Discussions on general chemistry and chemical engineering, organic chemistry, analytical chemistry, etc.

Hi everyone,

Again after long time i am here to clarify some facts on organic chemistry or chemistry. While i am reading orbital hybridization, it says hybrid orbitals have less energy than the individual S orbital. My doubt is does this means the hybrid orbitals are more closer to the nucleus than the S orbitals. The reason why ask this is i also learned that as distance increase from the nucleus the energy level of the orbital increase. Waiting for your reply guys.
Biosapien
Member

Posts: 124
Joined: 11 Mar 2015

Good question. Not all orbitals have spherical shape, so it's not straightforward to talk about "distance from nucleus".

The answer is that the hybrid orbitals will have an energy level (and if you want, distance from nucleus) that is somewhere between the lower and higher energy orbitals that are hybridizing. The overall energy of the hybrid orbitals, however, will be lower (under conditions that favor hybridization) than the overall energies of the un-hybridized orbitals (and hence why hybridization occurs - the system falls to a lower overall energy state).

If your hybridizing orbitals are s and p for example, the sp orbitals will be higher in energy (and distance) than the s orbital, and lower than the p orbitals. Total sp energy would be smaller than total s + p energy.

BioWizard

Posts: 12763
Joined: 24 Mar 2005
Location: United States
Blog: View Blog (3)
 Natural ChemE, Quantisierung liked this post

Thank you for the reply and clarification Biowiz. May I know does electron from P orbitals competes with electrons from S orbitals because of the reason that electron love to stay in lower energy orbitals.
Biosapien
Member

Posts: 124
Joined: 11 Mar 2015

Biosapien,

Quantum Mechanics is largely based off the Schrödinger equation. It's really stupidly hard to solve the Schrödinger equation, so we normally can't come up with exact solutions. At best, our idiot selves usually either simplify for analytical (mathematical) solutions or use computers to generate approximate numerical solutions. For analytical solutions, it's easiest to look at Hydrogen since it's got only one proton and one electron, making it a really simple atom.

When you solve the Schrödinger equation for Hydrogen, you get the atomic orbitals that they teach you about in Chemistry class (which are actually hydrogen-like atomic orbitals). These "orbitals" are literally just solutions to the Schrödinger equation if we assume that atoms are made up of a point-like nucleus with an electron. Other shapes aren't orbitals because they're not mathematically valid solutions to the Schrödinger equation.

Technically orbitals aren't the solid shapes that you usually see in pictures; those are simplifications. Orbitals are actually waves since quantum mechanics is all about waves. For example, the first S-orbital might be drawn like a sphere, but it's actually like the upper-left wave in this picture:
.
How do you draw that upper-left wave? As a sphere, if in three dimensions, e.g. as the S1 (upper-left again) image in this figure:
Note that the orbitals drawn in this picture don't actually have discrete boundaries, but rather just have areas which are more probable than others. Folks who draw orbitals discretely, e.g. the left-most image in this figure
,
are just drawing the areas with higher probablity distribution as being part of the orbital while ignoring areas with lower probability distribution. For this reason electrons aren't actually 100% confined to the areas shown in the solid drawings of orbitals.

But as stupidly complex as the Schrödinger equation is, it's got one really nice, clean property to it: solutions to it are linearly independent. This means that, if you have two waves which are a solution to the Schrödinger equation, then you can literally just add those two wave solutions together to get a third wave solution to the Schrödinger equation. See also:
Hybrid orbitals are solutions to the Schrödinger equation which can also be described as the sum of the solutions which we normally call atomic orbitals. Their shapes appear to be a mesh of the composing hybrid orbitals because it's just the two wave equations added together. Note that linear independence means that you can also subtract one solution from another to get a valid solution, which is where antibonding comes from.

