Percent s-character

In the foregoing treatment of the water molecule, which we shall use as an example, each of the two bond orbitals of the oxygen atom has been calculated to have 6 percent s character and 94 percent p character. Each of the two unshared-pair orbitals then has 44 percent s character and 56 percent p character. The maxima for the unshared-pair orbitals lie in directions making an angle of 142° with one another and such that their resultant is opposed to that for the two bond orbitals, which have their maxima at 93.5° with one another. The component for the four unshared-pair electrons is determined by the direction cosine —0.34, and that of the two bonding electrons of the oxygen atom by the direction cosine 0.68 hence the contribution of the four unshared-pair electrons to the dipole moment is just balanced by that of the two bonding electrons.18... 

A study of the hyperfine structure of the electron spin magnetic resonance spectrum, resulting from the interaction with the nuclear spins, has led to the conclusion9 that structure I contributes 65 percent and structure II35 percent, and that the odd electron occupies a 2pr orbital with 2.5 percent s character. 

The relation between hybridization and bond angle is simple for s-p hybrids. For two or more equivalent orbitals, the percent s character (5) or percent p character iP) is given by the relationship 27... 

The values of the Ag—P coupling constant were correlated with the percent s character in the Ag—P bond and also with the magnitude of the P—Ag—P bond angle.186... 

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. 

It relates to the 50 percent s-character of the spz linear combination that allows the excess density to screen the internuclear repulsion. Acetylene [2 x H(s)C(pz)2 s2], like N2 has two excess pairs in s-states to screen the nuclei and generate the bond order 3. 

The use of the newer empirical equations relating percent s character to and... 

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. 

Bond length decreases as the percent s-character increases (Section 1.1 OB). 

Increasing percent s-character Increasing carbanion stability... 

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 s-character in each case. 

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. 

Thus, acidity increases from CH3CH3 to CH2=CH2 to HC=CH as the negative charge of the conjugate base is stabilized by increasing percent s-character. Once again this is a specific example of a general trend. 

Hybridization effects The acidity of H-A increases as the percent s-character of A increases. 

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. 

Recall from Section 2.5D that the acidity of a C-H bond increases as the percent s-character of C increases. Thus, sp hybridized C-H bonds (having a C atom with 50% s-character) are more acidic than sp and sp hybridized C-H bonds (having C atoms with 33% and 25% s-character, respectively). 

The higher the percent s-character, the stronger the bond and the higher the wavenumber of absorption. 

Recall from Section 1.10B that increasing percent s-character decreases bond length... 

Based on hybridization, a Csp -Cgp bond should be shorter than a Csp -Cgp bond because it is formed from orbitals having a higher percent s-character. 

Hybridization effects Increasing the percent s-character in Pyridine is iess basic than piperidine, the orbitai with the lone pair decreases basicity. 

Percent s character in lone-pair hybrid, see Ref. (75JA4136). 

More recently Newton, Schulman and Manus (41) considered the relationship between experimental /(C-C) values in 12 different hydrocarbons and the product of the percent s character in the two bonding carbon atomic hybrids obtained from localization of INDO molecular orbitals (see Fig. 1). They found that... 

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