Wavefunction Equations Of Hybridized Orbitals

[LUFER]

Hi, I'm looking for the equations to perform the wave function of the orbitals, that way I can give a graph.
The problem I only encounter situations in basic energy transfer (valence).And also Schrodinger's theory for obtaining energy levels.


Task: Make wavefunctions of the sp2, sp3, sd3 hybridized orbitals respectively.

I need the wavefunction equations for the hybridized orbitals.

[exchemist]

7 hours ago, LUFER said:

Hi, I'm looking for the equations to perform the wave function of the orbitals, that way I can give a graph.
The problem I only encounter situations in basic energy transfer (valence).And also Schrodinger's theory for obtaining energy levels.


Task: Make wavefunctions of the sp2, sp3, sd3 hybridized orbitals respectively.

I need the wavefunction equations for the hybridized orbitals.

I'm not entirely sure what you are really asking. It's probably a language issue.

I doubt that you are being asked for the full algebraic expression for these wavefunctions. It seems more likely that you are being asked to show how linear combinations of appropriately chosen atomic orbitals generate the hybridised ones.

This link shows you how 4 sp3 hybrid orbitals can be constructed from s and 3 p orbitals: https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/05%3A_Bonding_in_polyatomic_molecules/5.2%3A_Valence_Bond_Theory_-_Hybridization_of_Atomic_Orbitals/5.2D%3A_sp3_Hybridization

You can do something analogous for the others. 

By all means come back if I have misunderstood.

[LUFER]

On 8/14/2021 at 5:46 AM, exchemist said:

I'm not entirely sure what you are really asking. It's probably a language issue.

I doubt that you are being asked for the full algebraic expression for these wavefunctions. It seems more likely that you are being asked to show how linear combinations of appropriately chosen atomic orbitals generate the hybridised ones.

This link shows you how 4 sp3 hybrid orbitals can be constructed from s and 3 p orbitals: https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/05%3A_Bonding_in_polyatomic_molecules/5.2%3A_Valence_Bond_Theory_-_Hybridization_of_Atomic_Orbitals/5.2D%3A_sp3_Hybridization

You can do something analogous for the others. 

By all means come back if I have misunderstood.

Exactly the algebraic expression for each case, that way I can generate the graph in matlab. I'm having to simulate, but lacking data in the equations.

[studiot]

7 hours ago, LUFER said:

Exactly the algebraic expression for each case, that way I can generate the graph in matlab. I'm having to simulate, but lacking data in the equations.

Can you post the full question ?

And also follow the rules by telling us what you have done so far?

I am with exchemist on this (+1) as what you seem to be asking is Nobel Prize level work.

A bit steep for SF Homework Help.

 

[exchemist]

8 hours ago, LUFER said:

Exactly the algebraic expression for each case, that way I can generate the graph in matlab. I'm having to simulate, but lacking data in the equations.

If you really need the full solutions, then these will involve the product of the Laguerre functions (expressing the radial wave function) and the spherical harmonics that express the angular dependence. You should be able to find these on the internet for 2s, 2p, 3d etc. but I confess I don't know where.  Here is an explanation of how they are used:

https://demonstrations.wolfram.com/HydrogenAtomRadialFunctions/

As you can see, it is not simple. I did an entire quantum chemistry supplementary subject at university without being required to master exact solutions. The reason is that exact solutions are only possible for the hydrogen atom. For multi-electron atoms and molecules one has to resort to approximate solutions anyway, so it is far better to focus on symmetry properties etc., rather than getting enmeshed in the algebra of analytical solutions.

So I'm still a bit sceptical that anyone expects you to get into all this. Like @studiotI too think it would be a good idea if can post the text of the task you have been assigned (if it is in English  - or maybe I could decipher it in French).

Edited August 17, 2021 by exchemist

[studiot]

6 hours ago, exchemist said:

The reason is that exact solutions are only possible for the hydrogen atom. For multi-electron atoms and molecules one has to resort to approximate solutions anyway,

 

I would go further than this, if @LUFER is really interested in discussing sp or any other hybridisation further.

