Q.
A metal $M$ reacts with sodium hydroxide to give a white precipitate $X$ which is soluble in excess of $NaOH$ to give $Y$. Compound $X$ is soluble in $HCl$ to form a compound $Z$. Identify $M, X, Y$ and $Z$.
M
X
Y
Z
(a)$\quad$
$Si$ $\quad$
$SiO_2$ $\quad$
$Na_2 SiO_3\quad$
$SiCl_4$ $\quad$
(b)
$Al$$\quad$
$Al(OH)_3$
$NaAlO_2$
$AlCl_3$ $\quad$
(c)
$Mg$$\quad$
$Mg(OH)_{3}$
$NaMgO_3$
$MgCl_2$ $\quad$
(d)
$Ca$$\quad$
$Ca(OH)_2$
$Na_2CO_3$
$NaHCO_3$ $\quad$
| M | X | Y | Z | |
|---|---|---|---|---|
| (a)$\quad$ | $Si$ $\quad$ | $SiO_2$ $\quad$ | $Na_2 SiO_3\quad$ | $SiCl_4$ $\quad$ |
| (b) | $Al$$\quad$ | $Al(OH)_3$ | $NaAlO_2$ | $AlCl_3$ $\quad$ |
| (c) | $Mg$$\quad$ | $Mg(OH)_{3}$ | $NaMgO_3$ | $MgCl_2$ $\quad$ |
| (d) | $Ca$$\quad$ | $Ca(OH)_2$ | $Na_2CO_3$ | $NaHCO_3$ $\quad$ |
The p-Block Elements
Solution: