Ground state for Sc

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804991762_4A
Posts: 47
Joined: Fri Apr 06, 2018 11:04 am

Ground state for Sc

Postby 804991762_4A » Tue May 01, 2018 11:16 pm

Can someone clarify how to get the ground state?
For example in Lecture the ground state for scandium is, Sc: [Ar] 3d¹ 4s², but how is that 4s^2 follows after 3d^1, and why is [Ar] present in the ground state?

Danielle Sumilang - 1F
Posts: 31
Joined: Fri Apr 06, 2018 11:05 am

Re: Ground state for Sc

Postby Danielle Sumilang - 1F » Wed May 02, 2018 12:00 am

Hello!

To answer your first question, once there is an electron in the 3d orbital (ex. 3d^1), the 3d orbital has less energy than the 4s orbital. Therefore, 3d is written before 4s.

To answer the second question, [Ar] is present because that is the last noble gas before the element. The last noble gas is used for writing the shorthand version of the electron configuration because it includes the previous orbitals (ex. 1s2, 2s2, 2p6, etc).

Alma Flores 1D
Posts: 64
Joined: Wed Nov 08, 2017 3:01 am

Re: Ground state for Sc

Postby Alma Flores 1D » Wed May 02, 2018 12:13 am

For your first question, electron configuration should be ordered primarily by increasing order of shells, the principle quantum number n. The ground state of scandium is [Ar] 3d¹ 4s² because 3 comes before 4.
For your second question, [Ar] is present in the ground state because it is the last noble gas before scandium. This is used when writing the shorthand electron configuration.


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