chemistry : class12 Halogen Mass Effect : Physical properties of haloalkanes : Haloalkanes and Haloarenes MCQ Questions and Answers
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Halogen Mass Effect
MCQ questions and answers.
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MCQ Questions
1. For the series CH3F, CH3Cl, CH3Br, and CH3I, which order of boiling points is expected primarily because the halogen mass increases the strength of London dispersion forces?
A.
CH3F < CH3Cl < CH3Br < CH3I
B.
CH3I < CH3Br < CH3Cl < CH3F
C.
CH3Cl < CH3F < CH3I < CH3Br
D.
CH3Br < CH3I < CH3F < CH3Cl
Correct Answer:
CH3F < CH3Cl < CH3Br < CH3I
Explanation:
With the same methyl group, replacing F by progressively heavier and more polarizable halogens increases London dispersion forces. Therefore, the boiling point generally rises from CH3F to CH3I.
2. Which explanation best accounts for the higher boiling point of chloroethane than fluoroethane, despite both compounds having the same carbon skeleton?
A.
Chloroethane forms strong hydrogen bonds with itself
B.
The chlorine atom has greater polarizability and produces stronger dispersion attractions
C.
Fluoroethane contains more carbon atoms
D.
The C–F bond is completely nonpolar
Correct Answer:
The chlorine atom has greater polarizability and produces stronger dispersion attractions
Explanation:
The heavier chlorine atom has a more easily deformable electron cloud than fluorine. This increases London dispersion forces between chloroethane molecules and raises its boiling point.
3. In the homologous halogen series R–F, R–Cl, R–Br, and R–I, where R is identical, which change is most directly associated with increasing halogen mass?
A.
A general increase in molar mass and polarizability
B.
A decrease in the number of electrons in the molecule
C.
Complete disappearance of dipole–dipole interactions
D.
A decrease in the size of the electron cloud around the halogen
Correct Answer:
A general increase in molar mass and polarizability
Explanation:
As the halogen changes from F to I, both molar mass and the number of electrons increase. The larger, more polarizable electron cloud strengthens dispersion forces and generally elevates boiling points.
4. Which compound is expected to have the highest boiling point among the following compounds with the same alkyl group, based chiefly on the halogen mass effect?
A.
1-Fluoropropane
B.
1-Chloropropane
C.
1-Bromopropane
D.
1-Iodopropane
Correct Answer:
1-Iodopropane
Explanation:
All four compounds have the same propyl group. Iodine is the heaviest and most polarizable halogen among the choices, giving 1-iodopropane the strongest dispersion forces and the highest expected boiling point.
5. A student claims that the boiling points of R–Cl, R–Br, and R–I must be identical because their carbon skeletons are identical. Which observation disproves this claim?
A.
R–I generally boils at a higher temperature than R–Cl
B.
R–Cl always has a higher melting point than R–I
C.
R–Br contains fewer carbon atoms than R–Cl
D.
R–Cl and R–I have identical molecular masses
Correct Answer:
R–I generally boils at a higher temperature than R–Cl
Explanation:
Changing only the halogen changes the molecular mass and polarizability. The iodine derivative generally experiences stronger dispersion attractions than the chlorine derivative, so its boiling point is higher.
6. For haloalkanes having the same alkyl group, which sequence best represents the expected trend in ease of liquefaction as the halogen becomes heavier?
A.
R–I > R–Br > R–Cl > R–F
B.
R–F > R–Cl > R–Br > R–I
C.
R–Cl > R–I > R–F > R–Br
D.
R–Br > R–F > R–I > R–Cl
Correct Answer:
R–I > R–Br > R–Cl > R–F
Explanation:
Heavier halogen atoms increase molecular polarizability and intermolecular dispersion forces. Consequently, the heavier derivatives generally condense more readily, giving the order R–I > R–Br > R–Cl > R–F.
7. Why does replacing chlorine by bromine in an otherwise unchanged haloalkane generally increase its boiling point?
A.
Bromine lowers the molecular mass but increases hydrogen bonding
B.
Bromine increases molecular mass and electron-cloud polarizability
C.
Bromine makes the molecule ionic in the liquid state
D.
Bromine eliminates all dipole interactions between molecules
Correct Answer:
Bromine increases molecular mass and electron-cloud polarizability
Explanation:
Bromine is heavier and has a more polarizable electron cloud than chlorine. The resulting stronger London dispersion forces require more thermal energy to separate the molecules, increasing the boiling point.
8. Consider equal-sized samples of 1-chlorobutane and 1-bromobutane. Which statement best reflects the halogen mass effect on their physical behavior?
A.
1-Bromobutane generally has stronger dispersion forces and a higher boiling point
B.
1-Chlorobutane has stronger dispersion forces because chlorine is more electronegative
C.
Both have identical boiling points because their carbon skeletons are the same
D.
1-Bromobutane has weaker intermolecular forces because bromine is less electronegative
Correct Answer:
1-Bromobutane generally has stronger dispersion forces and a higher boiling point
Explanation:
Electronegativity alone does not determine the boiling-point trend. The heavier bromine atom is more polarizable, so 1-bromobutane generally has stronger dispersion forces and a higher boiling point.
9. Which pair would provide the clearest comparison for studying the effect of halogen mass on boiling point while minimizing changes in carbon skeleton and branching?
A.
1-Chloropropane and 1-bromopropane
B.
1-Chloropropane and 2-chloropropane
C.
1-Chloropropane and 1-chlorobutane
D.
1-Chloropropane and chlorobenzene
Correct Answer:
1-Chloropropane and 1-bromopropane
Explanation:
These compounds have the same carbon skeleton and the same position of substitution; only the halogen changes. Their boiling-point difference can therefore be related most directly to the increased mass and polarizability of bromine.
10. The boiling points of CH3Cl, CH3Br, and CH3I increase in the order CH3Cl < CH3Br < CH3I. Which inference is most appropriate?
A.
The increase is mainly due to stronger intermolecular dispersion forces from the heavier halogens
B.
The carbon–halogen bond becomes ionic in the same order
C.
The molecules acquire additional carbon atoms in the same order
D.
Hydrogen bonding becomes progressively stronger from chlorine to iodine
Correct Answer:
The increase is mainly due to stronger intermolecular dispersion forces from the heavier halogens
Explanation:
The molecular framework remains CH3, while the halogen becomes heavier and more polarizable. Stronger dispersion forces therefore provide the main explanation for the increasing boiling points.
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