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The number of $\sigma$ and $\pi$-bonds between two carbon atoms in $CaC_2$ is

$\begin{array}{1 1}(a)\;\text{one }\pi\text{-bond and one }\sigma\text{ bond}\\(b)\;\text{two }\sigma\text{-bonds and one }\pi\text{ bond}\\(c)\;\text{two }\pi\text{-bonds and one }\sigma\text{ bond}\\(d)\;\text{three }\sigma\text{-bonds and no }\pi\text{ bond}\end{array}$

1 Answer

$CaC_2\rightarrow Ca^{2+}+C_2^{2-}$
$[C\equiv C]^{2-}$ (i.e) Two $\pi$-bonds and $\sigma$ -bond.
Hence (c) is the correct answer.
answered Mar 6, 2014 by sreemathi.v
 

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