[xiankai]

ok, i was puzzling on why the E.C. (electronic configuration) of Co³⁺ was different as compared to that of Cr, since both are isoelectronic.

Co³⁺ = [Ar] 4s2 3d4

Cr = [Ar] 4s1 3d5

ok so i had some thoughts:

1) is it possible that the E.C. of Co³⁺ that was given was that for an ion in a compound? because having other atoms/ions may distort its electron cloud, although in what way im not sure

2) maybe the increased nuclear charge of Co³⁺ as compared to Cr will distort the electron clouds, such that the 3d orbital for Co³⁺ is less energetic due to a smaller atomic radius and hence a greater nuclear attraction. this way, the electrons are arranged as such since the nuclear attraction of electrons to the 4s orbital is greater than the increased stability provided by a half-filled a d-subshell

but all these is just mere speculation... can anyone help me? thanks! 

[tamim83]

OK, the EC of cobalt³⁺ is [Ar] 3d⁶, cobalt will lose the 4s electrons and then one of the 3d electrons.  This would make it have the same number of valence electrons as Cr except the electrons are in slightly different places. 

However, here is my theory.  Perhaps, in the process of being oxidized the cobalt atom lose electrons so that it has the same electron configuration of Cr, since it slightly more energetically stable.  However, I really don't know if that is really the case. 

[Mitch]

Co3+ is not isoelectronic with Cr.

[xiankai]

so Co³⁺ is [Ar] 3d6 and not [Ar] 4s2 3d4..

i thought isoelectronic reffered to having the same number of electrons? if u meant the exact electronic configuration then i understand, but why isnt there a [Ar] 4s1 3d5 E.C instead of [Ar] 3d6, as an e[sup-[/sup] in a half-filled d subshell is more stable than in a fully-filled subshell?

im thinking in terms of energy stability here... but it doesnt seem to be the case here of course; are there any other factors?

thanks for all the replies so far...

[Mitch]

so Co3+ is [Ar] 3d6 and not [Ar] 4s2 3d4..

Yes

if u meant the exact electronic configuration then i understand

That is the correct meaning

why isnt there a [Ar] 4s1 3d5 E.C instead of [Ar] 3d6, as an e[sup-[/sup] in a half-filled d subshell is more stable than in a fully-filled subshell?

The simple answer is that the gain in exchange energy doesn't match the cost of placing an electron into a higher energy 4s subshell.

[xiankai]

hmm... i thought the 3d subshell was more energetic than the 4s subshell?

[Mitch]

The energy from the hydrogen atom in En = -NaRCh[1/n^2]

Where Na is avogadro's number, R rydberg constant, c speed of light, h plank's constant and n is the principle quantum number. Those with low n will be "stable" and be in lower energy, while those with higher n will be higher in energy. For multielectron atoms of course this changes a bit as seen from the fact that the transition metals do not all add in into the 3d, but this isn't because the 4s is so close in energy this is due to both pairing problems and the fact that the 3d orbitals are contracted closer to the nucleus than other d orbitals and thus are smaller and leading to a 3d contraction. This contraction forces electrons to be closer together than they would normally like to be, so they choose to occupy the higher energy 4s orbital instead of the lower lying 3d. Hope that helps. Mitch

[xiankai]

oh... that makes much sense... since the increased nuclear charge draws the orbitals closer to each other... ok thanks alot!