The one that is NOT suitable for the removal of permanent hardness of water is:

Water is said to be hard when it contains appreciable amounts of the ions $$\mathrm{Ca^{2+}}$$ and $$\mathrm{Mg^{2+}}$$. These ions can be present either as their bicarbonates or as their chlorides / sulphates.

We classify hardness into two types:

• Temporary hardness: caused mainly by $$\mathrm{Ca(HCO_3)_2}$$ and $$\mathrm{Mg(HCO_3)_2}$$.
• Permanent hardness: caused by $$$\mathrm{CaCl_2,\; CaSO_4,\; MgCl_2,\; MgSO_4}$$$ and similar salts.

Let us recall what each of the given treatments does.

Clark’s method (also called the lime-soda cold-lime process) removes temporary hardness by adding calculated amounts of lime, $$\mathrm{Ca(OH)_2}$$. We first state the reactions used:

$$$\mathrm{Ca(HCO_3)_2 + Ca(OH)_2 \;\longrightarrow\; 2\,CaCO_3 \downarrow + 2\,H_2O}$$$

$$$\mathrm{Mg(HCO_3)_2 + 2\,Ca(OH)_2 \;\longrightarrow\; 2\,CaCO_3 \downarrow + Mg(OH)_2 \downarrow + 2\,H_2O}$$$

The insoluble $$\mathrm{CaCO_3}$$ and $$\mathrm{Mg(OH)_2}$$ precipitate and can be filtered off, so the bicarbonate (temporary) hardness is removed. However, chlorides and sulphates of calcium and magnesium do not react with lime in this way, so permanent hardness remains untreated by Clark’s method.

Ion-exchange (zeolite) method works for both kinds of hardness. In the sodium-zeolite bed, the exchange reaction is

$$$\mathrm{Na_2Z + Ca^{2+} \;\longrightarrow\; CaZ + 2\,Na^{+}}$$$

which removes $$\mathrm{Ca^{2+}}$$ and $$\mathrm{Mg^{2+}}$$ regardless of their accompanying anions. Hence it is perfectly suitable for removing permanent hardness.

Calgon’s method uses sodium hexametaphosphate, $$\mathrm{(NaPO_3)_6}$$, popularly called Calgon. The anion $$\mathrm{[Na_2P_6O_{18}]^{2-}}$$ complexes $$\mathrm{Ca^{2+}}$$ and $$\mathrm{Mg^{2+}}$$:

$$$\mathrm{[P_6O_{18}]^{6-} + Ca^{2+} \;\longrightarrow\; [CaP_6O_{18}]^{4-}}$$$

This sequestration keeps the metal ions in solution but non-interfering, thus removing permanent hardness effectively.

Treatment with sodium carbonate (washing soda) also removes permanent hardness. Stating the precipitation reactions:

$$\mathrm{Ca^{2+} + CO_3^{2-} \;\longrightarrow\; CaCO_3 \downarrow}$$

$$\mathrm{Mg^{2+} + CO_3^{2-} \;\longrightarrow\; MgCO_3 \downarrow}$$

The insoluble carbonates settle out, and therefore chlorides and sulphates of $$\mathrm{Ca^{2+}}$$/$$\mathrm{Mg^{2+}}$$ are eliminated.

We can now compare:

• Clark’s method ⇒ removes only temporary hardness.
• The other three methods ⇒ remove permanent hardness very effectively.

Because the question asks for the method NOT suitable for removing permanent hardness, Clark’s method is the odd one out.

Hence, the correct answer is Option A.