(A) The reaction between $SnO_2$ and excess $HCl$ is as follows:$SnO_2 + 4HCl \rightarrow SnCl_4 + 2H_2O$Since $HCl$ is in excess,$SnCl_4$ reacts furt

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(A) The reaction between $SnO_2$ and excess $HCl$ is as follows:$SnO_2 + 4HCl \rightarrow SnCl_4 + 2H_2O$Since $HCl$ is in excess,$SnCl_4$ reacts further with $HCl$ to form a complex anion:$SnCl_4 + 2HCl \rightarrow H_2[SnCl_6]$This dissociates to give $[SnCl_6]^{2-}$ ions in the solution.