26. Which of the following best helps to explain why CCl4 is a liquid whereas CI4 is a solid when bo — Bonding Chemistry Question

STEPS:

1. Identify the molecular geometry and polarity of both molecules:
* Both carbon tetrachloride ($\text{CCl}_4$) and carbon tetraiodide ($\text{CI}_4$) consist of a central carbon atom covalently bonded to four halogen atoms.
* Based on VSEPR theory, both molecules possess a highly symmetrical tetrahedral geometry.
* Because of this perfect tetrahedral symmetry, the individual polar covalent bond dipoles ($\text{C–Cl}$ and $\text{C–I}$) pull symmetrically in opposite directions and completely cancel each other out.
* Therefore, both $\text{CCl}_4$ and $\text{CI}_4$ are nonpolar molecules with a net dipole moment of exactly zero.

2. Identify the primary intermolecular forces (IMFs) at play:
* Since both substances are nonpolar, they cannot undergo permanent dipole-dipole attractions or hydrogen bonding.
* Instead, the only intermolecular forces holding the molecules together in their condensed phases (liquid and solid) are London dispersion forces (LDFs).

3. Compare the factors that influence London dispersion forces:
* London dispersion forces arise from temporary, instantaneous dipoles created by the movement of electrons within a molecule's electron cloud.
* The strength of these temporary dipoles depends directly on the polarizability of the electron cloud—which is the ease with which the cloud can be distorted.
* Polarizability increases significantly with an increase in the size and total number of electrons in the electron cloud.

4. Compare the electron clouds of $\text{CCl}_4$ and $\text{CI}_4$:
* Iodine ($\text{I}$) is located much lower down Group 17 on the periodic table than chlorine ($\text{Cl}$) and possesses a much larger atomic radius and many more shell electrons.
* Consequently, the $\text{CI}_4$ molecule has a significantly larger and more polarizable electron cloud compared to the smaller $\text{CCl}_4$ molecule.

5. Relate IMF strength to the physical states observed at 25°C:
* Because $\text{CI}_4$ has a much more polarizable electron cloud, its London dispersion forces are dramatically stronger than those in $\text{CCl}_4$.
* At $25^\circ\text{C}$, the ambient thermal kinetic energy is high enough to overcome the weaker dispersion forces holding $\text{CCl}_4$ molecules together, allowing it to flow as a liquid.
* However, this same thermal energy is insufficient to overcome the much stronger dispersion forces in $\text{CI}_4$, keeping its molecules tightly locked in a structured lattice as a stable crystalline solid.
* This confirms Option D is the correct answer.

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WHY_OTHERS_WRONG:

• Option A is incorrect: This option falsely claims that $\text{CCl}_4$ has a larger dipole moment than $\text{CI}_4$. While chlorine is indeed more electronegative than iodine, the symmetrical tetrahedral arrangement causes the individual bond dipoles in both molecules to cancel out completely, yielding a net dipole moment of zero for both species.
• Option B is incorrect: This option falsely claims that $\text{CI}_4$ has a larger dipole moment than $\text{CCl}_4$. As nonpolar tetrahedral molecules, both have a net dipole moment of zero regardless of the relative electron repulsions in their individual bonds.
• Option C is incorrect: This option contains two fundamental chemical errors. First, it falsely claims that London dispersion forces are stronger in $\text{CCl}_4$ than in $\text{CI}_4$. Second, it incorrectly relates electronegativity to dispersion force strength; polarizability (which is larger in heavier halogens), not electronegativity, dictates the strength of London dispersion forces.