Question

rank the elements or compounds in the table below in decreasing order of their boiling points. that is, choose 1 next to the substance with the highest boiling point, choose 2 next to the substance with the next highest boiling point, and so on.
substance
a: cacl₂
b: (lewis structure: o=n−cl with lone pairs)
c: cl₂
d: (lewis structure: c=c with four cl atoms)
boiling point
1(highest) 2 4(lowest) 3 (for d)
try again
your answer is incorrect.

Explanation:

Step1: Identify Bond Types and Forces

• Substance A (\(CaCl_2\)): Ionic compound (metal + non - metal). Ionic bonds are strong, so ionic compounds have high boiling points due to strong electrostatic forces between ions.
• Substance B (Lewis structure: \(O = N - Cl\)): Covalent compound, polar (due to electronegativity differences in \(O - N\) and \(N - Cl\) bonds). Has dipole - dipole interactions and London dispersion forces.
• Substance C (\(Cl_2\)): Diatomic non - polar molecule. Only London dispersion forces (LDF). LDF strength depends on molar mass (or number of electrons). Molar mass of \(Cl_2=71\ g/mol\).
• Substance D (Lewis structure: \(C_2Cl_4\)): Covalent compound, non - polar (symmetrical structure). Has London dispersion forces. Molar mass of \(C_2Cl_4 = 2\times12+4\times35.5=24 + 142 = 166\ g/mol\).

Step2: Compare Boiling Point Factors

• Ionic compounds (\(CaCl_2\)) have much higher boiling points than covalent compounds because ionic bonds are stronger than intermolecular forces (dipole - dipole, LDF).
• Among covalent compounds:
• Polar compounds (Substance B) have dipole - dipole interactions in addition to LDF, so stronger than non - polar with only LDF of similar molar mass. But Substance D has a much higher molar mass than Substance B and Substance C.
• Molar mass comparison: \(M_{D}=166\ g/mol\), \(M_{B}\approx16 + 14+35.5 = 65.5\ g/mol\), \(M_{C}=71\ g/mol\). So LDF strength: \(D > C > B\) (since for non - polar, higher molar mass means stronger LDF; Substance B is polar, but its molar mass is less than \(Cl_2\) and much less than \(C_2Cl_4\)).

Step3: Rank Boiling Points

• Highest boiling point: \(CaCl_2\) (ionic, strong forces) → rank 1.
• Next: \(C_2Cl_4\) (non - polar, high molar mass, strong LDF) → rank 2.
• Then: \(Cl_2\) (non - polar, molar mass \(71\), LDF) → rank 3. Wait, no, wait: Substance B is polar (\(ONCl\)) with molar mass ~65.5, \(Cl_2\) is 71 (non - polar). The polar molecule (\(ONCl\)) has dipole - dipole which is stronger than LDF of \(Cl_2\) (since molar mass difference is small, dipole - dipole can be stronger). Wait, let's recalculate molar mass of \(ONCl\): \(O(16)+N(14)+Cl(35.5)=65.5\ g/mol\), \(Cl_2 = 71\ g/mol\). The dipole - dipole in \(ONCl\) vs LDF in \(Cl_2\): dipole - dipole is generally stronger than LDF for molecules of similar molar mass. So \(ONCl\) (Substance B) has stronger intermolecular forces than \(Cl_2\) (Substance C). Then \(C_2Cl_4\) (Substance D) has molar mass 166, so LDF is much stronger than both B and C. So correct order:

1. \(CaCl_2\) (A)
2. \(C_2Cl_4\) (D)
3. \(ONCl\) (B)
4. \(Cl_2\) (C)

Wait, the original table has A: \(CaCl_2\) (rank 1), C: \(Cl_2\) (rank 4), and we need to find rank 2 and 3. So rank 2 should be D (\(C_2Cl_4\)) and rank 3 should be B (\(ONCl\)).

Answer:

For rank 2: D (the substance with Lewis structure \(C_2Cl_4\)), for rank 3: B (the substance with Lewis structure \(O = N - Cl\))