JEE advanced problems on salt analysis

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Masterclass: 25 Ultra-Challenging JEE Advanced Problems on Salt Analysis

From the delicate control of the Common Ion Effect to the complex coordination chemistry of Nessler's Reagent and the Brown Ring Test. Master Qualitative Inorganic Analysis.

The Ultimate Detective Work!
Salt Analysis is not mere memorization of colors; it is the practical application of Ionic Equilibrium, Solubility Products ($K_{sp}$), and Coordination Chemistry. To conquer these JEE Advanced problems, you must understand exactly why certain buffers are used, how complexation masks precipitation, and the redox mechanics behind interference in tests.

Problem 1: Group I Separation (The Hot Water Anomaly)

A mixture containing $Ag^+$, $Pb^{2+}$, and $Hg_2^{2+}$ is treated with dilute $HCl$, yielding a white precipitate. When this precipitate is boiled with distilled water and filtered hot, the filtrate yields a yellow precipitate with $K_2CrO_4$. The residue on the filter paper turns black when treated with $NH_4OH$. Identify the chemical reactions confirming the exact identity of the ions.

View Solution

Strategy: Exploit the unique temperature-dependent solubility of Group I chlorides and the specific complexation/disproportionation reactions with Ammonia.

Step 1: The Group I Precipitate
Dilute $HCl$ precipitates all three Group I cations as insoluble chlorides: $AgCl$, $PbCl_2$, and $Hg_2Cl_2$ (all white).

Step 2: Hot Water Separation
Among these, Lead(II) chloride ($PbCl_2$) is uniquely highly soluble in boiling water but insoluble in cold water. Boiling and filtering removes $PbCl_2$ into the filtrate.
Adding Potassium chromate ($K_2CrO_4$) to the filtrate yields a brilliant yellow precipitate of Lead chromate, confirming $Pb^{2+}$.
$Pb^{2+} + CrO_4^{2-} \rightarrow \mathbf{PbCrO_4 \downarrow \text{ (Yellow)}}$

Step 3: Ammonia Treatment of the Residue
The remaining residue ($AgCl$ and $Hg_2Cl_2$) is treated with aqueous Ammonia ($NH_4OH$).
- $AgCl$ dissolves completely to form a soluble complex: $AgCl + 2NH_3 \rightarrow [Ag(NH_3)_2]Cl$.
- $Hg_2Cl_2$ undergoes a fascinating auto-redox (disproportionation) reaction with ammonia, yielding a mixture of finely divided elemental Mercury (black) and Mercury(II) amido chloride (white). The black color of the finely divided mercury dominates.
$Hg_2Cl_2 + 2NH_3 \rightarrow \mathbf{Hg \downarrow \text{ (Black)} + Hg(NH_2)Cl \downarrow \text{ (White)}} + NH_4^+ + Cl^-$

Final Answer: The hot water soluble fraction is $PbCl_2$ (confirmed by yellow $PbCrO_4$). The blackening of the residue confirms $Hg_2^{2+}$ due to its disproportionation into black $Hg$ metal and white $Hg(NH_2)Cl$ upon treatment with $NH_4OH$.

Problem 2: Group IIA vs IIB (Yellow Ammonium Sulfide)

A black precipitate from Group II analysis is suspected to be a mixture of $CuS$ and $As_2S_3$. To separate them, the precipitate is treated with Yellow Ammonium Sulfide (YAS). Which sulfide dissolves, what is the structure of the soluble complex formed, and why does the other sulfide remain insoluble?

View Solution

Strategy: Group II is divided into IIA (Copper group, insoluble in YAS) and IIB (Arsenic group, soluble in YAS). YAS acts as a sulfur-donating nucleophile capable of forming soluble thio-complexes with amphoteric-like highly charged metal sulfides.

Step 1: The Nature of Yellow Ammonium Sulfide (YAS)
YAS is ammonium polysulfide, $(NH_4)_2S_x$. It is a source of highly reactive polysulfide and sulfide ions, which act as strong Lewis bases (sulfur donors).

Step 2: Reaction with Group IIB ($As_2S_3$)
Arsenic(III) sulfide is amphoteric in the thio-system. It reacts with the excess sulfide ions in YAS to form a highly stable, water-soluble thioarsenite or thioarsenate complex (if oxidized by the polysulfide).
$As_2S_3 + 3S^{2-} \rightarrow 2[AsS_3]^{3-} \text{ (Soluble thioarsenite)}$
If oxidized by $S_x^{2-}$, it forms the thioarsenate complex: $\mathbf{[AsS_4]^{3-}}$.

Step 3: Reaction with Group IIA ($CuS$)
Copper(II) sulfide ($CuS$) is strongly basic in the thio-system and lacks the ability to expand its coordination with additional sulfur ligands to form stable soluble anions. Therefore, it completely fails to dissolve in YAS and remains as a black residue.

Final Answer: $As_2S_3$ dissolves by forming the soluble complex Thioarsenate $[AsS_4]^{3-}$. $CuS$ remains insoluble because it cannot form stable thio-complex anions.

Problem 3: Copper vs Cadmium (Complex Stability Trap)

A solution contains both $Cu^{2+}$ and $Cd^{2+}$ ions. Excess Potassium Cyanide ($KCN$) is added until the initial precipitates dissolve into colorless complexes. When Hydrogen Sulfide ($H_2S$) gas is passed through this solution, a yellow precipitate forms, but no black precipitate is observed. Explain the thermodynamic complexation kinetics causing this selective precipitation.

View Solution

Strategy: Both ions form soluble cyanide complexes. However, the stability constant ($K_f$) of the Copper complex is massively higher than that of the Cadmium complex, dictating the availability of free metal ions to exceed the solubility product ($K_{sp}$) of their respective sulfides.

Step 1: Complex Formation
Adding excess $KCN$ initially precipitates cyanides, which dissolve in excess to form complexes.
- Copper forms the extremely stable tetracyanocuprate(I) complex (Note: $Cu^{2+}$ is reduced to $Cu^+$ by cyanide, evolving cyanogen gas):
$2Cu^{2+} + 10CN^- \rightarrow \mathbf{2[Cu(CN)_4]^{3-}} + (CN)_2 \uparrow$
- Cadmium forms the tetracyanocadmate(II) complex:
$Cd^{2+} + 4CN^- \rightarrow \mathbf{[Cd(CN)_4]^{2-}}$

Step 2: Thermodynamic Stability
The $K_f$ for $[Cu(CN)_4]^{3-}$ is astronomically high ($\approx 10^{30}$). The complex dissociates so negligibly that the concentration of free $Cu^+$ ions in solution is effectively zero.
The $K_f$ for $[Cd(CN)_4]^{2-}$ is much lower ($\approx 10^{18}$). It dissociates enough to provide a small, but significant, concentration of free $Cd^{2+}$ ions.

