Chem 109 - General Chemistry - Spring 2011
Lecture Notes 25: 25 March
Electronic Configurations & Periodicity, cont.
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.)
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 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.
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.
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.)
- Barium iodide - lo & hi, thus ionic: Ba2+ + 2 I - = BaI2 (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!
© R A Paselk
Last modified 25 March 2011