1)Which of the following correctly lists the forces of attraction in order from weakest to strongest? A)Ionic bonding < dipole < hydrogen bonding < dispersion B)Dipole < dispersion < hydrogen bonding < ionic bonding C)Dispersion < dipole < hydrogen bonding < ionic bonding D)Dispersion < hydrogen bonding < dipole < ionic bonding E)Dipole < hydrogen bonding < ionic bonding < dispersion
2) How many hydrogen bonds are shown in the drawing below? A)0 B)1 C)2 D)5 E)6
3) Which drawing below represents what happens to the atoms/molecules of methanol (CH 3 OH) after it boils?
4) Which of the following correctly explains why hexane (C 6 H 14 ) has a higher boiling point than propane (C 3 H 8 )? A)Hexane has more bonds than propane, so its harder to break. B)Molecules of hexane are more strongly attracted to each other than molecules of propane. C)Hexane is heavier than propane, so gravity holds it together. D)Hexane is more polar than propane, holding it together better.
5) Which molecule(s) below is/are polar? A)CH4 B)CO 2 C)H 2 O D)SF 2 E)CF 4
6) Which picture below shows how two molecules of formaldehyde (CH 2 O) would likely arrange themselves near each other? A) B) C) D)
7) Which of the following explains why NaCl has a much higher melting point than HCl? A)NaCl has less electronegativity than HCl, so it is more strongly held together. B)The charges that attract NaCl particles are larger than the charges that attract HCl particles. C)NaCl is ionic and HCl is covalent, and ionic bonds are stronger than covalent bonds. D)NaCl has a greater mass than HCl, so its attraction is stronger.
8) How many hydrogen bonds are shown in the drawing below? A)0 B)2 C)6 D)8 E)18
9) Choose the best answer from below for this question: Which has a higher boiling point, CO 2 or CS 2 ? Why? A)CO 2 because it has a greater total electronegativity. B)CO 2 because it has stronger covalent bonds. C)CO 2 because it is polar while CS 2 is nonpolar. D)CS 2 because it is nonpolar while CO 2 is polar. E)CS 2 because it has stronger dispersion forces.