Examkracker Chemistry Question

[ratatat]

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Hi,

Regarding transition metal ions, EK says two things

1. Transition metals lose electrons from their s orbital first, and then they lose electrons from their d orbital.

2. Transition metal ions try to "even-out" their d orbitals, so each orbital has the same number of electrons.

I don't understand because cobalt for instance, ends up with 3d6 in its d-orbitals. I don't see how that 'evens-out' their d orbitals so that each orbital has the same number of electrons.

 

[G1SG2]

Hi,

Regarding transition metal ions, EK says two things

1. Transition metals lose electrons from their s orbital first, and then they lose electrons from their d orbital.

2. Transition metal ions try to "even-out" their d orbitals, so each orbital has the same number of electrons.

I don't understand because cobalt for instance, ends up with 3d6 in its d-orbitals. I don't see how that 'evens-out' their d orbitals so that each orbital has the same number of electrons.

1. is true. I think 2 may be referring to the deviations from predicted electron configuration for some transition metals. For example, for Cu we would predict [Ar] 3d^9 4s^2, but it's really [Ar] 4s^1 3d^10. Whenever you can partially (5 electrons) or fully (10 electrons) fill the d orbital so that all the orbitals can have the same number of electrons, it will happen.

 

[sv3]

1. is true. I think 2 may be referring to the deviations from predicted electron configuration for some transition metals. For example, for Cu we would predict [Ar] 3d^9 4s^2, but it's really [Ar] 4s^1 3d^10. Whenever you can partially (5 electrons) or fully (10 electrons) fill the d orbital so that all the orbitals can have the same number of electrons, it will happen.

I posted about this a while ago. I think there's only 4-5 elements that apply to rule 2. Rule 1 is a for sure thing though. I know TPR lists all those elements but not sure if Ek does. If you can find the earlier post they are all in there.

 

[ratatat]

Hey thanks, I found your thread.

Chromium, Copper, Molybdenum, Silver & Gold.

 

[sv3]

Hey thanks, I found your thread.

Chromium, Copper, Molybdenum, Silver & Gold.

yup those are specifically listed by TPR. If there are anymore i've never come across them and I did a ton of problems back then. I would say try to remember those if you can (two start with C, i always think of silver and gold together, and then the lone moly).

cheers

 

[Hemichordate]

For transition metals, do the d electrons count as valence electrons as well?

 

[thebillsfan]

wait, what exactly about those five elements should we understand? that they even our their d orbitals? what does that even mean? that they'll lose exactly enough electrons to have 5 d electrons? if thats the case, then why only those atoms?

a related question...I think I read the Cr and Cu have relatively equal energies for their 4s and 3d orbitals. usually the 4s electrons come off first, but with this, is it true that if they are ionized such that they have six valence electrons, then one s electron and one d electron will come off so that there is one electron in all 6 (1 s plus 5 d) orbitals?

 

[ratatat]

For transition metals, do the d electrons count as valence electrons as well?

Hmmm..I don't think so but i'm not exactly sure. I hate transition metals 😛

wait, what exactly about those five elements should we understand? that they even our their d orbitals? what does that even mean? that they'll lose exactly enough electrons to have 5 d electrons? if thats the case, then why only those atoms?

a related question...I think I read the Cr and Cu have relatively equal energies for their 4s and 3d orbitals. usually the 4s electrons come off first, but with this, is it true that if they are ionized such that they have six valence electrons, then one s electron and one d electron will come off so that there is one electron in all 6 (1 s plus 5 d) orbitals?

For those five elements, you should know that they form ions that have either 5 or 10 electrons in the d orbitals. They try to have either half full or "fully full" d orbitals.

But I still don't understand why Cr would form a 3+ ion then. How would losing three electrons make it have a d-shell with 5 or 10 electrons? 😕

 

[SuperSaiyan3]

😕

why would the s orbital electron come off FIRST?

Wouldn't the 3d orbital be at a HIGHER energy than 4s?

 

[ratatat]

😕

why would the s orbital electron come off FIRST?

Wouldn't the 3d orbital be at a HIGHER energy than 4s?

