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The electronic configuration of platinum is

$$\mathrm{[Xe] 4f^{14} 5d^9 6s^1}$$

and not $\mathrm{[Xe] 4f^{14} 5d^{10} 6s^0}$ or $\mathrm{[Xe] 4f^{14} 5d^8 6s^2}$.

I see this question has already been answered, but here is what Wikipedia says:

Other exceptions to Madelung's rule:
For example, in the transition metals, the 4s orbital is of a higher energy than the $\ce{3d}$ orbitals; and in the lanthanides, the $\ce{6s}$ is higher than the $\ce{4f}$ and $\ce{5d}$.

I don't understand the reason behind this.

Shouldn't it be that the $\ce{4s}$ orbital is of a lower energy than the $\ce{3d}$ orbitals (that's why it is filled first) and in the lanthanides, the $\ce{6s}$ is lower than the $\ce{4f}$ and $\ce{5d}$?

It also states that in heavier elements the inner electrons move with a speed approaching the speed of light. So why don't we consider the great amount of kinetic energy associated with the electron? If we consider this energy then the inner orbitals should not be filled first owing to their great amount of energy?

There is a list of exceptions arising from electronic configurations of lanthanides and transition series.

The exceptions in the transition series are explained by the fact that half and fully filled orbitals are more stable (although my text doesn't state why they are stable)?

So similarly do we have a rule that governs the electronic configuration of lanthanides? I believe for someone who is in his first year Hartree-Fock calculations are not a good thing to start with. So do I need to just memorize them all?

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A lot of questions in here.

It also states that in heavier elements the inner electrons move with a speed approaching the speed of light. So why don't we consider the great amount of kinetic energy associated with the electron? If we consider this energy then the inner orbitals should not be filled first owing to their great amount of energy?

In the Schrödinger Equation, $\hat H\Psi = E\Psi$, the Hamilton operator $\hat H$ is generally considered to consist of a kinetic term $\hat T$ and a potential term $\hat V$. In general, $\hat T$ is defined via an electron’s impulse $\vec p$ rather than its velocity $\vec v$, meaning that we end up with a term $$\hat T = -\frac{\hbar^2}{2m}\nabla^2$$ Within this operator, the kinetic energy of an electron is considered and since $\hat T$ forms part of $\hat H$, we use an electron’s kinetic energy in the original Schrödinger equation and thus it is considered in the calculations.

Using relativistic quantum mechanics will complicate the calculations but the general idea will remain the same.

The exceptions in the transition series are explained by the fact that half and fully filled orbitals are more stable (although my text doesn't state why they are stable)?

That’s … overly simplistic. In reality, it is a series of considerations we could make at the drawing board than when calculated leads to one electronic configuration being preferred over another. However, this question already covers that part.

Other exceptions to Madelung's rule:
For example, in the transition metals, the 4s orbital is of a higher energy than the 3d orbitals; and in the lanthanides, the 6s is higher than the 4f and 5d.

I don't understand the reason behind this.

Shouldn't it be that the 4s orbital is of a lower energy than the 3d orbitals (that's why it is filled first) and in the lanthanides, the 6s is lower than the 4f and 5d?

While your sentence is correct according to the aufbau principle or Madelung’s rule, this is no longer true when it gets to everything that’s not a neutral atom. Indeed, as soon as you turn scandium into a scandium(I) ion, the 4s orbital becomes less favourable than the 3d orbitals because the latter are still ‘closer to the nucleus’ and experience more stabilisation from an overall net positive charge. In general, once you go from neutral atoms to cations, the aufbau principle loses its relevance and a more ‘logical’ 3d-4s-4p order is favourable.

So similarly do we have a rule that governs the electronic configuration of lanthanides? I believe for someone who is in his first year Hartree-Fock calculations are not a good thing to start with. So do I need to just memorize them all?

Lanthanides tend to be rather subtle in their electronic effects if I recall my lecture correctly; as far as I am aware there is no explicit rule of any kind for them. But as for memorising: yes, you would have to do that.

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Electrons fill orbitals in the increasing energy order - 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p (starting from Hydrogen with 1 electron and ending at Oganesson with 108 electrons). While there are no exceptions for s-block and p-block elements, there are 2 expections for d-block (transition metals) - 1. No d9 rule (it goes from s2[f14]d8 to s1[f14]d10 to s2[f14]d10), 2. No s2d4 rule (from s2d3 to s1d5 to s2d5). As a result, d-block (transition metal) exceptions are 4(s1d10)+2(s1d5)=6 elements (Cu Ag Au Rg + Cr Mo). Note W/Sg (s2f14d4) are not exceptions.

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