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$PCl_5$ in solid state splits into :

$\begin{array}{1 1}(a)\;PCl_4^+\;and\;PCl_6^-&(b)\; PCl_6^+\;and\;PCl_4^-\\(c)\;PCl_4^+\;and\;Cl^-&(d)\;PCl_3\;and\;Cl_2\end{array}$

1 Answer

$2PCl_5\rightarrow PCl_4^++PCl_6^-$
Hence (a) is the correct answer.
answered Mar 5, 2014 by sreemathi.v
 

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