Orbitals have different energies since electronic charges like being near each other as described in Coulomb's law. I don't know how accurate this is as I haven't done Quantum Chemistry in quite a while, but conceptually you can think of an orbital's energy as being something like
$\large E=\large {\frac{-{q}_{\text{electron}}{q}_{\text{nucleus}}}{4{\pi}{\epsilon}_{0}}}{\int}_{x_{-\infty}}^{x_{+\infty}}{\int}_{y_{-\infty}}^{y_{+\infty}}{\int}_{z_{-\infty}}^{z_{+\infty}}{\frac{{{\Psi}{\left({x,y,z}\right)}}^2}{x^2+y^2+z^2}}{\text{d}}z{\text{d}}y{\text{d}}x$,
or, in an alternative notation,
$\large E=\large {\frac{-{q}_{\text{electron}}{q}_{\text{nucleus}}}{4{\pi}{\epsilon}_{0}}}{\int}{\left({\frac{{{\psi}{\left(r\right)}}}{r}}\right)}^{2}{\text{d}}V$,
i.e. as the energy of two separated electric charges (the electron and the nuclear of the atom) integrated over all space by the probability of the particle being in that spot in space (i.e. the square of the wave function).
• Effective nuclear charge, which replaces ${q}_{\text{nucleus}}$ when considering atoms that have multiple electrons (as opposed to hydrogen-like atoms).
• Table of values for effective nuclear charge. Note that Hydrogen has an effective nuclear charge of $1$ since it's the very definition of a Hydrogen-like atom.

Summary:
1. Atomic orbitals aren't actually solid shapes. Drawings showing orbitals as solid shapes are just simplifications highlighting the most probable locations within the orbital.
2. Hybrid orbitals are the sum of other orbitals. Their weird shapes are just the drawn solid approximations of the wave that results from adding the waves of the orbitals which are added together to form the hybrid orbital.
3. Orbitals closer to the nucleus separate electric charges (the electron and the protons in the nucleus) less, so their energy is lower.
4. We don't fully get quantum mechanics, so if this confuses you, it's okay.
Natural ChemE
Forum Moderator

Posts: 2754
Joined: 28 Dec 2009
 Quantisierung, Hendrick Laursen, BioWizard liked this post

Looks like Orthogonal and Non-Orthogonal Hybrids, W. Kutnelnigg (1988), has a pretty good discussion on constructing hybrid orbitals from atomic orbitals.

On the last paragraph of the second page, this paper credits Linus Pauling with describing hybrid orbitals as linear combinations of atomic orbitals in 1931.

Also I forgot to mention molecular orbitals. Molecular orbitals are the same thing as hybrid orbitals, i.e. the sum of several more basic orbitals, except orbitals from more than one atom are involved.

Finally, I'd note that I've been told weird, incorrect stuff in this field before. For example, in undergrad, the Chemistry department head and I got into several arguments over whether or not shielding was a derivable consequence of known quantum mechanical relationships. She argued that QM wasn't sufficient and that shielding was a separate, fundamental effect. To the best of my knowledge her belief not only contradicts accepted theory but was also entirely baseless. Still, she taught it every year in the Quantum half of Physical Chemistry.
Natural ChemE
Forum Moderator

Posts: 2754
Joined: 28 Dec 2009

For the benefit of onlookers, Khan Academy on Youtube has some pretty decent overview vids on Orbitals and Electron Configurations in their Chemistry Playlist.

I watched several of them a few months back, and found them very insightful.

Darby
Active Member

Posts: 1188
Joined: 14 Feb 2015
Location: Long Island, New York (USA)

Biosapien » 24 Jul 2015 10:50 am wrote:Thank you for the reply and clarification Biowiz. May I know does electron from P orbitals competes with electrons from S orbitals because of the reason that electron love to stay in lower energy orbitals.

Compete for what?

BioWizard

Posts: 12763
Joined: 24 Mar 2005
Location: United States
Blog: View Blog (3)

Competes to occupy the low energy level orbitals. means will they jump from P to S orbitals.
Thank you for profound explanation guys and reference.
Biosapien
Member

Posts: 124
Joined: 11 Mar 2015

Biosapien, yes, electrons can jump from P to S orbitals. This can release energy. See also: emission spectrum.

