How is it possible that 3d orbitals are more contracted than 4f?

In summary, the energy of an orbital depends on the quantum number n and the quantum number l. Contrary to the expectation that the 3d orbital should have a lower energy than the 4f orbital due to being more contracted, it is actually the other way around. This is because s orbitals have a higher probability of the electron being close to the nucleus, resulting in a lower energy. Additionally, relativistic effects play a role in the energy spectra of Hydrogen, known as the "fine structure." At higher n values, these effects become more significant in determining energy levels.
  • #1
Chemist20
86
0
just thinking... the energy of an orbital depends on:

a)quantum number n
b)quantum number l.

an orbital will be of less energy when it's more contracted. So technically, 3d should have a lower energy than 4f and hence be contracted. BUT IT'S THE OTHER WAY AROUND!

help?

thank you!
 
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  • #2
Chemist20 said:
an orbital will be of less energy when it's more contracted. So technically, 3d should have a lower energy than 4f and hence be contracted. BUT IT'S THE OTHER WAY AROUND!
I think you mean the 4s, not 4f, correct? Your reasoning works if you compare orbitals that are similar except for their n value, so if they have the same l value then the n will control how contracted they are and what their energy is. But when you also change the l value, you can no longer characterize the orbital by just its "size", there's also an issue about its "shape". It turns out that s orbitals have a probability of the electron being extremely close to the nucleus, so s orbitals have significantly lower energies then the other l states. As n rises, the energy differences between different n states becomes less than this dip in the s states.
 
  • #3
Ken G said:
I think you mean the 4s, not 4f, correct? Your reasoning works if you compare orbitals that are similar except for their n value, so if they have the same l value then the n will control how contracted they are and what their energy is. But when you also change the l value, you can no longer characterize the orbital by just its "size", there's also an issue about its "shape". It turns out that s orbitals have a probability of the electron being extremely close to the nucleus, so s orbitals have significantly lower energies then the other l states. As n rises, the energy differences between different n states becomes less than this dip in the s states.

Nope! I actually mean the 4f. the 4f are more contracted than 3d. And I don't see why! ;)
 
  • #4
If I remember correctly (and I'm vaguely remember something I read in an inorganic chemistry textbook several years ago), it has something to with relativistic effects.

I try to find the actual reference rather than just waving my hands.
 
  • #5
mjpam said:
If I remember correctly (and I'm vaguely remember something I read in an inorganic chemistry textbook several years ago), it has something to with relativistic effects.

I try to find the actual reference rather than just waving my hands.

You are right. The Hydrogen spectrum would be entirely dependent on n if relativistic effects are not taken into account.

Relativistic effects in the momentum-energy relation, the spin-orbit coupling, and the so called "Darwin term" account for the l (L, not I lol) dependence of the energy spectra of Hydrogen. The sum of these effects are usually called the "fine structure" of Hydrogen. They are called "fine structure" because these effects are alpha (~1/137) suppressed in comparison to non-relativistic effects (alpha being the fine structure constant). But, as Ken says, at the higher n's, the regular energy levels of Hydrogen get very close together, and these fine structure effects can become factors in which states have higher energy, etc.
 

1. Why are 3d orbitals more contracted than 4f orbitals?

The 4f orbitals have a higher principal quantum number (n) than the 3d orbitals, which means they are farther from the nucleus and experience less nuclear attraction. This allows the 4f orbitals to spread out more, resulting in a less contracted shape compared to the 3d orbitals.

2. How does nuclear charge affect the contraction of orbitals?

The higher the nuclear charge, the stronger the attraction between the nucleus and the electrons. This results in a more contracted shape for the orbitals closer to the nucleus, such as the 3d orbitals.

3. Can you explain the trend of contraction between 3d and 4f orbitals in the periodic table?

The contraction of 3d and 4f orbitals is due to the shielding effect of the inner electrons. As you move across a period in the periodic table, the number of inner electrons increases, causing more shielding of the outer electrons. This leads to a decrease in nuclear attraction and a more contracted shape for the 3d orbitals compared to the 4f orbitals.

4. Are there any exceptions to the trend of contraction between 3d and 4f orbitals?

Yes, there are a few exceptions to this trend. For example, the 4f orbitals in lanthanum (La) and actinium (Ac) are more contracted than the 3d orbitals due to their unique electronic configurations. Additionally, the 3d orbitals in copper (Cu) and chromium (Cr) are slightly less contracted than expected due to their half-filled and fully-filled subshells, respectively.

5. How does the shape of orbitals affect their reactivity?

The shape of orbitals can affect their reactivity by influencing the availability and orientation of electrons. More contracted orbitals, such as the 3d orbitals, have a higher electron density closer to the nucleus, making them less reactive compared to more spread out orbitals like the 4f orbitals. Additionally, the orientation of the orbitals can affect the likelihood of electron interactions and reactions with other atoms or molecules.

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