D Block electron arrangements

[druS]

I'm between study periods at the moment and have gone back to something that I noted as interesting during the last study period. Here is a quote: "...the transition metals have the general configuration [Ar]4s²3dⁿ except chromium (4s¹3d⁵) and copper (4s¹3d¹⁰). The reasons for these exceptions are complex"

[Zumdahl et al - Introductory Chemistry: A Foundation]  ps I had to edit the super scripts and trust I have it correct.

Digging through the various web based tutorials I can work out writing the electron configurations (with cheat notes) but so far nothing to explain why, just "it gets complex".  The half shell and complete shell being stable isn't really it either - why is that? OR at least what is the background? Why are the electron configurations for Cu and Cr the way they are?

It would be great if I could grasp, or at least be a little clearer, why and when Fe forms the cation iron(II) or iron(III). Let alone what is going on in something like Fe₃O_4. [I believe a mix of iron(II) and Iron(III) is happening but have no logic behind it.]

How does this relate to the concept of noble metals? Why are Cu, Ag and Au noble, but Zn, Cd and Hg not always considered noble?

 

 

 

 

[hypervalent_iodine]

19 minutes ago, druS said:

Digging through the various web based tutorials I can work out writing the electron configurations (with cheat notes) but so far nothing to explain why, just "it gets complex".  The half shell and complete shell being stable isn't really it either - why is that? OR at least what is the background? Why are the electron configurations for Cu and Cr the way they are?

 

Do you want to know why electrons are placed and lost first the 4s orbital, or specifically why Cu and Cr only have 1 electron in the 4s orbital or their ground state configuration? I am not an expert in this area of chemistry, but the intuitive / simple explanation in the case of Cu and Cr comes from considering stability of having a 3d⁴ / 3d⁹ configuration compared to a 3d⁵ / 3d¹⁰. The latter represents the more stable option, which is why you see 4s¹3d^5/10 for Cr and Cu, and not 4s²3d^4/9. I am curious to know why you say that stability is not the answer? It generally is with these things, but the exact cause of that stability can get fairly involved.

 

42 minutes ago, druS said:

It would be great if I could grasp, or at least be a little clearer, why and when Fe forms the cation iron(II) or iron(III). Let alone what is going on in something like Fe₃O_4. [I believe a mix of iron(II) and Iron(III) is happening but have no logic behind it.]

 

For iron oxidation states, I believe that you can use the same sort of logic. The two most commonly encountered states are iron(II) and iron(III). You need only look at the orbital diagrams or electronic configurations and compare to the above to guess as to why that might be. The compound you mention is a mix of oxidation states. It's not overly complicated, it just means that two of the iron atoms have given up 3 electrons, and the other has only given up 2 to form a stable structure. 

 

46 minutes ago, druS said:

How does this relate to the concept of noble metals? Why are Cu, Ag and Au noble, but Zn, Cd and Hg not always considered noble?

 

There are several definitions for noble metals, some are fairly arbitrary. Could you be more specific? 

[druS]

Hypervalent_iodine - thanks - i did expect responses that required more thinking on my part!

1. Can we start with why the d-block electron arrangement is 3d in the fourth period.

2. Thence to the specifics with Cr and Cu. In this case the stability issue seems like an observation rather than information behind why it is this way. Full valence electrons create a stable arrangement before we consider the d block. SOmething changes here where all of a sudden there is stability mid-period.

3. The exact cause can get fairly involved. I guess that I want to nudge that issue. Maybe I don't know enough to follow, but at the moment I haven't been able to find anything. Note that I have posited this thread under chemistry - I'm hoping to not end up too deep in quantum.

3. With Fe I need to look at the orbital diagrams or electronic configurations? Does that not take us back to my initial query? Perhaps though this is something that is not what I have learned yet? I'm comfortable with electron configuration up to Argon, and the step through d block doesn't look that hard. But maybe I'm missing something?

 

Cheers

 

Dru

 

[hypervalent_iodine]

14 hours ago, druS said:

1. Can we start with why the d-block electron arrangement is 3d in the fourth period.

Are you asking why is it 3 and not 4? This is to do with quantum numbers assigned to electrons, and what those numbers represent. Are you familiar with principal (n), azimuthal / angular (ℓ), and magnetic quantum numbers (m_ℓ) at all? In this example, the 3 represents the principal quantum number. At this energy level, there can be 3 values for ℓ, with each value essentially representing a type of orbital. For example, s orbitals are where ℓ = 0, p orbitals are described by ℓ = 1, d by ℓ = 2, and f by ℓ = 3.

