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## Q. 20.11

Estimate the dipole moment of the LiH molecule from the orbitals of Eq. (20.4-8):

$ψ_{1σ}=ψ_{1sLi}$          (20.4-8a)

$ψ_{2σ}=c_{2sp(1)Li}ψ_{2sp(1)Li} + c_{1sH}ψ_{1sH} =−0.47ψ_{2sp(1)Li }− 0.88ψ_{1sH}$          (20.4-8b)

## Verified Solution

The lithium nucleus has charge 3e . The two nonbonding electrons in the 1σ orbital contribute as though they were at the lithium nucleus. The probability density of an electron in the 2σ orbital is

$\left|ψ_{2σ}\right|^{2} = (0.47)^{2}ψ^{2}_{2sp(1)Li}+(0.47)(0.88)ψ_{2sp(1)Li}ψ_{1sH} + (0.88)^{2}ψ^{2}_{1sH}$

We neglect the second term in this expression because it is appreciably nonzero only in the overlap region. The third term will make its contribution to the integral in $Eq. (20.4-13):\left\langleμ\right\rangle =μ_{nuc} − e\sum\limits^{n_{e}}_{i=1}{\int{ψ_{i}(i)^{∗}r_{i}ψ_{i}(i)d^{3}r_{i} } } =μ_{nuc} − e\sum\limits^{n_{e}}_{i=1}{\int{r_{i}\left|ψ_{i}(i)\right|^{2}d^{3}r_{i}}}$ as though the electron were centered at the hydrogen nucleus. The 2sp(1) hybrid orbital does not have its center of charge exactly at the lithium nucleus, but we approximate the contribution of the first term as though it did. The net charge at the lithium nucleus is

$Q_{Li} = 3e − 2e − 2(0.47)^{2}e = 0.56e = 8.9 × 10^{−20} C$

The net charge at the hydrogen nucleus is

$Q_{H}= e − 2(0.88)^{2}e = −0.56e= −8.9 × 10^{−20} C$

with a bond length of $1.595 × 10^{−10} m,$

$μ ≈ (8.9 × 10^{−20} C)(1.595 × 10^{−10} m) = 1.42 × 10^{−29}Cm = 4.3 Debye$

This estimate of the dipole moment is about 60% as large as the value for a purely ionic bond, indicating a bond that is roughly 60% ionic in character. It agrees only roughly with the experimental value, 5.88 Debye.