4.16 - Intermolecular Forces - 2015

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| By Regentschemistry
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4.16 - Intermolecular Forces - 2015 - Quiz

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Questions and Answers
  • 1. 

    Determine the main type of intermolecular forces in CaO (aq).

    • A.

      London Dispersion Forces

    • B.

      Ion-Dipole Forces

    • C.

      Dipole-Dipole Forces

    • D.

      Hydrogen Bonds

    Correct Answer
    B. Ion-Dipole Forces
    Explanation
    Nonpolar Molecules (Symmetrical distribution of charge)=London Dispersion (weakest); Polar Molecules (Asymmetrical distribution of charge)=Dipole-Dipole Forces; Molecules with H-F, H-O, or H-N (because of big ∆EN)=Hydrogen Bonding;
    Saltwater (ionic compound and water)=Ion-Dipole Forces. Since CaO (aq) has an ionic compound, the CaO part, and water, the (aq) part, this qualifies as having ion-dipole forces.

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  • 2. 

    Determine the main type of intermolecular forces in C2H5OH.

    • A.

      London Dispersion Forces

    • B.

      Ion-Dipole Forces

    • C.

      Dipole-Dipole Forces

    • D.

      Hydrogen Bonds

    Correct Answer
    D. Hydrogen Bonds
    Explanation
    Nonpolar Molecules (Symmetrical distribution of charge)=London Dispersion (weakest); Polar Molecules (Asymmetrical distribution of charge)=Dipole-Dipole Forces; Molecules with H-F, H-O, or H-N (because of big ∆EN)=Hydrogen Bonding;
    Saltwater (ionic compound and water)=Ion-Dipole Forces. Since C2H5OH has O directly bonded to H somewhere in the molecule, based on the definition of hydrogen bonding, this qualifies as having hydrogen bonds.

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  • 3. 

    Determine the main type of intermolecular forces in pH3.

    • A.

      London Dispersion Forces

    • B.

      Ion-Dipole Forces

    • C.

      Dipole-Dipole Forces

    • D.

      Hydrogen Bonds

    Correct Answer
    C. Dipole-Dipole Forces
    Explanation
    Nonpolar Molecules (Symmetrical distribution of charge)=London Dispersion (weakest); Polar Molecules (Asymmetrical distribution of charge)=Dipole-Dipole Forces; Molecules with H-F, H-O, or H-N (because of big ∆EN)=Hydrogen Bonding; Saltwater (ionic compound and water)=Ion-Dipole Forces. Since PH3 is a polar molecule (it's asymmetrical overall, in terms of charge distribution) without H-F, H-O, or H-N, this qualifies as having dipole-dipole forces.

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  • 4. 

    Determine the main type of intermolecular forces in CCl4.

    • A.

      London Dispersion Forces

    • B.

      Ion-Dipole Forces

    • C.

      Dipole-Dipole Forces

    • D.

      Hydrogen Bonds

    Correct Answer
    A. London Dispersion Forces
    Explanation
    Nonpolar Molecules (Symmetrical distribution of charge)=London Dispersion (weakest); Polar Molecules (Asymmetrical distribution of charge)=Dipole-Dipole Forces; Molecules with H-F, H-O, or H-N (because of big ∆EN)=Hydrogen Bonding;
    Saltwater (ionic compound and water)=Ion-Dipole Forces. Since CCl4 is nonpolar (symmetrical distribution of charge overall), this qualifies as having London Dispersion Forces.

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  • 5. 

    Based on the ion shown, which side of water molecules should be touching the ion?

    • A.

      The partial negative end must be touching the Br^- ion, since like charges between partial negative end (O) in water and the negative ion attract.

    • B.

      The partial negative end must be touching the Br^- ion, since opposite charges between partial positive end (O) in water and the negative ion attract.

    • C.

      The partial positive end must be touching the Br^- ion, since opposite charges between partial positive end (H) in water and the negative ion attract.

    • D.

      None of the above

    Correct Answer
    C. The partial positive end must be touching the Br^- ion, since opposite charges between partial positive end (H) in water and the negative ion attract.
    Explanation
    The partial positive end of water molecules (H end) is attracted to the negative ion. The partial negative end of water molecules (O end) is attracted to the positive ion. This is because opposites attract.

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  • 6. 

    At standard pressure, NH3 has a boiling point of 240 K, AsH3 has a boiling point of 211 K, and pH3 has a boiling point of 186 K. What accounts for the higher boiling point of NH3?

    • A.

      NH3 has a higher boiling point, because it contains London forces.

    • B.

      NH3 has a higher boiling point, because it contains hydrogen bonds.

    • C.

      NH3 has a higher boiling point, because it contains dipole-dipole forces.

    • D.

      NH3 has a higher boiling point, because it contains ion-dipole forces.

