A metal \(M\) reacts with sodium hydroxide to give a white precipitate \(X\) which
is soluble in excess of \(\mathrm{NaOH}\) to give \(Y\). Compound \(X\) is soluble
in \(\mathrm{HCl}\) to form a compound \(Z\). Identify \(M, X, Y\) and \(Z\).
$$\begin{array}{lllc}
\boldsymbol{M} & {\boldsymbol{X}} & {\mathbf{Y}} & \mathbf{Z} \\
(a)\mathrm{Si} & \mathrm{SiO}_{2} & \mathrm{Na}_{2} \mathrm{SiO}_{3} &
\mathrm{SiCl}_{4} \\
(b)\mathrm{Al} & \mathrm{Al}(\mathrm{OH})_{3} & \mathrm{NaAlO}_{2} &
\mathrm{AlCl}_{3} \\
(c)\mathrm{Mg} & \mathrm{Mg}(\mathrm{OH})_{3} & \mathrm{NaMgO}_{3} &
\mathrm{MgCl}_{2} \\
(d)\mathrm{Ca} & \mathrm{Ca}(\mathrm{OH})_{2} & \mathrm{Na}_{2}
\mathrm{CO}_{3} & \mathrm{NaHCO}_{3}
\end{array}$$