Sp hybrid orbitals orthonormality

[josecuervo]

Homework Statement

Find the other sp hybrid orbital given Ψsp1 = 3Ψ2s + 4Ψ2pz using the orthonormal relationships.

Homework Equations

I know that you are supposed to use the orthonormal relationships like stated in the problem, and when finding the the second sp hybrid orbital for the normalized

Ψ[itex]_{1}[/itex]sp1 = [itex]\frac{1}{\sqrt{2}}[/itex](Ψ2s + Ψ2pz)

Ψ[itex]_{2}[/itex]sp1 = [itex]\frac{1}{\sqrt{2}}[/itex](Ψ2s - Ψ2pz)

and the integral for orthogonality and normalization is the standard:

[itex]\int[/itex] Ψ[itex]_{1}[/itex]sp1* Ψ[itex]_{2}[/itex]sp1=0

[itex]\int[/itex] Ψ[itex]_{1}[/itex]sp1* Ψ[itex]_{1}[/itex]sp1=1

The Attempt at a Solution

This is what I get and I'm not very confident about it. If someone could help me finish it or point me in the right direction I would appreciate it.

0=[itex]\int[/itex](3Ψ2s*+4Ψ2pz*)(c[itex]_{1}[/itex]Ψ2s+c[itex]_{2}[/itex]Ψ2pz)

Which I get reduces to

0= 3c[itex]_{1}[/itex] + 4c[itex]_2{}[/itex]

c[itex]_{2}[/itex] = -3/4c[itex]_{1}[/itex]

I then plugged in and used the relationship:

1= [itex]\int[/itex](c[itex]_{1}[/itex]Ψ2s* - 3/4c[itex]_{1}[/itex]Ψ2pz*)(c[itex]_{1}[/itex]Ψ2s - 3/4c[itex]_{1}[/itex]Ψ2pz)

which I get reduces to

1= -3/2c[itex]_{1}[/itex][itex]^{2}[/itex]

so

c[itex]_{1}[/itex]= +/- [itex]\sqrt{2/3}[/itex] and
c[itex]_{2}[/itex]= +/- [itex]\sqrt{6}[/itex]/4

Can someone double check me or point out any errors?

 

[mark726]

Thank you.





Your approach seems to be correct. The orthonormal relationships you have used are the correct ones for finding the second sp hybrid orbital. Your final values for c1 and c2 also seem to be correct. However, I would suggest double checking your calculations just to be sure. Also, it would be helpful to mention the values of c1 and c2 as well as the final orbital equation in your solution for clarity. Overall, good job on tackling this problem using the orthonormal relationships! Keep up the good work.