Step 3: Precipitation with $H_2S$
For a precipitate to form, the ionic product must exceed $K_{sp}$.
Because free $[Cu^+]$ is virtually zero, the ionic product NEVER exceeds the $K_{sp}$ of Copper sulfide. No black precipitate forms.
Because free $[Cd^{2+}]$ is present in sufficient quantity, the ionic product exceeds the $K_{sp}$ of $CdS$. A yellow precipitate of $CdS$ cleanly drops out.

Final Answer: The copper complex $\mathbf{[Cu(CN)_4]^{3-}}$ is highly stable, preventing $Cu_2S$ precipitation. The cadmium complex $\mathbf{[Cd(CN)_4]^{2-}}$ is less stable, yielding enough free $Cd^{2+}$ to exceed $K_{sp}$ and precipitate yellow $\mathbf{CdS}$.

Problem 4: Group III Precipitation (The Common Ion Effect)

In Group III analysis, a solid pinch of Ammonium Chloride ($NH_4Cl$) is strictly added before adding Ammonium Hydroxide ($NH_4OH$) to precipitate $Fe^{3+}$, $Al^{3+}$, and $Cr^{3+}$. What catastrophic analytical failure occurs if the $NH_4Cl$ is omitted, and how does the common ion effect mathematically prevent this?

View Solution

Strategy: We need to precipitate Group III hydroxides ($K_{sp} \approx 10^{-38}$) but strictly prevent the precipitation of Group IV/V hydroxides (like $Zn(OH)_2, Mg(OH)_2$ with $K_{sp} \approx 10^{-16}$). We must carefully throttle the $OH^-$ concentration.

Step 1: The Threat of High $OH^-$
$NH_4OH$ is a weak base, but it still produces enough $OH^-$ ions to easily exceed the relatively high $K_{sp}$ of Group IV and V metals like Zinc, Manganese, and Magnesium. If we add pure $NH_4OH$, all these metals will precipitate prematurely alongside Group III, ruining the separation scheme.

Step 2: The Common Ion Effect
$NH_4OH \rightleftharpoons NH_4^+ + OH^-$
By adding solid $NH_4Cl$, we flood the solution with $NH_4^+$ ions (a strong electrolyte). According to Le Chatelier's Principle (and the Common Ion Effect), this massive excess of $NH_4^+$ forcibly drives the weak base dissociation equilibrium drastically to the left.

Step 3: The Mathematical Result
This massive suppression lowers the $[OH^-]$ concentration to a microscopic level.
- This tiny $[OH^-]$ is still enough to exceed the incredibly tiny $K_{sp}$ of Group III hydroxides ($Fe(OH)_3$, $Al(OH)_3$), so they precipitate perfectly.
- However, this suppressed $[OH^-]$ is now mathematically too small to exceed the higher $K_{sp}$ values of Group IV/V metals, keeping them safely dissolved in solution.

Final Answer: Omitting $NH_4Cl$ causes premature precipitation of Group IV/V/VI metals (like $Zn, Mn, Mg$). The common ion ($NH_4^+$) drastically suppresses $NH_4OH$ dissociation, lowering $[OH^-]$ just enough to exclusively precipitate the low-$K_{sp}$ Group III hydroxides.

Problem 5: Prussian Blue vs Turnbull's Blue

In the specific tests for Iron, $Fe^{3+}$ reacts with Potassium ferrocyanide to give Prussian Blue, while $Fe^{2+}$ reacts with Potassium ferricyanide to give Turnbull's Blue. Modern X-ray crystallography reveals that both of these deeply colored precipitates are actually the exact same compound. Write its true chemical formula and explain the color origin via MO theory.

View Solution

Strategy: Analyze the oxidation states of iron inside and outside the coordination sphere.

Step 1: The Traditional View
Prussian Blue: $Fe^{3+}$ + $[Fe^{II}(CN)_6]^{4-} \rightarrow Fe_4^{III}[Fe^{II}(CN)_6]_3$
Turnbull's Blue: $Fe^{2+}$ + $[Fe^{III}(CN)_6]^{3-} \rightarrow Fe_3^{II}[Fe^{III}(CN)_6]_2$
Historically, they were thought to be distinct compounds.

Step 2: The Modern Reality
When $Fe^{2+}$ is mixed with Ferricyanide, an instant electron transfer (redox) occurs, oxidizing the $Fe^{2+}$ to $Fe^{3+}$ and reducing the Ferricyanide to Ferrocyanide. Thus, Turnbull's blue instantly converts into Prussian Blue. They are identical.
True Formula: $KFe^{III}[Fe^{II}(CN)_6]$ (or the insoluble polymeric form $\mathbf{Fe_4[Fe(CN)_6]_3}$).

Step 3: The Color Origin
The intense blue color is NOT due to a simple d-d transition. The structure contains adjacent Iron atoms in two different oxidation states ($+2$ and $+3$) bridged by cyanide ligands ($-C \equiv N-$). When struck by light, an electron is optically excited and transfers directly from the $Fe^{II}$ center through the cyanide bridge to the $Fe^{III}$ center. This is called Intervalence Charge Transfer (IVCT) or Metal-to-Metal Charge Transfer, which possesses an incredibly high molar absorptivity coefficient.

Final Answer: The true formula is $Fe_4[Fe(CN)_6]_3$. The intense color originates from Intervalence Charge Transfer (IVCT) between the adjacent $Fe^{II}$ and $Fe^{III}$ centers bridged by cyanide.

Problem 6: Chromyl Chloride Test Limitations

The Chromyl Chloride test is the definitive test for solid chloride salts. However, testing solid Silver Chloride ($AgCl$) or Mercuric Chloride ($HgCl_2$) with $K_2Cr_2O_7$ and concentrated $H_2SO_4$ yields absolutely no red vapors. Why does this test fail for these specific heavy metal chlorides?

View Solution

Strategy: The Chromyl Chloride test requires the generation of $HCl$ gas in situ. Evaluate the bond character of the starting chlorides to see if they can react with sulfuric acid to release $HCl$.

