# Tag Info

0

Directionality in a hybrid $p$ orbital comes exclusively from the directionality of the composite $p$ orbitals in the mix. From this, and the fact that the standard $p$ orbitals are mutually orthogonal, you can see that your claim is indeed true. In an $sp^{2}$ hybrid, we need to look at what $p$ orbitals are in the mix. Suppose, without loss of generality, ...

2

TL;DR Excitation of an electron typically conserves spin; that is to say, the spin must be left unchanged by the process. If the initial state has no net spin (one spin-up and one spin-down electron), then the final state should also have no net spin (one spin-up and one spin-down electron). This rule is reliable for small-ish atoms, but often breaks down ...

2

The picture shows a triplet excited state returning to the ground state by emitting a photon, i.e. phosphorescence. This can only happen if there is also an interaction that couples angular momentum change with the transition, such as spin orbit coupling. The electron has two quantum numbers; the spin $S=1/2$ but is not the spin of the electron that is ...

1

There is a general heuristic in quantum physics, often referred to jokingly as the totalitarian principle, that everything not forbidden is compulsory. That is, any process that can occur will occur, with some rate, probability, or cross-section, provided that it doesn't violate any conservation laws. An atom in an excited state can in most cases emit a ...

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 ...

14

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 ...

1

It sounds to me like your confusion is why these two definitions are describing the same thing. I will attempt to clarify that. My interpretations/slight rephrasings of these definitions are A probability distribution of an electron around a nucleus. A space occupied by the electron, where the probability is a maximum. (This is a bit different that what ...

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 ...

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 ...

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 ...

3

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 ...

Top 50 recent answers are included