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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%)

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

Solution & Explanation

### Related Formula mathrmNi^2+ + 2mathrmdmg^- rightarrow [mathrmNi(mathrmdmg)_2] downarrow quad text(Rosy Red Complex) ### Core Logic In basic medium (typically using ammonium hydroxide), nickel(II) ions react cleanly with dimethylglyoxime (textdmgH) 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 textNi^2+ ions. ### Pattern Recognition Nickel plus dmg in alkaline solution yields a signature rosy red chelate stabilized via symmetrical internal hydrogen bonds. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Principles of Qualitative Analysis

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Principles Of Qualitative Analysis

More Principles of Qualitative Analysis Previous-Year Questions

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. mathrmX = mathrmNH_3text and mathrmY = mathrmHgO
  • B. mathrmX = mathrmNH_3text and mathrmY = mathrmK_2mathrmHgI_4 + mathrmKOH
  • C. mathrmX = mathrmNH_4mathrmCltext and mathrmY = mathrmKOH
  • D. mathrmX = mathrmHCltext and mathrmY = mathrmNH_4mathrmCl

Solution

### Related Formula mathrmNH_4^+ + mathrmOH^- rightarrow mathrmNH_3uparrow + mathrmH_2mathrmO ### Core Logic 1. Treating an ammonium salt (containing mathrmNH_4^+) with mathrmNaOH releases Ammonia gas (mathrmX = mathrmNH_3): mathrmNH_4^+ + mathrmOH^- rightarrow mathrmNH_3uparrow + mathrmH_2mathrmO 2. Passing ammonia gas through Nessler's reagent (mathrmY = mathrmK_2[mathrmHgI_4] + mathrmKOH) produces a distinct brown precipitate (known as iodide of Millon's base): 2[mathrmHgI_4]^2- + mathrmNH_3 + 3mathrmOH^- rightarrow mathrmHgO cdot mathrmHg(NH_2)I downarrow (textbrown) + 7mathrmI^- + 2mathrmH_2mathrmO Hence, mathrmX is mathrmNH_3 and mathrmY is mathrmK_2mathrmHgI_4 + mathrmKOH. ### Pattern Recognition Classic confirmatory test for ammonium ion: heating with alkaline solutions yields mathrmNH_3 gas, which always produces a brown precipitate with alkaline Nessler's reagent. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Qualitative Salt Analysis Class 12 Chemistry: p-Block Elements
Q70 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. mathrmH_2mathrmS
  • B. mathrmCO_2
  • C. mathrmNH_3
  • D. mathrmCl_2

Solution

### Related Formula 2mathrmK_2[mathrmHgI_4] + mathrmNH_3 + 3mathrmKOH rightarrow mathrmHgOcdotmathrmHg(mathrmNH_2)mathrmI downarrow + 7mathrmKI + 2mathrmH_2mathrmO ### Core Logic Nessler's reagent (K_2[HgI_4] in alkaline solution) reacts explicitly with ammonia gas (NH_3) or ammonium ions to create an insoluble iodide of Millon's base (textHgOcdottextHg(textNH_2)textI). This characteristic complex showcases a signature brown structural coloration. ### Step 1: Finalization Thus, the unknown tracking gas 'X' is uniquely identified as ammonia (NH_3). ### Pattern Recognition Ammonia detection relies directly upon alkaline potassium tetraiodomercurate(II) resulting in brown Millon's base precipitations. ### Evaluation Rubric / Model Answer null ### 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 H_2SO_4, 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. textBoth Statement-I and Statement-II are false
  • B. textStatement-I is false but Statement-II is true
  • C. textStatement-I is true but Statement-II is false
  • D. textBoth Statement-I and Statement-II are true

Solution

### Related Formula Na_2S + H_2SO_4 rightarrow Na_2SO_4 + H_2S (CH_3COO)_2Pb + H_2S rightarrow PbS + 2CH_3COOH ### Core Logic When a salt containing sulphide is warmed with dilute H_2SO_4, H_2S gas is evolved. H_2S 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 H_2S gas, forming black PbS. Always watch out for the trap suffix: sulphite (SO_3^2-) vs sulphide (S^2-). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Qualitative Analysis Class 11 Chemistry: p-Block Elements
Q64 jee_main_2024_31_jan_morning Precipitate Colors
The compound that is white in color is
  • A. textammonium sulphide
  • B. textlead sulphate
  • C. textlead iodide
  • D. textammonium arsinomolybdate

Solution

### Core Logic Analyzing the colors of the given salts: Lead sulphate (PbSO_4) is white. Ammonium sulphide ((NH_4)_2S) is soluble/colorless in solution. Lead iodide (PbI_2) is bright yellow. Ammonium arsinomolybdate is yellow. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Salt Analysis

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