5 Steps to a 5 AP Chemistry

Bonding  163

Give yourself 1 point for each structure that is correct. The double bonds could be

between the nitrogen and any of the oxygens, not just the ones shown. Only one

double bond per structure is allowed.

ii. If you predicted the nitrite ion has the shorter bond length, you have earned 1 point.

The explanation must invoke resonance. You do not need to show all the resonance

structures. You need to mention that the double bond “moves” from one oxygen to

another. In the nitrite ion, each N–O bond is a double bond half the time and a single

bond the other half. This gives an average of 1.5 bonds between the nitrogen and each

of the oxygens. Similarly, for the nitrate ion, each N–O bond spends one-third of the

time as a double bond, and two-thirds of the time as a single bond. The average N–O

bond is 1.33. The larger the average number of bonds, the shorter the bond is. This

explanation will get you 1 point.

b. Give yourself 1 point for each correct Lewis structure.
BF 3 ClF 3

The BF 3 , with three bonding pairs and zero nonbonding pairs on the central atom, is

not polar. The ClF 3 , with five pairs about the central atom, is polar because of the two

lone pairs. Give yourself 1 point for this explanation.

c. You get 1 point for saying that the two compounds with the highest melting points are
ionic and the other compounds are molecular.
You get 1 point if you say that SrS is higher than KCl because the charges on the ions
in SrS are higher.

You get 1 point if you say H 2 O is higher than the lowest two because of hydrogen

bonding.

You get 1 point if you say H 2 S is higher than CH 4 because H 2 S is polar and CH 4 is

nonpolar.

There are a maximum of 11 points.

Second Free-Response Question
a. Silicon tetrafluoride is the only one of the three compounds that is not polar.

SiF 4 : four bonding SF 4 : four bonding XeF 4 : four bonding

pairs and no pairs and one pairs and two

lone pairs. lone pair. lone pairs.

Tetrahedral See-saw Square planar

F Xe

F

F

F

F S

F

F

F

F Si F

F

F

:F—B—F: :F











——Cl—F:

|

 



|

:F:



:F: