harrypotter said:
I thought I had finally gotten a handle on this but a question on a practice test proved me wrong.
Take Ti[22], the solutions say the electrons from 3d^2 will participate in a bond. Those are the valence electrons.
Now I looked up valence electrons and it was defined as electrons in the outermost shell. .. Wouldn't the electrons in the outermost shell be 4s because the principle quantum number of 4 is bigger than 3?
I'm pretty sure this is simple but I can't get it. Please help
QofQuimica said:
Both the 3d and the 4s electrons are valence electrons for titanium. You can't just go by the principle quantum number when you talk about transition metals like this, because the 3d and 4s orbitals are very close together in energy. In fact, they are close enough that 4s actually fills first before 3d does. (4s is slightly lower in energy than 3d.) Confusing the issue is the fact that once they are both filled with electrons, the 4s orbital becomes higher in energy than the 3d orbital is. So, the 4s fills before the 3d does, and the 4s empties before the 3d does also.
heymanooh1 said:
I have to disagree with the explanation that the 4s orbital is energetically lower than the 3d orbital. Clearly the 4s orbital is higher in energy because if neutral Titanium is singly ionized to a +1 oxidation state, the first electron removed is from the 4s orbital. This has been proven experimentally.
The real reason why Titanium fills in the 4s orbital first rather than the 3d can attributed to the penetration effect. Experimentally, when a "radial distribution v. distance from nucleus" for electrons is plotted, the 4s electron spends significanly more time i.e. penetrates closer to the nucleus compared to the 3d electron. Because of this effect, the 4s orbital is "preferred" first when filling in the orbitals for a transition metal such as Titanium. Also, the radial distribution plot confirms that the 4s electron in Titanium spends more time further from Titanium's nucleus compared to the 3d electron. This correlates with the experimental fact that the first ionization energy of neutral Titanium is due to the removal of the higher energy 4s electron and not the lower energy 3d electron. The consequence of Titanium being singly ionized to +1 is that the remaining 4s electron will drop down to the 3d orbital to give the following electron configuration Ti +1: [Ar] 3d3. If 4s were energetically lower than 3d, then the electron configuration for Ti +1 would be: [Ar] 4s1 3d2. This hasn't been shown experimentally.
Sorry to pile on heymanooh here, but QofQuimica is 100% right on the mark. Further, she presents an excellent way to look at the subject at the
level of the MCAT. Hence I apolgize for my rant in advance, as it is beyond the level of the MCAT and serves only to show that knowing a bunch of big
specialty science words doesn't help you at all on the MCAT and can confuse things. So, for the best explanation, see QofQuimica's comment that Foghorn referenced.
To heymanooh: What
experiment are you talking about? It's called
orbital theory because there is no experiment that can
truly see the elecrtons (so says Heisenberg). There are definitely electron topography diagrams as you reference, but those are not in lieu of the energy levels; those support the energy levels. As a neutral element, 4
s fills before 3
d, and this is seen by through the continued radial symmetry associated with the electron cloud as electrons are added beyond a nobel gas. The 4
s filling before 3
d supports the notion that the 4
s is more stable, and thus of lower energy. That is why chemists accept that 4
s is lower than 3
d. As the universally accepted theory, it's what you need to know for the MCAT.
When electrons are lost, they follow a different pattern. Once an electron is lost, shell contraction and the
very high potential for interactions with ligands occur, so atomic theory breaks down a bit. Your explanation involving what happens when Ti loses a 4
s electron to become Ti+ in terms of atomic orbitals is off. The second it becomes an ion, both the 4
s and 3
d levels shift due to what is known as the
Jahn-Teller effect. It basically states that orbitals can redistribute in such a way that the energy levels become degenerate. At its simplest level, it's the idea behind hybridization. Further, the theories we have for d-level electron distribution, are based on central metal cations in a ligated complex. If you look at a Tenabe-Sugano diagram for d-splitting associated with various ligands, you will see that the d-levels undergo energetic shifts. You will also note that in molecular orbital theory, which applies the minute the metal becomes a complexed ion, there is no longer
d-orbitals and
s-orbitals. However, the non-bonding orbitals correspond energetically to
d-levels and the bonding and anti-bonding correspond to hybrids of the
s-levels,
p-levels, and
d-levels. This means that the valence shell involves the
s-levels
AND the
d-levels,
exactly as QofQuimica said.
The bottomline for the MCAT:
- (1) The s-orbital fills before the d-orbital because it is of lower energy in a neutral metal
(2) The s-electrons are lost before the d-electrons because of the shift in energy levels associated with shell contraction lowers the d-level in energy more than it lowers the s-level in energy.
