Percent -character hybrid orbitals

In this discussion the orbital moments (along the orbital axes) have been assumed to be equal for bond orbitals and unshared-pair orbitals. Calculation of the moment (average value of cosine of angle with bond axis) for the %p hybrid orbitals leads to a value only half as great for the bond orbital as for the unshared-pair orbital. However, equality is found when the d-character (4 percent) and... 

Hybrid orbitals for nickel tet-racarbonyi have been discussed by G. Giacometti, J. Chem. Phys. 23, 2068 (1955), who reaches the conclusion that with dtp hybrid orbitals the bondB can have as much as 75 percent of double-bond character. 

Write the expression for a normalized hybrid orbital with 28 percent s character lying in the xy plane at an angle of 60° from the x six is. 

To understand why this is so, we must look at the atomic orbitals used to form each type of hybrid orbital. A single 2s orbital is always used, but the number of 2p orbitals varies with the type of hybridization. A quantity called percent s-character indicates the fraction of a hybrid orbital due to the 2s orbital used to form it. 

Why should the percent -character of a hybrid orbital affect the length of a C-H bond A 2s orbital keeps electron density closer to a nucleus compared to a 2p orbital. As the percent s-character increases, a hybrid orbital holds its electrons closer to the nucleus, and the bond becomes shorter and stronger. 

The higher the percent s-character of the hybrid orbital, the closer the lone pair is held to the nucleus, and the more stable the conjugate base. 

R groups increase the stability of an alkene because R groups are sp hybridized, whereas the carbon atoms of the double bond are sp hybridized. Recall from Sections 1.1 OB and 2.5D that the percent -character of a hybrid orbital increases from 25% to 33% in going from sp to sp. The higher the percent s-character, the more readily an atom accepts electron density. Thus, sp hybridized carbon atoms are more able to accept electron density and hybridized carbon atoms are more able to donate electron density. 

The bond dipole places a partial negative charge on the alkenyl carbon sjr) relative to the alkyl carbon because an sp hybridized orbital has greater percent -character (33%) than an sp hybridized orbital (25%). In a cis isomer, the two C ps-Cspi bond dipoles reinforce each other, yielding a small net molecular dipole. In a trans isomer, the two bond dipoles cancel. 

Even subtle differences that affect bond strength affect the frequency of an IR absorption. Recall from Section 1.10 that the strength of a C-H bond increases as the percent. v-character of the hybrid orbital on the carbon increases thus ... 

The C-O single bond of a carboxylic acid is shorter than the C—O single bond of an alcohol. This can be explained by looking at the hybridization of the respective carbon atoms. In the alcohol, the carbon is sp hybridized, whereas in the carboxylic acid the carbon is sp hybridized. As a result, the higher percent -character in the sp hybrid orbital shortens the C-O bond in the carboxylic acid. 

In molecules such as H2O, the composition of the four hybrid orbitals (two bonding and two lone pairs) on the O atom will be slightly different In such cases, the actual percent composition of each hybrid orbital will be dictated by the molecule s geometry. For sp" hybridized orbitals (n= I -3), the relationship between the bond angle (0) and the decimal percent s-character (S) is given by Equation (10.8). For hybrids that also contain some d-orbital character, the relationship between the bond angle and directionality of the hybrid orbitals is somewhat more complex and is not discussed in this textbook. 

Solution. All of the molecules have tetrahedral electron geometries and bent molecular geometries. The percent s-character calculated for the S-X bonding hybrid orbitals using Equation (10.8) are listed here. The percent s-character in each lone pair was calculated by using the formula 100 = 2S -I- IIL, where S is the percent s-character in the two S-X bonds and Z is the percent s-character in the two lone pairs. 

According to VBT, the hybridization on the central S atom is d sp where the S atom is forced to use two of its low-lying 3d orbitals in the formation of the hybrids. The percent d-character in a d sp hybrid is 33%. However, modern quantum mechanical calculations on the SF molecule indicate that there is only about 8% d-character in the S-F bonds, which is considerably less than the 33% expected for a (fisp hybrid. Clearly, while the d-orbitals seem to participate to a limited extent in... 

In general, the higher the percent of s-character in a hybrid orbital, the greater the deshielding effect on itself and adjacent atoms. 

The C-H coupling constants in tricyclo[3,1,0,0 ]hexane have been measured and compared with values derived from a semi-empirical relationship between /( C-H) and the calculated percent of s-character of the C-atom hybrid orbital in a... 

Note, however, that the hybridization of the carbon bearing the negative charge is different in each anion, so the lone pair of electrons occupies an orbital with a different percent -character in each case. A higher percent -character means a hybrid orbital has a larger fraction of the lower... 

Vinyl halides do not undergo Sn2 reactions in part because of the percent i -character in the hybrid orbital of the carbon atom in the C-X bond. The higher percent i -character in the sp hybrid orbital of the vinyl halide compared to the sp hybrid orbital of the alkyl halide (33% vs. 25%) makes the bond shorter and stronger. 

Remember shorter bonds are stronger bonds. A o bond formed from two sp hybridized C s is stronger than a o bond formed from two sp hybridized C s because the sp hybridized C orbitals have a higher percent -character. 

Bond length depends on hybridization and percent s -character. Bonds with a higher percent -character have smaller orbitals and are shorter. 

An sp hybrid orbital of carbon has approximately the same shape as an sp hybrid orbital. However, an sp hybrid orbital has 33% s character compared to 25% s character for an sp hybrid orbital. As the percent s character of hybrid orbitals increases, the electrons in the hybrid orbitals are closer to the nucleus. Therefore, the electrons in an sp hybrid orbital are closer to the nucleus than the electrons in an sp hybrid orbital. Increasing the percent s character of a hybrid orbital effectively increases the electronegativity of the carbon atom. 

Oxygen could form two O bonds to hydrogen to achieve a Lewis octet without hybridized orbitals. However, the overlap of a 2p orbital of oxygen with a Is orbital of hydrogen would not form as strong a bond as overlap of an sp hybrid orbital with a Is orbital. As we saw for ammonia, the percent s character of the sp hybrid results in a stronger bond and a more stable molecule. 

The acid ionization constants. A values, of hydrocarbons are extremely small, but there are substantial differences between various classes of hydrocarbons. The acidity of hydrocarbons is related to the hybridization of the carbon atom of the C—H bond. The acidity. A, of a carbon acid increases in the order sp < sp < sp. The order of acidities parallels the contribution of the lower energy of the 2s orbital to the hybrid orbitals in the a bond. The average distance of hybrid orbitals from the nucleus depends on the percent contribution of the s and p orbitals. For an sp hybrid orbital, the contribution of the 2s orbital is 25% because one 2s and three 2p orbitals contribute to the four hybrid orbitals. The contribution of the 2s orbital is 33% for an sp orbital and 50% for an sp hybrid orbital. Because an sp hybrid orbital has more s character than an sp or sp orbital, its electrons are located closer to the nucleus. Because the strength of an acid depends upon the stability of the conjugate base, a carbanion in which the negative charge is on an sp-hybridized carbon atom is more stable than a carbanion of an sp -hybridized or sp -hybridized carbon atom. 

The bond energy between common atoms increases in the order single < double < triple. This trend pardy reflects the effect of the closer approach of the O bonding electrons to the nucleus as the percent s character in the hybrid orbitals increases. However, the substantial increase in the carbon-carbon bond strength is largely a consequence of the increased number of bonds joining the carbon atoms. 

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