electronic config of Mn+ vs Fe2+

[rotten10]

Hey guys, I was hoping you all could help explain something to me...

A book I have says that Mn+ has electronic configuration [Ar] 4s1 3d5, which makes sense, because of the half-shell's extra stability. But it also says that Fe2+ has an electronic configuration of [Ar] 3d6 with an empty 4s, which doesn't make sense to me: they both have the same number of valence electrons, so why are the electrons found in different subshells? And wouldn't this configuration for the Fe2+ be higher in energy than one like Mn+'s?

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[Phantastic]

Fe2+ wouldn't necessarily be at higher energy Mn+, because they both have the same valence shell. (yes, the subshells differ)

The stability rules (Aufbau) are relevant to filling an electrons in, but I believe removal can take place in any(?) of the subshells of the valence shell.

 

[rotten10]

ok so that gets the right answer, filling in and taking out by different principles... I read somewhere else online that when your taking e-'s out, you take it from the subshell with the highest n (4s here).

so as far as the mcat goes, I think I'm set - thanks phantastic.

but conceptually, I dont understand why you would take out the e- from 4s before 3d, since (based on n+l) 4s is lower in energy than 3d? wouldn't you want to take out the highest E e- first? is it just that, physically, it spends more time farther from the nucleus (as n=4)? if nobody knows that cool, i'm just wondering

 

[Phantastic]

...but conceptually, I dont understand why you would take out the e- from 4s before 3d, since (based on n+l) 4s is lower in energy than 3d? wouldn't you want to take out the highest E e- first? is it just that, physically, it spends more time farther from the nucleus (as n=4)? if nobody knows that cool, i'm just wondering

You're welcome.

If you want the "hardcore mode" rule: All transition metal ions have a valence (n-1)d orbital that is lower energy than the valence ns orbital.

So for a transition metal ion like the one presented above, the 3d orbital is actually lower energy than the 4s. Remember, this applies to ions of the transition metals.

 

[loveoforganic]

Why does 4s fill prior to 3d if 3d is lower energy? Is it because you can get the spin paired electrons more quickly?

 

[Phantastic]

Why does 4s fill prior to 3d if 3d is lower energy? Is it because you can get the spin paired electrons more quickly?

As far as energy levels:

4s < 3d if and only if we are in the ground state in the gas phase of a transition metals. They're weird like that. When you create a transition metal ion by removing an electron, you necessarily will alter the effective nuclear charge (Zeff) in one way or another. This consideration, as well as electron-electron repulsion, "ligand fields" (for compounds--like coordination complexes) and a few other concepts way beyond the scope of the MCAT account for this observation that the s orbitals are empty. That is usually the case in the real world.

 

[loveoforganic]

Gotcha, (mostly), thanks 🙂

 

[Phantastic]

Sorry, those are concepts from inorganic chem special topic course on transition metals...so I have a hard time understanding it entirely myself...😀

 

[rotten10]

sweet answer, I didn't totally follow but I think I get the main idea there... anyway makes sense enough for me to be able to sleep easy now 🙂 thanks again