The dipole moments of CCl4, CHCl3 and CH4 are in the order

🧠 Compare dipole moments of three tetrahedral molecules with different substituents.

🗺️ Analyse each molecule:

$\mathrm{CH_4}$: Tetrahedral with four identical C–H bonds → all bond dipoles cancel by symmetry → μ = 0

$\mathrm{CCl_4}$: Tetrahedral with four identical C–Cl bonds → all bond dipoles cancel by symmetry → μ = 0

$\mathrm{CHCl_3}$: Tetrahedral with one C–H and three C–Cl bonds → asymmetric → bond dipoles do not cancel → μ ≠ 0 (≈ 1.01 D)

🗺️ Order:

$\mathrm{CH_4 = CCl_4 < CHCl_3}$

⚡ The substitution of one H by Cl (or one Cl by H) in an otherwise symmetric tetrahedron breaks the symmetry and creates a net dipole. Three C–Cl bonds pulling toward Cl and one C–H bond pulling oppositely gives a net resultant ≠ 0.

$\boxed{\text{Answer: } \mathrm{CH_4 = CCl_4 < CHCl_3}}$