The following are contributors to the resonance hybrid of propenal: (I) CH2=CH-CH=O, (II) ^+CH2-CH=CH-O^-, (III) ^-CH2-CH=CH-O^+ (carbon anion, oxygen cation, written with the oxygen sextet). Arrange them in decreasing order of contribution (stability).

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The following are contributors to the resonance hybrid of propenal: (I) $CH X_{2} = CH - CH = O$ , (II) $X_{+} X_{2}^{2 +} CH X_{2} - CH = CH - O X^{-}$ , (III) $X_{-} X_{2}^{2 -} CH X_{2} - CH = CH - O X^{+}$ (carbon anion, oxygen cation, written with the oxygen sextet). Arrange them in decreasing order of contribution (stability).

A $I > II > III$
B $III > II > I$
C $I > III > II$
D $II > I > III$

Answer: A

Structure I has no charge separation and all atoms have octets, so it contributes the most. Structure II separates charge but puts the negative charge on the electronegative oxygen, so it contributes moderately. Structure III places a negative charge on carbon and a positive charge on oxygen, with oxygen having only a sextet, which is the least favourable. The order is I > II > III.

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Answer: A

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