1. POLAR BONDS AND MOLECULES Ms. Withrow November 10, 2008

2. Polar Bonds  When involved in a bond, atoms of some elements attract the shared electrons to a greater extent than atoms of other elements – This property is called Electronegativity (EN)  The following chart is used to determine the electronegativities of each atom

4.  Based on the difference in electronegativities of atoms we can predict the type of bond that will form  Formula:  ∆EN = EN A – EN B  Chart:

5. Examples  Potassium Fluoride KF  ∆EN = EN F – EN K = 3.98 – 0.82 = 3.16  IONIC BOND  Two Oxygen Atoms O 2  ∆EN = EN O – EN O = 3.44 – 3.44 = 0  NON-POLAR COVALENT  Carbon Tetrachloride CCl 4  ∆EN = EN Cl – EN C = 3.16 – 2.55 = 0.61  POLAR COVALENT

6.  With respect to polar covalent bonds, the differences in electronegativity tell us about the sharing of electrons  Example: Carbon Tetrachloride (CCl 4 )  Cl has EN = 3.16  C has EN = 2.55  From this, we say that chlorine has stronger attraction for electrons than carbon  Thus, electrons will spend more time around the Cl than C

7.  This results in a slight separation of positive and negative charges which we call “partial charges” and represent them as δ + or δ -  Example: CCl 4  Chlorine with greater EN will have greater attraction of e- and thus will have partial negative charge δ -  Carbon with lower EN will have less attraction of e- and thus will have partial positive charge δ +  Shown as δ + C-Cl δ -

8.  When the bond is separated into partial positive and negative charges we call this bond a dipole bond  We represent dipole bonds with a vector arrow that points to the more electronegative atom  Example CCl 4 δ + C-Cl δ -

9. Examples  Remember to  Determine the bond type (by finding ∆EN)  Assign the partial charges  Place the dipole moment  Copper and Oxygen δ + C-O δ -  Carbon and Fluorine δ + C-F δ -

10. Polar Molecules  We use our information on polar bonds to predict whether molecules will be polar or non- polar  We also must know our VSEPR shapes in order to do this!!

11. Water H 2 0  Determine bond type  ∆EN = EN O – EN H = 3.44 – 2.20 = 1.24  Thus is POLAR COVALENT  Determine partial charges  O has higher EN and H has lower EN  Our partial charges are:  If we include the dipoles

12.  This is where VSEPR is important! -- You must know the shape of the molecule in order to determine it’s polarity  Water has two partially positive ends and one partially negative end  The two dipole arrows point in the same direction. If we add these together we can see the molecule will have an overall net dipole  Because the dipoles do not cancel each other a net dipole is produced and we say that the molecule is POLAR

13. Carbon Dioxide CO 2  Determine bond type  ∆EN = EN O – EN C = 3.44 – 2.55 = 0.89  Thus is POLAR COVALENT  Determine partial charges  O has greater EN than C  Our partial charges are:  If we include the dipoles

14.  The dipoles created in this molecule are pointing in opposite directions and thus will cancel each other  This molecule has no net dipole and therefore is said to be NON-POLAR

15.  Determine bond type  ∆EN = EN N – EN C = 3.04 – 2.55 = 0.49  Thus is slightly POLAR COVALENT  ∆EN = EN C – EN H = 2.55 – 2.20 = 0.35  Is also slightly POLAR COVALENT  Determine partial charges  N has greater EN than C – N will have δ -  C has greater EN than H – C will have δ - Hydrogen Cyanide HCN

16.  When we assign the dipoles  We see that they are both pointing the same direction  Thus they will not cancel, but will result in an overall net dipole  This molecule is said to be POLAR

17. Note the Difference!  When we had a linear molecule with the same atoms attached to the central atom the molecule was non-polar ex. CO 2  When we had a linear molecule with two different atoms attached to the central atom, the molecule was polar Ex. HCN  It is very important to look at the electronegativities associated with the atoms and not just the VSEPR shape

18. Sulfur Trioxide SO 3  Determine bond type  ∆EN = EN O – EN S = 3.44 – 2.58 = 0.86  Thus is POLAR COVALENT  Determine partial charges  O has greater EN than S  Our partial charges are:

19.  When we assign dipole arrows  All the dipoles are pulling away from the central atom  You may think that because there are three dipoles they will not cancel and will result in a polar molecule  This is not correct however!!

20.  Look at the horizontal and vertical components of the vectors (red and green arrows)  The red arrows will cancel  The green arrows can add together  This green arrow will cancel with the blue vector created by the top O  Therefore all dipole vectors will cancel in this molecule creating no net dipole and therefore the molecule is NON-POLAR

21.  Similar to our linear molecule, difference will occur when the atoms attached to the central atom are different  We must be sure to look at the electronegativities of each atom when comparing the dipole vectors  Ex. CCl 2 O  O has higher EN than Cl and will therefore have a greater dipole  The two dipoles from Cl will add together but they will still be less than that of O  Overall net dipole will result and thus molecule is POLAR

22. Ammonia NH 3  Determine bond type  ∆EN = EN N – EN H = 304 – 2.20 = 0.84  Thus is POLAR COVALENT  Determine partial charges  N has greater EN than H  Our partial charges are:

23.  Assign dipole vectors  The three vectors will add together to create an overall net dipole  This will result in a POLAR molecule

24. Carbon Tetrachloride CCl 4  Determine bond type  ∆EN = EN Cl – EN C = 3.16 – 2.55 = 0.61  Thus is POLAR COVALENT  Determine partial charges  Cl has greater EN than C  Our partial charges are:

25.  When we assign dipoles  We can see that all the dipoles are of the same magnitude because the EN differences are all the same  There are equal amounts of dipoles in opposite directions and thus they will all cancel  This results in no net dipole and therefore the molecule is NON-POLAR

26. Chloroform CHCl 3  Determine bond type  ∆EN = EN Cl – EN C = 3.16 – 2.55 = 0.61  Thus is POLAR COVALENT  ∆EN = EN C – EN H = 2.55 – 2.20 = 0.35  Thus is slightly POLAR COVALENT  Determine partial charges  Cl has greater EN than C  C has greater EN than H  Our partial charges are:

27.  Assign dipoles (blue arrows)  We can see that the dipoles to Cl will all add up to create the larger green dipole vector  This is opposite to the dipole vector created by H-C but does not have the same magnitude  Thus, it will not cancel and result in a net dipole  This molecule is POLAR

28. Summary of Polarity of Molecules  Linear:  When the two atoms attached to central atom are the same the dipoles will cancel, leaving no net dipole, and the molecule will be Non-Polar  When the two atoms are different the dipoles will not cancel, resulting in a net dipole, and the molecule will be Polar

29.  Bent:  The dipoles created from this molecule will not cancel creating a net dipole and the molecule will be Polar  Pyramidal:  The dipoles created from this molecule will not cancel creating a net dipole and the molecule will be Polar

30. Summary of Polarity of Molecules  Trigonal Planar:  When the three atoms attached to central atom are the same the dipoles will cancel, leaving no net dipole, and the molecule will be Non-Polar  When the three atoms are different the dipoles will not cancel, resulting in a net dipole, and the molecule will be Polar

31. Summary of Polarity of Molecules  Tetrahedral:  When the four atoms attached to the central atom are the same, the dipoles will cancel, leaving no net dipole, and the molecule will be Non-Polar  When the four atoms are different, the dipoles will not cancel, resulting in a net dipole, and the molecule will be Polar

32. Summary of Polarity of Molecules

33. Examples to Try  Determine whether the following molecules will be polar or non-polar  SI 2  CH 3 F  AsI 3  H 2 O 2