$Zn + NaOH \rightarrow Na _2 ZnO _2 \text {. }$
Mercury $(II)$iodide dissolves in excess of potassium iodide solution due to formation of
$HgI _2+2 KI \rightarrow K _2 HgI _4$
Tin $(IV)$ chloride is made by
$Sn +2 Cl _2 \rightarrow SnCl _4$
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$\begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}\,\,\,\,\,\,\,} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\mathop C\limits_7 {H_3} - \mathop C\limits_6 - \mathop C\limits_5 H = \mathop C\limits_4 H - \mathop C\limits_3 H - \mathop C\limits_2 \equiv \mathop C\limits_1 H} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$
is in the following sequence
