Only $ [Cr{{(N{{H}_{3}})}_{5}}Cl]\,S{{O}_{4}} $ give sulphate ions $ (SO_{4}^{2-}) $ in solution, hence only it will give white precipitate of barium sulphate with $ BaC{{l}_{2}}(aq). $ $ [Cr{{(N{{H}_{3}})}_{5}}Cl]\,S{{O}_{4}}+BaC{{l}_{2}}\xrightarrow{{}} $ $ [Cr{{(N{{H}_{3}})}_{5}}Cl]\,C{{l}_{2}}+\underset{\text{White}}{\mathop{BaS{{O}_{4}}\downarrow }}\, $