Products
Rewards
from HOLOOLY

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

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

HOLOOLY

HOLOOLY
TABLES

All the data tables that you may search for.

HOLOOLY
ARABIA

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

HOLOOLY
TEXTBOOKS

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

HOLOOLY
HELP DESK

Need Help? We got you covered.

## Q. 20.9

Using the fact that the $2s and 2p_{z}$ orbitals are normalized and orthogonal to each other, show that $c_{1} and c_{2}$ both equal $\sqrt{1/2}$ to normalize the hybrid orbitals.

## Verified Solution

$\int{ψ^{∗}_{2sp(1)}ψ_{2sp(1)}dq }=1$

$c^{2}_{1} \int{[−ψ_{2s}+ ψ_{2pz}][−ψ_{2s}+ ψ_{2pz}]dq } =1$

$c^{2}_{1} \int{ψ^{2}_{2s}dq} −2c^{2}_{1}\int{ψ_{2s}ψ_{2pz}dq} +c^{2}_{1} \int{ψ^{2}_{2pz}dq} =c^{2}_{1} (1-2×0+1)=2c^{2}_{1}=1$

$c_{1}=\frac{1}{\sqrt{2} }$

$\int{ψ^{∗}_{2sp(2)}ψ_{2sp(2)}dq }=1$

$c^{2}_{2} \int{[−ψ_{2s}− ψ_{2pz}]^{2}dq}=1$

$c^{2}_{2} \int{ψ^{2}_{2s} dq }+2c^{2}_{2}\int{ψ_{2pz}ψ_{2s}dq} +c^{2}_{2}\int{ψ^{2}_{2pz}dq} =c^{2}_{2}(1+0+1) =2c^{2}_{2} =1$

$c_{2}=\frac{1}{\sqrt{2} }$