constraints may be in order for linear and planar molecules, bond constraints are not satisfactory to guarantee the shape of the molecules during the described integration process. The reason for this is the absence of orthogonal forces that keep the molecular shape. In these instances, suitable primary atoms are chosen for the bond constraints, and the positions of the remaining secondary atoms are obtained simply by applying the vector constraints under the provision that the associated constraint forces are givea Vector constraints are of the form... 

Boron trifluoride is a trigonal planar molecule There are six electrons two for each B—F bond associated with the valence shell of boron These three bonded pairs are farthest apart when they are coplanar with F—B—F bond angles of 120°... 

Multiple bonds are treated as a single unit m the VSEPR model Formaldehyde is a trigonal planar molecule m which the electrons of the double bond and those of the two single bonds are maximally separated A linear arrangement of atoms m carbon diox ide allows the electrons m one double bond to be as far away as possible from the elec Irons m the other double bond... 

In the planar molecule BF3, in Figure 4.3(b), the C3 axis through B and perpendicular to the figure is the highest-fold axis and, therefore, the three planes of symmetry, perpendicular to the figure and through each of the B-F bonds, are labelled (t . The plane of the molecule is also a plane of symmetry and is labelled u , where /z stands for horizontal with respect to C3. 

Structure. Ethylene is a planar molecule with a carbon—carbon bond distance of 0.134 nm, which is shorter than the C—C bond length of 0.153 nm found in ethane. The C—H bond distance is 0.110 nm, and the bond angles are [Pg.432]

Carbon oxohalides are reactive gases or volatile liquids which feature planar molecules of C2t, symmetry they are isoelectronic with BX3 (p. 196) and the bonding is best described in terms of molecular orbitals spanning all 4 atoms rather than in terms of localized orbitals as... 

There is little experience with the von Niessen method, but for most molecules the remaining three schemes tend to give very similar LMOs. The main exception is systems containing both a- and vr-bonds, such as ethylene. The Pipek-Mezey procedure preserves the cr/yr-separation, while the Edmiston-Ruedenberg and Boys schemes produce bent banana bonds. Similarly, for planar molecules which contain lone pairs (like water), the Pipek-Mezey method produces one in-plane cr-type lone pair and one out-of-plane yr-type lone pair, while the Edmiston-Ruedenberg and Boys schemes produce two equivalent rabbit ear lone pairs. 

Cyclopropane, for example, must be a rigid, planar molecule because three points (the carbon atoms) define a plane. No bond rotation can take place around a cyclopropane carbon-carbon bond without breaking open the ring (Figure 4.1). 

Further evidence for the unusual nature of benzene is that all its carbon-carbon bonds have the same length—139 pm—intermediate between typical single (154 pm) and double (134 pm) bonds. In addition, an electrostatic potential map shows that the electron density in all six carbon-carbon bonds is identical. Thus, benzene is a planar molecule with the shape of a regular hexagon. All C-C—C bond angles are 120°, all six carbon atoms are sp2-hybridized. and each carbon has a p orbital perpendicular to the plane of the six-membered ring. 

The shape of the ethylene molecule has been learned by a variety of types of experiments. Ethylene is a planar molecule—the four hydrogen and the two carbon atoms all lie in one plane. The implication of this experimental fact is that there is a rigidity of the double bond which prevents a twisting movement of one of the CHj groups relative to the other. Rotation of one CHt group relative to the other—with the C—C bond as an axis—must be energetically restricted or the molecule would not retain this flat form. 

C21-0046. Borazine (B3 N3 Hg ) is a planar molecule analogous to benzene (Cg Hg ). Write the Lewis structure and describe the bonding of borazine. 

Nitromethylene anion (5) and aci-nitromethylene anion (7) are planar molecules. In the nitromethylene anion, there is an equivalence of the two NO bonds and a clear trend of the various bond angles toward the trigonal value of 120°. By contrast, the aci-nitromethylene anion shows the nonequivalence of the two NO bonds, a typical double bond character for the CN bond, and a larger deviation of all bond angles from 120°. 

If we now consider a planar molecule like BF3 (D3f, symmetry), the z-axis is defined as the C3 axis. One of the B-F bonds lies along the x-axis as shown in Figure 5.9. The symmetry elements present for this molecule include the C3 axis, three C2 axes (coincident with the B-F bonds and perpendicular to the C3 axis), three mirror planes each containing a C2 axis and the C3 axis, and the identity. Thus, there are 12 symmetry operations that can be performed with this molecule. It can be shown that the px and py orbitals both transform as E and the pz orbital transforms as A, ". The s orbital is A/ (the prime indicating symmetry with respect to ah). Similarly, we could find that the fluorine pz orbitals are Av Ev and E1. The qualitative molecular orbital diagram can then be constructed as shown in Figure 5.10. 

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