Tests for cations — flame test, borax bead, group reagent analysis

Question

How do we systematically identify cations in a salt mixture using flame tests, borax bead tests, and group reagent analysis? What is the logic behind the group separation scheme?

(JEE Main, CBSE 12 — qualitative analysis is a guaranteed practical/theory question in boards)

Solution — Step by Step

Heat a platinum wire in a Bunsen flame until no colour appears. Dip in concentrated HCl, then touch the salt, and hold in the flame. The colour tells you:

Cation	Flame Colour
$Na^+$	Persistent golden yellow
$K^+$	Violet (view through cobalt blue glass to filter Na)
$Ca^{2+}$	Brick red
$Sr^{2+}$	Crimson red
$Ba^{2+}$	Apple green
$Cu^{2+}$	Blue-green (with HCl: green)

Why does this work? Heating excites outer electrons to higher energy levels. When they fall back, they emit photons of specific wavelengths — each element has a unique emission spectrum.

Heat borax ( $Na_2B_4O_7 \cdot 10H_2O$ ) on a platinum loop until it forms a transparent glassy bead. Touch the bead to the salt and heat again.

Borax decomposes: $Na_2B_4O_7 \rightarrow 2NaBO_2 + B_2O_3$

The $B_2O_3$ (boric anhydride) reacts with metal oxides to form coloured metaborates:

Metal ion	Oxidising flame colour	Reducing flame colour
$Cu^{2+}$	Blue	Red (opaque)
$Co^{2+}$	Deep blue	Deep blue
$Cr^{3+}$	Green	Green
$Mn^{2+}$	Violet	Colourless
$Ni^{2+}$	Brown	Grey
$Fe^{3+}$	Yellow	Green

The group reagents are added in a specific order, exploiting differences in solubility product ( $K_{sp}$ ) values.

The systematic flowchart:

Group 0: Dissolve salt in water or dilute HCl.

Group I — Add dilute HCl: precipitates $PbCl_2$ , $HgCl_2$ (calomel), $AgCl$ (white precipitates of insoluble chlorides).

Group II — Pass $H_2S$ in acidic medium (pH ~0.5): precipitates $CuS$ , $PbS$ , $CdS$ , $As_2S_3$ , $Sb_2S_3$ , $SnS$ (sulphides insoluble even in acid).

Group III — Add $NH_4Cl + NH_4OH$ : precipitates $Fe(OH)_3$ (brown), $Al(OH)_3$ (white), $Cr(OH)_3$ (green) — hydroxides with low $K_{sp}$ .

Group IV — Pass $H_2S$ in basic medium (ammoniacal): precipitates $ZnS$ , $MnS$ , $NiS$ , $CoS$ — sulphides that need higher $S^{2-}$ concentration (basic pH).

Group V — Add $(NH_4)_2CO_3$ : precipitates $CaCO_3$ , $SrCO_3$ , $BaCO_3$ — insoluble carbonates.

Group VI — Remaining filtrate: $Mg^{2+}$ , $Na^+$ , $K^+$ , $NH_4^+$ — detected by specific tests.

flowchart TD
    A["Salt solution"] -->|"Add dil. HCl"| B{"Precipitate?"}
    B -->|"Yes: white ppt"| C["Group I: Pb²⁺, Hg₂²⁺, Ag⁺"]
    B -->|"No"| D["Pass H₂S in acidic medium"]
    D --> E{"Precipitate?"}
    E -->|"Yes: coloured sulphides"| F["Group II: Cu²⁺, Pb²⁺, Cd²⁺, As³⁺"]
    E -->|"No"| G["Add NH₄Cl + NH₄OH"]
    G --> H{"Precipitate?"}
    H -->|"Yes: hydroxides"| I["Group III: Fe³⁺, Al³⁺, Cr³⁺"]
    H -->|"No"| J["Pass H₂S in basic medium"]
    J --> K{"Precipitate?"}
    K -->|"Yes: sulphides"| L["Group IV: Zn²⁺, Mn²⁺, Ni²⁺, Co²⁺"]
    K -->|"No"| M["Add (NH₄)₂CO₃"]
    M --> N{"Precipitate?"}
    N -->|"Yes: carbonates"| O["Group V: Ca²⁺, Sr²⁺, Ba²⁺"]
    N -->|"No"| P["Group VI: Mg²⁺, Na⁺, K⁺, NH₄⁺"]

Why This Works

The entire scheme exploits one principle: selective precipitation based on $K_{sp}$ . By controlling pH and the concentration of the precipitating anion ( $Cl^-$ , $S^{2-}$ , $OH^-$ , $CO_3^{2-}$ ), we can precipitate one group at a time while keeping others in solution.

In acidic medium, $[S^{2-}]$ is very low (Le Chatelier pushes $H_2S \rightleftharpoons 2H^+ + S^{2-}$ backward), so only the most insoluble sulphides (Group II) precipitate. In basic medium, $[S^{2-}]$ increases, bringing down Group IV sulphides.

Common Mistake

Students often add group reagents out of order and then wonder why extra cations precipitate in the wrong group. The order is critical — if you skip adding HCl first, $Pb^{2+}$ will precipitate as $PbS$ in Group II instead of $PbCl_2$ in Group I. Always follow the sequence: I, II, III, IV, V, VI. The exam expects this discipline.

For flame test, the most common confusion is $Ca^{2+}$ (brick red) vs $Sr^{2+}$ (crimson). Remember: Calcium = Carrot colour (brick/orange-red), Sr = Scarlet Red (deeper crimson).

Question

Solution — Step by Step

Why This Works

Common Mistake

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