18. Which of the following Lewis electron-dot diagrams represents the molecule that is the most pola — Bonding Chemistry Question

STEPS:

1. Understand the dual factors governing molecular polarity:
A molecule's overall polarity (net dipole moment) depends on two fundamental concepts:
* Bond polarity: The electronegativity difference ($\Delta\chi$) between covalently bonded atoms, which determines the magnitude of charge separation ($\delta^+$ and $\delta^-$).
* Molecular geometry (shape): The spatial arrangement of those bonds (predicted by VSEPR theory), which determines whether the individual bond dipoles reinforce one another or cancel out symmetrically.

2. Analyze and eliminate symmetrical molecules (zero net dipole):
* Carbon disulfide ($\text{CS}_2$): Possesses a linear geometry ($180^\circ$ bond angle). Because it is completely symmetrical, the two polar $\text{C=S}$ bond dipoles pull in exactly opposite directions and completely cancel each other out. Thus, $\text{CS}_2$ has a net dipole moment of zero and is nonpolar.
* Boron trifluoride ($\text{BF}_3$): Possesses a trigonal planar geometry ($120^\circ$ bond angles). Despite having highly polar $\text{B–F}$ bonds, the three bond dipoles are arranged symmetrically in three directions and completely cancel each other out, making the molecule nonpolar.

3. Analyze the remaining asymmetric molecules (diatomic species):
* This leaves the two diatomic interhalogen compounds, $\text{ClF}$ and $\text{BrF}$.
* Diatomic molecules are inherently linear and asymmetrical, meaning their individual bond dipoles cannot be canceled by other bonds. Their molecular polarity is determined entirely by the polarity of their single covalent bond.

4. Compare electronegativity differences using periodic trends:
* Electronegativity measures an atom's ability to attract shared electrons in a covalent bond. Electronegativity increases going up and to the right on the periodic table.
* Fluorine ($\text{F}$) is the most electronegative element.
* Looking at Group 17 (the halogens), electronegativity decreases as you move down the group: $\text{F} > \text{Cl} > \text{Br}$. Therefore, bromine ($\text{Br}$) is less electronegative than chlorine ($\text{Cl}$).

5. Determine the molecule with the greatest dipole moment:
* Because $\text{Br}$ is less electronegative than $\text{Cl}$, the electronegativity difference between fluorine and bromine is greater than the difference between fluorine and chlorine:
$$\Delta\chi_{\text{Br–F}} > \Delta\chi_{\text{Cl–F}}$$
* This larger electronegativity difference in $\text{BrF}$ results in a more unequal sharing of electrons, producing larger partial charges ($\delta^+$ on $\text{Br}$ and $\delta^-$ on $\text{F}$) and a greater overall molecular dipole moment.
* Therefore, $\text{BrF}$ is the most polar molecule, matching Option B.

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WHY_OTHERS_WRONG:

• Option A is incorrect: While $\text{ClF}$ is a polar diatomic molecule, it is less polar than $\text{BrF}$. Because chlorine lies above bromine in Group 17, it is more electronegative than bromine. This makes the electronegativity difference in the $\text{Cl–F}$ bond smaller than that of the $\text{Br–F}$ bond, resulting in a smaller dipole moment.
• Option C is incorrect: Although the individual $\text{C=S}$ bonds are polar, the linear structural geometry of $\text{S=C=S}$ causes the two bond dipoles to point in opposite directions, canceling each other out to yield a nonpolar molecule.
• Option D is incorrect: While each of the three $\text{B–F}$ bonds is highly polar due to the large electronegativity difference between boron and fluorine, the symmetrical trigonal planar molecular geometry of $\text{BF}_3$ causes the three dipoles to completely cancel out, leaving the molecule nonpolar.