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This is a very fundamental question and for really understanding the "why" some advanced physics is involved. I will describe the process rather superficially. As you might know, the level energies of atoms and molecules can be calculated (in principle) using quantum mechanics. The simplest system is the hydrogen atom as it consists of a single ...


13

These orbitals represent the angular part of the wavefunction. The solution obtained directly from solving the Schrödinger equation produces equations containing complex numbers so cannot be drawn on normal $xyz$ axes and are hard to visualise. The angular part of the wavefunction is given by functions called Spherical Harmonics these usually are given the ...


9

I'd like to copy the answer by John Rennie to a similar question at Physics.SE, since it's much better than current answers here IMHO. Although it mostly speaks about binding energy, the same principle applies to the quanta of excitation energy. The mass of a hydrogen atom is $1.67353270 \times 10^{-27}$ kg. If you add the masses of a proton and electron ...


8

As @orthocresol points out, the key is that you need to compare the exchange energy before vs after the ionization process. Anything that is unchanged by ionization cannot affect ionization energy. This is a general idea for a lot of aspects of chemistry: close the thermodynamical circles carefully, since so much information can be gained from them. So, let ...


6

The very question you pose is addressed thoroughly in this open access work: https://www.nature.com/articles/ncomms9287 . The short answer to your question is that the electron density can be mapped using a technique akin to diffraction as described in their work. You mentioned a distinction between the electron density and the orbitals, and the orbital is ...


6

There is a misconception here. A p orbital is a 3D-function, and these functions don't have shapes, they have values at any point in space. If you describe an electron distribution with one of these functions, you can plot contours at a chosen value, and these contours have shapes. For example, you could choose a contour level such that an electron has a 90% ...


6

Does electron mass decrease when it changes its orbit? Essentially yes. If you add the mass of a free proton and a free electron you'll get a greater mass than that of a hydrogen atom. The mass difference will be equivalent to 13.6 eV which is the ionization energy of hydrogen. Now for any "practical" chemistry experiment the assumption is that ...


6

The wavefunction covers 3-dimensional space. Formally, it maps each point in space to a complex number. That is, it is a function that takes a point in 3-dimensional space as input and returns a complex number as output: $\Psi:\mathbb{R}^{3}\rightarrow \mathbb{C}$. There are many ways to represent 3-dimensional space: cartesian coordinates $(x,y,z)$, ...


5

Many things are only stable in their lowest energy state: electrons are no different Hold a ball in your hand. It is, in effect, in an excited state. Open your hand and the ball falls to the floor, without much effort or any push. Set the ball on the floor and it doesn't move. It is in its lowest energy state and won't move around unless given a push. Many ...


5

The question is misguided by the obsolete idea of shells and subshells as separated exclusive regions of the space around the nucleus. All orbitals largely overlaps and statistical distribution of electron density, (aside of being in the particular orbital) depends on the nucleus charge and overall electron configuration. It can be said, that distribution of ...


5

Be gentle with me, I am going to try to explain orbitals in a very simple way using many similes. I do this with primary school kids so many of the words I will use aren't literally accurate but I hope the overall picture is as accurate as possible (without knowing quantum math). Please fell free to make suggestions where the overall picture deviates from &...


5

Just to be sure: Note that orbitals(*) themselves have neither spin, neither charge. They are not real objects, but theoretical constructs of quantum atom models, that fit well the observations. Every atom has theoretically infinite number of orbitals ( 1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p, 4d, 4f etc), all but some unoccupied. For most scenarious, it is practical ...


5

Pauli's exclusion principle is a consequence of electrons being indistinguishable fermions. Fundamental particles are indistinguishable in that two of the same type differ only in a few properties but otherwise behave identically (compare that say to two apples - they are always in principle distinguishable because they differ in so many ways). Fermions ...


5

The values of $L$ come from the individual $\ell$ values, according to the Clebsch-Gordan series: $$L = |\ell_1 - \ell_2|, \ldots , \ell_1 + \ell_2.$$ where the dots indicate all "in-between" values with a step size of 1. The values of $S$ come from the individual $s$ values, in a similar fashion. Note that it is $s$, not $m_s$. Finally, the values ...


4

First of all, how can a "half" sigma bond exist? Usually, you expect double bonds to be shorter and stronger than the corresponding single bonds, and triple bonds even shorter and stronger. The OP already mentioned bond-orders of 1.5 that occur for conjugate double bond systems, and those have properties in between single and double bonds. Perhaps ...