Although you count up 3p, 4s, 3d....The 4s shell is still the highest energy level for transition metals, so those come off first. F-ed up, I know but that's what I've read. :|

 

[ratatat]

Hmmm..I don't think so but i'm not exactly sure. I hate transition metals 😛



For those five elements, you should know that they form ions that have either 5 or 10 electrons in the d orbitals. They try to have either half full or "fully full" d orbitals.

But I still don't understand why Cr would form a 3+ ion then. How would losing three electrons make it have a d-shell with 5 or 10 electrons? 😕

Bump, because I still don't understand how a Cr ion with 3+ has an even d-shell (5 or 10 electrons)

Thank you.

 

[Seraph 84]

It turns out that an empty s orbital has lower energy than an empty d orbital, so you fill the s orbital first. Also, a filled s orbital is higher than a filled d orbital, so when ionizing, you extract electrons from the s orbital before the d orbital.

Edit: To the post above me, a Cr 3+ ion will only have 3 d electrons in its valence shell. What you might be thinking of is the transition elements that are in the same column as Cr and Cu. Neutral Cr and Cu will have 5 d and 10 d electrons in their d subshells, respectively, because that is the most energetically stable state for those systems. If you draw out the orbital box diagrams for each, you can transfer one of the electrons from the 4s orbital to an unfilled 3d orbital, thus making all the 3d orbitals half filled or completely filled. This is more energetically stable than a completely filled 4s orbital and an only partially filled set of 3d orbitals. Just know that this exception only happens with transition elements that are in the same group as Cr and Cu.

Edit #2: Also, the electron configuration notation for cobalt is [Ar]4s^2, 3d^7. I'm not sure where you're getting only 6 d electrons, but it's wrong.

 

[ratatat]

It turns out that an empty s orbital has lower energy than an empty d orbital, so you fill the s orbital first. Also, a filled s orbital is higher than a filled d orbital, so when ionizing, you extract electrons from the s orbital before the d orbital.

Edit: To the post above me, a Cr 3+ ion will only have 3 d electrons in its valence shell. What you might be thinking of is the transition elements that are in the same column as Cr and Cu. Neutral Cr and Cu will have 5 d and 10 d electrons in their d subshells, respectively, because that is the most energetically stable state for those systems. If you draw out the orbital box diagrams for each, you can transfer one of the electrons from the 4s orbital to an unfilled 3d orbital, thus making all the 3d orbitals half filled or completely filled. This is more energetically stable than a completely filled 4s orbital and an only partially filled set of 3d orbitals. Just know that this exception only happens with transition elements that are in the same group as Cr and Cu.

Edit #2: Also, the electron configuration notation for cobalt is [Ar]4s^2, 3d^7. I'm not sure where you're getting only 6 d electrons, but it's wrong.

In regards to the Cr3+, I agree, it should have only 3 electrons in the d-orbital, but Examkrackers says otherwise:

"Ions are looking for symmetry....Transition metals try to 'even-out' their d orbitals, so each orbital has the same number of electrons"-EK

I don't see how that works.

Thanks for the reply though.

 

[Seraph 84]

In regards to the Cr3+, I agree, it should have only 3 electrons in the d-orbital, but Examkrackers says otherwise:

"Ions are looking for symmetry....Transition metals try to 'even-out' their d orbitals, so each orbital has the same number of electrons"-EK

I don't see how that works.

Thanks for the reply though.

EK is somewhat right, but that rule doesn't follow for every transition element - only those in the Cr/Cu groups. EK is known for errors and you have to pay to see their errata.

 

[ratatat]

EK is somewhat right, but that rule doesn't follow for every transition element - only those in the Cr/Cu groups. EK is known for errors and you have to pay to see their errata.

Yeah I see that Cr evens up as an element, but I don't see how it evens up as an ion.

 

[Hemichordate]

So do the d electrons count as valence electrons for transition metals?

 

[ratatat]

They do, but transitions lose s electrons first, then d.

 

[BigDolla]

Cobalt's atomic number is 27 so it's electron configuration would be [Ar]4s^2, 3d^7.

