Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 107	Fundamentals of Chemistry	Fall 2009
Lecture Notes: Elec Config Answers	© R. Paselk 2005
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• Spectroscopic notation:
  • H: 1s¹
  • He: 1s²
  • B: 1s² 2s² 2p¹
  • P: 1s² 2s² 2p⁶ 3s² 3p³
  • V: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d³

• Spectroscopic notation using the Noble gas core convention.
  • P: [Ne] 3s² 3p³ (instead of 1s² 2s² 2p⁶ 3s² 3p³)
  • V: [Ar] 4s² 3d³ (instead of 1s² 2s² 2p⁶ 3s² 3p³ 4s² 3d³)
  • U: [Rn] 7s²5f⁴ The savings here is really obvious! Note that if you use the Periodic chart that comes on exams you will fill the f's before the d's of a given period (this is not the case for the wall chart).

• Orbital Filling Diagrams:
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  • Notice how the electrons first fill into empty orbitals before they pair up
  • Now when we add one more electron it goes back to pair up with the first p electron.

• Ions: When an atom loses electrons we would expect it to lose its outermost electrons first. But which are outermost? Remember the "last added" electrons in the transition elements at in the d orbitals of the next outermost shell. Thus the d orbital electrons should not be the outermost electrons in an atom. Thus we will lose the s & p electrons first then the d electrons if any are present. If additional electrons are lost then we can go into the d shell. Examples:
  • Na⁺ = 1s² 2s² 2p⁶ 3s⁰ or [Ne] 3s⁰ (In both cases the 3s⁰ is usually not shown, I've shown it here for clarity.)
  • Cu²⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s⁰ 3d⁹ or [Ar] 4s⁰ 3d⁹ (In both cases the 3s⁰ is usually not shown, I've shown it here for clarity.)
  • Fe³⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s⁰ 3d⁵ or [Ar] 4s⁰ 3d⁵ (In both cases the 3s⁰ is usually not shown, I've shown it here for clarity.)
  • Br^- = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ or [Ar] 4s² 3d¹⁰ 4p⁶(Note that [Kr] is NOT an acceptable indication of electronic configuration - the outermost electrons must be shown..)

• Symmetry considerations:
  • Cr = [Ar] 4s¹3d⁵ instead of [Ar] 4s²3d⁴. This occurs because the s orbitals are already spherically symmetrical, whereas the d orbital set only becomes fully spherically symmetrical when all of the d orbitals are filled in the same way, in this case having one unpaired electron each. or when all of the d orbitals have two electrons each.
  • Cu = [Ar] 4s¹3d¹⁰ instead of [Ar] 4s²3d⁹. This occurs because the s orbitals are already spherically symmetrical, whereas the d orbital set only becomes fully spherically symmetrical when all of the d orbitals are filled in the same way, in this case all of the d orbitals have two electrons each.
  • Cu¹⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s⁰ 3d¹⁰ or [Ar]4s⁰ 3d¹⁰ This occurs because we are removing the outermost electron, the single electron in the 4s orbital. Note that without the symmetry filling of the 4d orbitals there would be two 4s electrons and we would then expect only Cu²⁺ as we see in the alkaline earths (Mg, Ca, etc.).
  • Zn²⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s⁰ 3d¹⁰ or [Ar]4s⁰ 3d¹⁰. Note that the electronic structure for the Zn(II) ion is the same as the Cu(I) ion. In this case however, Zn behaves like the alkaline earths, that is it only exhibits a 2+ ion.

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© R A Paselk

Last modified 20 September 2005