Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 109 - General Chemistry - Spring 2013

Lecture Notes 29: 15 April

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Molecular Geometry, cont.

VSEPR (Valence Shell Electron Pair Repulsion) Theory, cont.

Exceptions to the "Octet Rule." There are two categories of exceptions to the octet rule among representative elements:

1. Small atoms in Periods I & II which cannot accomodate or normally don't accomodate a full set of p-electrons. Thus hydrogen and helium have no ground state p-orbitals and so can only carry a maximum of 2 electrons (a duet). The first three atoms of Period II form a transition between hydrogen (duet) and carbon (octet). Thus lithium, because of its very small size, often forms covalent rather than ionic compounds, with a single bond (e.g. LiH). Similarly beryllium can form compounds with a quartet of electrons around it (e.g. BeH₂) and boron can accomodate a sextet (e.g. BH₃).
2. Atoms in the p-block of Periods 3 and higher can have "expanded valence shells" with 10 or 12 electrons in the outermost shell by using some of their "empty" d-orbitals to hold the extra electrons.

To help determine if the octet rule is followed recall Clark's Method (abbreviated) for determining bonding in covalent Lewis Structures: Add up all of the valence electrons in the structure (remember to add one electron for each negative charge, or subtract one for each positive charge) If e- = 6y + 2 where y = # atoms other than H, then octet rule is followed with single bonds only. If e- < 6y + 2 then probably have multiple bonding with the number of multiple bonds = /2 (remember a triple bond is 2 multiple bonds!). However, note the exceptions below with small atoms (H, Li, Be, and B). If e- > 6y + 2 then have an expanded valence shell. Note that if = 2, then pentavalent (10 electrons in the valence shell) , and if = 4, then hexavalent (12 electrons in the valence shell). If you can draw more than one structure, then chose the most symmetrical. If two or more structures are equally symmetrical, then you probably have resonance and should show all structures connected by double arrows.

Expanded valence shells

Representative atoms with empty d-shells can also have what are sometimes referred to as expanded valence shells. In these cases the d-orbitals also participate in bonding enabling more bonds to be formed. Two additional electronic geometries are possible:

• Trigonal bipyramidal with angles of 90° & 120° results when the valence shell is expanded to accomodate 10 electrons in five pairs (steric number = 5).
• Octahedral with angles of 90° results when the valence shell is expanded to accomodate 12 electrons in six pairs (steric number = 6).

These two electron pair geometries can lead to six new molecular geometries in addition to another way to make a linear molecule:

1. Trigonal bipyramidal with angles of 90° & 120° (PCl₅)
   
   
2. Seesaw with angles of 90° & 120° (SF₄)
   
   
3. T-shaped with angles of 90° (ClF₃)
   
   
4. Linear with angles of 180° (I₃^-)
   
   
5. Octahedral with angles of 90° (AsF₆^-)
   
   
6. Tetragonal pyramidal with angles of 90° (ICl₅)
   
   
7. Square planar with angles of 90° (XeF₄)
   
   

Polarity in Covalent Molecules

Polarity: So now we can predict bonding and shape in representative group molecules (and thus most biomolecules), how about electron density and thus charge distribution? Need two bits of information:

• Shape (based on VSEPR Theory)
• Electron distribution within a bond (based on electronegativity)

Examples:

Molecule	Geometry	Structure	Electronegativities	Bond Dipoles	Molecular Dipole	Model
Carbon monoxide	linear	C=O	ENC= 2.5, ENO= 3.5
Carbon dioxide	linear	O=C=O	ENC= 2.5, ENO= 3.5		None: two dipoles are of equal magnitude, but opposite in direction and cancel.
Water	bent		ENH= 2.1, ENO= 3.5
Ammonia	trigonal pyramidal		ENH= 2.1, ENN= 3.0
Ammonium ion	tetrahedral		ENH= 2.1, ENN= 3.0		None: four dipoles are symmetrically arranged to cancel each other out and give a spherically charged but non-polar ion.

Find whether Chlorine trifluoride (ClF₃) is polar.

1. Determine Lewis structure:

• Valence electrons = 7 + 3(7) = 28
• 6y + 2 = 24 + 2 = 26
• = 28 - 26 = 2 so two extra electrons 10 electrons around central atom:
  

2. Determine steric number (SN) = 3(bonded atoms) + 2(lone pairs) = 5

3. Determine Geometry:

• Electronic geometry = Trigonal bipyramidal:

• Since there are two lone pairs the trigonal bipyramidal electronic geometry give a T-Shape:
  
  

4. Determine Polarity:

• Is there a difference in the Electronegativity (EN) of the bonded atoms?
  • EN_Cl = 3.0, EN_F = 4.0
  • EN = 1.0, there is a difference, and
• The Cl-F bonds are polar, with a small negative charge on each Fluorine.
• One of the bonds is not opposed (the molecule is NOT symmetrical), it is polar.

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Last modified 15 April 2013