$Na_2CO_3$ is prepared by the Solvay process, but $K_2CO_3$ cannot be prepared by the same process because:

Observe the following statements:
Statement-$I$: Both $LiF$ and $CsI$ have low solubility in water.
Statement-$II$: Low solubility of $LiF$ in water is due to its high lattice enthalpy, and that of $CsI$ is due to its smaller hydration enthalpy of ions.

In the presence of air,washing soda loses $9$ molecules of water to form a monohydrate: $Na_2CO_3 \cdot 10H_2O \xrightarrow{\text{In open air}} Na_2CO_3 \cdot H_2O + 9H_2O$. This process is known as:

$A$ metal $M$ on heating in nitrogen gas gives $Y$. $Y$ on treatment with $H_2O$ gives a colourless gas which when passed through $CuSO_4$ solution gives a blue colour. $Y$ is

Match the elements given in Column-$I$ with the colour they impart to the flame given in Column-$II$.

Column-$I$	Column-$II$
$A$. $Cs$	$1$. Apple green
$B$. $Na$	$2$. Violet
$C$. $K$	$3$. Brick red
$D$. $Ca$	$4$. Yellow
$E$. $Sr$	$5$. Crimson red
$F$. $Ba$	$6$. Blue

The atomic radii of the alkali metals follow the order