[monicad95]

Hi, I'm in my first year of pharmacology and in Mcmurry's textbook (the one with the shell on it, I think it's edition 1?) page 12, it says that the reason that hybridization occurs is because with the sp3 orbital, the two lobes of it are unsymmetrical (one bigger lobe and one smaller lobe), and this allows it to overlap more effectively with another orbital when forming a bond.

However, this is the only place where I've seen that explanation being mentioned.
I thought the reason was because when you mix the s and the p orbitals (which have different energy levels), the hybridisation forms 4 equivalent orbitals all of an average energy (between the s and the p energy levels) which means that it's a lower energy overall and more stable so can form more stable bonds??

[Enthalpy]

Hi monicad95, welcome here!
Just a note before you get more orthodox answers:

Electrons do what's best for them. Descriptions as orbital combinations is just what we humans would have liked them to do so we can understand their behaviour.

Electrons interact strongly within an atom or a molecule. Though, the only algebraic solutions we have is for the lone hydrogen atom (something rare on Earth), so we pretend to misuse these solutions for atoms and molecules of several electrons and nuclei. This demands strong updates to the model here and there.

Linear algebra tells that the wave of any electron caught by a nucleus is a weighed sum of orbitals, but this holds for one nucleus. Chemical bonds between atoms are commonly and usefully described as sums and differences of the orbitals of individual atoms, but this is an approximation that would hold if the atoms were far apart - not very accurate in a bond.

As schrödinger's equation is linear, any weighed sum of solutions is a solution... but orbitals are special solutions: stationary ones, where the wave's modulus doesn't change over time, only the phase does. A weighed sum of orbitals having exactly the same energy, say 2px and 2py, is still an orbital - but if the energies differ, even a little bit as between 2s and 2p, then the waves' phases rotate at different paces, and the weighed sum is not stationary, is not an orbital, because the locations where the 2s and 2p add or subtract vary over time. In this sense, a weighed sum like sp3 is not an orbital.

Carbon uses to bind with several more atoms, and the resulting orbitals span each over all neighbours, instead of over one atom pair or one "bond".

With that in mind, it's clearer that bonds and hybrid orbitals are an approximation. They shouldn't be stretched too far trying to explain everything. Though, molecular orbitals are less accessible to reasoning and I haven't seen them used to predict on the paper reaction mechanisms for instance (a theory still to be created?), while bonds and hybrid orbitals are a fertile model, so these must be known.

[mjc123]

Sit down and take a deep breath. Hybridisation doesn't happen. Hybridisation isn't a physical thing, but a mathematical convenience. Any set of orbitals can also be described mathematically by a suitable set of linear combinations of those orbitals, the overall energy being unchanged. So if you have a 2s and three 2p orbitals, you can also describe them mathematically using a set of four symmetry-equivalent orbitals we call sp³. These orbitals have, as you say, one large projecting lobe, and are disposed tetrahedrally, so they are convenient for describing the bonding of carbon to 4 other atoms. They are not convenient for describing a trigonal planar bonding situation, as in ethylene, so there we use a different combination - three sp² orbitals and one p orbital.

I thought the reason was because when you mix the s and the p orbitals (which have different energy levels), the hybridisation forms 4 equivalent orbitals all of an average energy (between the s and the p energy levels) which means that it's a lower energy overall and more stable so can form more stable bonds??

This is not true. (sp³)⁴ has a higher energy than the ground state of 2s²2p. It has exactly the same energy as 2s¹2p³, precisely because the sp³ orbitals are linear combinations of 2s + 3*2p orbitals, and the linear combination doesn't change the total energy - it's just a different mathematical description of the same thing. The energy required to promote an electron from 2s to 2p is more than compensated by the formation of two extra bonds.

[monicad95]

Thanks for your reply

When you say that "(sp3)4 has a higher energy than the ground state of 2s22p. It has exactly the same energy as 2s12p3" but then you also say that " The energy required to promote an electron from 2s to 2p is more than compensated by the formation of two extra bonds", so then shouldn't (sp3)4 aka 2s12p3 have a lower energy than 2s22p?
Sorry if this was a stupid question!
Thanks  

[Enthalpy]

Any set of orbitals can also be described mathematically by a suitable set of linear combinations of those orbitals, the overall energy being unchanged. So if you have a 2s and three 2p orbitals, you can also describe them mathematically using a set of four symmetry-equivalent orbitals we call sp³.

We have a disagreement here. The linear combination of orbitals is commonly used, but that's not justified by linear algebra. The linear combination would be an orbital only if the energies were exactly the same - not for 2s and 2p. Calling sp³ an orbital is already an approximation, not math - a useful approximation.