Generally the universe is "cold", i.e. low on energy. Systems within the universe tend to minimize their energy because they're bleeding their energy to colder parts of the universe. It's kinda like mixing up marbles of different colors in a bowl; the mixed states are more common as there are more possible combinations. Ditto for low-energy configurations: stochastic processes tend to lead to Markov-esque equilibria in which energy is distributed across systems in the universe, effectively leading hot systems toward their ground state. The movement toward equilibrium in a very complex statistical ensemble is largely indistinguishable from energy minimization at the macroscopic level (noting that even subatomic systems can be "macroscopic" in this sense), so we use energy minimization as an (often very good) approximation. See also: Statistical Mechanics.

Not all parts of the universe are "cold". For example, in the interior of stars, electrons generally fail to remain in their ground states (e.g. the 1s orbital in a Hydrogen atom), and they're actually ionized away, forming plasma.

Natural ChemE
Forum Moderator

Posts: 2754
Joined: 28 Dec 2009

Thank you Nature chemE, I understand a little on what you explained that's because lack of my profound knowledge in this, but your reply makes me more curious to learn all these aspects of chemistry.
Biosapien
Member

Posts: 124
Joined: 11 Mar 2015

Nice post Natural ChemE. Just a few observations.

1. Can you explain this diagram? It seems that it is ambiguous.

a) Second derivative greater than zero therefore T is a minimum, of course from fundamental calculus. Viceversa.
b) E = T +V, if T decreases V increases (E-V<0), yeah from the law of conservation of energy.
c) what is the meaning of a wavefunction greater than zero?. Viceversa. it does make sense?

I think the diagram corresponds to something very specific, and it is not a general approach.

2. This is incorrect. The energy of orbitals comes from the eigen problem and they are found in the diagonal matrix elements of an adiabatic Hamiltonian.

Natural ChemE » July 24th, 2015, 12:33 pm wrote:
Orbitals have different energies since electronic charges like being near each other as described in Coulomb's law. I don't know how accurate this is as I haven't done Quantum Chemistry in quite a while, but conceptually you can think of an orbital's energy as being something like
$E={\frac{-{q}_{\text{electron}}{q}_{\text{nucleus}}}{4{\pi}{\epsilon}_{0}}}{\int}_{x_{-\infty}}^{x_{+\infty}}{\int}_{y_{-\infty}}^{y_{+\infty}}{\int}_{z_{-\infty}}^{z_{+\infty}}{\frac{{{\Psi}{\left({x,y,z}\right)}}^2}{\sqrt{x^2+y^2+z^2}}}{\text{d}}z{\text{d}}y{\text{d}}x$,

Quantisierung
Member

Posts: 78
Joined: 30 Sep 2011

Natural ChemE » July 24th, 2015, 2:20 pm wrote:
Also I forgot to mention molecular orbitals. Molecular orbitals are the same thing as hybrid orbitals, i.e. the sum of several more basic orbitals, except orbitals from more than one atom are involved.

Hi Natural ChemE, it is me again.

I have to say something. Years ago, when I was more active in this forum, I noticed several posts of yours saying the same idea of this quote.

It is OK, the construction of molecular orbitals starts from atomic orbital basis as you know; but they are not just a superposition as you are saying, like hybrid orbitals; if true, the life of a theoretician would be wonderful. The approximations are not also untrue because they start from hydrogen-like orbitals; well, as said, they are approximations; no theory is powerful enough to provide all the information we require.

Let's see:

1. Hartree-Fock theory.

The electron wavefunctions are single Slater determinants. Molecular orbitals are constructed from them. Orbitals are itetarively modified under variational principle so that the ground state energy is found.

2. Kohn-Sham DFT

Actually, it shares many mechanistic details as Hartree-Fock; but they focus in the construction of a ground state density from orbitals that minimizes the energy.

3. Moller-Plesset Perturbation theory.

Orbitals are Taylor expansions of an pertubed Hamiltonian.

4. Coupled Cluster, Configuration Iteration, etc.

The electron wavefunctions are many Slater determinants at once. These are the so-called post Hartree-Fock theories and they have a very high degree of accuracy.

In all of those theories, the initial molecular orbitals constructed from (not always) hydrogen-like orbitals are iteratively optimized to reach a convergence criterion. The final molecular orbitals have nothing to do with hydrogen like.

