Electronic config. of Cr and Fe2+

[20090900]

Is this the correct electronic config. of Cr? --> [Ar] 4s1 3d5

Assuming this is correct, why isn't this the electronic config. for Fe2+?

In DAT destroyer, it says the config; is blah blah 3p6 3d6 (so both electrons from the 4s orbital have been removed, why isn't just 1 removed like in that of Cr, doesn't it just want to achieve a 1/2 filled shell?)

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[jefff]

i believe this is the reasoning, if someone can confirm or deny this it would be appreciated...

Cr is in the ground state and when we "make elements" and fill orbitals we follow the (n+&#8467 rule

(n+&#8467 rule: (fill lower energy orbital's first)
~FILLING: electrons fill orbitals with lower energy (n+&#8467 before filling orbital's with higher energy (n+&#8467…
~if both the (n+&#8467 values are the same then the lower n value gets filled first
4s = (n+&#8467 = (4 +0) = 4
3d = (n+&#8467 = (3 + 2) = 5
~~prior to being filled the 4s orbital has lower energy than the 3d orbital

we know that once the exception transition metals (Cr,Mo and Cu,Ag) are formed and in their ground state, they prefer to have the max number of half-filled or fully-filled orbitals
so they "promote" an electron from the 4s orbital into a 3d orbital which creates an ideal situation (either half-filled or fully-filled orbitals)


sɐǝɹǝɥʍ

Fe2+ is an ion and has lost 2 electrons

while the (n+&#8467 rule enabled us to see which orbital had higher energy, in these cases its only prior to them being filled…

~LEAVING: in cation formation electrons leave orbital's of higher energy prior to leaving orbital's of lesser energy
~~once filled the 4s orbital is higher energy than the 3d orbital (4s orbital will be the outermost orbital)

 

[thejamespark]

i believe this is the reasoning, if someone can confirm or deny this it would be appreciated...

Cr is in the ground state and when we “make elements” and fill orbitals we follow the (n+&#8467 rule

(n+&#8467 rule: (fill lower energy orbital’s first)
~FILLING: electrons fill orbitals with lower energy (n+&#8467 before filling orbital’s with higher energy (n+&#8467…
~if both the (n+&#8467 values are the same then the lower n value gets filled first
4s = (n+&#8467 = (4 +0) = 4
3d = (n+&#8467 = (3 + 2) = 5
~~prior to being filled the 4s orbital has lower energy than the 3d orbital

we know that once the exception transition metals (Cr,Mo and Cu,Ag) are formed and in their ground state, they prefer to have the max number of half-filled or fully-filled orbitals
so they “promote” an electron from the 4s orbital into a 3d orbital which creates an ideal situation (either half-filled or fully-filled orbitals)


sɐǝɹǝɥʍ

Fe2+ is an ion and has lost 2 electrons

while the (n+&#8467 rule enabled us to see which orbital had higher energy, in these cases its only prior to them being filled…

~LEAVING: in cation formation electrons leave orbital’s of higher energy prior to leaving orbital’s of lesser energy
~~once filled the 4s orbital is higher energy than the 3d orbital (4s orbital will be the outermost orbital)

very confusing wording imo.

 

[jefff]

very confusing wording imo.

how about this...

once filled the 4s orbital is higher energy than the 3d orbital (4s orbital will be the outermost orbital)

therefore when making a cation you first take from the 4s orbital

 

[jefff]

you seemed as if you already knew the electron config for exception transition elements

but you need to understand thats only when making a "ground state" transition element (that is one of the exceptions...)


and you need to understand when a cation is formed from a ground state transition element, the 4s orbital (ONCE FILLED) is higher in energy than a 3d orbital, it will act as if its the OUTER most orbital/electron cloud

...again, im pretty confident on this but if someone could confirm or deny these statements it would be appreciated

 

[thejamespark]

so because it is a transition element (Cr), normally it would be [Ar]4s2 3d4 -> [Ar] 3d6
since the 4s orbital (once filled) is in higher energy level than the 3d orbital.

 

[jefff]

so because it is a transition element (Cr), normally it would be [Ar]4s2 3d4 -> [Ar] 3d6
since the 4s orbital (once filled) is in higher energy level than the 3d orbital.

nooo, it would "promote" 1 electron from the 4s orbital to the 3d

[Ar] 4s1 3d5
this satisfies Hunds rule



the electron configuration for the ION Fe2+ is what is "different"....when they form cations they loose electrons from the 4s (higher energy) orbital

 

[jefff]

if you're still confused, explain whats confusing you

 

[20090900]

Ok i think i get it. Does this apply for ALL cations? Do you know if this applies to anions?

 

[hchc]

Ok i think i get it. Does this apply for ALL cations? Do you know if this applies to anions?

I read through this and the better explanation is. That Cr is one of the exceptions in filling orbitals along with Mo, Cu, Ag, Au. They only fill their S orbital with 1 electron then completely fill their D orbitals. So that is why Cr is [Ar]4s1 3d5.

Because Fe is Not an exception it will fill it's orbitals normally, s first then d in this case since it's a transition element.
Fe = [Ar]4s2 3d6
now remove 2 electrons =
Fe 2+ = [Ar]3d6
since you remove electrons from the highest orbital first for all the transition atoms. And you remove the electron that you last put in for the other atoms.