Step 1: The Standard Mechanism
In a standard test (e.g., with $NaCl$), the conc. $H_2SO_4$ displaces the chloride to form $HCl$ gas:
$NaCl + H_2SO_4 \rightarrow NaHSO_4 + \mathbf{HCl}$
The $HCl$ then attacks the chromic acid formed from the dichromate to yield volatile, red Chromyl Chloride gas ($CrO_2Cl_2$).

Step 2: The Failure of Heavy Metal Chlorides
Chlorides of Silver ($AgCl$), Lead ($PbCl_2$), Mercury ($Hg_2Cl_2$, $HgCl_2$), and Tin ($SnCl_4$) possess significant covalent character (due to high polarization, Fajans' rules). Because their $M-Cl$ bonds are tightly covalent, they do not ionize easily, and they completely resist attack by concentrated $H_2SO_4$.

Step 3: Conclusion
Because no $HCl$ gas is generated in the first step, the subsequent condensation reaction with chromic acid cannot occur, and no $CrO_2Cl_2$ vapors are evolved.

Final Answer: Heavy metal chlorides like $AgCl$ and $HgCl_2$ possess high covalent character. They fail to react with concentrated $H_2SO_4$ to yield the necessary $HCl$ intermediate, completely breaking the reaction chain.

Problem 7: The Brown Ring Test Paradox

In the Brown Ring test for nitrates, a brown complex is formed at the junction of two liquids. Determine the exact formula of this complex. Furthermore, state the highly unusual oxidation state of Iron in this complex and the nature of the Nitric Oxide ligand that causes it.

View Solution

Strategy: Trace the redox reaction between nitrate and Ferrous sulfate. Evaluate the charge transfer between the Iron center and the NO ligand.

Step 1: The Redox Generation
Concentrated $H_2SO_4$ converts nitrate ($NO_3^-$) to nitric acid, which is then reduced by the excess Iron(II) sulfate ($Fe^{2+}$) into Nitric Oxide ($NO$) gas.
$3Fe^{2+} + NO_3^- + 4H^+ \rightarrow 3Fe^{3+} + \mathbf{NO} + 2H_2O$

Step 2: Complex Formation
The $NO$ gas traps itself in the remaining unreacted aqueous iron hexaaqua complex: $[Fe(H_2O)_6]^{2+}$. It displaces one water molecule to form the brown complex.
Formula: $[Fe(H_2O)_5(NO)]^{2+}$ (Pentaaquanitrosoiron(I) ion).

Step 3: The Oxidation State Paradox
Initially, the Iron is $Fe^{2+}$ and $NO$ is neutral. However, the $NO$ molecule acts as a non-innocent ligand. It contains an unpaired electron in an anti-bonding $\pi^*$ orbital. Upon coordination, $NO$ transfers this single electron entirely to the Iron atom.

Step 4: Charge Assignment
- $NO$ loses an electron to become the Nitrosonium cation ($NO^+$).
- $Fe^{2+}$ gains an electron to drop to the highly unusual $+1$ oxidation state ($Fe^+$).
The complex has $3$ unpaired electrons ($3d^7$ high spin), leading to a magnetic moment of $\approx 3.87 \text{ B.M.}$, which experimentally proves this unique $+1$ oxidation state.

Final Answer: The formula is $[Fe(H_2O)_5(NO)]^{2+}$. The NO ligand acts as a positive $NO^+$ ion (Nitrosonium), forcing the Iron into the highly anomalous $+1$ oxidation state.

Problem 8: Group IV Separation (Alkaline H₂S)

In Group II, we passed $H_2S$ gas in an acidic medium ($HCl$). For Group IV ($Zn^{2+}, Mn^{2+}, Ni^{2+}, Co^{2+}$), we pass $H_2S$ in an alkaline medium ($NH_4OH$). Why is the switch to an alkaline medium absolutely mandatory to precipitate Group IV sulfides?

View Solution

Strategy: Evaluate the solubility products ($K_{sp}$) of the metal sulfides and apply Le Chatelier's Principle to the dissociation of $H_2S$.

Step 1: Dissociation of $H_2S$
$H_2S \rightleftharpoons 2H^+ + S^{2-}$
$H_2S$ is a very weak acid, producing only a tiny amount of $S^{2-}$ ions.

Step 2: Group II (Acidic Medium)
Group II sulfides ($CuS, HgS, PbS$) have astronomically low $K_{sp}$ values ($\approx 10^{-35}$ to $10^{-50}$). Even with the $HCl$ adding common $H^+$ ions and fiercely suppressing the $S^{2-}$ concentration, the microscopic amount of $S^{2-}$ left is still more than enough to exceed their tiny $K_{sp}$ and precipitate them.

Step 3: Group IV (Alkaline Medium Requirement)
Group IV sulfides ($ZnS, MnS, NiS, CoS$) have much higher $K_{sp}$ values ($\approx 10^{-15}$ to $10^{-20}$). The suppressed $S^{2-}$ concentration from the acidic medium is mathematically insufficient to reach their $K_{sp}$. They will not precipitate.
By adding $NH_4OH$ (alkaline medium), the $OH^-$ ions instantly neutralize the $H^+$ ions from the $H_2S$ dissociation. By continuously removing the product ($H^+$), Le Chatelier's principle heavily drives the $H_2S$ equilibrium forward to the right, creating a massive spike in $[S^{2-}]$ concentration. This high concentration easily exceeds the $K_{sp}$ of Group IV metals, precipitating them completely.

Final Answer: Group IV sulfides have higher $K_{sp}$ values. An alkaline medium removes $H^+$ ions, driving $H_2S$ dissociation forward. This massively increases the $\mathbf{[S^{2-}]}$ concentration required to exceed their solubility products.

Problem 9: The DMG Test for Nickel

Dimethylglyoxime (DMG) is a highly specific reagent for $Ni^{2+}$, yielding a brilliant cherry-red precipitate in a slightly ammoniacal medium. Draw the structure of the $Ni(DMG)_2$ complex. What unique intramolecular force provides this specific complex with its extraordinary stability?

View Solution

Strategy: Evaluate the coordination chemistry of DMG. It is a bidentate ligand. Check the stereochemistry of the resulting square planar complex.

Step 1: The Ligand
Dimethylglyoxime is a bidentate ligand. In the slightly basic medium, it loses one proton to become the monoanion ($DMG^-$). It coordinates to the Nickel via its two nitrogen atoms.

