Answer
See the explanation
Work Step by Step
Let's go through the given compounds and explain their polarity:
1. $CF_{4}$ (Carbon Tetrafluoride):
- $CF_{4}$ is a nonpolar compound because it has a tetrahedral molecular geometry with four equivalent C-F bonds.
- In a tetrahedral arrangement, the bond dipole moments cancel out due to the symmetry of the molecule, resulting in a net dipole moment of zero.
2. $XeF_{4}$ (Xenon Tetrafluoride):
- $XeF_{4}$ is also a nonpolar compound due to its square planar molecular geometry.
- The bond dipole moments in the square planar arrangement cancel out, leading to a net dipole moment of zero.
3. $SF_{4}$ (Sulfur Tetrafluoride):
- $SF_{4}$ is a polar compound because it has a distorted tetrahedral molecular geometry.
- In the distorted tetrahedral arrangement, the bond dipole moments do not cancel out completely, resulting in a net dipole moment.
4. $CO_{2}$ (Carbon Dioxide):
- $CO_{2}$ is a nonpolar compound because it has a linear molecular geometry.
- The two C=O bond dipole moments cancel out due to the symmetry of the molecule, leading to a net dipole moment of zero.
5. COS (Carbonyl Sulfide):
- COS is a polar compound because it has a bent molecular geometry.
- The bond dipole moments in the bent arrangement do not cancel out, resulting in a net dipole moment.
In summary:
- $CF_{4}$ and $XeF_{4}$ are nonpolar compounds because their molecular geometries (tetrahedral and square planar, respectively) result in the cancellation of bond dipole moments, leading to a net dipole moment of zero.
- $SF_{4}$ is a polar compound due to its distorted tetrahedral molecular geometry, where the bond dipole moments do not cancel out completely.
- $CO_{2}$ is a nonpolar compound because of its linear molecular geometry, where the bond dipole moments cancel out.
- COS is a polar compound due to its bent molecular geometry, where the bond dipole moments do not cancel out.
