Principles of Qualitative Analysis

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

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).

:$$\text{PbCrO}_4 + 4\,\text{NaOH (hot, excess)} \;\longrightarrow\; \text{Na}_2[\text{Pb(OH)}_4] + \text{Na}_2\text{CrO}_4$$
$[\text{Pb(OH)}_4]^{2-}$ is a dianionic complex with coordination number = 4.

$N{i}^{2+}$ 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.

The greenish-yellow gas is chlorine (Cl₂).

MnO₂ + H₂SO₄ (conc.) generates nascent oxidizing conditions and can oxidize chloride ions (Cl⁻) to chlorine gas.

Reaction: 2Cl⁻ → Cl₂ (greenish-yellow gas)

Among the options, only NH₄Cl contains chloride ions.

So, NH₄Cl → Cl⁻ → Cl₂ (gas formed)

The greenish-yellow gas is chlorine (Cl₂).

MnO₂ + H₂SO₄ (conc.) generates nascent oxidizing conditions and can oxidize chloride ions (Cl⁻) to chlorine gas.

Reaction: 2Cl⁻ → Cl₂ (greenish-yellow gas)

Among the options, only NH₄Cl contains chloride ions.

So, NH₄Cl → Cl⁻ → Cl₂ (gas formed)

Correct qualitative tests:
•${\mathrm{Cu}}^{2+}$+ ${\mathrm{K}}_4\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]$ → chocolate brown ppt
• ${\mathrm{Fe}}^{3+}$+ ${\mathrm{K}}_4\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]$→ Prussian blue
•$Z{n}^{2+}$ + ${\mathrm{K}}_4\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]$ (neutralised) → white ppt

The sodium carbonate extract is first acidified with dilute nitric acid (HNO₃).
This step removes interfering ions like carbonate (CO₃²⁻) and sulfite by converting them into gases (CO₂, SO₂).
Then silver nitrate (AgNO₃) is added to detect halide ions.

If halide ions are present, they form precipitates:

Cl⁻ + Ag⁺ → AgCl (white precipitate)
Br⁻ + Ag⁺ → AgBr (pale yellow precipitate)
I⁻ + Ag⁺ → AgI (deep yellow precipitate)

The question clearly says: Pale yellow precipitate

So: Not Cl⁻ (white), Not I⁻ (deep yellow)

This strongly indicates Br⁻

Confirmation using NH₄OH solubility:

AgCl → dissolves easily
AgBr → dissolves with difficulty (sparingly soluble)
AgI → does not dissolve at all

Given: Pale yellow precipitate, dissolves with difficulty in NH₄OH

Only AgBr satisfies both conditions.

The acidic radical present is: Bromide ion (Br⁻)

The sodium carbonate extract is first acidified with dilute nitric acid (HNO₃).
This step removes interfering ions like carbonate (CO₃²⁻) and sulfite by converting them into gases (CO₂, SO₂).
Then silver nitrate (AgNO₃) is added to detect halide ions.

If halide ions are present, they form precipitates:

Cl⁻ + Ag⁺ → AgCl (white precipitate)
Br⁻ + Ag⁺ → AgBr (pale yellow precipitate)
I⁻ + Ag⁺ → AgI (deep yellow precipitate)

The question clearly says: Pale yellow precipitate

So: Not Cl⁻ (white), Not I⁻ (deep yellow)

This strongly indicates Br⁻

Confirmation using NH₄OH solubility:

AgCl → dissolves easily
AgBr → dissolves with difficulty (sparingly soluble)
AgI → does not dissolve at all

Given: Pale yellow precipitate, dissolves with difficulty in NH₄OH

Only AgBr satisfies both conditions.

The acidic radical present is: Bromide ion (Br⁻)

MgCl₂, AlCl₃, CuCl₂ are water soluble at room temperature.

AgCl, Hg₂Cl₂ are sparingly soluble in water

PbCl₂ is soluble in hot water.

${\left(N{H}_4\right)}_{2}S$ 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

${\mathrm{Fe}}^{+2}$ ions undergoes hydrolysis, therefore while preparing aqueous solution of ferrous sulphate and ammonium sulphate in water dilute sulphuric acid is added to prevent hydrolysis of ferrous sulphate.

${\mathrm{Fe}}^{+2}$ ions undergoes hydrolysis, therefore while preparing aqueous solution of ferrous sulphate and ammonium sulphate in water dilute sulphuric acid is added to prevent hydrolysis of ferrous sulphate.

To identify Ba²⁺ & Ca²⁺

Reagent (NH₄)₂CO₃ + NH₄Cl is used BaCO₃ & CaCO₃ are obtained as precipitates

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

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

The ligand NO converts to NOS^–.

On heating borax:

Na₂B₄O₇ → 2NaBO₂ + B₂O₃

So:
X = NaBO₂
Y = B₂O₃

CuSO₄ reacts with B₂O₃ (non-luminous flame, oxidizing condition):

Forms copper metaborate:
CuSO₄ + B₂O₃ (Y) ⟶ Cu(BO₂)₂ + SO₃

Z = Cu(BO₂)₂

Oxidation state of Cu = +2

In reducing flame (with carbon):

Cu(BO₂)₂ gets reduced to Cu₂O:
Cu(BO₂)₂ + 2NaBO₂ + C ⟶ 2CuBO₂(Q) + N a 2 B 4 O 7 + C O

Q = CuBO₂

Oxidation state of Cu = +1

Assertion (A):

The gas liberated on warming a salt with dilute H₂SO₄ turns a piece of paper dipped in lead acetate black — confirmatory test for sulphide ion.