Step 2: Coordination Geometry
Two $DMG^-$ ligands coordinate to a single $Ni^{2+}$ ion, forming a neutral $Ni(DMG)_2$ complex. Because $Ni^{2+}$ is a $d^8$ system and DMG is a strong-field chelating ligand, it forces electron pairing to form an inner-orbital $dsp^2$ square planar geometry.

Step 3: The Intramolecular Lock
Because the complex is perfectly flat, the $O-H$ group of one DMG ligand ends up exactly adjacent to the $O^-$ group of the other DMG ligand. This perfect spatial alignment allows the formation of two incredibly strong, symmetrical Intramolecular Hydrogen Bonds bridging the two ligands across the top and bottom of the complex.

Step 4: Conclusion
These hydrogen bonds lock the entire flat complex together into a massive macrocyclic ring. This extra thermodynamic "chelate-like" stability makes the precipitate highly insoluble and definitive for Nickel.

Final Answer: The complex is a square planar $Ni(DMG)_2$ molecule. Its phenomenal stability and insolubility are derived from the formation of two highly stable Intramolecular Hydrogen Bonds that physically lock the two DMG ligands together.

Problem 10: Vogel's Test for Cobalt

Cobalt(II) salts react with solid Ammonium Thiocyanate ($NH_4SCN$) in the presence of Amyl Alcohol or Ether to produce a beautiful deep blue coloration in the organic layer. Identify the formula of the complex responsible for the blue color and explain why extraction into an organic solvent is necessary.

View Solution

Strategy: Assess the complex formed between $Co^{2+}$ and $SCN^-$. Consider the geometry of Cobalt complexes and their solubility traits.

Step 1: Complex Formation
$Co^{2+}$ reacts with a large excess of thiocyanate ions to form a tetracoordinated complex.
$Co^{2+} + 4SCN^- \rightleftharpoons \mathbf{[Co(SCN)_4]^{2-}}$
Unlike the pink octahedral hexaaqua complex, this tetrathiocyanatocobaltate(II) complex adopts a tetrahedral geometry. Due to intense d-d transitions in tetrahedral symmetry (lacking a center of inversion, breaking Laporte selection rules), the complex is intensely deep blue.

Step 2: The Solvent Extraction (Why Ether/Amyl Alcohol?)
The formation of $[Co(SCN)_4]^{2-}$ is a highly reversible equilibrium. In purely aqueous solutions, the massive concentration of water easily outcompetes the $SCN^-$ ligands, pushing the equilibrium back to the pale pink $[Co(H_2O)_6]^{2+}$ complex.
By adding an immiscible organic solvent like Amyl alcohol or Ether, the uncharged ion pair $(NH_4)_2[Co(SCN)_4]$ preferentially dissolves and extracts into the organic layer. This physical removal traps the complex, protecting it from water hydrolysis and concentrating the brilliant blue color for easy observation.

Final Answer: The complex is the tetrahedral $[Co(SCN)_4]^{2-}$. Organic extraction is necessary because the complex is unstable in water; the non-polar solvent traps it, preventing aqueous hydrolysis back to the pink hexaaqua ion.

Problem 11: Group V Separation (The Chromate vs Sulfate Choice)

In Group V, the cations $Ba^{2+}$, $Sr^{2+}$, and $Ca^{2+}$ are precipitated together as carbonates. To separate them, they are dissolved in Acetic Acid. Potassium Chromate ($K_2CrO_4$) is then added. Why does only Barium precipitate as Barium Chromate, while Strontium and Calcium chromates remain in solution?

View Solution

Strategy: Evaluate the solubility products ($K_{sp}$) of the chromates in conjunction with the mildly acidic environment provided by Acetic Acid.

Step 1: The Role of Acetic Acid
Acetic acid is a weak acid. It provides a moderate concentration of $H^+$ ions. These protons react with the added Chromate ions, establishing an equilibrium with Dichromate:
$2CrO_4^{2-} + 2H^+ \rightleftharpoons Cr_2O_7^{2-} + H_2O$
This equilibrium significantly lowers the concentration of free $CrO_4^{2-}$ ions in the solution.

Step 2: $K_{sp}$ Comparisons
- Barium Chromate ($BaCrO_4$) has a very low $K_{sp}$ ($\approx 10^{-10}$).
- Strontium Chromate ($SrCrO_4$) has a much higher $K_{sp}$ ($\approx 10^{-5}$).
- Calcium Chromate is highly soluble.

Step 3: The Selective Precipitation
Because the $CrO_4^{2-}$ concentration has been suppressed by the Acetic acid buffer, the Ionic Product $[Ba^{2+}][CrO_4^{2-}]$ is still large enough to comfortably exceed the tiny $K_{sp}$ of $BaCrO_4$, causing it to precipitate as a yellow solid.
However, this suppressed chromate concentration is mathematically too low to exceed the much larger $K_{sp}$ of $SrCrO_4$. Thus, Strontium (and Calcium) safely remain dissolved in the solution for later tests.

Final Answer: The weak Acetic Acid suppresses the $[CrO_4^{2-}]$ concentration. This suppressed concentration is only high enough to exceed the very low $K_{sp}$ of $\mathbf{BaCrO_4}$, effectively separating Barium from the more soluble Strontium and Calcium chromates.

Problem 12: Nessler's Reagent Mechanics

Nessler's reagent is the ultimate test for Ammonium ions ($NH_4^+$), yielding a brown precipitate. What is the chemical composition of Nessler's reagent? Draw the structure of the brown precipitate, known historically as the "Iodide of Millon's Base."

View Solution

Strategy: Identify the complex salt used and trace the substitution/condensation reaction with ammonia in an alkaline medium.

Step 1: Composition of the Reagent
Nessler's reagent is an alkaline solution of Potassium tetraiodomercurate(II).
Formula: $K_2[HgI_4]$ dissolved in $KOH$ or $NaOH$.

Step 2: The Reaction
When an ammonium salt reacts with the strong alkali ($KOH$), it liberates free Ammonia gas ($NH_3$).
The $NH_3$ then attacks the $[HgI_4]^{2-}$ complex. A complex substitution and condensation reaction occurs, linking two Mercury atoms via a Nitrogen bridge and an Oxygen bridge (from the alkaline solvent).

