Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: Each of the following options contains a set of four molecules. Identify the option(s) where all four molecules possess permanent dipole moment at room temperature.

The invisible world of molecules is constantly engaged in a microscopic tug-of-war. When atoms bond, they don't always share electrons equally. The more electronegative atom pulls the electron cloud towards itself, creating a bond dipole. But does this mean the entire molecule is polar? Not necessarily! The ultimate judge is symmetry.

In this problem, we are tasked with finding the options where all four molecules possess a permanent dipole moment. Let's break down the geometry and symmetry of each molecule using VSEPR theory.

Vector Addition

A molecule has a permanent dipole moment ($\mu eq 0$) if the vector sum of all its individual bond dipoles and lone pair moments is non-zero. If the molecule is perfectly symmetric, these vectors cancel out perfectly, resulting in a non-polar molecule ($\mu =0$).

The Symmetric Squad

Let's look at the first set: ${\text{BeCl}}_2$, ${\text{CO}}_2$, ${\text{BCl}}_3$, and ${\text{CHCl}}_3$.

- ${\text{BeCl}}_2$ and ${\text{CO}}_2$: Both of these molecules are perfectly linear. The two bond dipoles point in exactly opposite directions and cancel each other out. ($\mu =0$) - ${\text{BCl}}_3$: This molecule is trigonal planar. The three B-Cl bonds are separated by $120^{\circ }$, and their vector sum is zero. ($\mu =0$) - ${\text{CHCl}}_3$: While this molecule is tetrahedral, the presence of one hydrogen and three chlorines breaks the symmetry, making it polar.

Since ${\text{BeCl}}_2$, ${\text{CO}}_2$, and ${\text{BCl}}_3$ are non-polar, Option A is incorrect.

The Asymmetric Group

Now, let's examine the second set: ${\text{SO}}_2$, ${\text{C}}_6{\text{H}}_5\text{Cl}$, ${\text{H}}_2\text{Se}$, and ${\text{BrF}}_5$.

- ${\text{SO}}_2$: Sulfur dioxide has a lone pair on the central sulfur atom, giving it a bent shape. The bond dipoles do not cancel. ($\mu eq 0$) - ${\text{C}}_6{\text{H}}_5\text{Cl}$: Chlorobenzene has a single highly polar C-Cl bond that is not balanced by any opposite bond. ($\mu eq 0$) - ${\text{H}}_2\text{Se}$: Just like water, hydrogen selenide has a bent geometry due to two lone pairs on selenium. ($\mu eq 0$) - ${\text{BrF}}_5$: Bromine pentafluoride has a square pyramidal geometry with one lone pair. The asymmetry guarantees a net dipole moment. ($\mu eq 0$)

All four molecules are polar! Option B is a correct answer.

The Mixed Bag

Let's check the third set: ${\text{BF}}_3$, ${\text{O}}_3$, ${\text{SF}}_6$, and ${\text{XeF}}_6$.

- ${\text{BF}}_3$: As we saw with ${\text{BCl}}_3$, this is trigonal planar and perfectly symmetric. ($\mu =0$) - ${\text{SF}}_6$: Sulfur hexafluoride is perfectly octahedral. All six S-F bonds cancel each other out. ($\mu =0$)

Since we already found non-polar molecules, we don't need to analyze further. Option C is incorrect.

The Final Polar Set

Finally, let's look at the last set: ${\text{NO}}_2$, ${\text{NH}}_3$, ${\text{POCl}}_3$, and ${\text{CH}}_3\text{Cl}$.

- ${\text{NO}}_2$: Nitrogen dioxide has an unpaired electron, giving it a bent shape. ($\mu eq 0$) - ${\text{NH}}_3$: Ammonia has a lone pair on nitrogen, creating a trigonal pyramidal shape. The N-H bond dipoles add up to a net dipole. ($\mu eq 0$) - ${\text{POCl}}_3$ and ${\text{CH}}_3\text{Cl}$: Both have a tetrahedral electron geometry, but the attached atoms are not identical. This lack of symmetry means the bond dipoles cannot cancel. ($\mu eq 0$)

All four molecules in this set are polar! Option D is also a correct answer.

By carefully drawing the VSEPR structures and analyzing the symmetry, we can confidently conclude that the sets in Option (B) and Option (D) consist entirely of polar molecules. Always remember: symmetry is the ultimate deciding factor for molecular polarity!