Chemistry Q, from Destroyer

[Svart Aske]

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Should be an easy one for the GC-whiz.

#84.
Which species has the most unpaired electrons?
a) Na
b) I
c) Fe
d) Cu
e) Cr

We can safely rule Na and I out. These are the electron configurations according to Destroyer:
Fe: 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d6
Cu: 1s2, 2s2, 2p6, 3s2, 3p6, 4s1, 3d10
Cr: 1s2, 2s2, 2p6, 3s2, 3p6, 4s1, 3d5

This shows that Cr is the correct choice, with 6 unpaired electrons. My question is why do we fill up the 3d subshells of Cu and Cr before the 4s subshell, but we don't do the same with Fe?

 

[jbravo12]

My guess would be that the ground states of both Cr and Cu are more volatile than Fe...

Cr has 4/5 orbitals filled (1 electron each), so adding one more electron to that empty orbital will add substantial stability.

Cu is only 1 electron away from completeing it's entire d subshell, so that's pretty urgent.

Fe, however, has all orbitals somewhat occupied and is too far from completing it's subshell, so it won't have enough drive to go to the excited state.

This is how I look at it. Hope it helps.

 

[Svart Aske]

That's what I'm thinking too, especially with regard to Cu, but even with filling the 4s subshell first you still get 1 unpaired electron in one of the 3d orbitals. As for Cr, filling that one electron in 3d subshell creates 6 unpaired electrons, which I'd imagine to be less stable. Is there some type of rule of thumb that can help us with these types of problems?

 

[zuma35]

The rule of thumb is that if you have an electron configuration that ends in nd9 or nd4, the d gets the electon and is pushed up to 10 and 5. This has been shown from experiments.