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# $CF_2Cl_2+hv\rightarrow CF_2Cl+A,CFCl_3+hv\rightarrow CFCl_2+B$.what is A?

$\begin{array}{1 1}(a)\;Cl\\(b)\;Cl_2\\(c)\;Cl_3\\(d)\;FCl\end{array}$

$CF_2Cl_2+hv\rightarrow CF_2Cl+Cl$
Hence A is $Cl$
Hence (a) is the correct option.