Salt Analysis

In Gries–Ilosvay test for nitrite ion $\left(\text{NO}_2^-\right)$, the reagents used are sulphanilic acid and $\alpha$-naphthylamine.

• In acidic medium, nitrite ion forms nitrous acid $HNO_2$, which diazotizes sulphanilic acid to give a diazonium salt.

• This diazonium salt then couples with $\alpha$-naphthylamine to form a coloured azo dye (usually reddish/pink).

So, Statement I is true and Statement II is also true.

Correct option: D (Both Statement I and Statement II are true)

In classical group analysis, Ba²⁺ and Ca²⁺ belong to Group V and are precipitated by (NH₄)₂CO₃ (in presence of NH₄Cl and NH₄OH) as their carbonates:

• Ba²⁺ → BaCO₃ (white ppt)

• Ca²⁺ → CaCO₃ (white ppt)

Correct option: D) carbonate & carbonate

$\mathrm{x}$ is a metal chloride that dissolves in water at room temperature (water soluble).

Examples of such chlorides are $ \mathrm{MgCl}_2$, $ \mathrm{AlCl}_3$, and $ \mathrm{CuCl}_2$.

$\mathrm{y}$ is a metal chloride that dissolves only a little in water at room temperature (sparingly soluble).

Examples of such chlorides are $ \mathrm{AgCl}$ and $ \mathrm{Hg}_2 \mathrm{Cl}_2$.

$\mathrm{z}$ is a metal chloride that does not dissolve well in cold water but dissolves in hot water.

An example of such a chloride is $ \mathrm{PbCl}_2$.

(1) Carboxylic acid gives efferve scence with sodium bicarbonate solution

(2) Phenolic-OH gives voilet coloured complex with Neutral $\mathrm{FeCl}_3$.

(3) Alcoholic-OH gives Red colour with cerric ammonium Nitrate.

(4) When alkaline $\mathrm{KMnO}_4$ reacts with an unsaturated compound (Alkene or alkyne) the purple colour of $\mathrm{KMnO}_4$ solution disappears, indicating positive test for unsaturation.

Concept - Those compounds which are more acidic than $\mathrm{H}_2 \mathrm{CO}_3$ can gives effervescence of $\mathrm{CO}_2$ with aq. $\mathrm{NaHCO}_3$.

Release $\mathrm{CO}_2 \uparrow$ gas with aq. NaHCO3

$\Rightarrow[\text { A.S. }]_{\text {Comp. }}>[\text { A.S. }]_{\mathrm{H}_2 \mathrm{CO}_3}$

$\rightarrow$ Option 1, 2 and 3 gives effervescence of $\mathrm{CO}_2$ gas with $\mathrm{NaHCO}_3$

$\rightarrow$ Option (4) Not gives $\mathrm{CO}_2$ gas with $\mathrm{NaHCO}_3$.

Mohr's salt is chemically represented as $ \mathrm{FeSO}_4 \cdot\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4 \cdot 6 \mathrm{H}_2 \mathrm{O} $.

To understand its composition, Mohr's salt dissociates into the following ions:

Fe$^ {2+} $ (ferrous ion)

2 NH$_4^+ $ (ammonium ions)

2 SO$_4^{2-} $ (sulfate ions)

Thus, Mohr's salt contains ferrous, ammonium, and sulfate ions, but also has water molecules that hydrate these ions.

Using Kohlrausch's Law of independent ionic mobility, the molar conductance at infinite dilution for an electrolyte is the sum of the ionic conductances of its constituent ions:

$ \lambda_{\mathrm{m}}^{\infty}(\text{Mohr's salt}) = \mathrm{x}_1 + 2\mathrm{x}_2 + 2\mathrm{x}_3 $

Here:

$ x_1 $ represents the molar conductance of the ferrous ion (Fe$^ {2+} $),

$ x_2 $ represents the molar conductance of the ammonium ion (NH$_4^+$),

$ x_3 $ represents the molar conductance of the sulfate ion (SO$_4^{2-}$).

