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Polarisation of electrons in acrolein may be written as :

$\begin{array}{1 1}(a)\;CH_2^{\delta -}=CH-CH^{\delta +}=O\\(b)\; CH_2^{\delta -}=CH-CH=O^{\delta +}\\(c)\; CH_2^{\delta -}=CH^{\delta +}-CH=O\\(d)\; CH_2^{\delta +}=CH-CH=O^{\delta -}\end{array}$

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$CH_2^{\delta +}=CH-CH=O^{\delta -}$
Because here,'O' is more electronegative ,hence it should take negative charge.
Hence (d) is the correct answer.
answered Mar 10, 2014 by sreemathi.v

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