Chem 109 - General Chemistry - Spring 2013

Lecture Notes 25: 29 March

Sodium Demonstration

Electronic Configurations & Periodicity, cont.

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 are in the d orbitals of the next outermost shell. The 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: look at Periodic Chart and figure out configurations for Na ion, Ni 2+ ion and Fe 2+ ion,

**Periodic Table of the Elements**
	IA	IIA											IIIA	IVA	VA	VIA	VIIA	VIIIA
H																		He
2	Li	Be											B	C	N	O	F	Ne
3	Na	Mg	IIIB	IVB	VB	VI	VIIB	VIIIB			IB	IIB	Al	Si	P	S	Cl	Ar
4	K	Ca	Sc	Ti	V	Cr	Mn	Fe	Co	Ni	Cu	Zn	Ga	Ge	As	Se	Br	Kr
5	Rb	Sr	Y	Zr	Nb	Mo	Tc	Ru	Rh	Pd	Ag	Cd	In	Sn	Sb	Te	I	Xe
6	Cs	Ba	Lu	Hf	Ta	W	Re	Os	Ir	Pt	Au	Hg	Tl	Pb	Bi	Po	At	Rn
	s¹	s²	d¹	d²	d³	d⁴	d⁵	d⁶	d⁷	d⁸	d⁹	d¹⁰	p¹	p²	p³	p⁴	p⁵	p⁶

Na⁺ = 1s² 2s² 2p⁶ 3s⁰ or [Ne] 3s⁰ (In both cases the 3s⁰ is usually not shown, I've shown it here for clarity.)
Ni²⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s⁰ 3d⁸ or [Ar] 4s⁰ 3d⁸ (In both cases the 4s⁰ is usually not shown, I've shown it here for clarity.)
Fe²⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s⁰ 3d⁵ or [Ar] 4s⁰ 3d⁶ (In both cases the 4s⁰ is usually not shown, I've shown it here for clarity.)

Symmetry considerations

It turns out that symmetry is a strong driving force in nature and symmetry considerations are a powerful tool for predicting how nature operates. This is important in predicting electronic configurations because when two electronic energy levels are close to each other, as in the 3d orbitals (highest energy in the 3 shell) and the 4s orbitals (lowest energy in the 4 shell), symmetry considerations can result in an electron preferring to "fill" the 3d orbital set, making it symmetrical, instead of going to the already symmetrical 4s orbital. This can be done in two ways: we can put one electron in each of the five d orbitals giving a spherical half-filled d orbital set, or we can put 2 electrons in each orbital. Examples: look at Periodic Chart and figure out configurations for Cr, Cu, Cu +1 ion, Zn +2 ion and Fe +3 ion,

**Periodic Table of the Elements**
	IA	IIA											IIIA	IVA	VA	VIA	VIIA	VIIIA
H																		He
2	Li	Be											B	C	N	O	F	Ne
3	Na	Mg	IIIB	IVB	VB	VI	VIIB	VIIIB			IB	IIB	Al	Si	P	S	Cl	Ar
4	K	Ca	Sc	Ti	V	Cr	Mn	Fe	Co	Ni	Cu	Zn	Ga	Ge	As	Se	Br	Kr
5	Rb	Sr	Y	Zr	Nb	Mo	Tc	Ru	Rh	Pd	Ag	Cd	In	Sn	Sb	Te	I	Xe
6	Cs	Ba	Lu	Hf	Ta	W	Re	Os	Ir	Pt	Au	Hg	Tl	Pb	Bi	Po	At	Rn
	s¹	s²	d¹	d²	d³	d⁴	d⁵	d⁶	d⁷	d⁸	d⁹	d¹⁰	p¹	p²	p³	p⁴	p⁵	p⁶

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.
Fe³⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s⁰ 3d⁵ or [Ar] 4s⁰ 3d⁵ (Note can lose 3d⁵ to give the symmetrical 3d⁵ arrangement.)

Paramagnetism

Atoms and compounds are paramagnetic when they have unpaired electrons. Recall that electron spin can be thought of as electrons behaving as tiny magnets, and the arrows we use in orbital filling diagrams corresponding to the direction of the magnetic poles.

As a result, atoms such as oxygen and iron will be paramagnetic and be attracted to a magnetic field. For example, oxygen has two unpaired electrons as seen in the orbital filling diagram: , so it is paramagnetic. Similarly, iron is paramagnetic due to its unpaired d electrons: