1. List all types of IMFs that would occur in each of the following and circle t

Question

1. List all types of IMFs that would occur in each of the following and circle the strongest force present. a. CHiCF b. CCl d. NaBr c. O2 e. (CHs)sN 2. Rank the following compounds from weakest IMF to strongest IMF using the less than f. CH:OH symbol, and list the strongest force present for each compound. a. CaCO3 CH4 CH3OH CH3OCH3 b. HF MgO HC1 H2 c. SO2 PCl3 F2 FeCl 3. Rank the following compounds from weakest IMF to strongest IM, using the less than symbol, list the strongest force present for each compound, and justify your answer if the same force is present in more than one compound a. H2S 2 N2 H20 b. H2Se H2S H2Po H2Te 4. Rank the following compounds from lowest to highest boiling point using the less than symbol, and justify your answer. CH CH OHCH CH2OH CH3CH CH3CH3CH CH b. Br Clh 12

Explanation / Answer

1)

CH3CF3 ------- london dispersion , dipole dipole

CCl4 ---------- london dispersion

O2 ------------- london dispersion

NaBr -------- ionic , london dispersion force

(CH3)3N ------- dispersion , dipole dipole

CH3OH --------- hydrogen bonding , dipole dipole , dispersion

2)

a) CH4 < CH3OCH3 < CH4<CaCO3

CaCO3— ion-ion attractions.
CH4— London dispersion forces
CH3OH— hydrogen bonding
CH3OCH3— dipole-dipole attractions

b) H2<HCl< HF < MgO

H2 ------ london dispersion force

HCl ------- dipole dipole

HF ------- hydrogen bonding

MgO ---- ionic bond

c) F2< SO2 < PCl3 < FeCl2

F2 ------ london dispersion

  SO2 ----- dipole dipole

PCl3 ----- dipole dipole

FeCl2 ---- ionic bond force

3)  

N2 < I2 < H2S < H2O

N2 and I2 are nonpolar, so they only have dispersion forces; I2 has stronger forces because it
is larger . H2S has dipole-dipole, so it is stronger than I2 .H2O has hydrogen bonding, so it is stronger than H2S  

H2S < H2Se < H2Te < H2Po

These compounds are all the same shape. Although H2S is slightly more polar than the
others, it is not very polar so it has very weak dipole-dipole forces. Therefore, the difference
in dispersion forces are more important for these compounds. H2Po is the largest and,
therefore, has the strongest dispersion forces.

4) a) Ch3CH2CH3CH3CH2CH3 < CH3CH2OHCH3CH2OH < CH3OCH2o3CH3OCH3

becuase of strong hydrogen bonding present in CH3CH2OHCH3CH2OH has more boiling point compare to others so it has mor boiling point in CH3CH2CH3CH3CH2CH3 only dispersion force is presnt so it has less boiling point

b) Cl2< Br2< I2

Chlorine has the lowest boiling point as compared to other two (bromine and iodine) because Br and I is larger than Cl, therefore Br2 AND I2 has stronger intermolecular foces (Van der Waals) compared to Cl2. Among Br2 and I2 , Bromine has lower boiling point as compared to iodine  

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