Lufer, you do realise you have only asked half a question?

If we have solutions to the Schrodinger equation as s and p etc orbitals, why do we need hybridisation ?

Under what conditions will hybrids arise, they don't happen by themselves?

Edited August 17, 2021 by studiot

[exchemist]

4 hours ago, studiot said:

 

I would go further than this, if @LUFER is really interested in discussing sp or any other hybridisation further.

Lufer, you do realise you have only asked half a question?

If we have solutions to the Schrodinger equation as s and p etc orbitals, why do we need hybridisation ?

Under what conditions will hybrids arise, they don't happen by themselves?

I think he has been asked to graph them, to see how the shapes arise from the linear combinations, or something.  

Edited August 17, 2021 by exchemist

[studiot]

1 hour ago, exchemist said:

I think he has been asked to graph them, to see how the shapes arise from the linear combinations, or something.  

'Or something'  is correct.

 

For instance take the sd3 hybrid in the OP.

What s and what d orbitals are involved, and why.

In say potassium permanganate why are the perfectly good and valid solutions of Schrodinger changed in the manganese atom ?

 

This is what I mean by half a question.

Hybridisation arises because the environment of the atom (ie the orbitals in the rest of the atoms in the compound) make it energetically favourable.

Manganese atoms do not wander around on their own configured in the sd3 condition.

 

So we need the full information to be able to help.

 

And the OP (who has been back since I last posted) seems reluctant to help us to help himself.

Edited August 17, 2021 by studiot

[LUFER]

On 8/17/2021 at 5:32 PM, exchemist said:

I think he has been asked to graph them, to see how the shapes arise from the linear combinations, or something.  

Hello after having done a search if I can find the equations and I'll give you their source. The only thing I couldn't find was this one in the annex. Please help me know what this value or name is.

https://www.clutchprep.com/chemistry/practice-problems/143395/the-s-orbital-is-spherical-nondirectional-because-check-all-that-apply-a-the-ang

 

https://www.learnpick.in/prime/documents/ppts/details/4225/atomic-structure-and-quantum-numbers

https://slideplayer.com.br/slide/1595583/

http://www.ensinoadistancia.pro.br/ead/qg/aula-7/aula-7.html

https://pbs.twimg.com/media/EWxsOfKWoAIk6eq?format=jpg&name=large

https://web1.eng.famu.fsu.edu/~dommelen/quantum/style_a/hyd.html

 

ρ=r/α      ?

 

 

[exchemist]

9 hours ago, LUFER said:

Hello after having done a search if I can find the equations and I'll give you their source. The only thing I couldn't find was this one in the annex. Please help me know what this value or name is.

https://www.clutchprep.com/chemistry/practice-problems/143395/the-s-orbital-is-spherical-nondirectional-because-check-all-that-apply-a-the-ang

 

https://www.learnpick.in/prime/documents/ppts/details/4225/atomic-structure-and-quantum-numbers

https://slideplayer.com.br/slide/1595583/

http://www.ensinoadistancia.pro.br/ead/qg/aula-7/aula-7.html

https://pbs.twimg.com/media/EWxsOfKWoAIk6eq?format=jpg&name=large

https://web1.eng.famu.fsu.edu/~dommelen/quantum/style_a/hyd.html

 

ρ=r/α      ?

 

 

Obrigado! 😀

According to my old Atkins QM book, ρ = r.(μe²/hbar²4πε₀) for the hydrogen atom, so that seems to fit. What is the problem?   

[studiot]

I still say that hybrid orbitals only occur in bonds, not free atoms.

And the actual hybrid varies depending upon the actual atoms concerned.

Here are some simple examples of the maths.

 

[LUFER]

5 hours ago, exchemist said:

Obrigado! 😀

According to my old Atkins QM book, ρ = r.(μe²/hbar²4πε₀) for the hydrogen atom, so that seems to fit. What is the problem?   

ρ = r / α

[exchemist]

39 minutes ago, LUFER said:

ρ = r / α

Er, I've just confirmed what ρ is.What are you asking?   

3 hours ago, studiot said:

I still say that hybrid orbitals only occur in bonds, not free atoms.

And the actual hybrid varies depending upon the actual atoms concerned.

Here are some simple examples of the maths.

 

My understanding of the VB model is that you make hybridised atomic orbitals from the atomic orbitals of the free atom and you then attribute bonding to the overlap of these hybridised atomic orbitals with those of neighbouring atoms. This is what your book is saying. 

Our questioner is engaged in constructing graphs, using some maths software, of these hybridised atomic orbitals, that's all. 

 

[LUFER]

32 minutes ago, exchemist said:

Er, acabei de confirmar o que é ρ. O que você está perguntando?   

Meu entendimento do modelo VB é que você faz orbitais atômicos hibridizados a partir dos orbitais atômicos do átomo livre e então atribui a ligação à sobreposição desses orbitais atômicos hibridizados com aqueles de átomos vizinhos. Isso é o que seu livro está dizendo. 

Nosso questionador está empenhado em construir gráficos, usando algum software de matemática, desses orbitais atômicos hibridizados, só isso. 

 

Because I'm asking about p, I think it's a constant. That I have neither p-value nor r-value.

[exchemist]

1 hour ago, LUFER said:

Because I'm asking about p, I think it's a constant. That I have neither p-value nor r-value.

Well no, it is proportional to r, surely? a is a constant, but ρ =r/a, right?   

By the way, are you in Brazil or Portugal?

Edited August 20, 2021 by exchemist

[studiot]

3 hours ago, exchemist said:

My understanding of the VB model is that you make hybridised atomic orbitals from the atomic orbitals of the free atom and you then attribute bonding to the overlap of these hybridised atomic orbitals with those of neighbouring atoms. This is what your book is saying. 

Our questioner is engaged in constructing graphs, using some maths software, of these hybridised atomic orbitals, that's all. 

Not exactly, no.

The free atom does indeed possess the s an p orbitals. I already said that.

When this free atom enter a bond it does not reorganise its orbitals into hybrid ones as a preparatory process.
This a a theoretical and mathematical device for us to transform the s and p orbitals into the hybrid ones, just as you say is outlined in the text.

But the text also has different hybridisation equations depending upon the environment as it shows with methane and ethylene.

Nor does the carbon in methane deconstruct its hybrids to reform the s and p orbitals then form  different hybrids when the methan molecule enter a chemical reaction where say one hydrogen is replaced by a chlorine.

So you cannot get away from the environment ie the other atoms in the molecule.

This is just as you cannot solve Schrodinger without the boundary conditions.

 

[LUFER]

2 hours ago, exchemist said:

Well no, it is proportional to r, surely? a is a constant, but ρ =r/a, right?   

By the way, are you in Brazil or Portugal?

but sorry, do you have the p-value?
I don't know if translation fails. r i know the radius, but the p don't know.
I don't know if it's constant.

[exchemist]

1 hour ago, LUFER said:

but sorry, do you have the p-value?
I don't know if translation fails. r i know the radius, but the p don't know.
I don't know if it's constant.

 ρ = r.(μe²/hbar²4πε₀). So it is r multiplied by a constant, made up of these various physical and mathematical quantities. 

2 hours ago, studiot said:

Not exactly, no.

The free atom does indeed possess the s an p orbitals. I already said that.

When this free atom enter a bond it does not reorganise its orbitals into hybrid ones as a preparatory process.
This a a theoretical and mathematical device for us to transform the s and p orbitals into the hybrid ones, just as you say is outlined in the text.

But the text also has different hybridisation equations depending upon the environment as it shows with methane and ethylene.

Nor does the carbon in methane deconstruct its hybrids to reform the s and p orbitals then form  different hybrids when the methan molecule enter a chemical reaction where say one hydrogen is replaced by a chlorine.

So you cannot get away from the environment ie the other atoms in the molecule.

This is just as you cannot solve Schrodinger without the boundary conditions.

 

Yes of course. But none of this is the subject of this thread.