# Molecular Geometry, cont.

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

### Exceptions to the "Octet Rule" cont.

1. 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 pyramidal:

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

4. Determine Polarity:

• Is there a difference in the Electronegativity (EN) of the bonded atoms?
• ENCl = 3.0, ENF = 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.

# Energy of Formation for Ionic Compounds

It turns out that the transfer of an electron from a metal to a non-metal will not generally provide enough energy to favor the process. So how is it that these are in fact favorable reactions?

Let's look at the energy of the process by breaking it into steps and looking at the enthapies of formation starting with free atoms (the reality will be somewhat more complex since we would start with solid metal and molecules, each of which must first react to give free atomic state, but the results are similar). Of course we can get away with this because we are looking at state functions, which as we saw before are pathway independent!

 Ionization Energy Na Na+ + e- H = +495 kJ/mol Electron Affinity Energy Cl + e- Cl- H = -348 kJ/mol Total H = + 147 kJ/mol However, this value is for the free ions. If we allow them to come together by coulombic attraction into a crystal lattice a large additional amount of energy is released: Lattice Energy Na+(g) + Cl-(g) NaCl(s) H = - 449 kJ/mol Overall H = - 302 kJ/mol

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