The Permanent hardness of water is due to compound $A$ . One of the method to remove permanent hardness of water is $B$ whereas the temporary hardness of water is due to compound $X$ . Boiling this compound converts $X$ to compound $Y$ . $A, B, X, and Y,$ respectively can be:
(Given: $M = M g, C a$ )

M C l_{2}

, by boiling method,

C a (H C O_{3})_{2}

and

C a (O H)_{2}

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M C l_{2}

, Calgon's method,

M g (H C O_{3})_{2}

and

M g (O H)_{2}

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M (H C O_{3})_{2}

, ion- exchange method,

M g (H C O_{3})_{2}

and

M g C O_{3}

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M S O_{4}

, Clark's method,

C a (H C O_{3})_{2}

and

C a O

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Solution

The correct option is B $M C l_{2}$ , Calgon's method, $M g (H C O_{3})_{2}$ and $M g (O H)_{2}$
The permanent hardness of water is due to soluble salts of magnesium and calcium in the form of chlorides and sulphates in water.

One of the methods to remove the permanent hardness of water is Calgon's method. In this method, sodium hexametaphosphate, also known as Calgon, is added to hard water and the following reaction takes place:
$N_{6} P_{6} O_{18} \to 2 N a^{+} + N a_{4} P_{6} O_{18}^{2 -} M^{2 +} + N a_{4} P_{6} O_{18}^{2 -} \to [N a_{2} M P_{6} O_{18}]^{2 -} + 2 N a^{+}$
Where, $M = M g, C a$
The complex anion keeps the $M g^{2 +}$ and $C a^{2 +}$ ions in the solution.

Temporary hardness of water is due to $M g (H C O_{3})_{2}$ and $C a (H C O_{3})_{2}$ .

During boiling, the soluble $M g (H C O_{3})_{2}$ is converted into insoluble $M g (O H)_{2}$ and $C a (H C O_{3})_{2}$ is converted into insoluble $C a C O_{3}$ . These precipitates can be removed by filtration. The filtrate thus obtained will be soft water.
$M g (H C O_{3})_{2} h e a t i n g - --- \to M g (O H)_{2} ↓ + 2 C O_{2} ↑$
$C a (H C O_{3})_{2} h e a t i n g - --- \to C a C O_{3} ↓ + H_{2} O + 2 C O_{2} ↑$

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