4

Which representation you would use depends on what kind of information you would want to represent. The p-orbitals (and all orbitals with a finite angular momentum quantum number $l$, for that matter) have a spatially dependent phase: The nodal planes in your bottom representation separate the parts of the p-orbitals that have opposite phases. In other words,...


4

Although the general trend $\mathrm{s} < \mathrm{p} < \mathrm{sp}^n$ makes sense, these magic numbers $1.73$, $1.93$, $1.99$, and $2.00$ seem to have just been pulled out of a hat. If these are really "approximate strengths of bonds", a good book would justify these by showing which bonds they use to come up with these numbers. Just to show ...


4

These rules go back to a publication by Russel S. Drago criticizing the use of the VSEPR model in Introductory Chemistry. As an alternative, he suggests a set of heuristics. First, if there are no lone pairs, use essentially the same rules as VSEPR. How to treat lone pairs is described below: Here we shall consider central atoms with eight or less electrons ...


4

The Bohr Model tried, quite successfully for its time, to model the energy states of an electron. This model has turned out inadequate, as it cannot answer question like yours. There have been more refined models (where your question isn't possible), but they all have their drawbacks. Not an answer to what you were asking, but some kind matching your ...


4

Stationary wavefunctions are solutions of the time-dependent Schrödinger equation $$\hat{H}\psi=i\hbar\frac{\partial \psi}{\partial t} $$ for which the energy $E$ is constant (it being - like the Hamiltonian - otherwise generally time-dependent) so that $$\begin{align}\hat{H}\psi&=E\psi\\&=i\hbar\frac{\partial \psi}{\partial t}\end{align} $$ which ...


4

In multielectronic atoms we have a relatively large difference between $s$ and $d$ orbitals when they have the same $n$ quantum number, or in terms of the actual quantum mechanics when they have the same total number of nodes in the wavefunction (this being what we label as $n-1$). But in the case of transition metals the $s$ orbital that mixes in with the $...


3

The answer comes in several layers, as Martin has alluded to. I'll try and give a short summary: each point goes slightly deeper than the previous one. For an isolated atom, the labelling of the orbitals is arbitrary. That is to say, the $p_x$, $p_y$, and $p_z$ orbitals are all interchangeable. More formally, this reflects the isotropy of the system, which ...


3

I found an old slide from when I was teaching. These are plots of the 2s and 2p orbitals for hydrogen (assuming 1 electron). The radial wave function is on the left and the probability density as a function of distance from the nucleus is on the right ($4\pi r^{2}R(r)dr$ if you understand the calculus). You can see that there's a blob of density that comes ...


3

This video nicely explains your question. The essence is that paulis exclusion principle states that two electrons(or generally fermions) cannot occupy the same quantum state. This is because electrons are fundamentally indistinguishable,but their wavefunctions are not. So there must be some quantum factors which allows to distinguish between their quantum ...


3

The discussion surrounding SF6 is often centered around two opposing hypotheses: (1) Hypothesis # 1: SF6 actually obeys the octet rule, because the sulfur atom has a net positive charge and some of the S-F bonds are ionic in nature. For example, four single S-F bonds and two ionic S - F bonds. The six equal S-F bond lengths are then explained as a resonance ...


3

Here is the meaning of the words ""vulnerable attack" and "negative region of space". The nucleus of the six atoms (2 C and 4 H) are situated in the same plane, which is usually defined as horizontal. The electrons of the 4 H atoms and the three first electrons from each Carbon atom are all together included in sigma bonds, and their ...


3

This collapsing electron orbit issue was a stumble stone for the last major classical Rutherford model of an atom. According to the classical electrodynamics, an electron accelerated by a central force should emit radiation and finally collapse on the nucleus. The Bohr model of an atom was the first major quantum atomic model, addressing this electron ...


3

In the isolated atom, yes, they will have the same energy. Dealing with d orbitals, you are in the field of transition metal chemistry. As correctly said by Buck Thorn, the degeneracy can be removed by external electromagnetic field. In the case of transition metals, it can be extensively seen in the crystal field theory. The ligands act as a source of ...


3

(Take any axis , the result would be same) Comparision of Energies : 2p electrons are higher in energy than 2s ones due to screening effects that result from electron-electron interactions. On average, the 2s electrons will be at a slightly greater distance from the nucleus than the 2p electrons. However, the 2s electrons have a higher probability of being ...


3

Whether you say if $\ce{F2}$ hybridises or not you reach the same answer - the molecule is linear. As indicated in the comments, hybridisation isn't an accurate reflection of how bonding actually works, however it is a good enough approximation to determine bond angles for simple molecules like ammonia. Note: to add some detail on your comment of the lone ...


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