Well, also just keep in mind that orbital is an abstract concept. It would be better to have available a pure electron density in theoretical chemistry, just as extracted from experiments.

Quantisierung
Member

Posts: 78
Joined: 30 Sep 2011

Quantisierung,

It's good to see you back!

1. I used a diagram from Wikipedia for the pictures on it. The equations weren't relevant to the topic. However, if you're interested in the equations on that diagram, I'd suggest:
1. clicking on the diagram (it's a link) to view the source;
2. making a post about the parts you're interested in.
2. The Hamiltonian is a mathematical tool that holds descriptive information about the relevant physics. In this case, it would include Coulombic interactions since Coulombic interactions are the relevant physics. Technically kinetic energy and gravity are also involved, but they're not the major conceptual point.
• Yup, as I'd noted in the second paragraph of that post, I was speaking in conceptual terms to introduce the topic. As you correctly point out, there are computational methods that are able to consider the many-body interactions. The focus of the post wasn't on these topics because they don't really help folks to understand stuff like why orbitals have different energies or the shapes that they do.
Natural ChemE
Forum Moderator

Posts: 2754
Joined: 28 Dec 2009
 Quantisierung liked this post

Hi Natural ChemE, I agree in most of you say; however, I just add a correction here

Natural ChemE » September 14th, 2015, 12:05 am wrote:
The Hamiltonian is a mathematical tool that holds descriptive information about the relevant physics. In this case, it would include Coulombic interactions since Coulombic interactions are the relevant physics. Technically kinetic energy and gravity are also involved, but they're not the major conceptual point.

The Hamiltonian is, for a molecular system:

H = T + Vee + Vne

T = Kinetic operator
Vee = electron-electron potential
Vne = nuclear attraction potential

What you say about Coulomb interactions is commonly described for Vne. However, Vee is a challenging task for theoreticians.

If we know the wavefunction of electron i, let's name it wf(i), then we can know its orbital energy as follows

<wf(i) | H | wf (i)> = E(i)

With this I mean that orbital energy E(i) is not only Coulomb energy, but it also includes kinetic and electron-electron potentials. Moreover, an adiabatic Hamiltonian means a diagonilized matrix where the diagonal contains all the E(j) of the system.

PS
Gravitiy is not considered in traditional quantum mechanics. Observe that the Hamiltonian does not include anything related to a "gravity potential"

Quantisierung
Member

Posts: 78
Joined: 30 Sep 2011
 Natural ChemE liked this post

Quantisierung,

Yup, as you'd said,
${{H}_{\text{atom}}}=T+{{V}_{{\text{e}},{\text{e}}}}+{{V}_{{\text{n}},{\text{e}}}}$
where
• ${{H}_{\text{atom}}}$ is the commonly used Hamiltonian description for electrons in an atom;
• $T$ is kinetic energy;
• ${{V}_{{\text{e}},{\text{e}}}}$ holds electron-to-electron potential contribution;
• ${{V}_{{\text{n}},{\text{e}}}}$ holds nucleus-to-electron potential contribution.
This commonly used relationship neglects gravity because:
1. it's believed to be an extremely, extremely small contribution;
2. we don't really understand gravity at the quantum level yet (quantum gravity).
Because we're looking at hydrogen-like atoms (i.e. there's only one electron), ${{V}_{{\text{e}},{\text{e}}}}=0$(Note 1), so
${{H}_{\text{hydrogen-like atom}}}=T+{{V}_{{\text{n}},{\text{e}}}}$.
The electron has kinetic energy as it's moving around inside of the orbital (kinda; I guess this gets fuzzy in Quantum). However, the idea that electrons move around in orbitals isn't particularly useful to conceptualizing the differences between orbitals. This leaves us with
${{H}_{\text{conceptual hydrogen-like atom}}}{\left[{\approx}\right]}{{V}_{{\text{n}},{\text{e}}}}$,
meaning that we're pretty much just concerned with the Coulombic interactions for our explanatory purposes.

