The coagulating power of electrolytes having ions Na$$^+$$, Al$$^{3+}$$ and Ba$$^{2+}$$ for arsenic sulphide sol increases in the order,

First, we recall the Hardy-Schulze rule, which says:

“The coagulating (precipitating) power of an electrolyte for a given lyophobic colloid is directly proportional to the valency of the ion that has a charge opposite to that of the colloidal particles.”

Arsenic sulphide sol, $$\text{As}_2\text{S}_3$$, is formed by passing hydrogen sulphide through an arsenious acid solution. The colloidal particles acquire sulphide ions on their surface, so the sol particles carry a negative charge. Therefore, the ions that will neutralise this negative charge and bring about coagulation are the positively charged ions (cations) present in the electrolytes we are comparing.

We have three cations:

$$\text{Na}^+,\qquad \text{Ba}^{2+},\qquad \text{Al}^{3+}$$

Their valencies are:

$$\text{Na}^+ : +1,\qquad \text{Ba}^{2+} : +2,\qquad \text{Al}^{3+} : +3$$

Applying the Hardy-Schulze rule, a cation with a higher positive charge neutralises the negative charge on the colloidal particles more efficiently, so its coagulating power is greater. Hence the order of coagulating power must increase with valency:

$$+1\;(\text{Na}^+) \;<\; +2\;(\text{Ba}^{2+}) \;<\; +3\;(\text{Al}^{3+})$$

Writing this explicitly, we obtain:

$$\text{Na}^+ \;<\; \text{Ba}^{2+} \;<\; \text{Al}^{3+}$$

This sequence corresponds to Option D in the given list.

Hence, the correct answer is Option D.