In the Group analysis of cations, Ba2+ & Ca2+ are precipitated respectively as

Q: In the Group analysis of cations, Ba2+ & Ca2+ are precipitated respectively as

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. Group V cations (Ba2+, Sr2+, Ca2+) always precipitate as Carbonates upon adding (NH4)2CO3 in an ammoniacal medium. Class 11 Chemistry: Salt Analysis

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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 $NH₄^+$ $\mathrm{NH}_4^+$) with $NaOH$ $\mathrm{NaOH}$ releases Ammonia gas ( $X = NH₃$ $\mathrm{X} = \mathrm{NH}_3$):

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 \cdot Hg(NH₂)I (brown) + 7I^- + 2H₂O

Hence, $X$ $\mathrm{X}$ is $NH₃$ $\mathrm{NH}_3$ and $Y$ $\mathrm{Y}$ is $K₂HgI₄ + KOH$ .

Pattern Recognition

Classic confirmatory test for ammonium ion: heating with alkaline solutions yields $NH₃$ $\mathrm{NH}_3$ 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. $H₂S$ $\mathrm{H}_2\mathrm{S}$
B. $CO₂$ $\mathrm{CO}_2$
C. $NH₃$ $\mathrm{NH}_3$
D. $Cl₂$ $\mathrm{Cl}_2$

Solution

Related Formula

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

Core Logic

Nessler's reagent ( $K₂[HgI₄]$ $\text{K}_2[\text{HgI}_4]$ in alkaline solution) reacts explicitly with ammonia gas ( $NH₃$ $\text{NH}_3$) 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 ( $NH₃$ $\text{NH}_3$).

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

Q67 jee_main_2024_29_january_evening Identification of Cations

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

A. sodium nitroprusside
B. neutral $FeCl₃$ $\mathrm{FeCl}_3$
C. meta-dinitrobenzene
D. dimethyl glyoxime

Solution

Related Formula

Ni²⁺ + 2dmg^- arrow [Ni(dmg)₂] (Rosy Red Complex)

Core Logic

In basic medium (typically using ammonium hydroxide), nickel( $II$ $II$) ions react cleanly with dimethylglyoxime ( $dmgH$ $\text{dmgH}$) 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 $Ni²⁺$ $\text{Ni}^{2+}$ 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

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 $H₂SO₄$ $H_2SO_4$, $H₂S$ $H_2S$ gas is evolved. $H₂S$ $H_2S$ gas turns lead acetate paper black due to the formation of lead sulphide ( $PbS$ $PbS$), not lead sulphite. Thus, Statement I is correct, but Statement II is incorrect.

Pattern Recognition

Lead acetate paper test is standard for $H₂S$ $H_2S$ gas, forming black $PbS$ $PbS$. Always watch out for the trap suffix: sulphite ( $SO₃²⁻$ $SO_3^{2-}$) vs sulphide ( $S²⁻$ $S^{2-}$).

Chapter Mix

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

In the Group analysis of cations, Ba²⁺$Ba^{2+}$ & Ca²⁺$Ca^{2+}$ are precipitated respectively as

Solution & Explanation

Core Logic

Pattern Recognition

Chapter Mix

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In the Group analysis of cations, $Ba²⁺$ $Ba^{2+}$ & $Ca²⁺$ $Ca^{2+}$ are precipitated respectively as