TG EAMCET 2 May 2025 Shift 2 Paper

(A) Mg(HCO3)2⟶Mg(OH)2↓+2CO2↑
This reaction shows the decomposition of magnesium bicarbonate upon heating (boiling) to form insoluble magnesium hydroxide and carbon dioxide. This process removes temporary hardness from water and is achieved by Boiling.
So, (A) matches with (III) Boiling.
(B) M2++Na4P6O18‌2−⟶[Na2MP6O18]2−+2Na+
In this reaction, hard water ions ( M2+, typically Ca2+ or Mg2+ ) react with sodium hexametaphosphate ( Na4P6O18‌2− ), commonly known as Calgon. The Calgon forms a soluble complex with the metal ions, thereby sequestering them and preventing them from causing hardness. This is the principle behind Calgon's method.
So, (B) matches with (IV) Calgon's method.
(C) Ca(HCO3)2+Ca(OH)2⟶2CaCO3↓+2H2O
This reaction shows calcium bicarbonate reacting with a calculated amount of slaked lime (Ca(OH)2) to precipitate insoluble calcium carbonate. This method is specifically known as Clark's method for removing temporary hardness.
So, (C) matches with (I) Clark's method.
(D) 2NaZ+Ca2+(aq)⟶2Na++CaZ(Z= Zeolite )
This reaction illustrates the exchange of calcium ions ( Ca2+ ) from hard water with sodium ions ( Na+) from sodium zeolite (NaZ). Zeolites are ion-exchange resins that replace hardness-causing ions with non-hardness-causing ions. This is the basis of the lon exchange method (also known as the Zeolite process or Permutit process).
So, (D) matches with (II) Ion exchange method.
Combining the matches:
A - III
B - IV
C-I
D - II
Comparing this with the given options:
Option A: A-III, B-IV, C-I, D-II (Matches our findings)
Option B: A-IV, B-II, C-I, D-III
Option C: A-III, B-IV, C-II, D-I
Option D: A-II, B-IV, C-I, D-III
The correct option is A.
The final answer is A-III, B-IV, C-I, D-II.