Humboldt State University ® Department of Chemistry

Richard A. Paselk

	General Chemistry	Spring 2015
Exercise: Hybrid Molecular Orbitals	© R. Paselk 2015
	Discussion Modules
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Hybrid Molecular Orbitals

Periodic Table of the Elements

IA

IIA

IIIA

IVA

VA

VIA

VIIA

VIIIA

H

He

2

Li

Be

B

C

N

O

F

Ne

3

Na

Mg

IIIB

IVB

VB

VI

VIIB

VIIIB

IB

IIB

Al

Si

P

S

Cl

Ar

4

K

Ca

Sc

Ti

V

Cr

Mn

Fe

Co

Ni

Cu

Zn

Ga

Ge

As

Se

Br

Kr

5

Rb

Sr

Y

Zr

Nb

Mo

Tc

Ru

Rh

Pd

Ag

Cd

In

Sn

Sb

Te

I

Xe

6

Cs

Ba

Lu

Hf

Ta

W

Re

Os

Ir

Pt

Au

Hg

Tl

Pb

Bi

Po

At

Rn

1. Phosphorus pentachloride, PCl₅.

• Determine if the molecules follows the octet rule with single bonds:
  • Looking at the Periodic Table, P is in group V so it has 5 valence electrons while Cl is in group VII so each has 7 valence electrons,
  • valence electrons = 5 + 5(7) = 40
• For Clark's rules we see that there are 1(P) plus 5(Cl) atoms or a total of 6 atoms other than hydrogen,
  • 6y + 2 = 6(6) + 2 = 38
  • Comparing the number of valence electrons needed for eight valance electrons around each atom (6y +2) to the actual we find = 40 - 38 = 2, we have two extra electrons around the central atom!
  • This compound has an expanded valence shell with 10 electrons in 5 pairs.
• Lewis Structure
• The Steric Number (SN) = 5 so must be 5 bonding orbitals, can only get 4 with s and p so need to use a d as well: dsp³ = Trigonal bipyramidal electronic geometry.
• No lone pairs, so molecular geometry = electronic geometry.
  
  = =

 

Polarity:

• P and Cl have significantly different Electronegativity values (2.1& 3.0), so the P-Cl bonds will be polar, with Cl partially negative.
• Of course the equatorial chlorine-iodine bonds will cancel since they are completely symmetrically arranged.
• The axial chlorine-iodine bonds will also cancel since they are opposed, thus the molecule is non-polar.

2. Determine the Geometry of Sulfur tetrafluoride, SF₄.

Valence electrons.
• Looking at the Periodic Table, S is in group VI so it has 6 valence electrons while F is in group VII so each has 7 valence electrons,
• valence electrons = 6 + 4(7) = 34
• For Clark's rules we see that there are 1(S) and 5(F) atoms or a total of 5 atoms other than hydrogen,
  • 6y + 2 = 6(5) + 2 = 32
  • Comparing the number of valence electrons needed for eight valance electrons around each atom (6y +2) to the actual we find = 34 - 32 = 2, we have two extra electrons around the central atom!
  • This compound has an expanded valence shell with 10 electrons in 5 pairs.
• Lewis Structure
• SN = 5 so must be 5 orbitals, one s three p and one d: dsp³ = Trigonal bipyramidal electronic geometry.
• One lone pair, so molecular geometry = Seesaw shaped: =

Polarity:

• S and F have significantly different Electronegativity values (2.5 & 4.0), so the S-F bond will be polar, with F partially negative.
• Of course the axial sulfur-fluorine bonds will cancel since they completely oppose each other.
• However, the two equatorial bonds are unbalanced, making Sulfur tetafluoride a polar molecule, as seen in the diagram below:

 

3. Determine the Geometry of Arsenic hexafluoride ion, AsF₆^-.

• Valence electrons.
  • Looking at the Periodic Table, As is in group V so it has 5 valence electrons while F is in group VII with 7 electrons each, and we have one extra electron because of the -1 charge,
  • valence electrons = 5 + 6(7) + 1= 48
• For Clark's rules we see that there are 1(Cl) and 3(F) atoms or a total of 4 atoms other than hydrogen,
  • 6y + 2 = 6(7) + 2 = 44
  • Comparing the number of valence electrons needed for eight valance electrons around each atom (6y +2) to the actual we find = 48 - 44 = 4, we have four extra electrons around the central atom!
  • This compound has an expanded valence shell with 12 electrons in 6 pairs.
• Lewis Structure
• The Steric Number (SN) = 6 so must be 6 orbitals, one s three p and two d: d²sp³ = Octahedral electronic geometry.
• No lone pairs so the Molecular Geometry is identical to the electronic geometry, and the molecule is Octahedral:

 

4. Determine the Geometry of Iodine pentachloride, ICl₅.

• Valence electrons.
  • Looking at the Periodic Table, both Cl and I are in group VII with 7 electrons each,
  • valence electrons = 6(7) = 42
• For Clark's rules we see that there are 1(I) and 5(Cl) atoms or a total of 6 atoms other than hydrogen,
  • 6y + 2 = 6(6) + 2 = 38
  • Comparing the number of valence electrons needed for eight valance electrons around each atom (6y +2) to the actual we find = 42 - 38 = 4, we have four extra electrons around the central atom!
  • This compound has an expanded valence shell with 12 electrons in 6 pairs.
• Lewis Structure
• The Steric Number (SN) = 5 bonded atoms + 1 lone pairs = 6 so must be 6 orbitals, one s three p and two d: d²sp³ = Octahedral electronic geometry.
  
• Since there is one lone pair the Molecular Geometry is different from the electronic geometry, and the molecule is tetragonal pyramidal:

=

Polarity:

• I and Cl have somewhat different Electronegativity values (2.5 & 3.0), so the I-Cl bond will be polar, with Cl partially negative.
• Of course the equatorial chlorine-iodine bonds will cancel since they are completely symmetrically arranged.
• However, the remaining bond is unbalanced, making Iodine pentachloride a polar molecule, as seen in the diagram below:

In addition to these exercises you should familiarize yourself with the text materials.

Hybrid Molecular Orbitals

© R A Paselk

Last modified 20 April 2015