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.
table columns: substance (a: f₂, b: lewis structure (h-c-c-o-c-h with o double bond), c: lewis structure (n≡n-o with o lone pairs), d: lewis structure (h-c-c-n-c-h with o double bond and n lone pairs)), chemical symbol/formula/lewis structure, boiling point (select dropdowns: a has f₂, b has h-c-c-o-c-h structure, c has n≡n-o structure, d has h-c-c-n-c-h structure)

Explanation:

To solve the problem of ranking the substances (F₂, B, C, D) by boiling point, we analyze the intermolecular forces (IMFs) present in each:

Step 1: Identify IMFs for each substance

• Substance A (\( \boldsymbol{\ce{F2}} \)): Nonpolar diatomic molecule. Only London dispersion forces (LDF) (weakest IMF).
• Substance B: Contains a carbonyl (\( \ce{C=O} \)) and ether (\( \ce{C-O-C} \)) group. Polar molecule (due to electronegative O), so it has dipole-dipole interactions (stronger than LDF) and LDF.
• Substance C: Contains a triple bond (\( \ce{N≡N} \)) and a polar \( \ce{N-O} \) bond. Polar molecule, so dipole-dipole interactions and LDF. Also, \( \ce{N≡N} \) is nonpolar, but the \( \ce{N-O} \) bond creates polarity.
• Substance D: Contains a carbonyl (\( \ce{C=O} \)) and an amine (\( \ce{C-N} \)) group with a lone pair on N. Polar molecule, and the N atom can form hydrogen bonds (strongest IMF, since H is bonded to N, and O is electronegative) + dipole-dipole + LDF.

Step 2: Relate IMFs to boiling point

Boiling point is determined by IMF strength:

• Hydrogen bonding (D) > Dipole-dipole (B, C) > London dispersion (A).

Now compare B and C (both dipole-dipole):

• Substance B: Formula is \( \ce{CH3COOCH3} \) (ethyl acetate, molar mass ~74 g/mol).
• Substance C: Formula is \( \ce{N2O} \) (nitrous oxide, molar mass ~44 g/mol).

LDF strength increases with molar mass (more electrons, larger electron cloud). Since B has a higher molar mass than C, B has stronger LDF (in addition to dipole-dipole) than C.

Step 3: Final ranking (highest to lowest boiling point)

1. D (hydrogen bonding + dipole-dipole + LDF)
2. B (dipole-dipole + LDF, higher molar mass than C)
3. C (dipole-dipole + LDF, lower molar mass than B)
4. A (\( \ce{F2} \), only LDF)

Answer:

1. D
2. B
3. C
4. A

(If selecting the next highest after D, choose B; after B, choose C; after C, choose A.)