Solved: CHEMISTRY Question for NEET

NEET CHEMISTRYMedium

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

CH $_4$ > H $_2$ S > NH $_3$ > HF

H $_2$ S > NH $_3$ > HF > CH $_4$

NH $_3$ > HF > CH $_4$ > H $_2$ S

HF > NH $_3$ > H $_2$ S > CH $_4$

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.
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).
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).
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 $_3$ .
Conclusion: The decreasing order is HF > NH $_3$ > H $_2$ S > CH $_4$ .

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