Match the words in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer. Reset Help hydrogen bonding influences Given the molecules phosgene (Cl2 CO) and formaldehyde (H2CO). has the higher reduced polarity and strongerboiling point mainly due to dispersion forces phosgene (C2 CO) its greater molar mass and Given the molecules sulfur dioxide (SO2) and sulfur trioxide (SOs). has the higher boiling point mainly due to stronger dispersion forces sulfur dioxide (SO2) molecule shape and stronger dipole-dipole interactions sulfur trioxide (SOs) formaldehyde (H2CO)
Solution
1. Phosgene has the higher boiling point than formaldehyde. This is mainly due to phosgene\'s (Cl2CO) greater molar mass and stronger dispersion forces.
Explanation:
For molecules that do not participate in hydrogen bonding, the majority of the attraction between those molecules is due to London dispersion forces. Even though dipole-dipole interactions are stronger, they apply to specific orientations of the interacting molecules. The ability to form an instantaneous dipole is referred to as polarizability. Molecules that are more polarizable, owing to their size and number of electrons, experience greater intermolecular attractions because those interactions are much less dependent on the molecular orientations.
2. Sulfur trioxide (SO3) has a higher boiling point than sulfur dioxide (SO2) mainly due to its greater molar mass and stronger dispersion forces.
Explanation:
There can be multiple types of interactions that occur between molecules, but comparing the boiling points of substances should mostly be based on polarizability (with the exception of hydrogen bonding). Both sulfur trioxide and phosgene would experience much greater attractions due to London dispersion forces than sulfur dioxide and formaldehyde, respectively. The contributions of dipole-dipole interactions in sulfur dioxide and formaldehyde are too insignificant in both cases.
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Match the words in the left column to the appropriate blanks in the se.docx
1. Match the words in the left column to the appropriate blanks in the sentences on the right. Make
certain each sentence is complete before submitting your answer. Reset Help hydrogen bonding
influences Given the molecules phosgene (Cl2 CO) and formaldehyde (H2CO). has the higher
reduced polarity and strongerboiling point mainly due to dispersion forces phosgene (C2 CO) its
greater molar mass and Given the molecules sulfur dioxide (SO2) and sulfur trioxide (SOs). has
the higher boiling point mainly due to stronger dispersion forces sulfur dioxide (SO2) molecule
shape and stronger dipole-dipole interactions sulfur trioxide (SOs) formaldehyde (H2CO)
Solution
1. Phosgene has the higher boiling point than formaldehyde. This is mainly due to phosgene's
(Cl2CO) greater molar mass and stronger dispersion forces.
Explanation:
For molecules that do not participate in hydrogen bonding, the majority of the attraction between
those molecules is due to London dispersion forces. Even though dipole-dipole interactions are
stronger, they apply to specific orientations of the interacting molecules. The ability to form an
instantaneous dipole is referred to as polarizability. Molecules that are more polarizable, owing
to their size and number of electrons, experience greater intermolecular attractions because those
interactions are much less dependent on the molecular orientations.
2. Sulfur trioxide (SO3) has a higher boiling point than sulfur dioxide (SO2) mainly due to its
greater molar mass and stronger dispersion forces.
Explanation:
There can be multiple types of interactions that occur between molecules, but comparing the
boiling points of substances should mostly be based on polarizability (with the exception of
hydrogen bonding). Both sulfur trioxide and phosgene would experience much greater attractions
due to London dispersion forces than sulfur dioxide and formaldehyde, respectively. The
contributions of dipole-dipole interactions in sulfur dioxide and formaldehyde are too
insignificant in both cases.