Step 3: The Iodide of Millon's Base
Reaction: $2K_2[HgI_4] + NH_3 + 3KOH \rightarrow \mathbf{HgO \cdot Hg(NH_2)I \downarrow} + 7KI + 2H_2O$
The brown precipitate is a polymeric coordination compound. Its core structural unit consists of an $NH_2$ group bridging two $Hg$ atoms, which are also bridged by an Oxygen atom. Formulaically, it is written as $H_2N-Hg-O-Hg-I$.

Final Answer: Nessler's reagent is alkaline $K_2[HgI_4]$. The brown precipitate (Iodide of Millon's Base) has the formula $HgO \cdot Hg(NH_2)I$ (or $H_2N-Hg-O-Hg-I$).

Problem 13: Sodium Carbonate Extract Preparation

Before performing tests for most anions, the unknown salt is heavily boiled with Sodium Carbonate ($Na_2CO_3$) to prepare the "Sodium Carbonate Extract". Why is this grueling extraction step necessary? What fatal analytical errors occur if tests are performed directly on the water extract of the salt?

View Solution

Strategy: Evaluate the interference of heavy metal cations with standard anion test reagents (like $AgNO_3$ or $BaCl_2$).

Step 1: The Threat of Cation Interference
Many qualitative tests for anions rely on precipitation (e.g., adding $AgNO_3$ to test for halides). If the original salt contains heavy metal cations (like $Pb^{2+}, Cu^{2+}, Fe^{3+}$), these cations will wildly interfere. They might form their own insoluble precipitates with the test reagents, or their intense colors (like blue $Cu^{2+}$ or yellow $Fe^{3+}$) will completely mask the color of the intended precipitate.

Step 2: The Action of Sodium Carbonate
Boiling the salt with excess $Na_2CO_3$ forces a double displacement reaction. Almost all heavy metal cations react with the carbonate to form highly insoluble Metal Carbonates (or basic carbonates) which precipitate out as a solid sludge.
$MX_2 + Na_2CO_3 \rightarrow \mathbf{MCO_3 \downarrow} + 2NaX$

Step 3: The Pristine Filtrate
The mixture is filtered. The heavy metals are left behind on the filter paper. The filtrate (the extract) now contains only the pure Sodium salts of the original anions (e.g., $NaCl, Na_2SO_4$). Because Sodium ($Na^+$) is a universal spectator ion that never precipitates or causes color interference, the extract can be safely used to test for anions without any fear of false positives.

Final Answer: The boiling extracts the anions into soluble Sodium salts while precipitating out all heavy metal cations as insoluble Carbonates. This physically removes the cations, preventing their precipitation and color interferences during subsequent anion testing.

Problem 14: Differentiating Nitrate and Nitrite

Both Nitrate ($NO_3^-$) and Nitrite ($NO_2^-$) ions give a positive Brown Ring test. Describe a definitive chemical test using dilute acids to instantly distinguish between a solid nitrite salt and a solid nitrate salt.

View Solution

Strategy: Nitrite is the salt of a weak, highly unstable acid (Nitrous acid). Nitrate is the salt of a strong, stable acid (Nitric acid). Their reactivity towards mild, dilute acids is drastically different.

Step 1: Action of Dilute Acid on Nitrite
When dilute $HCl$ or dilute $H_2SO_4$ is added to a solid Nitrite salt, it instantly forms Nitrous acid ($HNO_2$). Nitrous acid is highly unstable at room temperature and violently disproportionates to yield Nitric Oxide ($NO$) and Nitric Acid.
The colorless $NO$ gas immediately contacts the oxygen in the air and oxidizes into Nitrogen Dioxide ($NO_2$), a deeply reddish-brown, pungent, suffocating gas.
Reaction: $2NO + O_2 \rightarrow \mathbf{2NO_2 \uparrow \text{ (Brown)}}$

Step 2: Action of Dilute Acid on Nitrate
When dilute $HCl$ or $H_2SO_4$ is added to a solid Nitrate salt, it forms Nitric acid ($HNO_3$). Nitric acid is highly stable and does not decompose in cold, dilute conditions. Therefore, no gas is evolved, and no reaction is observed.

Note: Nitrates require concentrated $H_2SO_4$ and intense heating (often with copper turnings) to force the evolution of brown $NO_2$ gas.

Final Answer: Add dilute $HCl$ or $H_2SO_4$. Nitrite salts will instantly evolve pungent, reddish-brown $NO_2$ gas in the cold. Nitrate salts will show absolutely no reaction.

Problem 15: The Layer Test for Bromides and Iodides

A solution contains a mixture of Bromide ($Br^-$) and Iodide ($I^-$) ions. To identify them, Chlorine water ($Cl_2$ in water) is added dropwise, followed by extraction with Chloroform ($CHCl_3$). Describe the exact sequence of color changes observed in the organic layer as excess Chlorine water is added, and explain the redox chemistry.

View Solution

Strategy: Chlorine is a stronger oxidizing agent than both Bromine and Iodine. It will oxidize the heaviest, most easily oxidized halide first (Iodide), and then move to Bromide. The solubility of halogens in chloroform provides the colors.

Step 1: First Oxidation (Iodide)
Because $I^-$ is a stronger reducing agent than $Br^-$, the added Chlorine water selectively oxidizes the Iodide ions to free Iodine ($I_2$) first.
$Cl_2 + 2I^- \rightarrow 2Cl^- + I_2$
The free non-polar $I_2$ molecules preferentially dissolve into the heavy Chloroform layer at the bottom of the test tube, turning the organic layer a brilliant Violet/Purple.

Step 2: Excess Chlorine (Destruction of Iodine)
As more $Cl_2$ water is added, the excess Chlorine acts as a powerful oxidant, further oxidizing the violet $I_2$ molecules into colorless Iodic Acid ($HIO_3$).
$I_2 + 5Cl_2 + 6H_2O \rightarrow 2HIO_3 \text{ (colorless)} + 10HCl$
The violet color in the chloroform layer completely fades away.

Step 3: Second Oxidation (Bromide)
With all iodine destroyed, the excess Chlorine now attacks the remaining Bromide ions, oxidizing them to free Bromine ($Br_2$).
$Cl_2 + 2Br^- \rightarrow 2Cl^- + Br_2$
The free $Br_2$ dissolves into the chloroform layer, turning it a distinct Orange/Brown color.

