Question

Question 11.28 When metal X is treated with sodium hydroxide, a white precipitate (A) is obtained, which is soluble in excess of NaOH to give soluble complex (B). Compound (A) is soluble in dilute HCl to form compound (C). The compound (A) when heated strongly gives (D), which is used to extract metal. Identify (X), (A), (B), (C) and (D). Write suitable equations to support their identities.

Class XI The p-Block Elements Page 325

Answer 1

When metal 'X' is treated with sodium hydroxide gives white precipitate which dissolves in excess of NaOH to give soluble complex 'B' . It means Metal is nothing it is Aluminum.
see reactions :
Al( 'X') + 3NaOH ------>Al(OH)3 ↓(white ppt)(compound A) + 3Na+
Al(OH)3 + NaOH (excess) ----> 2Na+[Al(OH)4]–

now compound 'A' e.g., Al(OH)3 is soluble in dilute HCl to form Compound 'C'
e.g., Al(OH)3 + dil.HCl ----> AlCl3 (C)+ 3H2O

now the compound 'A' is heated strongly gives compound 'D' which is used to extract metal.
e.g., Al(OH)3 -----> Al2O3 + 3H2O

so, finally
metal 'X' = Al
compound 'A' = Al(OH)3
compound 'B' = 2Na+[Al(OH)4]–
compound 'C' = AlCl3
compound 'D' = Al2O3

Answer 2

Hi

Here is your answer,

Since metal X on treatment with sodium hydroxide gives white precipitate which dissolves in excess of NaOH to give soluble complex (B), therefore, the metal X is Al.

  Al + 2NaOH -----> Al(OH)₃↓ + 2Na⁺
(X)                            white ppt. (A)
 
Al(OH)₃ + NaOH ------->  2Na⁺[Al(OH)₄]⁻
(A)             (excess)       Sodium tetra hydroxoaluminate (III)

Al(OH)₃ +3HCl(aq) --------> AlCl + 3H₂O
(A)                                         (C)
                     
   2Al(OH)  -------->  Al₂O₃  + 3H₂O
     (A)                        (D)


Hope it helps you !