Among the following compounds, the one which shows highest dipole moment is

The dipole moment $$\mu$$ of a hetero-atomic bond is the product of the magnitude of charge separation $$q$$ and the bond length $$r$$, i.e. $$\mu = q\,r$$ (expressed in Debye, D).
A larger electronegativity difference increases $$q$$, while a longer bond increases $$r$$. Hence, to compare dipole moments we must examine the combined effect of both factors.

For the tetrahedral molecules $$CH_3X$$ (where $$X = F, Cl, Br, I$$) the three nearly coplanar C-H bonds have dipoles that almost cancel one another. The net molecular dipole therefore points almost entirely along the C-X bond. Consequently,

$$\mu(CH_3X) \approx \mu(\text{C-}X) = q_{\,\text{C-}X}\; r_{\,\text{C-}X} \quad -(1)$$

Let us analyse both factors moving down the halogen group:

Electronegativity difference (controls $$q$$)
$$\Delta\chi(\text{C-}X):\; F \gt Cl \gt Br \gt I$$
Thus $$q_{\,\text{C-}F} \gt q_{\,\text{C-}Cl} \gt q_{\,\text{C-}Br} \gt q_{\,\text{C-}I}$$.

Bond length (controls $$r$$)
C-F (1.39 Å) < C-Cl (1.78 Å) < C-Br (1.94 Å) < C-I (2.14 Å)

Putting the two trends together:

• From F → Cl the decrease in $$q$$ is modest, whereas the increase in $$r$$ is substantial (≈0.39 Å).
• From Cl → Br → I both factors ($$q$$ decreases and $$r$$ increases only slightly) lower the product $$q\,r$$.

Experimental dipole moments corroborate this assessment:

• $$\mu(CH_3F) = 1.85\;\text{D}$$
• $$\mu(CH_3Cl) = 1.87\;\text{D}$$
• $$\mu(CH_3Br) = 1.79\;\text{D}$$
• $$\mu(CH_3I) = 1.64\;\text{D}$$

The slight fall in charge separation from F to Cl is more than compensated by the longer C-Cl bond, giving $$CH_3Cl$$ the highest dipole moment of the four molecules.

Hence, among the compounds listed, the molecule with the greatest dipole moment is $$\mathbf{CH_3Cl}$$.

Option A which is: $$CH_3Cl$$