Since the value for ℓ under a given value for n can equal anything from 0 - n-1 (eg. if n = 2, ℓ can equal 0 or 1), there must be 3 possible types of orbitals that can be described by n = 3. Those are the s (ℓ = 0), the p (ℓ = 1) and d (ℓ=2). This is not possible when n = 2, because the highest possible value for ℓ is 1, which represents the p orbitals. Thus, the first time the d orbitals can appear is at n = 3, which is why you first row of transition metals have electronic configurations of 3d rather than 4, despite where they show up on the table. 

I hope that wasn't too confusing.

 

14 hours ago, druS said:

Thence to the specifics with Cr and Cu. In this case the stability issue seems like an observation rather than information behind why it is this way. Full valence electrons create a stable arrangement before we consider the d block. SOmething changes here where all of a sudden there is stability mid-period.

 

I'll try and explain this a bit better. Firstly, note that the 4s and 3d energies are quite close together in Cr and Cu. I am unsure of why this is exactly. Probably something to do with nuclear charge. Consider chromium first. Pairing electrons costs energy because of Coulombic repulsion. This is why when we fill orbitals, we don’t start pairing until all the degenerate orbitals have an unpaired electron in them first. Because we fill electrons in order of energy, because the 4s and 3d are so close together, and because it costs more energy to pair an electron in the 4s than it is to place another unpaired electron into the 3d orbital, chromium exists as 3d⁵ 4s¹. 

Another explanation comes from effective nuclear charge and nuclear screening caused by the existence of s electrons. The nature of s orbitals means that when you place electrons in them, you place them close to the nucleus. This effectively screens or creates a barrier between the nucleus and the electrons in the d orbitals, and reduces the effective charge felt by them. If you were to only place one electron into the s orbital, there would be less screening, and thus the 5 or 10 electrons left over in the Cr / Cu d orbitals feel more charge from the nucleus and are more tightly held by it compared to if there were 2 electrons in the 4s. Of course, this phenomenon would also be true for all the other d block elements so I guess it begs the question of why do they have 2 s electrons and not 1. The 4s orbital is still lower in energy than the 3d, so the reason is likely because in other elements that energy gap is too high to be worth it. I believe that as you fill up the d orbitals, the energy difference between placing all the electrons in the d orbitals and one in the s instead of 2 does decrease. In any case, this effect isn’t as strong or prominent in all atoms which is why chromium and copper are exceptions to the rule and not the rule. 

 

14 hours ago, druS said:

I'm hoping to not end up too deep in quantum.

Unfortunately, that's probably the only area fit to fully give you a full answer as to 'why.' 

14 hours ago, druS said:

With Fe I need to look at the orbital diagrams or electronic configurations? Does that not take us back to my initial query? Perhaps though this is something that is not what I have learned yet? I'm comfortable with electron configuration up to Argon, and the step through d block doesn't look that hard. But maybe I'm missing something?

Well, yes. Consider sodium as a simple example of what I am getting at. It has a configuration of [Ne] 3s¹. When it ionises, it becomes Na⁺ and adopts a [Ne] configuration. We don't see Na²⁺ because [Ne] is a very stable configuration, certainly more so than [He]2s²2p⁵. It is similar for Fe(II) and Fe(III), even if it isn't as simplistic, in the sense that the things you get out the most can be surmised to be the things that are the more stable in some sense. I realise that still doesn't answer you, but it's as good as I've got. What is very important to note is that firstly, when talking of Fe(II) and Fe(III), we are no longer speaking about neutral elemental species with degenerate orbitals, and more importantly, d electron counting is only a formalism. Not all complexes are able to be described very well by using this system, which is why things like crystal field and ligand field theory exist. I think it is probably too much to get into, but if you are interested I did find a related question with some great answers here: https://www.researchgate.net/post/Why_CuII_is_more_stable_than_CuI

 

[druS]

Loving it - plenty for me to cogitate here. Give me a moment or two.

Thanks so far hyper