Tetrahedral shape hybrid orbitals

The hybrid orbitals, constructed by the combination of atomic orbitals, can be represented in a similar way (Fig. 1.5). Only the two sp-hybrid orbitals are shown but the trigonal sp - and tetrahedral sp hybrid orbitals have similar shapes. In each case the large bonding lobes have the same sign, taken as positive. 

The concept of hybridization explains how carbon forms four equivalent tetrahedral bonds but not why it does so. The shape of the hybrid orbital suggests the answer. When an 5 orbital hybridizes rvith three p orbitals, the resultant sp3 hybrid orbitals are unsyimmetrical about the nucleus. One of the two... 

The VSEPR notation for the Cl2F+ ion is AX2E3. According to Table 11.1, molecules of this type exhibit an angular molecular geometry. Our next task is to select a hybridization scheme that is consistent with the predicted shape. It turns out that the only way we can end up with a tetrahedral array of electron groups is if the central chlorine atom is sp3 hybridized. In this scheme, two of the sp3 hybrid orbitals are filled, while the remaining two are half occupied. 

Figure 11.13 Hybridization of 5- and p- atomic orbitals, (a) Linear sp hybrid, from one s- and one p-orbital. (b) sp2 hybrid, from one s- and two p-orbitals, with a plane triangular shape, (c) sp3 hybrid, from one 5-orbital and the three p-orbitals, which has a tetrahedral shape in three dimensions.

The shape of the CH4 molecule is tetrahedral. A tetrahedral orientation of equal bonds (which are formed from the overlap of the identical sp3 hybrid orbitals and the hydrogen Is orbitals) gives a bond angle of 109.5° (Figure 6). 

The simplest compounds to consider here are ammonia and water. It is apparent from the above electronic configurations that nitrogen will be able to bond to three hydrogen atoms, whereas oxygen can only bond to two. Both compounds share part of the tetrahedral shape we saw with 5/ -hybridized carbon. Those orbitals not involved in bonding already have their full complement of electrons, and these occupy the remaining part of the tetrahedral array (Figure 2.21). These electrons are not inert, but play a major role in chemical reactions we refer to them as lone pair electrons. 

The VSEPR theory assumes that the four electrons from the valence shell of the carbon atom plus the valency electrons from the four hydrogen atoms form four identical electron pairs which, at minimum repulsion, give the observed tetrahedral shape. To rationalize the tetrahedral disposition of four bond-pair orbitals with those of the 2s and three 2p atomic orbitals of the carbon atom, sp3 hybridization is invoked. 

It is important to realize that methane is not tetrahedral because carbon has sp3, hybrid orbitals. Hybridization is only a model—a theoretical way of describing the bonds that are needed for a given molecular structure. Hybridization is an interpretation of molecular shape shape is not a consequence of hybridization. 

In an atom, the hybridization of s and p orbitals to form sp orbitals provides electron probability areas where bonds can form to make a molecule more stable than if the bonding had occurred in the individual s and p orbitals. The sp orbitals have one large lobe and one small lobe and are aligned along x, y, and z coordinates so that four sp orbitals, called sp3 orbitals because they are made of one s and three p orbitals, result in a tetrahedral-shaped arrangement. When there are three sp orbitals, made of one s and two p orbitals, called sp orbitals, the molecular has a triangular-planar shape. If there is bonding in two sp orbitals, made of one s and one p orbital, a linear molecule results. 

Two molecular fragments are isolobal if the number, symmetry properties, shapes, and approximate energies of their frontier orbitals are the same. They may or may not also be isoelectronic. For example, the HB and HC fragments are isolobal (but not isoelectronic), whereas the H2C and H2N moieties are both isolobal and isoelectronic. These relationships are illustrated in Fig. 1-23. As shown there, the symbol is used to express isolobality. Also shown in Fig. 1-23 is the idea that we may choose to picture the isolobality in more than one way. Thus, for the HB and HC fragments, we can either envision one sp hybrid orbital whose axis is colinear with the H—B or H—C bond, plus two p orbitals perpendicular to this axis, or we may envision a state of full hybridization, where the frontier orbitals in each case are three of the four in a set of sp3 tetrahedral hybrids. The orbitals whose similarity is critical in determining isolobality are called the frontier orbitals. 