Therefore, statement II should include the correct expression $ x_1 + 2x_2 + 2x_3 $ to accurately describe the molar conductance for Mohr's salt.

$\mathrm{NH}_3$ is identify by $\mathrm{K}_2\left[\mathrm{HgI}_4\right]+\mathrm{KOH}$

The primary standard is a highly pure stable compound with a known exact composition that can be accurately weighed and dissolved to creat a solution of known concentration.

$\mathrm{NaOH}$ is hygroscopic and can't be used.

$\mathrm{FeSO}_4 \cdot 6 \mathrm{H}_2 \mathrm{O}$ is unstable and can be easily oxidised.

$\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is hygroscopic and can't be used.

$\underset{\substack{\text { Sodium } \\ \text { sulphide }}}{\mathrm{Na}_2 \mathrm{S}}+\underset{\substack{\text { Sodium nitro } \\ \text { Prusside }}}{\mathrm{Na}_2\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right]}\rightarrow \underset{\substack{\text { Violet Colour } \\ }}{\mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}\right]}$

$2 \mathrm{CuSO}_4+\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \xrightarrow{\mathrm{CH}_3 \mathrm{COOH}}\begin{aligned} &\mathrm{Cu}_2\left[\mathrm{Fe}(\mathrm{CN})_6\right]+2 \mathrm{~K}_2 \mathrm{SO}_4\\ &\text { (Chocolate brown ppt.) } \end{aligned}$

$4 \mathrm{FeCl}_3+3 \mathrm{~K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \longrightarrow \begin{aligned} &\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3+12 \mathrm{KCl}\\ &\text { (Prussian Blue ppt.) } \end{aligned}$

$3 \mathrm{ZnCl}_4+2 \mathrm{~K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \xrightarrow{\mathrm{NH}_4 \mathrm{OH}}\underset{\text { (White or bluish white ppt.) }}{\mathrm{K}_2 \mathrm{Zn}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2+6 \mathrm{KCl}}$

We want to identify which among the listed sulfides are yellow in color:

$\mathbf{(NH_4)_2S}$

Ammonium sulfide is typically encountered as an aqueous solution that can be colorless to slightly yellowish in solution form. It is not known as a distinct “yellow sulfide” compound in the same sense as solid colored sulfides like orpiment (see below).

$\mathbf{PbS}$ (lead(II) sulfide)

$ \text{PbS} $ (Galena) is well-known to be black or grayish‐black in color.

$\mathbf{CuS}$ (copper(II) sulfide)

$ \text{CuS} $ is black (covellite).

$\mathbf{As_2S_3}$ (arsenic(III) sulfide)

$ \text{As}_2\text{S}_3 $ (orpiment) is characteristically yellow.

$\mathbf{As_2S_5}$ (arsenic(V) sulfide)

$ \text{As}_2\text{S}_5 $ is also typically a yellow sulfide.

Hence, among the options given,

$\mathbf{(D)}\;As_2S_3$ and

$\mathbf{(E)}\;As_2S_5$

are the two yellow inorganic sulfides.

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Correct Choice

$ \boxed{\text{(D) and (E) only.}} $

$\mathrm{CoS} \xrightarrow{\text { Aquaregia }} \mathrm{CoCl}_2 \xrightarrow[2 \cdot \mathrm{AcOH}]{\mathrm{1.KNO}_2 / \mathrm{NH}_4 \mathrm{OH}} \mathrm{K}_3\left[\mathrm{Co}\left(\mathrm{NO}_2\right)_6\right] \downarrow \text { yellow }$

Let’s analyze each statement in the context of the classical (and standard) Lassaigne’s test:

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Statement (I)

“Nitrogen, sulphur, halogen, and phosphorus present in an organic compound are detected by Lassaigne's Test.”

This is true.

The Lassaigne’s test is used to detect N, S, X (halogens), and can also be adapted for P in organic compounds.

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Statement (II)

“The elements present in the compound are converted from covalent form into ionic form by fusing the compound with Magnesium in Lassaigne's test.”

This is false, because:

In the classical Lassaigne’s test, one fuses the organic compound with sodium metal (Na), not magnesium.

