VSEPR & Molecular Geometry practice questions

Question 1

Which of the following correctly identifies the molecular geometry and the hybridization of the central atom in a molecule of $\text{XeF}_4$?

Accepted Answer

Square planar and $sp^3d^2$

Answer

Square planar and $sp^3d$

Answer

Tetrahedral and $sp^3$

Answer

See-saw and $sp^3d$

Question 2

A student compares the bonding around each carbon atom in ethene, $\text{H}_2\text{C}=\text{CH}_2$, with the bonding around each carbon atom in ethyne, $\text{HC}\equiv\text{CH}$.

Which row correctly identifies the hybridization of each carbon atom in the two molecules?

Accepted Answer

Ethene carbon: $sp^2$; ethyne carbon: $sp$

Answer

Ethene carbon: $sp^2$; ethyne carbon: $sp^3$

Answer

Ethene carbon: $sp^3$; ethyne carbon: $sp$

Answer

Ethene carbon: $sp^3$; ethyne carbon: $sp^3$

Question 3

A student draws the Lewis diagram shown below for xenon tetrafluoride ($\text{XeF}_4$). Which of the following claims best describes the predicted $\text{F}-\text{Xe}-\text{F}$ bond angles in $\text{XeF}_4$ and provides the correct justification?

Accepted Answer

The $\text{F}-\text{Xe}-\text{F}$ bond angles are equal to $90^\circ$ because the two lone pairs on the central $\text{Xe}$ atom occupy positions $180^\circ$ apart, exerting symmetrical repulsions above and below the square plane.

Answer

The $\text{F}-\text{Xe}-\text{F}$ bond angles are less than $90^\circ$ because lone pair-bonding pair repulsions are stronger than bonding pair-bonding pair repulsions, compressing the bonds toward each other.

Answer

The $\text{F}-\text{Xe}-\text{F}$ bond angles are less than $90^\circ$ because six electron domains around the central $\text{Xe}$ atom create steric crowding that distorts the square planar geometry.

Answer

The $\text{F}-\text{Xe}-\text{F}$ bond angles are equal to $90^\circ$ because the central $\text{Xe}$ atom forms four equivalent bonds with zero lone pairs in its valence shell.

Question 4

Propyne, $\text{CH}_3\text{C}\equiv\text{CH}$, is an alkyne used as a specialty fuel and chemical building block. The structural formula of propyne is shown below. Which of the following correctly identifies the hybridization of the triply bonded carbon atoms and the total number of $\pi$ bonds in one molecule of propyne?

Accepted Answer

$sp$ hybridization and 2 $\pi$ bonds

Answer

$sp^2$ hybridization and 1 $\pi$ bond

Answer

$sp^2$ hybridization and 2 $\pi$ bonds

Answer

$sp$ hybridization and 3 $\pi$ bonds

Question 5

Sulfur hexafluoride, $\text{SF}_6$, is a dense, colorless gas widely used as an electrical insulator in high-voltage equipment. Based on VSEPR theory, which of the following best predicts the molecular geometry and ideal bond angles around the central sulfur atom in an $\text{SF}_6$ molecule?

Accepted Answer

Octahedral geometry with bond angles of $90^\circ$ and $180^\circ$

Answer

Seesaw geometry with bond angles of less than $90^\circ$ and less than $120^\circ$

Answer

Trigonal bipyramidal geometry with bond angles of $90^\circ$ and $120^\circ$

Answer

Square planar geometry with bond angles of $90^\circ$ and $180^\circ$

Question 6

A student uses molecular-modeling software to compare the arrangements of electron domains around the central atoms in three molecules.

| Molecule | Bonding domains | Lone-pair domains |
|---|---:|---:|
| $\text{CH}_4$ | 4 | 0 |
| $\text{NH}_3$ | 3 | 1 |
| $\text{H}_2\text{O}$ | 2 | 2 |

Which choice correctly ranks the average bond angles from greatest to least and explains the ranking?

Accepted Answer

$\text{CH}_4 > \text{NH}_3 > \text{H}_2\text{O}$, because repulsions involving lone-pair domains are stronger than bonding-domain–bonding-domain repulsions and compress the bond angles.

Answer

$\text{CH}_4 > \text{NH}_3 > \text{H}_2\text{O}$, because bonding domains occupy more space than lone-pair domains and therefore exert stronger repulsions.

Answer

$\text{H}_2\text{O} > \text{NH}_3 > \text{CH}_4$, because repulsions involving lone-pair domains are stronger and therefore increase the angles between the bonding domains.

Answer

$\text{NH}_3 > \text{CH}_4 > \text{H}_2\text{O}$, because the three bonding domains in $\text{NH}_3$ can spread farther apart than the four bonding domains in $\text{CH}_4$.

Question 7

A student uses a Lewis diagram and VSEPR theory to construct the molecular-domain model of $\text{SF}_4$ shown. Each $\text{SF}$ bond is polar because $\text{F}$ is more electronegative than $\text{S}$.

Which prediction about the molecular polarity of $\text{SF}_4$ is consistent with the model?

Accepted Answer

The $\text{SF}_4$ molecule is polar because its seesaw arrangement is asymmetric, so the four $\text{SF}$ bond-dipole vectors do not sum to zero.

Answer

The $\text{SF}_4$ molecule is nonpolar because its four identical $\text{SF}$ bond dipoles cancel in pairs.

Answer

The $\text{SF}_4$ molecule is nonpolar because a lone pair is not a bond dipole and therefore cannot affect molecular polarity.

Answer

The $\text{SF}_4$ molecule is polar because every molecule containing polar bonds has a nonzero molecular dipole.

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