The correct order of dipole moments for molecules NH$$_3$$, H$$_2$$S, CH$$_4$$ and HF is:

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1. **CH$$_4$$ (Methane):** It has a symmetrical tetrahedral geometry. The vector sum of the four C-H bond moments cancels out completely, resulting in a net dipole moment of **0 D**.
2. **HF (Hydrogen Fluoride):** Fluorine is the most electronegative element (EN = 4.0). The electronegativity difference between H and F is very high, and the molecule is linear. This results in the highest dipole moment among the group ($$\approx 1.78$$ D).
3. **NH$$_3$$ (Ammonia):** Nitrogen is more electronegative (EN = 3.04) than Sulphur. The molecule is pyramidal. The orbital dipole of the lone pair reinforces the resultant dipole of the N-H bonds, leading to a high dipole moment ($$\approx 1.47$$ D).
4. **H$$_2$$S (Hydrogen Sulphide):** Sulphur (EN = 2.58) is less electronegative than Nitrogen and Fluorine. While bent, the lower electronegativity difference results in a smaller dipole moment ($$\approx 0.95$$ D) compared to HF and NH$$_3$$.
5. **Conclusion:** The decreasing order is HF > NH$$_3$$ > H$$_2$$S > CH$$_4$$.

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