True.

When a sulphide salt (like FeS, ZnS, etc.) is treated with dilute H₂SO₄, hydrogen sulphide gas (H₂S) is evolved:

$FeS + H_{2}S{O}_4 \stackrel{}{\to } FeS{O}_4 + H_{2}S$↑

The evolved H₂S gas reacts with lead acetate paper to form black lead sulphide (PbS).

So this is indeed a confirmatory test for sulphide ion (S²⁻).

False.

The black compound formed is lead sulphide (PbS), not lead sulphite (PbSO₃).

Correct Answer: A

Assertion is true but the reason is false.

Sulphide ion reacts with sodium nitroprusside in alkaline medium to form a purple coloured complex.
This test confirms $S^{2-}$ ion.

$N{i}^{2+}$ 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.

$\text{Cu}\stackrel{\text{dil. HCl }}{\to }$ no reaction

$\mathrm{CuS}{\mathrm{O}}_4\stackrel{\mathrm{KCN}}{\to }\underset{\text{perfect complex}}{{\left[\mathrm{Cu}{\left(\mathrm{CN}\right)}_4\right]}^{3-}}$

$\stackrel{{\mathrm{H}}_2\mathrm{S}}{\to }$ no sulphide ppt

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.

$$\begin{gathered} C{u}^{2+}+K_4\left[Fe(CN{)}_6\right]\to \text{ Brown } \\ \text{ precipitate of }C{u}_2\left[Fe(CN{)}_6\right] \\ F{e}^{3+}+K_4\left[Fe(CN{)}_6\right]\to \text{ Prussian blue } \\ \text{ precipitate of }F{e}_4{\left[Fe(CN{)}_6\right]}_3 \\ Z{n}^{2+}+K_4\left[Fe(CN{)}_6\right]\to \text{ White precipitate } \\ \text{ of }Z{n}_2\left[Fe(CN{)}_6\right] \\ C{a}^{2+}+K_4\left[Fe(CN{)}_6\right]\to \text{ White precipitate } \\ \text{ of }\left(C{a}_2\left[Fe(CN{)}_6\right]\right) \end{gathered}$$

Iodine of millon's base Brown

2K₂Hgl₄ + NH₃ + 3KOH$\to$HgO.Hg(NH₂)I + 7KI + 2H₂O

On heating borax:

Na₂B₄O₇ → 2NaBO₂ + B₂O₃

So:
X = NaBO₂
Y = B₂O₃

CuSO₄ reacts with B₂O₃ (non-luminous flame, oxidizing condition):

Forms copper metaborate:

Z = Cu(BO₂)₂

Oxidation state of Cu = +2

In reducing flame (with carbon):

Cu(BO₂)₂ gets reduced to Cu₂O:

Q = Cu₂O

Oxidation state of Cu = +1

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.

$\text{Cu}\stackrel{\text{dil. HCl }}{\to }$ no reaction

$\mathrm{CuS}{\mathrm{O}}_4\stackrel{\mathrm{KCN}}{\to }\underset{\text{perfect complex}}{{\left[\mathrm{Cu}{\left(\mathrm{CN}\right)}_4\right]}^{3-}}$

$\stackrel{{\mathrm{H}}_2\mathrm{S}}{\to }$ no sulphide ppt

Sulphide ion reacts with sodium nitroprusside in alkaline medium to form a purple coloured complex.
This test confirms $S^{2-}$ ion.

Correct qualitative tests:
•${\mathrm{Cu}}^{2+}$+ ${\mathrm{K}}_4\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]$ → chocolate brown ppt
• ${\mathrm{Fe}}^{3+}$+ ${\mathrm{K}}_4\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]$→ Prussian blue
•$Z{n}^{2+}$ + ${\mathrm{K}}_4\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]$ (neutralised) → white ppt

The oxidation number of iron in the compound formed during the brown ring test for the $N{O}_3^{-}$ ion, the complex compound [Fe(H₂O)₅NO]SO₄ , is +1.

$N{i}^{2+}$ 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.

(A) (NH₄)₂S (Ammonium sulphide)
Exists mainly as an aqueous solution, not as a solid precipitate.
It may appear colorless to pale yellow due to impurities (like polysulphides).
But it is not a solid sulphide precipitate.
In qualitative analysis, only insoluble coloured precipitates are considered.

Not counted as yellow sulphide.

(B) PbS (Lead sulphide)
Forms a black precipitate.

Not yellow

(C) CuS (Copper sulphide)
Forms a black precipitate.

Not yellow

(D) As₂S₃ (Arsenic trisulphide)
Forms a bright yellow precipitate.
Classic Group II qualitative analysis compound.

Yellow sulphide

(E) As₂S₅ (Arsenic pentasulphide)
Forms a yellow precipitate.

Yellow sulphide

To identify Ba²⁺ & Ca²⁺

Reagent (NH₄)₂CO₃ + NH₄Cl is used BaCO₃ & CaCO₃ are obtained as precipitates