Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 109 - General Chemistry - Spring 2015

Lecture Notes 26: 1 April

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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,

• Na⁺, note process: Na = 1s² 2s² 2p⁶ 3s¹ Na⁺ = 1s² 2s² 2p⁶ 3s⁰ + e^–
  • 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⁺, show process as above,
  • 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⁺ , show process as above,
  • 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,

• 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.
  • Show expected Orbital filling diagram , then transform to symmetrical correct Orbital filling diagram.
• 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.
  • Show expected Orbital filling diagram , then transform to symmetrical correct Orbital filling diagram.
• 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.).
  • Show expected Orbital filling diagram for Cu, then transform to diagram for Cu¹⁺.
• 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.
  • Show expected Orbital filling diagram for Zn, then transform to diagram for Zn²⁺.
• Fe³⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s⁰ 3d⁵ or [Ar] 4s⁰ 3d⁵ (Note can lose an electron from 3d⁶to give the symmetrical 3d⁵ arrangement.)
  • Show expected Orbital filling diagram for Fe, then transform to diagram for Fe³⁺, talk about Fe²⁺ vs Fe³⁺ and how diagram "predicts" both.

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:

Fe
	1s	2s	2p	3s	3p	4s	3d

Whereas zinc, with the d shell filled, is not paramagentic since it has no unpaired electrons:

Zn
	1s	2s	2p	3s	3p	4s	3d

Chemical Bonds

Chemical bonds are the strongest forces that exist between atoms. They are the forces that hold atoms together in molecules and atoms or ions together in solids. We will look at other weak bonds and forces later.

The two most important and common strong bond types in chemistry are ionic bonds and covalent bonds, a third bond type, found in metallic solids, will be discussed later.

Electronegativity

So how do we determine whether two atoms will form an ionic or a covalent bond? Use a new property - electronegativity (EN). Electronegativity is a periodic measure of how electrons are shared by atoms with the highest value for F and the lowest for Cs. There are a couple of ways of determining EN's:

• Look up values on table.
  • You should memorize values for Period 2, Li (1.0) to F (4.0) in steps of 0.5 and Hydrogen (2.1)
• Use the high, low, intermediate approximation:
  • all metals are low
  • the most electronegative of the non-metals are high (N, O, F, S, Cl, Br, I)
  • all other elements are intermediate.

Bond Type

So how do we use this to predict whether a bond is covalent or ionic?

• For numbers, use a difference of 1.7 to distinguish ionic ( EN> 1.7) and covalent ( EN< 1.7).
• Easier to use the hi, lo, intermediate system where we assign approximately hi EN, lo EN or intermediate EN values:
  • If combine hi + lo, then ionic
  • Otherwise, covalent (hi + hi, lo + lo, hi + inter., lo + inter., inter. + inter.)

    Examples:

    • Barium iodide - lo & hi, thus ionic: Ba²⁺ + 2 I ^- = BaI_{2 (s)}
    • Carbon disulfide - intermediate & hi, thus covalent: CS₂
    • Arsenic triiodide - intermediate & hi, thus covalent: AsI₃
    • Hydrogen selenide - intermediate & intermediate, thus covalent: H₂Se
    • Oxygen difluoride - hi & hi, thus covalent: OF₂

  • Note that in each case the positive ion or lower EN element is written first!

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Last modified 1 April 2015