    Correct Answer
    B. NH3 has a higher boiling point, because it contains hydrogen bonds.
    Explanation
    Nonpolar Molecules (Symmetrical distribution of charge)=London Dispersion (weakest); Polar Molecules (Asymmetrical distribution of charge)=Dipole-Dipole Forces; Molecules with H-F, H-O, or H-N (because of big ∆EN)=Hydrogen Bonding; Saltwater (ionic compound and water)=Ion-Dipole Forces (strongest). NH3 has H-N, so it has hydrogen bonding. AsH3 and PH3 are polar molecules (asymmetrical distribution of charge) without H-O, N-H, and H-F, so they have dipole-dipole forces. Hydrogen bonding is stronger than dipole-dipole forces are, so NH3 has the highest boiling point, because it has the strongest IMF's out of the three molecules.

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  • 7. 

    At standard pressure, HF has a higher boiling point than HCl, HBr, or H2 does. What accounts for the higher boiling point of HF?

    • A.

      HF has a higher boiling point, because it contains London forces.

    • B.

      HF has a higher boiling point, because it contains hydrogen bonds.

    • C.

      HF has a higher boiling point, because it contains dipole-dipole forces.

    • D.

      HF has a higher boiling point, because it contains ion-dipole forces.

    Correct Answer
    B. HF has a higher boiling point, because it contains hydrogen bonds.
    Explanation
    Nonpolar Molecules (Symmetrical distribution of charge)=London Dispersion (weakest); Polar Molecules (Asymmetrical distribution of charge)=Dipole-Dipole Forces; Molecules with H-F, H-O, or H-N (because of big ∆EN)=Hydrogen Bonding;
    Saltwater (ionic compound and water)=Ion-Dipole Forces (strongest). HF has H-F, so it has hydrogen bonding. HCl and HBr are polar molecules (asymmetrical distribution of charge) without H-F, H-N, or H-O, so they have dipole-dipole forces. H2 is a nonpolar molecule (symmetrical distribution of charge), so it has London forces. Hydrogen bonding in H-F is the strongest IMF (stronger than dipole-dipole and London forces), so HF will have the highest boiling point, because it has the strongest IMF's of the four molecules.

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  • 8. 

    Compounds like CuO and NaHCO3 are soluble in water. Compare the boiling point of a solution of CuO to the boiling point of water at standard pressure.

    • A.

      The solution of CuO has a higher boiling point than water does.

    • B.

      Water has a higher boiling point than the CuO solution does.

    Correct Answer
    A. The solution of CuO has a higher boiling point than water does.
    Explanation
    Ionic Compound + Water=Ion-Dipole Forces (strongest IMF's); H-O, H-N, H-F=Hydrogen Bonds (Weaker than Ion-dipole forces); CuO in water is an ionic compound in water, so it has ion-dipole forces. On the other hand, water has H-O, so it has hydrogen bonding. Ion-dipole has stronger IMF's and, therefore, a higher boiling point.

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  • 9. 

    A mixture contains hydrogen fluoride and ammonia. Identify a strong intermolecular force of attraction between hydrogen fluoride and ammonia.

    • A.

      Dipole-Dipole Forces

    • B.

      Hydrogen Bonding

    • C.

      Ion-Dipole Forces

    • D.

      London Dispersion Forces

    Correct Answer
    B. Hydrogen Bonding
    Explanation
    H-O, H-N, H-F=Hydrogen Bonds (Weaker than Ion-dipole forces). Both hydrogen fluoride (HF) and ammonia (NH3) contain H-F, H-O, or H-N bonds, so the IMF's between them are hydrogen bonds.

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  • 10. 

    Rank the following molecules in order of increasing boiling point: NaF (aq), H2S, HF, CO2.

    • A.

      NaF (aq) < H2S < HF < CF4

    • B.

      H2S < CO2 < HF < NaF (aq)

    • C.

      CO2 < H2S < HF < NaF (aq)

    • D.

      CO2 < HF < H2S < NaF (aq)

    Correct Answer
    C. CO2 < H2S < HF < NaF (aq)
    Explanation
    Nonpolar Molecules (Symmetrical distribution of charge)=London Dispersion (weakest IMF's, lowest boiling point); Polar Molecules (Asymmetrical distribution of charge)=Dipole-Dipole Forces; Molecules with H-F, H-O, or H-N (because of big ∆EN)=Hydrogen Bonding; Saltwater (ionic compound and water)=Ion-Dipole Forces (strongest IMF's, highest boiling point). CO2 is nonpolar (symmetrical distribution of charge), so it has London forces. H2S is a polar molecule (asymmetrical distribution of charge) without H-F, H-O, or H-N, so it has dipole-dipole forces. HF has F-H in its structure, so it has hydrogen bonding. NaF (aq) has an ionic compound, NaF, dissolved in water (aq), so it has ion-dipole forces. In order, London forces are weakest, dipole-dipole forces are stronger, hydrogen bonds are even stronger than dipole-dipole, and ion-dipole forces are strongest.

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  • 11. 

    Rank the following molecules in order of increasing boiling point: CH3Cl, CH3OH, Na2O (aq),  CF4.

    • A.

