Hybrid orbital number

Problem 3.1 Determine the hybrid orbital number (HON) of the five C-containing intermediates tabulated below and give the hybrid state of the C atom. Unpaired electrons do not require an HO and should not be counted in your determination. M... 

Number of atoms + lone pairs (X + E) Hybridization Number of hybrid orbitals Number ofp orbitals Geometry (bond angle) Example"... 

Hunsdiecker reaction, 341 Hybridization, 17 Hybrid orbital number, 17, 18, 32 Hybrid, resonance, 24 Hydration of cyclohexane derivatives, 191 Hydrazine, 4 Hydride shift, 93 Hydroboration, 95 Hydroboration-oxidation, 258, 270 Hydrocarbons, cyclic, 162 unsaturated, 87 Hydrogenation of alkenes, 57 Hydrogen bond, 22 Hydroperoxides in ethers, 284 Hydroquinone, 430 Hydroxy acids, 344... 

Electron arrangement around the central atom Hybrid orbitals (number) Angle(s) between a-bonds Example(s) underlined atom]... 

Hybrid Orbitals Number of Hybrid Orbitals in Set Component Orbitals Angle Between Orbitals Arrangement of Hybrid Orbitals in Set... 

Ethylene is planar with bond angles close to 120° (Figure 2 15) therefore some hybridization state other than sp is required The hybridization scheme is determined by the number of atoms to which carbon is directly attached In sp hybridization four atoms are attached to carbon by ct bonds and so four equivalent sp hybrid orbitals are required In ethylene three atoms are attached to each carbon so three equivalent hybrid orbitals... 

Phosphorus and sulfur are the third-row analogs of nitrogen and oxygen, and the bonding in both can be described using hybrid orbitals. Because of their positions in the third row, however, both phosphorus and sulfur can expand their outer-shell octets and form more than the typical number of covalent bonds. Phosphorus, for instance, often forms five covalent bonds, and sulfur occasionally forms four. 

Trends in Au-X and Au-P bond lengths in complexes (PPh3) AuX should be noted (Table 4.10) the Au-Cl bond length varies more with changes in coordination number than does the Au-P bond and is, therefore, more sensitive to the decrease in s character as the hybrid orbitals used by gold change from sp to sp3. 

Table 3.2 summarizes the relation between electron arrangement and hybridization type. No matter how many atomic orbitals we mix together, the number of hybrid orbitals is always the same as the number of atomic orbitals with which we started ... 

Electron arrangement Number of atomic orbitals Flybridization of the central atom Number of hybrid orbitals... 

Complexes of d- and /-block metals can be described in terms of hybridization schemes, each associated with a particular shape. Bearing in mind that the number of atomic orbitals hybridized must be the same as the number of hybrid orbitals produced, match the hybrid orbitals sp1d, sp fd , and sp d3f to the following shapes (a) pentagonal bipyramidal ... 

Next we count the number of lone pairs on the carbon atom. There are no lone pairs on the carbon atom. (If you are not sure how to tell that there are no lone pairs there, go back to Chapter 1 and review the section on counting lone pairs.) Now we take the sum of the attached atoms and the number of lone pairs—in this case, 3 + 0 = 3. Therefore, three hybridized orbitals are being used here. That means that we have mixed two p orbitals and one s orbital (a total of three orbitals) to get three equivalent sp orbitals. Thus, the hybridization is sp. Let s take a closer look at how this works. 

So here s the rule Just add the number of bonded atoms to the number of lone pairs. The number you get teUs you how many hybridized orbitals you need according to the following ... 

An inner atom with a steric number of 4 has tetrahedral electron group geometry and can be described using S p hybrid orbitals. 

Both inner atoms have steric numbers of 4 and tetrahedral electron group geometry, so both can be described using s p hybrid orbitals. All four hydrogen atoms occupy outer positions, and these form bonds to the inner atoms through 1 s-s p overlap. The oxygen atom has two lone pairs, one in each of the two hybrid orbitals not used to form O—H bonds. 

The steric number of an inner atom uniquely determines the number and type of hybrid orbitals. 

A molecule with a steric number of 6 requires six hybrid orbitals arranged in octahedral geometry. In Chapter 9, sulfur hexafluoride appears as the primary example of a molecule with a steric number of 6 (Figure ). Six equivalent orbitals for sulfur can be constmcted for the inner sulfur atom by combining the 3. S, the three 3 p,... 

With a steric number of 5, chlorine has trigonal bipyramidal electron group geomehy. This means the inner atom requires five directional orbitals, which are provided hymsp d hybrid set. Fluorine uses its valence 2 p orbitals to form bonds by overlapping with the hybrid orbitals on the chlorine atom. Remember that the trigonal bipyramid has nonequivalent axial and equatorial sites. As we describe in Chapter 9, lone pairs always occupy equatorial positions. See the orbital overlap view on the next page. 

Each carbon atom has a steric number of 2, indicating that acetylene is a linear molecule and that sp hybrid orbitals can be used to construct the bonding orbital framework. Figure 10-24a shows the a bonding system of acetylene. 

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