[mjc123]

the ground state of 2s22p

I should of course have said 2s²2p².

so then shouldn't (sp3)4 aka 2s12p3 have a lower energy than 2s22p?

Not in the isolated carbon atom. The energy is lowered by forming bonds with other atoms.

Calling sp3 an orbital is already an approximation, not math - a useful approximation.

I think you're probably right. It's a long time since I went through this stuff in detail. And even I sometimes find it convenient to talk as if sp³ orbitals were real things!

[Corribus]

We have a disagreement here. The linear combination of orbitals is commonly used, but that's not justified by linear algebra. The linear combination would be an orbital only if the energies were exactly the same - not for 2s and 2p. Calling sp³ an orbital is already an approximation, not math - a useful approximation.

This doesn't make sense to me. Linear combination of atomic orbitals on the same nucleus are what hybridized orbitals are. The energies of the orbitals don't have to be the same to take a linear combination. We do this all the time for LCAO treatment of molecular orbitals. Hybrid orbitals are no different, except that evaluating the integrals to determine the resulting energies is easier due to the orthonormality of the atomic orbitals involved.

Maybe I'm just misunderstanding your point?

[Irlanur]

but orbitals are special solutions: stationary ones,

I am not sure this is true. It might depend on who you ask, but I think that orbitals are even less than that. They are just the one-electron-functions in a product or Slater-determinant ansatz. In the HF formalism, you then look for the orbitals which minimize the energy (i.e. orbitals that make a Lagrangian, including the orthonormality as Lagrangian multipliers, stationary. they are NOT stationary solutions of the real Schrödinger Equation (these would be the exact solutions) ). You can calculate these orbitals numerically, without any introduction of atomic orbitals or hybridisation whatsoever. These is however computationally very demanding and often inconveniant. So the usual approach is to introduce a Basis-Set (this is where LinAlg becomes important), which leads to the Roothaan-Hall equations, the basis of many Computational Chemistry approaches.

[Enthalpy]

As soon as one attributes a definite energy to an orbital - often the case - or defines an orbital as a proper function of the Hamiltonian, it's a stationary wave function. But true, for instance the Slater-type "orbitals" are no proper functions
http://en.wikipedia.org/wiki/Slater-type_orbital

I wish they wouldn't be called "orbital", but it is the habit:
http://en.wikipedia.org/wiki/Atomic_orbital
"Atomic orbitals can be the hydrogen-like "orbitals" which are exact solutions to the Schrödinger equation for a hydrogen-like "atom" (i.e., an atom with one electron). Alternatively, atomic orbitals refer to functions that depend on the coordinates of one electron (i.e. orbitals) but are used as starting points for approximating wave functions that depend on the simultaneous coordinates of all the electrons in an atom or molecule."

I know it's common practice to compute orbitals as linear combinations of orbital moduli only, but it is my opinion up to now that this does not result from the linearity of Schrödinger's equation. The exp(i2pi*Et/h) is  a part of the wave function and can't be omitted when expressing any trapped electron's wave function as a linear combination of orbitals. It's even the beat at F=(E₂-E₁)/h of orbitals with energies E₂ and E₁ that makes the electron wobble and emit or absorb a photon.

So the LCAO method gives sensible results by forgetting the individual phase terms, and there must be good reasons for it, be these exact or approximate, but the linearity of Schrödinger's equation is too short an argument. Wiki's article omits this difficulty
http://en.wikipedia.org/wiki/Linear_combination_of_atomic_orbitals

In the case of several atoms in a molecule, the situation is clearer: an orbital at one atom is not a solution of Schrödinger's equation at the other since the nuclei's positions differ, so the equation's linearity is no sufficient reason to take a linear combination of the atomic orbitals.

My guess up to now is that many non-obvious approximations are necessary to the method, and most textbooks conceal that under "linearity" and "basis".

[Irlanur]

Hmm... As I understand it, if a function is stationary under the Hamiltonian, it is an Eigenfunction (if it's not an Eigenfunction, it could be expressed as a linear combination of Eigenfunctions with different Energies -> the phase changes incoherently and the total function is not stationary).

So if we knew stationary solutions of the TRUE hamiltonian, we would know its exact solutions. that's not the case. A Determinant or product of One-electron-functions (=orbitals) cannot be the true solution of a many-electron system, that's prevented by the electronic repulsion terms in the hamiltonian. Every Operator, including the Hamiltonian, expressed in the basis of one-electron-functions, is incomplete.

that's my understanding of it.