Complete the statements in COLUMN A with the options in COLUMN B. Write only A or B next to the question numbers (1.1.1 to 1.1.7) in the ANSWER BOOK, for example 1.1.8. B.

White precipitate formed, insoluble in excess NaOH.

$$Pb^{2+} + 2OH^- \to Pb(OH)_2 \downarrow$$

No dissolution on adding excess $NaOH$.
Inference: $Pb^{2+}$, $Hg_2^{2+}$, $Bi^{3+}$, $Cu^{2+}$, $Cd^{2+}$, $Fe^{3+}$, $Al^{3+}$, $Cr^{3+}$, $Zn^{2+}$ (but pattern matches group II).

White precipitate, insoluble in excess.

$$Pb(OH)_2 \text{ insoluble in } NH_4OH$$

Confirms group II cation.

Black precipitate.

$$Pb^{2+} + H_2S \to PbS \downarrow + 2H^+$$

Inference: $Pb^{2+}$.

Cation in A: $Pb^{2+}$

Blue precipitate formed, insoluble in excess NaOH.

$$Cu^{2+} + 2OH^- \to Cu(OH)_2 \downarrow$$

No dissolution on adding excess $NaOH$.

Precipitate dissolves in excess to give deep blue solution.

$$Cu(OH)_2 + 4NH_3 \to [Cu(NH_3)_4]^{2+} + 2OH^-$$

Inference: $Cu^{2+}$.

Cation in B: $Cu^{2+}$

Reddish-brown precipitate, insoluble in excess NaOH and $NH_4OH$.

$$Fe^{3+} + 3OH^- \to Fe(OH)_3 \downarrow$$

Insoluble in excess reagents.

Prussian blue precipitate.

$$3K^+ + [Fe(CN)_6]^{4-} + Fe^{3+} \to Fe_4[Fe(CN)_6]_3 \downarrow$$

Cation in C: $Fe^{3+}$

White precipitate, soluble in excess NaOH.

$$Zn^{2+} + 2OH^- \to Zn(OH)_2 \downarrow$$ $$Zn(OH)_2 + 2OH^- \to [Zn(OH)_4]^{2-}$$

White precipitate dissolves in excess.

$$Zn(OH)_2 + 4NH_3 \to [Zn(NH_3)_4]^{2+} + 2OH^-$$

Cation in D: $Zn^{2+}$

Final answers (Q11):
A: Pb^{2+}
B: Cu^{2+}
C: Fe^{3+}
D: Zn^{2+}