Chem 109 - General Chemistry - Spring 2011

Lecture Notes 25: 25 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:

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.)
Fe³⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s⁰ 3d⁵ or [Ar] 4s⁰ 3d⁵ (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:

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.

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!