I really like it that you'd brought up an actual formulation for the ${{H}_{\text{atom}}}$. It's a great addition to this thread.

Notes:
1. Here zero $\left(0\right)$ is the zero-operator, i.e. "zero times", because the Hamiltonian is itself an operator. Readers who don't get this point can see it as a normal zero for the purposes of this post.
Natural ChemE
Forum Moderator

Posts: 2754
Joined: 28 Dec 2009
 Quantisierung liked this post

Quantisierung » September 12th, 2015, 1:02 pm wrote:Nice post Natural ChemE. Just a few observations.

1. Can you explain this diagram? It seems that it is ambiguous.

a) Second derivative greater than zero therefore T is a minimum, of course from fundamental calculus. Viceversa.
b) E = T +V, if T decreases V increases (E-V<0), yeah from the law of conservation of energy.
c) what is the meaning of a wavefunction greater than zero?. Viceversa. it does make sense?

I think the diagram corresponds to something very specific, and it is not a general approach.

1. Responses:
1. Yup, you're right.
• Kinetic energy, $T=E-V$, can be negative when tunneling, i.e. passing through a region of space that classical mechanics would've forbidden it being in. I believe that there's a better way of looking at this, but a quick Google search suggests that it's non-standard (I can't find it anywhere).
• If we integrate the square of a wave function over all space, it must equal one:
$\int_{\text{space}}{{\Psi}^{2}}{\text{d}}V=1$.
This means that the probability of finding the particle described by that wave function somewhere in all of space is exactly 100%. Also, since any subspace must have a probably between 0 and 1, the square can't be negative at any point (since that'd lead to a negative probably at that point, and a greater-than-100% probably in the rest of the universe). However, that's the square of the wave function. The wave function itself is a complex number, and its real and imaginary components may be negative.
In this picture, the blue regions are where the wave function is negative.

Generally, because wave functions are smooth waves, you'll get signs flipping (negative to positive or vice versa) between neighboring regions. For example, the 7s orbital:
.
Just to note it, the locations that aren't colored red or blue still have non-zero values of the wave function; they're just relatively close to zero except at the exact nodes between sign flips. This isn't just the gaps between the colored shells, but also radiating out to infinity since the s-shells never really "stop", but rather go to zero as distance from the nucleus goes to infinity,
${\text{lim}}_{r{\rightarrow}{\infty}}{\left({{\Psi}{\left(r\right)}}^2\right)}{\rightarrow}0$.

In the actual universe, usually there will be other potentials as we go out to infinity. These would cause the wave function to decline far more rapidly when they exceed the electron's energy, though the electron may yet still tunnel into those regions (according to current models when taken to their unvalidated extremes).
Natural ChemE
Forum Moderator

Posts: 2754
Joined: 28 Dec 2009
 Quantisierung liked this post

Hi again Natural ChemE; I'm back

From here, even for explanatory purposes, the kinetic contribution should not be neglected.

Natural ChemE » September 15th, 2015, 3:29 pm wrote:
Because we're looking at hydrogen-like atoms (i.e. there's only one electron), ${{V}_{{\text{e}},{\text{e}}}}=0$, so
${{H}_{\text{hydrogen-like atom}}}=T+{{V}_{{\text{n}},{\text{e}}}}$.
The electron has kinetic energy as it's moving around inside of the orbital (kinda; I guess this gets fuzzy in Quantum). However, the idea that electrons move around in orbitals isn't particularly useful to conceptualizing the differences between orbitals. This leaves us with
${{H}_{\text{conceptual hydrogen-like atom}}}{\left[{\approx}\right]}{{V}_{{\text{n}},{\text{e}}}}$,
meaning that we're pretty much just concerned with the Coulombic interactions for our explanatory purposes.

First, electrons in orbitals are indeed moving, the proof is the existence of electron density

therefore they have an associated kinetic energy and angular momentum (first and second quantum numbers respectively).

Second; I am going to demonstrate how important the kinetic energy is. Let's see.

Consider the 1s solution to the Schrödinger equation for the hydrogen atom.

$\Large \psi_{1s} = \Large \frac {1}{sqrt\pi} (\frac {Z}{a_0})^{3/2} e^{-Zr/a_0}$

This orbital expression can be translated into the bra-ket notation, which is more easy to write:

$\Large \psi_{1s} = |1s >$

and

$\Large (\psi_{1s})* = <1s|$

From text books, you can find that

$\Large E_{1s} = \Large \left\langle 1s \middle|\hat H\middle| 1s \right\rangle$

$\Large E_{1s} = \frac{-Z^2 e^2} {8 \pi \epsilon_0 a_0}$

However, if you are able also to solve the potential energy you shoud obtain

$\Large V_{1s} = \Large \left\langle 1s \middle|\hat V\middle| 1s \right\rangle$

$\Large V_{1s} = \Large \left\langle 1s \middle| 1/r \middle| 1s \right\rangle$

$\Large V_{1s} = \frac{-Z^2 e^2} {4 \pi \epsilon_0 a_0}$

Carefully pay attention in this last result, it means that

$\Large E_{1s} = \Large \frac {1}{2} V_{1s}$

This result is known as the quantum virial theorem. It has important consequences. You already know that

E = V + T

Therefore

T = E - V = (V/2) - V = - V/2

or well

0 = 2T + V

What is the point?

The point is that kinetic energy is not negligible. The energy of the 1s level is -0.5 atomic units, that means:

$\Large E_{1s} = -0.5 a.u.$
$\Large V_{1s} = -1.0 a.u.$
$\Large T_{1s} = 0.5 a.u.$

Observe, that 0.5 a.u. = 314 kcal/mol, a very large quantity that, if ignored, it will lead you to unphysical results when studying for instance dissociation of hydrogen or other chemical reactions.

As a final remark, some quantum theories work preferably minimizing the kinetic energy as well; like DFT.

To sum up, Coulomb interactions are as important as kinetic contributions, and of course, other more complex terms like exchange and correlation (Vee, not avaiable for hydrogen atom, as you already mentioned)

Kindest,
JP

Quantisierung
Member

Posts: 78
Joined: 30 Sep 2011
 Natural ChemE liked this post

Natural ChemE, for your second post, I agree with all your explanation.

In my case, it created some confusion since I am usually interested in total probability density, as you already described, and total average values:

$\Large \left\langle\psi\|\psi\right\rangle = 1$

$\Large \left\langle\psi\middle|\hat O\middle|\psi\right\rangle = o$

Those integrals must be done through the entire domain. Therefore, for me it does not matter if the wavefunction is in phase A or phase B (as you colored as red and blue showing the electron wavefunction of orbital 7s for instance) since the integration through the entire domain lead to the same result for the molecular property "o" associated to operator $\hat O$.

In other words, if I have the bra then I make the vector product from the right. Viceversa if I have the ket. Observe that for the wavefunction the bra represents pahse A (in the picture, it is the red region); and the ket is phase B (blue region).

Therefore where T or V is a minimum for me it is not important, since the electron is distributed through the entire orbital; and the probability of finding it at some specific region (at local maximum or minimum points) is low.

Quantisierung
Member

Posts: 78
Joined: 30 Sep 2011
 Natural ChemE liked this post

Quantisierung,

Earlier you'd commented that the energy-of-an-orbital equation didn't look quite right:
Quantisierung » September 12th, 2015, 1:02 pm wrote:2. This is incorrect. The energy of orbitals comes from the eigen problem and they are found in the diagonal matrix elements of an adiabatic Hamiltonian.

Natural ChemE » July 24th, 2015, 12:33 pm wrote:
Orbitals have different energies since electronic charges like being near each other as described in Coulomb's law. I don't know how accurate this is as I haven't done Quantum Chemistry in quite a while, but conceptually you can think of an orbital's energy as being something like
$E={\frac{-{q}_{\text{electron}}{q}_{\text{nucleus}}}{4{\pi}{\epsilon}_{0}}}{\int}_{x_{-\infty}}^{x_{+\infty}}{\int}_{y_{-\infty}}^{y_{+\infty}}{\int}_{z_{-\infty}}^{z_{+\infty}}{\frac{{{\Psi}{\left({x,y,z}\right)}}^2}{\sqrt{x^2+y^2+z^2}}}{\text{d}}z{\text{d}}y{\text{d}}x$,

. I just wanted to show how it follows from what you've posted.