The purpose is indeed to convert elements from their covalent forms (e.g., C–N, C–S, C–X bonds) into their ionic forms (e.g., NaCN, Na₂S, NaX, etc.). But the metal used is sodium, not magnesium.

Hence, while the overall description of “converting covalent to ionic form” is correct, the statement incorrectly specifies “magnesium” instead of “sodium,” making it false as written.

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Conclusion

Statement (I) is True.

Statement (II) is False.

Therefore, the correct choice is:

(B) Statement I is true but Statement II is false.

(A) Borax bead test

$\mathrm{MSO}_4 \xrightarrow[\Delta]{\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7} \mathrm{M}\left(\mathrm{BO}_2\right)_2 \rightarrow \mathrm{MBO}_2 \rightarrow \mathrm{M}$

(B) Charcoal cavity test

$\mathrm{MSO}_4 \xrightarrow[\Delta]{\mathrm{Na}_2 \mathrm{CO}_3} \mathrm{MCO}_3 \rightarrow \mathrm{MO} \rightarrow \mathrm{M}$

(C) Cobalt nitrate test

$\mathrm{MCO}_3 \rightarrow \mathrm{MO} \xrightarrow[+\Delta]{\mathrm{CO}\left(\mathrm{NO}_3\right)_2} \mathrm{CoO} \cdot \mathrm{MO}$

(D) Flame test

$\mathrm{MCO}_3 \rightarrow \mathrm{MCl}_2 \rightarrow \mathrm{M}^{2+}$

So, $\mathrm{A} \rightarrow$ (III), $\mathrm{B} \rightarrow$ (IV), $\mathrm{C} \rightarrow$ (I), $\mathrm{D} \rightarrow$ (II).

The $\mathrm{Br}^{-}$ ion present in the salt gives pale yellow ppt. of $\mathrm{AgBr}$ with $\mathrm{AgNO}_3$, insoluble in dil $\mathrm{HNO}_3$ but partially soluble in aq. $\mathrm{NH}_4 \mathrm{OH}$.

$\mathrm{Br}^{-}+\mathrm{AgNO}_3 \rightarrow \underset{\text { Pale yellow }}{\mathrm{AgBr} \downarrow}+\mathrm{NO}_3^{-}$

$\mathrm{AgBr}+2 \mathrm{NH}_4 \mathrm{OH} \rightarrow \underset{\text { Partially soluble }}{\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right] \mathrm{Br}}+2 \mathrm{H}_2 \mathrm{O}$

$\begin{array}{lll} \mathrm{A} & \rightarrow \mathrm{Al}^{3+}(\mathrm{III}) \\ \mathrm{B} & \rightarrow \mathrm{Sr}^{2+}(\mathrm{IV}) \\ \mathrm{C} & \rightarrow \mathrm{Mn}^{2+}(\mathrm{I}) \\ \mathrm{D} & \rightarrow \mathrm{Pb}^{2+}(\mathrm{II}) \end{array}$

In qualitative analysis, the precipitation of the iron group (III) requires careful control of conditions to ensure selective precipitation of the desired ions. Ammonium chloride is added before ammonium hydroxide to achieve this control effectively.

Let's analyze the options to understand why ammonium chloride is used:

Option A: decrease concentration of $\mathrm{OH}^{-}$ ions

This option is correct. Adding ammonium chloride ($\mathrm{NH}_4\mathrm{Cl}$) introduces a common ion effect, where the concentration of $\mathrm{OH}^{-}$ ions is reduced due to the formation of $\mathrm{NH}_4\mathrm{OH}$. This reaction can be represented as:

$\mathrm{NH}_4^+ + \mathrm{OH}^- \leftrightarrow \mathrm{NH}_4\mathrm{OH}$

By reducing the concentration of $\mathrm{OH}^{-}$ ions, it prevents the premature precipitation of hydroxides of metals that might otherwise not selectively precipitate in the desired group.