One s and three p orbitals hybridize to form four sp orbitals. According to VSEPR, a tetrahedral shape minimizes repulsion between the orbitals. 

What are the hybrid orbitals in a molecule with a tetrahedral shape ... 

FIGURE 6.43 Shapes and relative orientations of the four sp hybrid orbitals in CH4 pointing at the corners of a tetrahedron with the carbon atom at its center. The "exploded view" at the bottom shows the tetrahedral geometry. 

Figure 6.43 shows the shape and orientation of these four orbitals, pointing toward the vertices of a tetrahedron, which has the carbon atom at its center. The bottom image in Figure 6.43 shows an exploded view in which the orbitals have been displaced from one another to show the tetrahedral geometry. Each hybrid orbital can overlap a Is orbital of one of the hydrogen atoms to give an overall tetrahedral structure for CH4. 

In the nitrogen atom, the same hybridisation of the atomic orbitals occurs and so four sp hybrid orbitals are formed. These are distributed around the central atom in the same tetrahedral manner as in carbon. Thus, suggest the shape of the ammonia molecule. 

The s and p orbitals in a carbon atom combine into four hybridized orbitals that repel each other in a shape much like tliat of four balloons tied together. Carbon takes this tetrahedral shape because it only has six electrons which fill the the s but only two of the p orbitals. 

Plan We note, as in Sample Problem 11.1, the shape around each central atom to postulate the type of the hybrid orbitals, paying attention to the multiple bonding of the C and O bond. Solution In Sample Problem 10.8, we determined the shapes around the three central atoms of acetone tetrahedral around each C of the two CH3 (methyl) groups and trigonal planar around the middle C atom. Thus, the middle C has three sp orbitals and one unhybridized p orbital. Each of the two methyl C atoms has four sp orbitals. Three of these sp orbitals overlap the U orbitals of the H atoms to form a bonds the fourth overlaps an sp orbital of the middle C atom. Thus, two of the three sp orbitals of the middle C form CT bonds to the other two C atoms. 

Figure 8.19 A carbon atom s 2s and 2p electrons occupy the hybrid sp orbitals. Notice that the hybrid orbitals have an intermediate amount of potential energy when compared with the energy of the original s and p orbitals. According to VSEPR theory, a tetrahedral shape minimizes repulsion between the hybrid orbitals in a CH4 molecule. 

We have seen that the tetrahedral shape of methane is consistent with a bonding model based on sp hybrid orbitals on carbon. We should not conclude, however, that the geometry is a result of sp hybridization. We... 

It can be shown that the sp hybrids are the only four hybrids that can be formed from the s and the three p orbitals under the condition that the hybrids are orthogonal and have the same shape. The hybridization model might therefore be used to predict the tetrahedral shape of these molecules, but in a much less direct manner than the VSEPR model. 

This recaptured much of the intuitive picture of bonding because now atomic orbitals could be sketched and bonds drawn where these orbitals overlapped. The device offered an explanation for bonding in many molecules, and where it failed— as in the tetrahedral bonding of carbon—Pauling showed that a hybrid atomic orbital system sufficed Just as two ripples on a pond can come together to form a differently shaped wave, one s and three p orbitals can mix to form four hybrid orbitals extended tetrahedrally in four directions in space. 

The electronic configuration of a carbon atom in its ground state, established unambiguously by spectroscopic measurements as s 2s 2p precludes a simple rationalization of the tetrahedral bonding at carbon. Pauling suggested that the four atomic orbitals (2s, 2px, 2py, 2p ) are replaced by a set of four equivalent hybrid orbitals designated sp. The approximate shapes of these orbitals are shown in Fig. 

As you review each type of hybrid orbital in the following subsections, take note of (1) the number and types of atomic orbitals that were combined to make the hybrid orbitals, (2) the number of orbitals that remain uncombined, and (3) the three-dimensional arrangement in space of the hybrid orbitals and any uncombined p orbitals. In particular, you will find that these three-dimensional arrangements will retain the names (tetrahedral, trigonal planar, linear) and bond angles (109.5°, 120°, and 180°) used to describe the shapes of molecules in our section on VSEPR (Section 1.3). 

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