As you'd noted:
Quantisierung » September 20th, 2015, 10:07 am wrote:This orbital expression can be translated into the bra-ket notation, which is more easy to write:

$\Large \psi_{1s} = |1s >$

and

$\Large (\psi_{1s})* = <1s|$

From text books, you can find that

$\Large E_{1s} = \Large \left\langle 1s \middle|\hat H\middle| 1s \right\rangle$

$\Large E_{1s} = \frac{-Z^2 e^2} {8 \pi \epsilon_0 a_0}$

Here I'd note that
$\Large E_{1s} = \Large \left\langle 1s \middle|\hat H\middle| 1s \right\rangle$,
when translated into a Hilbert space (i.e. what most folks think of), becomes something like
$\Large E_{1s} = \Large \int \left({{\psi_{1s}}^{2} \hat H }\right)$,
which, as you note, evaluates to
$\Large E_{1s} = \frac{-Z^2 e^2} {8 \pi \epsilon_0 a_0}$.

This is pretty much exactly what the post referring to
$E={\frac{-{q}_{\text{electron}}{q}_{\text{nucleus}}}{4{\pi}{\epsilon}_{0}}}{\int}_{x_{-\infty}}^{x_{+\infty}}{\int}_{y_{-\infty}}^{y_{+\infty}}{\int}_{z_{-\infty}}^{z_{+\infty}}{\frac{{{\Psi}{\left({x,y,z}\right)}}^2}{\sqrt{x^2+y^2+z^2}}}{\text{d}}z{\text{d}}y{\text{d}}x$
was talking about. Except, to modify the notation to match, it'd be more like
$E=\large {\frac{-{Z}^{2}{e}^{2}}{4{\pi}{\epsilon}_{0}}}{\int}{\frac{{{\psi}^{2}}}{\sqrt{x^2+y^2+z^2}}}$,
noting that
$\large {\hat H}=\large {{\frac{-{Z}^{2}{e}^{2}}{4{\pi}{\epsilon}_{0}}}{\frac{1}{\sqrt{x^2+y^2+z^2}}}}$
in this case, as we'd previously discussed, i.e. it's Coulomb's law.

Generally bra-ket notation isn't really meaningful to folks except those who already know Quantum, so I think that integrals can be more helpful to introduce the topic.

The main point here was to address the OP's question about the orbital's location versus its energy:
Biosapien » July 19th, 2015, 11:55 pm wrote:Hi everyone,

Again after long time i am here to clarify some facts on organic chemistry or chemistry. While i am reading orbital hybridization, it says hybrid orbitals have less energy than the individual S orbital. My doubt is does this means the hybrid orbitals are more closer to the nucleus than the S orbitals. The reason why ask this is i also learned that as distance increase from the nucleus the energy level of the orbital increase. Waiting for your reply guys.
I think that showing an orbital's energy using an integration of its wave function's square* by Coulomb's law is the most fruitful way to answer the question.
* Technically it's not the square of the wave function, but rather its inner dot product with its own complex conjugate. This reduces to squaring if the wave function were entirely real, but in general it can be complex, i.e. have an imaginary component. Still, I think that this factoid is more confusing to someone just learning Quantum than it is instructive, except as a side note like this.
Natural ChemE
Forum Moderator

Posts: 2754
Joined: 28 Dec 2009
 Quantisierung liked this post

Natural ChemE » July 24th, 2015, 12:33 pm wrote:$E={\frac{-{q}_{\text{electron}}{q}_{\text{nucleus}}}{4{\pi}{\epsilon}_{0}}}{\int}_{x_{-\infty}}^{x_{+\infty}}{\int}_{y_{-\infty}}^{y_{+\infty}}{\int}_{z_{-\infty}}^{z_{+\infty}}{\frac{{{\Psi}{\left({x,y,z}\right)}}^2}{\sqrt{x^2+y^2+z^2}}}{\text{d}}z{\text{d}}y{\text{d}}x$
Eww! Sorry folks, I made a stupid mistake!

