You must be familiar with the fact that melting point is essentially the same thing as freezing point. What is meant by this is that the same temperature will melt something (if we\'re considering an endothermic reaction) as it will freeze something (exothermic reaction). After you\'ve reached this clarity, simply observe all of the compounds and check for which ones have low melting points and then work your way up. Now the question is, how can we identify compounds with low melting points? Remember, if a compound is polar (charged, or is a dipole, like water), it will have a high boiling point (it will be very tough to boil). So compounds that have a high boiling point will most likely have high melting points as well (generally, not universally). For NaBr, Na+ and Br- and the constituents and they certainly have an electronegativity difference (uneven charge distribution in the electron density implies a dipole). We have to look at calcium nitrate next before determining the priority of NaBr. Ca(NO3)2 is obviously going to have charged constituents. The Ca will be positively charged and the nitrate groups which are oxyANIONs will be negatively charged. Since there are MORE oxygens in calcium nitrate, this compound will be more electronegative and will form a stronger dipole compared to NaBr (Br is NOT as electronegative as oxygen). Greater electronegativity implies stronger dipole in a compound and this means higher boiling point, which means higher melting point, which means higher freezing point compared to NaBr. HCl is an acid and although there are H atoms, HCl CANNOT form Hydrogen bonds (h-bonding occurs only for N-H, O-H, and F- H. Cl is obviously NOT one of the constituents that can form hydrogen bonding and therefore will have a LOW boiling point, meaning low freezing point as well. HCl falls at the very bottom of our list. Ethanol is polar (because of the free electrons from O--H from the sp3 orbital). Hydrogen bonding IMMEDIATELY implies high boiling point, so EtOH will have a relatively large melting point, and thus freezing point. Now that I\'ve laid out the information as thoroughly as possible. Simply arrange them to form your freezing point trend list given the four compounds. Should be rather straightforward. Solution You must be familiar with the fact that melting point is essentially the same thing as freezing point. What is meant by this is that the same temperature will melt something (if we\'re considering an endothermic reaction) as it will freeze something (exothermic reaction). After you\'ve reached this clarity, simply observe all of the compounds and check for which ones have low melting points and then work your way up. Now the question is, how can we identify compounds with low melting points? Remember, if a compound is polar (charged, or is a dipole, like water), it will have a high boiling point (it will be very tough to boil). So compounds that have a high boiling point will most likely have high melting poin.

1. You must be familiar with the fact that melting point is essentially the same thing as
freezing point. What is meant by this is that the same temperature will melt something (if we're
considering an endothermic reaction) as it will freeze something (exothermic reaction). After
you've reached this clarity, simply observe all of the compounds and check for which ones have
low melting points and then work your way up. Now the question is, how can we identify
compounds with low melting points? Remember, if a compound is polar (charged, or is a dipole,
like water), it will have a high boiling point (it will be very tough to boil). So compounds that
have a high boiling point will most likely have high melting points as well (generally, not
universally). For NaBr, Na+ and Br- and the constituents and they certainly have an
electronegativity difference (uneven charge distribution in the electron density implies a dipole).
We have to look at calcium nitrate next before determining the priority of NaBr. Ca(NO3)2 is
obviously going to have charged constituents. The Ca will be positively charged and the nitrate
groups which are oxyANIONs will be negatively charged. Since there are MORE oxygens in
calcium nitrate, this compound will be more electronegative and will form a stronger dipole
compared to NaBr (Br is NOT as electronegative as oxygen). Greater electronegativity implies
stronger dipole in a compound and this means higher boiling point, which means higher melting
point, which means higher freezing point compared to NaBr. HCl is an acid and although there
are H atoms, HCl CANNOT form Hydrogen bonds (h-bonding occurs only for N-H, O-H, and F-
H. Cl is obviously NOT one of the constituents that can form hydrogen bonding and therefore
will have a LOW boiling point, meaning low freezing point as well. HCl falls at the very bottom
of our list. Ethanol is polar (because of the free electrons from O--H from the sp3 orbital).
Hydrogen bonding IMMEDIATELY implies high boiling point, so EtOH will have a relatively
large melting point, and thus freezing point. Now that I've laid out the information as
thoroughly as possible. Simply arrange them to form your freezing point trend list given the four
compounds. Should be rather straightforward.
Solution
You must be familiar with the fact that melting point is essentially the same thing as
freezing point. What is meant by this is that the same temperature will melt something (if we're
considering an endothermic reaction) as it will freeze something (exothermic reaction). After
you've reached this clarity, simply observe all of the compounds and check for which ones have
low melting points and then work your way up. Now the question is, how can we identify
compounds with low melting points? Remember, if a compound is polar (charged, or is a dipole,
like water), it will have a high boiling point (it will be very tough to boil). So compounds that
have a high boiling point will most likely have high melting points as well (generally, not
universally). For NaBr, Na+ and Br- and the constituents and they certainly have an

2. electronegativity difference (uneven charge distribution in the electron density implies a dipole).
We have to look at calcium nitrate next before determining the priority of NaBr. Ca(NO3)2 is
obviously going to have charged constituents. The Ca will be positively charged and the nitrate
groups which are oxyANIONs will be negatively charged. Since there are MORE oxygens in
calcium nitrate, this compound will be more electronegative and will form a stronger dipole
compared to NaBr (Br is NOT as electronegative as oxygen). Greater electronegativity implies
stronger dipole in a compound and this means higher boiling point, which means higher melting
point, which means higher freezing point compared to NaBr. HCl is an acid and although there
are H atoms, HCl CANNOT form Hydrogen bonds (h-bonding occurs only for N-H, O-H, and F-
H. Cl is obviously NOT one of the constituents that can form hydrogen bonding and therefore
will have a LOW boiling point, meaning low freezing point as well. HCl falls at the very bottom
of our list. Ethanol is polar (because of the free electrons from O--H from the sp3 orbital).
Hydrogen bonding IMMEDIATELY implies high boiling point, so EtOH will have a relatively
large melting point, and thus freezing point. Now that I've laid out the information as
thoroughly as possible. Simply arrange them to form your freezing point trend list given the four
compounds. Should be rather straightforward.