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Principles of Qualitative Analysis appeared 7 times across 3 years — 0.8% of Chemistry. This question is from Identification of Cations.

Year 2026 2025 2024 Total
Questions 1 1 5 7

A reagent which gives brilliant red precipitate with Nickel ions in basic medium is

Solution & Explanation

Related Formula
Ni²⁺ + 2dmg^- arrow [Ni(dmg)₂] (Rosy Red Complex)
Core Logic

In basic medium (typically using ammonium hydroxide), nickel(II) ions react cleanly with dimethylglyoxime (dmgH) to generate a stable, neutral coordination complex containing structural hydrogen bonding. This complex isolates out of solution as a highly signature brilliant rosy red precipitate.

Step 1: Finalization

Therefore, dimethyl glyoxime is universally utilized for the quantitative and qualitative determination of Ni²⁺ ions.

Pattern Recognition

Nickel plus dmg in alkaline solution yields a signature rosy red chelate stabilized via symmetrical internal hydrogen bonds.

Chapter Mix

Class 12 Chemistry: Principles of Qualitative Analysis

Reference Study Guides

More Principles of Qualitative Analysis Previous-Year Questions

Q62 jee_main_2026_24_january_evening Analysis of Group V Cations
In the Group analysis of cations, Ba²⁺ & Ca²⁺ are precipitated respectively as
  • A. sulphide & sulphide
  • B. hydroxide & carbonate
  • C. carbonate & carbonate
  • D. chromate & sulphide

Solution

Core Logic

In qualitative inorganic analysis, both Ba²⁺ and Ca²⁺ belong to Group V cations. To identify Ba²⁺ & Ca²⁺, the group reagent (NH₄)₂CO₃ + NH₄Cl (with NH₄OH) is used. As a result, they precipitate out as insoluble carbonates: BaCO₃ & CaCO₃.

Pattern Recognition

Group V cations (Ba²⁺, Sr²⁺, Ca²⁺) always precipitate as Carbonates upon adding (NH₄)₂CO₃ in an ammoniacal medium.

Chapter Mix

Class 11 Chemistry: Salt Analysis

Q37 jee_main_2025_07_april_morning Identification of Ammonium Ion
When a salt is treated with sodium hydroxide solution it gives gas X. On passing gas X through reagent Y a brown coloured precipitate is formed. X and Y respectively, are:
  • A. X = NH₃ and Y = HgO
  • B. X = NH₃ and Y = K₂HgI₄ + KOH
  • C. X = NH₄Cl and Y = KOH
  • D. X = HCl and Y = NH₄Cl

Solution

Related Formula
NH₄^+ + OH^- arrow NH₃ + H₂O
Core Logic
  • Treating an ammonium salt (containing NH₄^+) with NaOH releases Ammonia gas (X = NH₃):
NH₄^+ + OH^- arrow NH₃ + H₂O
  • Passing ammonia gas through Nessler's reagent (Y = K₂[HgI₄] + KOH) produces a distinct brown precipitate (known as iodide of Millon's base):
2[HgI₄]²⁻ + NH₃ + 3OH^- arrow HgO · Hg(NH₂)I (brown) + 7I^- + 2H₂O

Hence, X is NH₃ and Y is K₂HgI₄ + KOH.

Pattern Recognition

Classic confirmatory test for ammonium ion: heating with alkaline solutions yields NH₃ gas, which always produces a brown precipitate with alkaline Nessler's reagent.

Chapter Mix

Class 11 Chemistry: Qualitative Salt Analysis Class 12 Chemistry: p-Block Elements

Q jee_main_2024_29_january_evening Identification of Anions and Gases
On passing a gas, 'X', through Nessler's reagent, a brown precipitate is obtained. The gas 'X' is
  • A. H₂S
  • B. CO₂
  • C. NH₃
  • D. Cl₂

Solution

Related Formula
2K₂[HgI₄] + NH₃ + 3KOH arrow HgO (NH₂)I + 7KI + 2H₂O
Core Logic

Nessler's reagent (K₂[HgI₄] in alkaline solution) reacts explicitly with ammonia gas (NH₃) or ammonium ions to create an insoluble iodide of Millon's base (HgO (NH₂)I). This characteristic complex showcases a signature brown structural coloration.

Step 1: Finalization

Thus, the unknown tracking gas 'X' is uniquely identified as ammonia (NH₃).

Pattern Recognition

Ammonia detection relies directly upon alkaline potassium tetraiodomercurate(II) resulting in brown Millon's base precipitations.

Chapter Mix

Class 12 Chemistry: Principles of Qualitative Analysis

Q61 jee_main_2024_30_jan_morning Test for Sulphide Ion
Given below are two statements: Statement-I: The gas liberated on warming a salt with dil H₂SO₄, turns a piece of paper dipped in lead acetate into black, it is a confirmatory test for sulphide ion. Statement-II: In statement-I the colour of paper turns black because of formation of lead sulphite. In the light of the above statements, choose the most appropriate answer from the options given below:
  • A. Both Statement-I and Statement-II are false
  • B. Statement-I is false but Statement-II is true
  • C. Statement-I is true but Statement-II is false
  • D. Both Statement-I and Statement-II are true

Solution

Related Formula
Na₂S + H₂SO₄ arrow Na₂SO₄ + H₂S (CH₃COO)₂Pb + H₂S arrow PbS + 2CH₃COOH
Core Logic

When a salt containing sulphide is warmed with dilute H₂SO₄, H₂S gas is evolved. H₂S gas turns lead acetate paper black due to the formation of lead sulphide (PbS), not lead sulphite. Thus, Statement I is correct, but Statement II is incorrect.

Pattern Recognition

Lead acetate paper test is standard for H₂S gas, forming black PbS. Always watch out for the trap suffix: sulphite (SO₃²⁻) vs sulphide (S²⁻).

Chapter Mix

Class 11 Chemistry: Qualitative Analysis Class 11 Chemistry: p-Block Elements

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)