It's
$\large E=\large {\frac{-{q}_{\text{electron}}{q}_{\text{nucleus}}}{4{\pi}{\epsilon}_{0}}}{\int}_{x_{-\infty}}^{x_{+\infty}}{\int}_{y_{-\infty}}^{y_{+\infty}}{\int}_{z_{-\infty}}^{z_{+\infty}}{\frac{{{\Psi}{\left({x,y,z}\right)}}^2}{x^2+y^2+z^2}}{\text{d}}z{\text{d}}y{\text{d}}x$,
and not
$\large E=\large {\frac{-{q}_{\text{electron}}{q}_{\text{nucleus}}}{4{\pi}{\epsilon}_{0}}}{\int}_{x_{-\infty}}^{x_{+\infty}}{\int}_{y_{-\infty}}^{y_{+\infty}}{\int}_{z_{-\infty}}^{z_{+\infty}}{\frac{{{\Psi}{\left({x,y,z}\right)}}^2}{\sqrt{x^2+y^2+z^2}}}{\text{d}}z{\text{d}}y{\text{d}}x$.
I was taking the square root to move out of spherical coordinates into the normal $\left(x,y,z\right)$ coordinate system folks are more used to, but that was silly since Coulomb's law is by $\large \frac{1}{r^2}$, not $\large \frac{1}{r}$. The square cancels out the square root, i.e.
$\large {\frac{1}{r^2}}=\large {\frac{1}{{\sqrt{\left(x^2+y^2+z^2\right)}}^2}}=\large {\frac{1}{x^2+y^2+z^2}}$.

Natural ChemE
Forum Moderator

Posts: 2754
Joined: 28 Dec 2009
 Quantisierung liked this post

*Gasp* NCE!

I was going to point it out, but you beat me to it by one second. And two minutes. And three hours. And four days. Ok fine I wasn't going to open my quantum chemistry notes for this. But yeah, it's always 1/r2.

BioWizard

Posts: 12763
Joined: 24 Mar 2005
Location: United States
Blog: View Blog (3)
 Natural ChemE, Quantisierung liked this post

Hi guys,

One question to both, Biowizard and Natural.

Are you sure the operator is 1/r2 instead of 1/r?

This is indeed the nucleus-electron distance, I'd rather write |R-r|; being R the position of the nucleus and r the position of the electron, moreover vertical lines | | stand for geometrical distance.

https://en.wikipedia.org/wiki/Molecular_Hamiltonian

Finally, the original question by Biosapien was immediately answered by Biowizard, in his/her first post.

Quantisierung
Member

Posts: 78
Joined: 30 Sep 2011
 Natural ChemE liked this post

Quantisierung,

I made that mistake in the OP. It's one of those
$\large{\int{\frac{1}{x^2}}{\text{d}}x=-\frac{1}{x}}+c$
things.

So Coulomb force is proportional to $r^{-2}$. To get energy, you integrate the force over the distance of displacement relative to some reference (generally infinity). Conceptually integrating a force over distance like this means that you're getting the total work needed to move against that force from the start point to the end point, i.e. the energy.

Wikipedia uses $r^{-1}$ since they're writing the Coulomb energy found after integration. Since the equation uses the force within the integrand, it's Coulomb force, so $r^{-2}$.

But, yeah, I keep changing my mind on how to word posts and sometimes I get stuff like that mixed up!

PS - I've been pulling tons of all-nighters and I'm not thinking too clearly right now. I suspect that there may be an error in integrating Coulomb force and the wave function together; perhaps they should be separated into the product of two integrations, with the Coulombic one reducing down to the normal expression for Coulombic energy? I can update later once I've had some sleep (maybe Monday-ish since I'll be out-of-town this weekend).
Natural ChemE
Forum Moderator

Posts: 2754
Joined: 28 Dec 2009

Benefit a lot，you can go on
smfood
Forum Neophyte

Posts: 17
Joined: 02 Feb 2015