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Need help with atomic oribitals.Help please

In the 3d series : Scandium

Configuration: 3d1 4s2

Why isn't it 3d2 4s1?
Since half filled orbitals ( in this case 4s1 ) are more stable?
Fully filled sub-shells(orbitals) are more stable than half filled, added to that in the first configuration all electrons are paired in 4s, so more stability.
Since half filled orbitals
( in this case 4s1 ) are
more stable?

You are having some misinterpretation here, they are more stable than an irregular/asymmetrical sub-shell

BTW , next time ask on a chemistry/science forum
Last edited on
Thanks so much.

I did, but it's been 3 days. No one gets replied to. I've posted here beofre and get instant help :)
np
What forum you posted earlier ?
Yes it takes extra energy to put two electrons together, and if 4s could, it would shove that second electron away somewhere (it does just that in Cr, which is 3d5 4s1 and Cu, which is 3d10 4s1, and there are a lot more examples with 5s), but those are exceptions. Under the normal Aufbau principle, 4s has to be filled completely before 3d.
I might be wrong here Cubbi, but what i learned was that when an orbital consist of two electrons it has less energy as two electrons with opposite spins counteract each other's influence , and the exceptional configuration of Cr ,Cu,Pd etc. is because half filled sub-shells are more stable , here talking about the (n-1)d orbital which is either half -filled or fully-filled in the exceptional case.
Spin is not charge, opposite spins do not attract.
There was a decent article on transition element electron configs, a couple years ago in J. Chem. Educ, i'll get it when I have a moment (for once a Lounge thread about something I like)
There we go, J. Chem. Educ., 2010, 87 (4), pp 444–448

To summarize, the energy of the 3d orbital is lower than 4s, but only starting with Sc: that's known as d-orbital collapse (usually explained by imperfect shielding by the core orbitals)
K  : 3p << 4s < 4p << 3d
Ca : 3p << 4s < 3d <  4p
Sc+: 3p << 3d < 4s <  4p

(this continues: the more the nuclear charge, the deeper is 3d under 4s, just not as dramatically). f orbitals similarly collapse after group 3

From this alone, transition elements should be 3dn4s0, and that's what they are in real chemistry (in compounds), but ground states of most unbound, neutral transition metal atoms in vacuum have the 4s occupied by one or two electrons because 3d is compact while 4s is diffuse (in a free atom): it becomes energetically favorable to shift one or even two electrons from the 3d shell into the slightly higher energy 4s, where the electronic repulsion is much smaller.

Awesome example from the article: V+[3d4] + e- -> V0[3d3][4s2]

that is known as d-s electron repulsion. Naturally, as you go to bigger shells, this grows weaker, so out of 9 elements in Y-Ag, 6 don't manage to push the electrons out of 4d to a full 5s2.

Other factors such as d-d electron repulsion, spin-orbit and spin-spin splitting come into play: Ni is d9s1 and the next lowest configuration, d8s2 is a whole 100kJ/mol higher, but its lowest term is only 2kJ/mol higher, and spin-orbit splitting makes them overlap by 2 kJ/mol. On heavier atoms, you get relativistic effects in, too.

in compounds, of course, the massively diffuse outer s gets destroyed by the nearby atoms, and all electrons pull back into d, where they belong (even on neutral atoms)
Thank you Cubbi, I guess what they teach at high school is half-truth and incomplete .
Depending on your point of view, that may be sufficient to provide pseudo-explanations to help memorize a few basic facts without overloading you with quantum mechanics.
Cubbi wrote:
Depending on your point of view, that may be sufficient to provide pseudo-explanations to help memorize a few basic facts without overloading you with quantum mechanics.

That's what I always hated about schools , they literally *lie* in a sense.
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