Write the resonance structures of CH 2 = CH – CHO and arrange them in order of decreasing stability.
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Structure (I) is most stable since each C and O atom has an octet of electrons and none of these atoms carries any charge. Structures (II and III) both involve separation of charge and hence both are less stable than structure (I). However, structure (II is more stable than structure (III) since it carries a – ve charge on the more electronegative O atom and + ve charge on the less electronegative C atom while in structure (III), the more electronegative O atom carries the + ve charge while the less electronegative C atom carries the – ve charge. Thus, the decreasing order of stability is I > II > III.
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