spdf notation qn (1 Viewer)

fx82au

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i literally have no idea what a block is, thus i cant answer the q
 

011235

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The d-block compromises groups 3-12. The x-block has elements with the highest energy subshell of x. I'm assuming that by ending they're listing not in energy order but in numerical order, so 4s1 would be the only one in the fourth shell; hence, we only need to look at elements 21-30, the 3d block (as there is no 4p or 4d subshells).
But that doesn't seem to make much sense! The Aufbau principle says that the 3d subshell should be filled AFTER the 4s subshell, so having a 4s1 in the 3d block shouldn't be possible.
(I'm not certain this quesiton is in syllabus, actually), but anyways, there's actually one exception - the 3d and 4s orbitals have a very similar energy, meaning that having 4s1 is in two cases (Cr and Cu) more electronically stable than 4s2.
(i might be getting the details a little wrong here) In the case of Cr this is because the d-subshell has 5 orbitals, and symmetry = stability. 3d5 is much more stable than 3d4. Ditto for Cu and 3d9 vs 3d10 - this can happen because of the small difference in energy between 3d and 4s. Anything further than this is well outside of what you need to know.
Because of these two exceptions the answer should be B.

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fx82au

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The d-block compromises groups 3-12. The x-block has elements with the highest energy subshell of x. I'm assuming that by ending they're listing not in energy order but in numerical order, so 4s1 would be the only one in the fourth shell; hence, we only need to look at elements 21-30, the 3d block (as there is no 4p or 4d subshells).
But that doesn't seem to make much sense! The Aufbau principle says that the 3d subshell should be filled AFTER the 4s subshell, so having a 4s1 in the 3d block shouldn't be possible.
(I'm not certain this quesiton is in syllabus, actually), but anyways, there's actually one exception - the 3d and 4s orbitals have a very similar energy, meaning that having 4s1 is in two cases (Cr and Cu) more electronically stable than 4s2.
(i might be getting the details a little wrong here) In the case of Cr this is because the d-subshell has 5 orbitals, and symmetry = stability. 3d5 is much more stable than 3d4. Ditto for Cu and 3d9 vs 3d10 - this can happen because of the small difference in energy between 3d and 4s. Anything further than this is well outside of what you need to know.
Because of these two exceptions the answer should be B.

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great explanation, thanks! but doesnt this sound like a very niche question - you either know the exceptions or you dont.
 

someth1ng

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(i might be getting the details a little wrong here) In the case of Cr this is because the d-subshell has 5 orbitals, and symmetry = stability. 3d5 is much more stable than 3d4. Ditto for Cu and 3d9 vs 3d10 - this can happen because of the small difference in energy between 3d and 4s. Anything further than this is well outside of what you need to know.
There are many reasons that can be argued. For Cr specifically, a common one is about pairing electrons (i.e. electron repulsion), which is part of it. There are a total of 6 orbitals to fill with 6 electrons (1 s orbital, 5 d orbitals). It just happens that pairing the electrons in the s orbital takes more energy than just placing the electron in an empty d orbital.

A better explanation is that the 4s and 3d orbitals are affected differently by increasing nuclear charge (because they are distinct shapes and will experience shielding differently). Specifically, the energies of the d orbitals decreases faster than the s orbital as Z increases. It just happens that the energy levels cross over at Cr and Cu.

On this topic, you might hear people referring to completely or partially filled shells having extra stability, but there's nothing magical about it. If you look at the 4d and 5d periods, the electron configurations are a complete mess with some being half-filled, some fully filled etc. It's just a memory trick to remember it for the 3d elements (which are the most common elements). The reality is, electrons just go where the energy is lowest and it's not exactly predictable.

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