According to the Aufbau Principle orbitals are going to be filled 1s, 2s 2p, 3s, 3p, 4s, 3d, 4p... Based on this principle some transition metals will shift electrons around in order to get half filled orbits because it is more stable.

EX. Cromium Z=24
Cr configuration should be [Ar]4s^2, 3d^4. It takes 10 electrons to fill a d orbital and 5 electrons to make a half filled orbital. So Cr will promote an electron from the 4s^2 orbital and put it in the 3d orbital to make:
[Ar]4s^1, 3d^5

 

[BerkReviewTeach]

In regards to the Cr3+, I agree, it should have only 3 electrons in the d-orbital, but Examkrackers says otherwise:

"Ions are looking for symmetry....Transition metals try to 'even-out' their d orbitals, so each orbital has the same number of electrons"-EK

I don't see how that works.

Thanks for the reply though.

This is a case of oversimplification, problem in the interest of minimizing the text. That rule (if it really is one as opposed to being an excpetion another rule) explains half-filled and filled d-shell configurations for Cu, Mo, W, Cu, and Ag. In those cases, neutral elements have electronic configurations of s^1d^5 or s^1d^10, as was well explained by Pookiez88 and BigDolla.

In the case of Cr3+, the electronic configuration is d^3 (no s electrons remain). The thing you must keep in mind is that Cr3+ cannot be found as an isolated ion. It is solvated (complexed) by ligands. When the ligands attach, they do so in a octahedral fashion, with the 6 ligands aligning on the x-, y-, and z-axes. This stabilizes the dxy, dxz, and dyz orbitals, making them of lower energy than the other two d-orbitals. This is referred to as d-shell splitting. Because those three orbitals are of lower energy than the other two d-orbitals (dz2 and dx2-y2), there is one electron in each of the three lower-level d-orbitals. If the ligands bind the cation tightly, the energy gap between the lower d-orbitals and higher d-orbitals becomes significant, and the lower level will completely fill before electrons go into the higher d-orbitals. This is know as low spin filling.

This is borderline beyond the scope of the MCAT, although it is found in general chemistry textbooks. Look up crystal field theory for a more detailed explanation. The wickepedia link is pretty good.

 

[ratatat]

This is a case of oversimplification, problem in the interest of minimizing the text. That rule (if it really is one as opposed to being an excpetion another rule) explains half-filled and filled d-shell configurations for Cu, Mo, W, Cu, and Ag. In those cases, neutral elements have electronic configurations of s^1d^5 or s^1d^10, as was well explained by Pookiez88 and BigDolla.

In the case of Cr3+, the electronic configuration is d^3 (no s electrons remain). The thing you must keep in mind is that Cr3+ cannot be found as an isolated ion. It is solvated (complexed) by ligands. When the ligands attach, they do so in a octahedral fashion, with the 6 ligands aligning on the x-, y-, and z-axes. This stabilizes the dxy, dxz, and dyz orbitals, making them of lower energy than the other two d-orbitals. This is referred to as d-shell splitting. Because those three orbitals are of lower energy than the other two d-orbitals (dz2 and dx2-y2), there is one electron in each of the three lower-level d-orbitals. If the ligands bind the cation tightly, the energy gap between the lower d-orbitals and higher d-orbitals becomes significant, and the lower level will completely fill before electrons go into the higher d-orbitals. This is know as low spin filling.

This is borderline beyond the scope of the MCAT, although it is found in general chemistry textbooks. Look up crystal field theory for a more detailed explanation. The wickepedia link is pretty good.

Wow, great reply. Thank you very much.

 

[ratatat]

Cobalt's atomic number is 27 so it's electron configuration would be [Ar]4s^2, 3d^7.

According to the Aufbau Principle orbitals are going to be filled 1s, 2s 2p, 3s, 3p, 4s, 3d, 4p... Based on this principle some transition metals will shift electrons around in order to get half filled orbits because it is more stable.

EX. Cromium Z=24
Cr configuration should be [Ar]4s^2, 3d^4. It takes 10 electrons to fill a d orbital and 5 electrons to make a half filled orbital. So Cr will promote an electron from the 4s^2 orbital and put it in the 3d orbital to make:
[Ar]4s^1, 3d^5

Thanks a lot man.