Final Answer: The organic layer first turns Violet (due to $I_2$ formation). Upon adding excess $Cl_2$, the violet fades as $I_2$ is oxidized to colorless $HIO_3$, and the layer then turns Orange/Brown as $Br_2$ is liberated.

Problem 16: Sulfate vs Sulfite (Barium Precipitation)

Both Sulfate ($SO_4^{2-}$) and Sulfite ($SO_3^{2-}$) ions yield a white precipitate upon the addition of Barium Chloride ($BaCl_2$) solution. How can you chemically distinguish between the two precipitates using dilute Hydrochloric acid?

View Solution

Strategy: Evaluate the salts formed. Barium sulfate is a salt of a strong acid. Barium sulfite is a salt of a weak acid. Their solubilities in dilute mineral acids differ radically.

Step 1: The Precipitates
- $SO_4^{2-} + Ba^{2+} \rightarrow BaSO_4 \downarrow$ (White)
- $SO_3^{2-} + Ba^{2+} \rightarrow BaSO_3 \downarrow$ (White)

Step 2: Acid Addition (Sulfite)
Barium sulfite is the salt of a weak acid (Sulfurous acid, $H_2SO_3$). When dilute $HCl$ is added, the $H^+$ ions strongly protonate the sulfite anion to form sulfurous acid, which instantly decomposes into water and Sulfur Dioxide ($SO_2$) gas.
$BaSO_3 + 2HCl \rightarrow BaCl_2 \text{ (soluble)} + H_2O + \mathbf{SO_2 \uparrow}$
The precipitate dissolves completely, evolving a gas with the suffocating odor of burning sulfur.

Step 3: Acid Addition (Sulfate)
Barium sulfate is the salt of a strong acid ($H_2SO_4$). Dilute $HCl$ cannot protonate the sulfate anion effectively. The precipitate is incredibly stable and remains completely insoluble.

Final Answer: Add dilute $HCl$. The Barium Sulfite precipitate will dissolve, evolving pungent $SO_2$ gas. The Barium Sulfate precipitate is completely inert and will remain insoluble.

Problem 17: The Phosphate Test (Ammonium Molybdate)

The definitive test for Phosphate ions ($PO_4^{3-}$) involves heating the extract with concentrated Nitric Acid and adding Ammonium Molybdate, yielding a canary yellow precipitate. Write the formula of this massive coordination complex. Why is the addition of Nitric acid mandatory?

View Solution

Strategy: The test relies on forming a massive heteropoly acid complex. The acidic medium is required to drive the dehydration and condensation of molybdate units around the central phosphorus.

Step 1: The Reaction
Phosphate ions react with Ammonium Molybdate in the presence of $HNO_3$ to form an intricate, massive cage structure where 12 molybdate units surround a central phosphate tetrahedron.

Step 2: The Formula
$PO_4^{3-} + 3NH_4^+ + 12MoO_4^{2-} + 24H^+ \rightarrow \mathbf{(NH_4)_3[PMo_{12}O_{40}] \downarrow} + 12H_2O$
The canary yellow precipitate is Ammonium Phosphomolybdate.

Step 3: The Role of Nitric Acid
The formation of this heteropoly complex requires a massive influx of protons ($24H^+$ per molecule) to strip oxygen atoms from the molybdate anions as water, forcing the molybdates to condense and link together around the phosphorus. Without a highly acidic medium, the condensation simply cannot occur. Nitric acid is used specifically because, unlike $HCl$ or $H_2SO_4$, the nitrate ion is a poor ligand and does not interfere with the complexation.

Final Answer: The yellow precipitate is Ammonium Phosphomolybdate: $(NH_4)_3[PMo_{12}O_{40}]$. Nitric acid is mandatory to provide the massive concentration of $H^+$ ions required to drive the dehydration-condensation of the molybdate cages.

Problem 18: The Borate Green Flame Test

To test for Borate ions ($BO_3^{3-}$), the solid salt is mixed with concentrated $H_2SO_4$ and Ethanol ($C_2H_5OH$), then ignited. A beautiful green-edged flame burns. Identify the highly volatile compound that burns with this green flame, and write its formation reaction.

View Solution

Strategy: Acid-catalyzed esterification occurs between the borate (which forms boric acid) and the alcohol.

Step 1: Formation of Boric Acid
The strong $H_2SO_4$ protonates the borate salt to yield Orthoboric Acid ($H_3BO_3$).
$2Na_3BO_3 + 3H_2SO_4 \rightarrow 2H_3BO_3 + 3Na_2SO_4$

Step 2: Esterification
Boric acid reacts with Ethanol in a classic esterification reaction, heavily catalyzed and dehydrated by the concentrated $H_2SO_4$.
$H_3BO_3 + 3C_2H_5OH \xrightarrow{H_2SO_4} \mathbf{B(OC_2H_5)_3 \uparrow} + 3H_2O$

Step 3: The Green Flame
The product, Triethyl Borate, is a highly volatile ester. As the ethanol burns, the triethyl borate vaporizes and burns alongside it. Boron atoms uniquely emit a characteristic, brilliant green light when electronically excited in a flame.

Final Answer: The volatile compound is Triethyl Borate [$B(OC_2H_5)_3$]. It is formed via the acid-catalyzed esterification of the generated boric acid with ethanol.

Problem 19: Amphoterism in Group III (Al vs Fe)

Both $Al^{3+}$ and $Fe^{3+}$ precipitate as gelatinous hydroxides ($Al(OH)_3$ white, $Fe(OH)_3$ brown) upon addition of $NH_4OH$. If an excess of strong Sodium Hydroxide ($NaOH$) is added to these precipitates, the $Al(OH)_3$ completely dissolves while the $Fe(OH)_3$ remains as a solid. Explain the chemical property causing this separation and write the formula of the soluble complex.

View Solution

Strategy: Evaluate the position of the metals on the periodic table. Aluminum is on the metalloid boundary, making its oxide/hydroxide amphoteric. Iron is a true transition metal.

Step 1: Iron(III) Hydroxide
$Fe(OH)_3$ is strictly basic. It will neutralize acids, but it has no ability to react with or accept additional $OH^-$ ions from a strong base. It completely ignores the excess $NaOH$ and remains insoluble.

Step 2: Aluminum(III) Hydroxide (Amphoterism)
$Al(OH)_3$ is amphoteric. While it acts as a base against acids, in the presence of a very strong base like $NaOH$, it acts as a Lewis acid. The electron-deficient Aluminum atom accepts an additional Hydroxide ligand to expand its coordination sphere.