      CF4 < CH3OH < CH3Cl < Na2O (aq)

    • B.

      CF4 < CH3Cl < CH3OH < Na2O (aq)

    • C.

      CF4 < Na2O (aq) < CH3Cl < CH3OH

    • D.

      CF4 < Na2O (aq) < CH3OH < CH3Cl

    Correct Answer
    B. CF4 < CH3Cl < CH3OH < Na2O (aq)
    Explanation
    Nonpolar Molecules (Symmetrical distribution of charge)=London Dispersion (weakest IMF's, lowest boiling point); Polar Molecules (Asymmetrical distribution of charge)=Dipole-Dipole Forces; Molecules with H-F, H-O, or H-N (because of big ∆EN)=Hydrogen Bonding; Saltwater (ionic compound and water)=Ion-Dipole Forces (strongest IMF's, highest boiling point). CF4 is nonpolar (symmetrical distribution of charge), so it has London forces. CH3Cl is a polar molecule (asymmetrical distribution of charge) without H-F, H-O, or H-N, so it has dipole-dipole forces. CH3OH has O-H in its structure, so it has hydrogen bonding. Na2O (aq) has an ionic compound, Na2O, dissolved in water (aq), so it has ion-dipole forces. In order, London forces are weakest, dipole-dipole forces are stronger, hydrogen bonds are even stronger than dipole-dipole, and ion-dipole forces are strongest.

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  • 12. 

    Which element combines with hydrogen to form a compound with the strongest hydrogen bonding between its molecules?

    • A.

      Sulfur

    • B.

      Tellurium

    • C.

      Selenium

    • D.

      Oxygen

    Correct Answer
    D. Oxygen
    Explanation
    H-O, H-N, H-F=Hydrogen Bonds (Weaker than Ion-dipole forces). H and O makes an H-O bond, so the IMF's between the molecules are hydrogen bonding.

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  • 13. 

    Substance A has a boiling point of 13 degrees Celsius, Substance B has a boiling point of 20 degrees Celsius, Substance C has a boiling point of 23 degrees Celsius, Substance D has a boiling point of 12 degrees Celsius. Which substance has the strongest intermolecular forces?

    • A.

      Substance A

    • B.

      Substance B

    • C.

      Substance C

    • D.

      Substance D

    Correct Answer
    C. Substance C
    Explanation
    Higher Boiling Point=Stronger Intermolecular Forces.

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  • 14. 

    At standard pressure, HF and NH3 have higher boiling points than HBr and CH4 do. What accounts for the higher boiling point of HF?

    • A.

      NH3 and HF have higher boiling points, because they contain London Forces.

    • B.

      NH3 and HF have higher boiling points, because they contain hydrogen bonds.

    • C.

      NH3 and HF have higher boiling points, because they contain dipole-dipole forces.

    • D.

      NH3 and HF have higher boiling points, because they contain ion-dipole forces.

    Correct Answer
    B. NH3 and HF have higher boiling points, because they contain hydrogen bonds.
    Explanation
    Nonpolar Molecules (Symmetrical distribution of charge)=London Dispersion (weakest); Polar Molecules (Asymmetrical distribution of charge)=Dipole-Dipole Forces; Molecules with H-F, H-O, or H-N (because of big ∆EN)=Hydrogen Bonding;
    Saltwater (ionic compound and water)=Ion-Dipole Forces (strongest). NH3 and HF have hydrogen bonding because they have H-F/H-O/H-N bonded. HBr is polar, so it's just dipole-dipole. CH4 is nonpolar, so it's London Dispersion Forces. Hydrogen Bonds are stronger IMF's than dipole-dipole and London Forces are, so NH3 and HF have higher boiling points as a result.

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  • 15. 

    At standard pressure, H2Se and CO have higher boiling points than CF4 and CO2 do. What accounts for the higher boiling point of HF?

    • A.

      H2Se and CO have higher boiling points, because they contain London Forces.

    • B.

      H2Se and CO have higher boiling points, because they contain hydrogen bonds.

    • C.

      H2Se and CO have higher boiling points, because they contain dipole-dipole forces.

    • D.

      H2Se and CO have higher boiling points, because they contain ion-dipole forces.

    Correct Answer
    C. H2Se and CO have higher boiling points, because they contain dipole-dipole forces.
    Explanation
    Nonpolar Molecules (Symmetrical distribution of charge)=London Dispersion (weakest); Polar Molecules (Asymmetrical distribution of charge)=Dipole-Dipole Forces; Molecules with H-F, H-O, or H-N (because of big ∆EN)=Hydrogen Bonding
    Saltwater (ionic compound and water)=Ion-Dipole Forces (strongest). H2Se and CO have dipole-dipole forces because they are polar molecules without H-F/H-O/H-N. CF4 and CO2 are both nonpolar, so they have London Dispersion Forces. Dipole-Dipole Forces are stronger IMF's than London Forces are, so H2Se and CO have higher boiling points as a result.

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  • Mar 22, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Dec 07, 2015
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    Regentschemistry
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