Identification of Sets of Molecules Possessing Permanent Dipole Moments

To determine which set(s) contain only molecules with a non-zero permanent dipole moment ($\mu \neq 0$), we analyze the molecular geometry and symmetry of each molecule in the given options:

Analysis of Option (A):

1. $\text{BeCl}_2$: Linear geometry ($\text{sp}$ hybridization) with symmetrical distribution of charge. The bond dipoles cancel each other out ($\mu = 0$).
2. $\text{CO}_2$: Linear geometry ($\text{sp}$ hybridization). Symmetrical structure leads to cancellation of dipole moments ($\mu = 0$).
3. $\text{BCl}_3$: Trigonal planar geometry ($\text{sp}^2$ hybridization). Symmetrical structure leads to cancellation of dipole moments ($\mu = 0$).
4. $\text{CHCl}_3$: Distorted tetrahedral geometry ($\text{sp}^3$ hybridization). Asymmetrical bond polarities result in a non-zero net dipole moment ($\mu \neq 0$).

Since three of the molecules have a zero dipole moment, Option (A) is incorrect.

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Analysis of Option (B):

1. $\text{NO}_2$: Bent (angular) geometry due to an odd, unpaired electron on nitrogen. Asymmetric distribution of charge gives $\mu \neq 0$.
2. $\text{NH}_3$: Trigonal pyramidal geometry ($\text{sp}^3$ hybridization with one lone pair). Net dipole moment points towards the lone pair ($\mu \neq 0$).
3. $\text{POCl}_3$: Distorted tetrahedral structure ($\text{sp}^3$ hybridization with a double-bonded oxygen). Difference in electronegativities and bond polarities gives $\mu \neq 0$.
4. $\text{CH}_3\text{Cl}$: Distorted tetrahedral structure ($\text{sp}^3$ hybridization). Asymmetry in bond dipoles gives $\mu \neq 0$.

All four molecules possess a non-zero permanent dipole moment. Option (B) is correct.

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Analysis of Option (C):

1. $\text{BF}_3$: Trigonal planar geometry ($\text{sp}^2$ hybridization). Symmetrical vectors cancel out ($\mu = 0$).
2. $\text{O}_3$: Bent geometry due to a lone pair on the central oxygen atom ($\mu \neq 0$).
3. $\text{SF}_6$: Regular octahedral geometry ($\text{sp}^3\text{d}^2$ hybridization). Highly symmetrical, so individual $S-F$ bond dipoles cancel completely ($\mu = 0$).
4. $\text{XeF}_6$: Distorted octahedral geometry (capped octahedron) due to the presence of a stereochemically active lone pair ($\mu \neq 0$).

Since $\text{BF}_3$ and $\text{SF}_6$ have $\mu = 0$, Option (C) is incorrect.

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Analysis of Option (D):

1. $\text{SO}_2$: Bent (V-shaped) geometry ($\text{sp}^2$ hybridization with one lone pair). Asymmetrical geometry gives $\mu \neq 0$.
2. $\text{C}_6\text{H}_5\text{Cl}$: Chlorobenzene has a polar $C-Cl$ bond and an asymmetrical structure ($\mu \neq 0$).
3. $\text{H}_2\text{Se}$: Bent (V-shaped) geometry ($\text{sp}^3$ hybridization with two lone pairs). Asymmetrical structure gives $\mu \neq 0$.
4. $\text{BrF}_5$: Square pyramidal geometry ($\text{sp}^3\text{d}^2$ hybridization with one lone pair). Net axial dipole moment gives $\mu \neq 0$.

All four molecules possess a non-zero permanent dipole moment. Option (D) is correct.

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Conclusion:

The options containing sets of molecules where all four possess a non-zero permanent dipole moment at room temperature are B and D.