Step 3: Complex Formation
$Al(OH)_3 + NaOH \rightarrow \mathbf{Na[Al(OH)_4] \text{ (Soluble)}}$
The formation of the soluble Sodium Tetrahydroxoaluminate(III) complex cleanly pulls the aluminum back into the aqueous phase, leaving the brown iron precipitate behind for easy physical separation.

Final Answer: $Al(OH)_3$ is amphoteric and dissolves in excess strong base by forming the soluble coordination complex Sodium Tetrahydroxoaluminate ($Na[Al(OH)_4]$). $Fe(OH)_3$ is strictly basic and cannot form such complexes.

Problem 20: Tin(II) vs Tin(IV) Redox Test

A solution is suspected to contain Tin(II) ions ($Sn^{2+}$). Mercuric Chloride ($HgCl_2$) solution is added dropwise. Initially, a stark white precipitate forms. Upon continuous addition of the Tin(II) solution, the white precipitate turns completely grey/black. Write the sequential redox equations explaining this color shift.

View Solution

Strategy: Tin(II) is a powerful reducing agent, desperate to reach the highly stable Tin(IV) state. It will sequentially reduce Mercury from $+2$ to $+1$, and finally to metallic $0$.

Step 1: First Reduction (White Precipitate)
The $Sn^{2+}$ donates two electrons to two $Hg^{2+}$ ions. The Tin is oxidized to $Sn^{4+}$. The Mercury is reduced to the $+1$ state, precipitating as insoluble Mercurous Chloride (Calomel), which is stark white.
$Sn^{2+} + 2HgCl_2 \rightarrow Sn^{4+} + \mathbf{Hg_2Cl_2 \downarrow \text{ (White)}} + 2Cl^-$

Step 2: Second Reduction (Grey/Black Precipitate)
If the reducing agent ($Sn^{2+}$) is in excess, it continues its attack. It reduces the solid white $Hg_2Cl_2$ further, pushing the Mercury all the way down to the $0$ oxidation state.
$Sn^{2+} + Hg_2Cl_2 \rightarrow Sn^{4+} + \mathbf{2Hg \downarrow \text{ (Grey/Black)}} + 2Cl^-$
The resulting finely divided elemental Mercury metal appears grey/black.

Final Answer: Tin(II) acts as a strong reducing agent. It first reduces $HgCl_2$ to white Mercurous Chloride ($Hg_2Cl_2$). Excess Tin(II) further reduces this to finely divided elemental Mercury ($Hg$), which is grey/black.

Problem 21: Bismuth Hydrolysis

A salt solution dissolved in concentrated $HCl$ is poured into a large beaker of pure distilled water. Instantly, a massive cloudy white precipitate forms. This precipitate dissolves upon adding concentrated $HCl$ back in. This is characteristic of Bismuth ($Bi^{3+}$) and Antimony ($Sb^{3+}$). Write the chemical equation for Bismuth and identify the name of the precipitate.

View Solution

Strategy: Highly charged cations of Group VA are susceptible to massive hydrolysis. Adding water drives the equilibrium toward the insoluble oxo-salt.

Step 1: The Reversible Hydrolysis
Bismuth chloride ($BiCl_3$) is stable only in highly acidic solutions. When heavily diluted with pure water, the massive excess of $H_2O$ acts as a nucleophile, stripping away chlorine atoms and replacing them with oxygen to form a basic salt.
$BiCl_3 + H_2O \rightleftharpoons \mathbf{BiOCl \downarrow \text{ (White)}} + 2HCl$

Step 2: Le Chatelier's Principle
- Pouring into water floods the reactant side, pushing the equilibrium hard to the right, precipitating Bismuth Oxychloride ($BiOCl$).
- Adding concentrated $HCl$ floods the product side with $H^+$ and $Cl^-$, violently driving the equilibrium back to the left, dissolving the precipitate back into soluble $BiCl_3$.

Final Answer: $BiCl_3 + H_2O \rightleftharpoons \mathbf{BiOCl \downarrow} + 2HCl$. The white precipitate is Bismuth Oxychloride. It forms due to rapid hydrolysis upon dilution.

Problem 22: Iron(III) Thiocyanate Complex (Blood Red)

To distinguish between $Fe^{2+}$ and $Fe^{3+}$, Ammonium Thiocyanate ($NH_4SCN$) is used. $Fe^{2+}$ yields no color, while $Fe^{3+}$ yields an incredibly intense blood-red coloration. What is the formula of the complex responsible, and why is this test famously used in "fake blood" magic tricks?

View Solution

Strategy: Evaluate the coordination complex formed. The intense color is due to a charge transfer band.

Step 1: Complex Formation
The $Fe^{3+}$ ion coordinates rapidly with the thiocyanate ligand ($SCN^-$). Unlike many coordination complexes that go all the way to 6 ligands, in standard dilute analytical tests, the mono-thiocyanato or di-thiocyanato complex dominates visually.
$Fe^{3+} + SCN^- \rightarrow \mathbf{[Fe(SCN)]^{2+}}$ (or strictly $[Fe(H_2O)_5(SCN)]^{2+}$)

Step 2: The Color Origin
This complex features a massive Ligand-to-Metal Charge Transfer (LMCT) band. An electron is optically excited from the ligand's molecular orbitals directly into the d-orbitals of the $Fe^{3+}$. LMCT transitions have extraordinarily high molar absorptivities, meaning even a microscopic trace of $Fe^{3+}$ produces a violently dark, opaque, blood-red color.

Step 3: Magic Trick Utility
Because it is instant, works in water, and looks exactly like blood, magicians can rub colorless $KSCN$ on their arm and use a plastic knife dipped in colorless $FeCl_3$ to "cut" themselves, instantly generating the red liquid.

Final Answer: The complex is the Pentaaquathiocyanatoiron(III) ion ($[Fe(H_2O)_5(SCN)]^{2+}$). The intense color is caused by a highly allowed Ligand-to-Metal Charge Transfer (LMCT).

Problem 23: Master Deduction 1 (The Anion Cascade)

A white salt X dissolves in water. Addition of $BaCl_2$ yields a white precipitate Y. Precipitate Y is completely insoluble in dilute $HCl$. When solid X is heated with concentrated $H_2SO_4$ and $MnO_2$, a pungent, greenish-yellow gas Z is evolved. Deduce the identities of the anion in X, the precipitate Y, and the gas Z.

