As Sb$$_2$$S$$_3$$ is a negative sol, so, Al$$_2$$(SO$$_4$$)$$_3$$ will be the most effective coagulant due to higher charge density on Al$$^{3+}$$ in accordance with Hardy-Schulze rule.

Order of effectiveness of cations: Al$$^{3+}$$ > Ca$$^{2+}$$ > Na$$^{+}$$ > NH$$^{+}_{4}$$