chemistry : class12 Halogen Atom Effect : Physical properties of haloalkanes : Haloalkanes and Haloarenes MCQ Questions and Answers
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Halogen Atom Effect
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MCQ Questions
1. For the homologous series CH3X, where X is F, Cl, Br, or I, which order of boiling points is expected primarily because of increasing halogen polarizability and molecular mass?
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:
Down the halogen group, molecular mass and electron-cloud polarizability increase, strengthening London dispersion forces. Consequently, the boiling points of methyl halides generally increase from CH3F to CH3I.
2. Which methyl halide is expected to have the greatest density at room temperature, considering the effect of replacing hydrogen by progressively heavier halogen atoms?
A.
Methyl fluoride
B.
Methyl chloride
C.
Methyl bromide
D.
Methyl iodide
Correct Answer:
Methyl iodide
Explanation:
The density of haloalkanes generally increases with the increasing atomic mass of the halogen atom. Among the methyl halides, methyl iodide therefore has the greatest density.
3. Which statement best explains why methyl iodide has a much higher boiling point than methyl fluoride, although both compounds contain only one carbon atom?
A.
Methyl iodide forms strong hydrogen bonds with itself
B.
The larger iodine atom gives methyl iodide much greater polarizability and stronger dispersion forces
C.
The C–F bond is nonpolar whereas the C–I bond is ionic
D.
Methyl fluoride has a greater molecular mass than methyl iodide
Correct Answer:
The larger iodine atom gives methyl iodide much greater polarizability and stronger dispersion forces
Explanation:
Iodine has a much larger and more easily distorted electron cloud than fluorine. The resulting stronger London dispersion forces raise the boiling point of methyl iodide.
4. Which physical-property combination is most reasonable for methyl chloride and methyl bromide at ordinary room temperature?
A.
Both are solids because the C–X bond is highly polar
B.
Methyl chloride is a gas, whereas methyl bromide is much closer to condensation because of its higher mass and polarizability
C.
Methyl chloride is a liquid, whereas methyl bromide is a solid because bromine forms hydrogen bonds
D.
Both have identical physical states because they have the same carbon skeleton
Correct Answer:
Methyl chloride is a gas, whereas methyl bromide is much closer to condensation because of its higher mass and polarizability
Explanation:
Replacing chlorine by the heavier, more polarizable bromine atom increases intermolecular dispersion forces and raises the boiling point. Thus methyl bromide is less volatile than methyl chloride.
5. For 1-chloropropane, 1-bromopropane, and 1-iodopropane, which property is expected to increase in the stated order?
A.
Volatility
B.
Vapour pressure at the same temperature
C.
Boiling point
D.
Ease of evaporation
Correct Answer:
Boiling point
Explanation:
The sequence Cl, Br, I represents increasing halogen mass and polarizability. Stronger dispersion forces produce progressively higher boiling points, while volatility and vapour pressure decrease.
6. A student compares chloromethane and fluoromethane. Which prediction is most consistent with the halogen atom effect on physical properties?
A.
Fluoromethane has the higher boiling point because the C–F bond is stronger
B.
Chloromethane has the higher boiling point because chlorine is more polarizable
C.
Both have the same boiling point because their molecular formulas contain one carbon atom
D.
Fluoromethane has the higher boiling point because fluorine has the highest electronegativity
Correct Answer:
Chloromethane has the higher boiling point because chlorine is more polarizable
Explanation:
Boiling point depends strongly on intermolecular attractions, especially dispersion forces, not simply on bond strength or electronegativity. Chlorine has greater polarizability and mass than fluorine, giving chloromethane the higher boiling point.
7. Which sequence correctly represents the expected increase in liquid density for the corresponding 1-haloalkanes, R–X, when the alkyl group R is kept unchanged?
A.
R–I < R–Br < R–Cl < R–F
B.
R–F < R–Cl < R–Br < R–I
C.
R–Br < R–F < R–I < R–Cl
D.
R–Cl < R–I < R–F < R–Br
Correct Answer:
R–F < R–Cl < R–Br < R–I
Explanation:
With the same alkyl group, replacing fluorine by progressively heavier halogens increases the mass per molecule substantially. This generally produces an increasing density from alkyl fluoride to alkyl iodide.
8. Why is an alkyl iodide generally less volatile than the corresponding alkyl chloride?
A.
The alkyl iodide has weaker intermolecular forces because iodine is less electronegative
B.
The alkyl iodide has stronger dispersion forces because its iodine atom is larger and more polarizable
C.
The alkyl chloride has stronger hydrogen bonding through chlorine
D.
The alkyl iodide has fewer electrons than the alkyl chloride
Correct Answer:
The alkyl iodide has stronger dispersion forces because its iodine atom is larger and more polarizable
Explanation:
Lower volatility means a lower tendency to enter the vapour phase. The larger iodine atom increases electron-cloud polarizability and dispersion attractions, raising the boiling point relative to the corresponding alkyl chloride.
9. Which comparison correctly identifies the dominant reason that bromoethane has a higher boiling point than fluoroethane?
A.
Bromoethane has a larger, more polarizable bromine atom and stronger dispersion forces
B.
Fluoroethane has no permanent dipole moment
C.
Bromoethane forms intermolecular hydrogen bonds through bromine
D.
The C–F bond contributes more molecular mass than the C–Br bond
Correct Answer:
Bromoethane has a larger, more polarizable bromine atom and stronger dispersion forces
Explanation:
Although fluoroethane contains a strongly polar C–F bond, bromoethane has much greater molecular mass and polarizability. Its stronger dispersion forces lead to the higher boiling point.
10. In a set of structurally similar haloalkanes, which change would generally produce the greatest increase in boiling point due specifically to the halogen atom effect?
A.
Replacing iodine with fluorine
B.
Replacing bromine with chlorine
C.
Replacing chlorine with iodine
D.
Replacing iodine with bromine
Correct Answer:
Replacing chlorine with iodine
Explanation:
The change from chlorine to iodine produces a large increase in atomic size, electron count, and polarizability. These factors strengthen dispersion forces more substantially than the other listed substitutions.
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