98: Lanthanum metal forms cations with a charge of 3+. Consider the following observations about the chemistry of lanthanum: When lanthanum metal is exposed to air, a white solid (compound A) is formed that contains lanthanum and one other element. When lanthanum metal is added to water, gas bubbles are observed and a different white solid (compound B) is formed. Both A and B dissolve in hydrochloric acid to give a clear solution. When either of these solutions is evaporated, a soluble white solid (compound C) remains. If compound C is dissolved in water and sulfuric acid is added, a white precipitate (compound D) forms, (a) Propose identities for the substances A, B, C, and D. (b) Write net ionic equations for all the reactions described. (c) Based on the preceding observations, what can be said about the position of lanthanum in the activity series (Table)? Table Activity Series of Metals in Aqueous Solution
Solution: Here, we are going to identify the compounds A, B, C and D and also write the net ionic equations for all the reactions described. Step1: When, lanthanum is exposed to air, lanthanum oxide is formed. The balanced molecular equation for the reaction is: 4Ln(s) + 3O (g) --2--> 2Ln O 2 3 The compound (A) thus formed is Ln O 2 3 Step2: When lanthanum metal is added to water, lanthanum hydroxide is formed. The balanced molecular equation for the reaction is: 2Ln(s) + 6H O(l) -2---> 2Ln(OH) (s) + 3H (g) 3 2 The compound (B) thus formed is Ln(OH) 3 Step3: Both the compounds A and B dissolves in HCl as shown below: Ln O2(s3+ 6HCl(aq) -------> 2LnCl (aq) + 3H O 3 2 Ln(OH) (s) +3HCl(aq) -------> LnCl (aq) + 3H O 3 2 The net ionic equation for the above reactions are: Ln O (s) + 6H ------> 2Ln + 3H O 3+ 2 3 2 Ln(OH) (s) +3H ------> Ln + 3H O 3+ 2 On evaporation of LnCl , solid LnCl3s formed. Thus, t3 compound C is LnCl . 3 Step4: When lanthanum chloride is dissolved in water and sulfuric acid is added, lanthanum sulfate is formed. The reaction involved in the process is: LnCl (s3+ 3H O(l) -2--> Ln(OH) (aq) + 3HCl 3 2Ln(OH) (s) +3H SO ------2 Ln4SO ) (s) + 6H2 4 3 2 The net ionic equation for the above reactions are: LnCl (s) + 3H ------> Ln + 3HCl 3+ 3 2Ln + 3SO 4-------> Ln (SO2) (s4 3 Thus, the compound D is Ln (SO ) . 2 4 3 Step5: c) Since, lanthanum is readily reactive with most of the solvents, so it should be placed high in the reactivity series. -------------------------
