Polar Bonds and Molecules

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 = ENA – ENB • Chart:

5. Examples • Potassium Fluoride KF • ∆EN = ENF – ENK = 3.98 – 0.82 = 3.16 • IONIC BOND • Two Oxygen Atoms O2 • ∆EN = ENO – ENO = 3.44 – 3.44 = 0 • NON-POLAR COVALENT • Carbon Tetrachloride CCl4 • ∆EN = ENCl – ENC = 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 (CCl4) • 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+ orδ- • Example: CCl4 • 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-Clδ-

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 CCl4 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 H20 • Determine bond type • ∆EN = ENO – ENH = 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 Bent shape according to VSEPR

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 CO2 • Determine bond type • ∆EN = ENO – ENC = 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 Linear shape according to VSEPR

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. Hydrogen Cyanide HCN • Determine bond type • ∆EN = ENN – ENC = 3.04 – 2.55 = 0.49 • Thus is slightly POLAR COVALENT • ∆EN = ENC – ENH = 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-

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. CO2 • 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 SO3 • Determine bond type • ∆EN = ENO – ENS = 3.44 – 2.58 = 0.86 • Thus is POLAR COVALENT • Determine partial charges • O has greater EN than S • Our partial charges are: Trigonal Planar shape according to VSEPR

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. CCl2O • 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 NH3 • Determine bond type • ∆EN = ENN – ENH = 304 – 2.20 = 0.84 • Thus is POLAR COVALENT • Determine partial charges • N has greater EN than H • Our partial charges are: Pyramidal shape according to VSEPR

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 CCl4 • Determine bond type • ∆EN = ENCl – ENC = 3.16 – 2.55 = 0.61 • Thus is POLAR COVALENT • Determine partial charges • Cl has greater EN than C • Our partial charges are: Tetrahedral shape according to VSEPR

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 CHCl3 • Determine bond type • ∆EN = ENCl – ENC = 3.16 – 2.55 = 0.61 • Thus is POLAR COVALENT • ∆EN = ENC – ENH = 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: Tetrahedral shape according to VSEPR

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 • SI2 • CH3F • AsI3 • H2O2