In comparison to the zeolite process for the removal of permanent hardness, the synthetic resin method is

We first recall what happens in the zeolite, also called the permutit, process. Natural or artificial zeolite is represented ideally by the formula $$\text{Na}_{2}\text{Z}$$ where $$\text{Z}^{2-}$$ is the aluminosilicate framework. When hard water containing $$\text{Ca}^{2+}$$ and $$\text{Mg}^{2+}$$ ions passes through the bed, the following cation‐exchange takes place:

$$\text{Na}_{2}\text{Z}\;+\;\text{Ca}^{2+}\;\longrightarrow\;\text{CaZ}\;+\;2\text{Na}^{+}$$

$$\text{Na}_{2}\text{Z}\;+\;\text{Mg}^{2+}\;\longrightarrow\;\text{MgZ}\;+\;2\text{Na}^{+}$$

So, in the zeolite method only cations are exchanged; the accompanying anions such as $$\text{HCO}_{3}^{-}$$, $$\text{SO}_{4}^{2-}$$, $$\text{Cl}^{-}$$ remain in solution. After the exchange, although $$\text{Ca}^{2+}$$ and $$\text{Mg}^{2+}$$ are removed, the water still contains the anions, now accompanied by harmless $$\text{Na}^{+}$$ ions. Thus the zeolite process softens water, but it does not produce completely de-ionised water.

Now we look at the synthetic resin, or ion-exchange, method. In this technique two different resins are employed consecutively:

(i) A cation-exchange resin, generally denoted $$\text{R}\!-\!\text{SO}_{3}\text{H}$$, reacts as

$$\text{R}\!-\!\text{SO}_{3}\text{H}\;+\;\text{M}^{n+}\;\longrightarrow\;\bigl(\text{R}\!-\!\text{SO}_{3}\bigr)_{n}\text{M}\;+\;n\text{H}^{+}$$

where $$\text{M}^{n+}$$ stands for all cations present (not only $$\text{Ca}^{2+}$$ and $$\text{Mg}^{2+}$$, but also $$\text{Na}^{+},\; \text{K}^{+},\; \text{Fe}^{3+}$$, etc.).

(ii) Immediately after that, the water passes through an anion-exchange resin, commonly written $$\text{R}\!-\!\text{N}\bigl(\text{CH}_{3}\bigr)_{3}\text{OH}$$, which functions according to

$$\text{R}\!-\!\text{N}\bigl(\text{CH}_{3}\bigr)_{3}\text{OH}\;+\;\text{A}^{-}\;\longrightarrow\;\text{R}\!-\!\text{N}\bigl(\text{CH}_{3}\bigr)_{3}\text{A}\;+\;\text{OH}^{-}$$

with $$\text{A}^{-}$$ representing all anions present ($$\text{Cl}^{-},\;\text{SO}_{4}^{2-},\;\text{HCO}_{3}^{-},\;\dots$$). The $$\text{H}^{+}$$ ions from the first resin and the $$\text{OH}^{-}$$ ions from the second resin combine to give $$\text{H}_{2}\text{O}$$, so the final output is nearly completely de-ionised water.

Because the synthetic resin method eliminates both cations and anions, the degree of purification it achieves is much higher than that of the zeolite process, which eliminates only cations. Therefore we say that, in comparison with the zeolite process, the synthetic resin method is more efficient since it can exchange both kinds of ions.

The statement that exactly conveys this idea among the given choices is:

“more efficient as it can exchange both cations as well as anions.”

Hence, the correct answer is Option B.