In context with the industrial preparation of hydrogen from water gas $$(CO + H_2)$$, which of the following is the correct statement?

The mixture obtained after passing steam over red‐hot coke is called water gas and has the composition $$CO + H_2$$. To use this stream as an industrial source of pure hydrogen, the carbon monoxide must be removed. The standard industrial route is the water-gas shift reaction.

Step 1 : Water-gas shift reaction
$$CO + H_2O \;(\text{steam}) \;\xrightarrow{\text{Fe-Cr or Cu-Zn catalyst, 400 °C}}\; CO_2 + H_2$$
The catalyst converts the poisonous CO into $$CO_2$$ while producing one extra mole of $$H_2$$.

Step 2 : Removal of $$CO_2$$
The resulting mixture $$(CO_2 + H_2)$$ is passed through an alkaline solution such as $$NaOH$$, $$K_2CO_3$$ or monoethanolamine, where $$CO_2$$ dissolves/gets neutralised:
$$CO_2 + 2\,NaOH \rightarrow Na_2CO_3 + H_2O$$
Hydrogen, being almost insoluble, escapes and is collected as the final pure product.

Hence, the correct industrial procedure is described by Option D.

Why the other statements are incorrect

• Option A: CO and $$H_2$$ have very similar densities (1.25 g L⁻¹ vs 0.09 g L⁻¹ at STP), so fractional separation is impractical on an industrial scale.
• Option B: CO can indeed form a complex with $$CuCl$$ in concentrated $$HCl$$, but this method is restricted to laboratory analysis of CO, not bulk industrial purification.
• Option C: Palladium can occlude (absorb) large amounts of $$H_2$$, yet Pd is far too expensive for tonnage production; moreover, desorption and recovery steps are uneconomical.

Therefore, the only viable industrial sequence is that given in Option D.

Final answer: Option D which is: CO is oxidised to $$CO_2$$ with steam in the presence of a catalyst followed by absorption of $$CO_2$$ in alkali.