- #1
josecuervo
- 18
- 0
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?