We are determined to provide the latest solutions related to all subjects FREE of charge!

Please sign up to our reward program to support us in return and take advantage of the incredible listed offers.

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program


Advertise your business, and reach millions of students around the world.


All the data tables that you may search for.


For Arabic Users, find a teacher/tutor in your City or country in the Middle East.


Find the Source, Textbook, Solution Manual that you are looking for in 1 click.


Need Help? We got you covered.

Chapter 20

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.