Principles of Qualitative Analysis

Yellow compounds: CdS (cadmium sulfide, yellow pigment) and PbCrO₄ (lead chromate, chrome yellow).

Non-yellow compounds: PbS (black), CuS (black), ZnS (cold) (white).

Thus, the correct option is (B) (a) and (e).

${\left(N{H}_4\right)}_{2}S$ is typically yellow.

PbS is black
CuS is black
$A{s}_2{S}_3$ is yellow in colour
$A{s}_2{S}_5$ is yellow in colour

Al³⁺ → Group III

Cu²⁺ → Group II

Ba²⁺ → Group V

Co²⁺ → Group IV

Mg²⁺ → Group VI

When precipitating Fe³⁺ as Fe(OH)₃, we add NH₄Cl before NH₄OH because:

• NH₄Cl dissociates to give NH₄⁺ ions.

• The NH₄⁺ ions suppress the ionization of NH₄OH (common-ion effect).

• Thus, the concentration of OH⁻ decreases, preventing excess OH⁻ in solution.

This is important because:

• If OH⁻ concentration becomes too high, other cations (like Zn²⁺, Al³⁺) might also precipitate.

• Lower OH⁻ ensures selective precipitation of Fe(OH)₃.

Hence, NH₄Cl is added to control (reduce) OH⁻ concentration, giving a clean precipitate.

✔ A is correct.

${\mathrm{Na}}_2\mathrm{S}+{\mathrm{Na}}_2\left[\mathrm{Fe}\left(\mathrm{CN}{)}_5\mathrm{NO}\right]\to {\mathrm{Na}}_4\right[\mathrm{Fe}\left(\mathrm{CN}{)}_5\mathrm{NOS}\right]$

Purple solution $[{\mathrm{Fe}}^{+2}<\leftarrow :\mathrm{N}{\le }_{\mathrm{S}}^{\mathrm{O}}]$

The ligand NO converts to NOS^–.

Yellow compounds: CdS (cadmium sulfide, yellow pigment) and PbCrO₄ (lead chromate, chrome yellow).

Non-yellow compounds: PbS (black), CuS (black), ZnS (cold) (white).

Thus, the correct option is (B) (a) and (e).

$N{i}^{2+}$ has $d^8$ electronic configurations which is unaffected by the nature of ligand, and gives a violet colour in nonluminous flame under hot condition in borax bead test.