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Allyl isocynide has

$ (a)\;9 \sigma \; and\; 4\pi\;bonds \\(b)\;8 \sigma \; and\; 5 \pi\;bonds \\ (c)\; 9 \sigma, 3 \pi \; and\; 2\;non-bonded\; electrons \\ (d)\; 8 \sigma, 3 \pi \; and\; 4\;non-bonded\; electrons $

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$CH_2 = CH = CH_2 - N \equiv C$
In it we find $5 \;C-H(\sigma\;bonds),1 \;C-C(\sigma\; bond), 1\;C-N(\sigma\; bond),1\;C=C(1 \sigma \;and \;1 \pi \;bonds), !\;N \equiv (1 \sigma \;and \;2 \pi \;bonds)$
ie a total of $9 \sigma \;and \;3 \pi \;bonds$
Two non- bonded electrons are also present on C-atom (of $N \equiv C:)$
$ 9 \sigma, 3 \pi \; and\; 2\;non-bonded\; electrons$
Hence c is the correct answer.
answered Feb 21, 2014 by meena.p

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