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Read the passage on resonance: delocalisation of pi and lone-pair electrons (atoms fixed in position) stabilises a molecule; a resonance form is less stable when an electronegative atom carries a positive charge or when like charges sit on adjacent atoms. For vinyl chloride the three resonating forms are (I) CH2=CH-Cl with a lone pair on Cl, (II) the carbanion-type form:CH2-CH=Cl, and (III) the carbocation-type form (+)CH2-CH=Cl. Arrange them in decreasing order of stability.

1. I > II > III
2. II > III > I
3. III > II > I
4. I > III > II

Correct answer: I > II > III

Solution

Form I is neutral and uncharged, hence most stable. Forms II and III both involve C=Cl with chlorine bearing a positive charge (less favourable). Form II places the negative charge on carbon while III places a positive charge on carbon and is the least stable contributor here, giving I > II > III.

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