View Solution

Strategy: The Barium test isolates sulfates/sulfites. The $HCl$ solubility test differentiates them. The $MnO_2$ oxidation isolates the specific halide.

Step 1: The Barium Test
A white precipitate with $BaCl_2$ that is insoluble in $HCl$ is the definitive proof of a Sulfate ($SO_4^{2-}$). (If it had dissolved with $SO_2$ gas, it would be a sulfite).
Therefore, precipitate Y is Barium Sulfate ($BaSO_4$), and the salt contains the Sulfate anion.

Step 2: The Halide Oxidation
Heating a salt with conc. $H_2SO_4$ and an oxidizing agent ($MnO_2$) is the classic test for halides. The strong acid generates $HX$, which the $MnO_2$ oxidizes to the free halogen gas ($X_2$).
The gas evolved is greenish-yellow and pungent. This perfectly matches the physical description of Chlorine gas ($Cl_2$). (Bromine would be red-brown, Iodine would be violet).
Therefore, the salt also contains the Chloride ($Cl^-$) anion.

Step 3: Conclusion
The salt is a mixed salt or a double salt containing both Chloride and Sulfate (e.g., $KCl \cdot MgSO_4$ or similar, though usually questions imply identifying the active ions). The anion pool contains both $Cl^-$ and $SO_4^{2-}$.

Final Answer: The anions are Sulfate ($SO_4^{2-}$) and Chloride ($Cl^-$). Precipitate Y is $BaSO_4$. Gas Z is Chlorine ($Cl_2$).

Problem 24: Master Deduction 2 (The Cation Cascade)

An acidic solution of salt M yields no precipitate with $H_2S$. Upon boiling off the $H_2S$, adding a drop of $HNO_3$, and introducing $NH_4Cl$ followed by $NH_4OH$, a gelatinous white precipitate N forms. Precipitate N dissolves entirely in excess $NaOH$. When the resulting solution is heated with solid $NH_4Cl$, the white precipitate N miraculously reappears. Deduce the cation in M.

View Solution

Strategy: Follow the Group separation scheme. Group II is absent. Group III yields a white amphoteric hydroxide. Verify the reprecipitation chemistry.

Step 1: Group Separation
- No ppt with acidic $H_2S$: Group II ($Cu, Hg, Pb, Cd, As, Sb, Sn$) is absent.
- Gelatinous white ppt with $NH_4OH/NH_4Cl$: This is the classic signal for Group III Aluminum ($Al^{3+}$). The precipitate N is $Al(OH)_3$. (Iron would be brown, Chromium would be green).

Step 2: Amphoterism
$Al(OH)_3$ is amphoteric. It dissolves in excess strong base ($NaOH$) to form the soluble Sodium tetrahydroxoaluminate complex:
$Al(OH)_3 + NaOH \rightarrow \mathbf{Na[Al(OH)_4]}$ (Soluble).

Step 3: The Reprecipitation Trick
Solid $NH_4Cl$ acts as a weak acid in solution ($NH_4^+ \rightleftharpoons NH_3 + H^+$). When heated with the highly basic $Na[Al(OH)_4]$ solution, the $NH_4^+$ ions aggressively neutralize the $OH^-$ ligands holding the complex together, pulling them off as water and evolving ammonia gas.
$Na[Al(OH)_4] + NH_4Cl \xrightarrow{\Delta} \mathbf{Al(OH)_3 \downarrow} + NaCl + NH_3 \uparrow + H_2O$
The destruction of the complex forces the Aluminum hydroxide to reprecipitate, definitively proving Aluminum.

Final Answer: The cation is Aluminum ($Al^{3+}$). Precipitate N is $Al(OH)_3$. It dissolves in $NaOH$ to form $Na[Al(OH)_4]$, which is destroyed by the acidic $NH_4Cl$ to reprecipitate $Al(OH)_3$.

Problem 25: The Ultimate Salt Identification

A pale green crystalline salt S dissolves in water.
1) The solution decolorizes acidified $KMnO_4$.
2) Addition of $BaCl_2$ yields a white precipitate insoluble in $HCl$.
3) Heating the solid salt with $NaOH$ yields a pungent gas that turns Nessler's reagent brown.
4) Adding $K_3[Fe(CN)_6]$ (Ferricyanide) to the salt solution yields a deep blue precipitate.
Deduce the complete, exact chemical formula of the double salt S.

View Solution

Strategy: Break down each individual test to identify the specific cations and anions present. A pale green double salt has a very famous identity in titration chemistry.

Step 1: Test 2 (The Anion)
A white precipitate with $BaCl_2$ that is insoluble in $HCl$ definitively proves the presence of the Sulfate ion ($SO_4^{2-}$).

Step 2: Test 3 (The First Cation)
Heating a salt with strong base ($NaOH$) to yield a pungent gas that turns Nessler's reagent brown is the absolute definitive test for Ammonia ($NH_3$). This proves the salt contains the Ammonium ion ($NH_4^+$).

Step 3: Test 1 & 4 (The Second Cation)
- Decolorizing acidified $KMnO_4$ means the salt contains a strong reducing agent.
- Reacting with Potassium Ferricyanide to yield a deep blue precipitate (Turnbull's Blue) definitively proves the presence of the Iron(II) ion ($Fe^{2+}$). The $Fe^{2+}$ easily reduces the $KMnO_4$. The pale green color of the original crystals is also highly characteristic of hydrated $Fe^{2+}$ salts.

Step 4: Assembling the Salt
The salt contains $Fe^{2+}$, $NH_4^+$, and $SO_4^{2-}$. This is a classic double salt used extensively in redox titrations because it resists air-oxidation much better than plain ferrous sulfate.

Final Answer: The salt is Ferrous Ammonium Sulfate (Mohr's Salt). Its exact formula is $FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O$.

Mastering the Analytical Matrix

Congratulations on conquering these 25 ultra-challenging problems in Qualitative Inorganic Analysis! Salt analysis is the ultimate test of your inorganic chemistry knowledge. You cannot rely on isolated facts; you must integrate solubility product principles, complexation thermodynamics (like Nessler's reagent or Cyanide complexes), and intense redox chemistry (like Chromyl Chloride or Brown Ring tests) simultaneously. Always follow the rigid hierarchy of Group separations to ensure no false-positives slip through. Keep honing your deductive logic, and visit Chemca.in for more elite masterclasses!

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