Option B: increase concentration of $\mathrm{Cl}^{-}$ ions

While it is true that ammonium chloride increases the concentration of $\mathrm{Cl}^{-}$ ions, this is not the primary reason for its addition in this context. The focus is more on regulating $\mathrm{OH}^{-}$ ion concentration.

Option C: prevent interference by phosphate ions

This is not relevant in the context of adding ammonium chloride, as the main idea behind adding ammonium chloride is to control the $\mathrm{OH}^{-}$ ion concentration rather than dealing directly with phosphate ions.

Option D: increase concentration of $\mathrm{NH}_4^+$ ions

While ammonium chloride does increase the concentration of $\mathrm{NH}_4^+$ ions, which in turn helps in maintaining the $\mathrm{OH}^{-}$ concentration at a lower level, this is again a secondary consideration to the primary goal of controlling the $\mathrm{OH}^{-}$ ion concentration to ensure selective precipitation.

Therefore, the correct answer is:

Option A: decrease concentration of $\mathrm{OH}^{-}$ ions

Lead sulphate-white

Ammonium sulphide-soluble

Lead iodide-Bright yellow

Ammonium arsinomolybdate-yellow

$\mathrm{Na}_2 \mathrm{S}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{Na}_2 \mathrm{SO}_4+\mathrm{H}_2 \mathrm{S}$

$\mathrm{{(C{H_3}COO)_2}Pb + {H_2}S \to \mathop {PbS}\limits_{Black\,lead\,sulphide} + 2C{H_3}COOH}$

$\begin{aligned} & \mathrm{Fe}^{2+} \xrightarrow[\text { Conc. } \mathrm{H}_2 \mathrm{SO}_4]{\mathrm{H}^{+}} \mathrm{Fe}^{+3} \\ & \mathrm{Fe}^{+3} \xrightarrow{- \text { SCN }} \mathrm{Fe}(\mathrm{SCN})_3 \text { (blood red colour) } \end{aligned}$

Appearance of blood red colour indicates presence of both nitrogen and sulphur.

In the wet test for detection of Group 5 cations, including Ba²⁺, Ca²⁺, and Sr²⁺, ammonium carbonate ((NH₄)₂CO₃) is used as a group reagent. The reaction with barium ions (Ba²⁺) can be written as :

$\mathrm{Ba}^{2+} + \left(\mathrm{NH}_4\right)_2\mathrm{CO}_3 \rightarrow \underset{\text{(white precipitate)}}{\mathrm{BaCO}_3 \downarrow} + 2\mathrm{NH}_4^+$

Upon adding ammonium carbonate to the solution containing Ba²⁺ ions, a white precipitate of barium carbonate (BaCO₃) is formed, which can be used to identify the presence of barium ions in the sample.

The mixture of salts gives a dark brown ring when treated with ferrous sulphate in acidic medium. This is a characteristic reaction of nitrate ions (NO$_3^-$). The brown ring is due to the formation of a complex between NO and Fe$^{2+}$.

Additionally, the mixture gives a deep red color when treated with neutral ferric chloride (FeCl$_3$), which then disappears upon boiling to form a red-brown precipitate. This is a classic reaction of acetate ions (CH$_3$COO$^-$) with Fe$^{3+}$. Initially, the acetate ions form a complex with ferric ions, leading to the deep red color. Upon boiling, hydroxide ions displace some acetate ions in the complex, resulting in the red-brown precipitate.

Therefore, the mixture contains acetate (CH$_3$COO$^-$) and nitrate (NO$_3^-$) ions.

A confirmatory test for $Pb^{2+}$ ions involves the formation of a specific compound that indicates the presence of lead ions in a solution.

Let's go through the options:

Option A: Lead chromate ($\mathrm{PbCrO}_4$) forms a yellow precipitate, which is a confirmatory test for $Pb^{2+}$ ions.

Option B: Lead iodide ($\mathrm{PbI}_2$) forms a yellow precipitate, which is a confirmatory test for $Pb^{2+}$ ions.

Option C: Lead sulfate ($\mathrm{PbSO}_4$) forms a white precipitate, which is a confirmatory test for $Pb^{2+}$ ions.

Option D: Lead nitrate ($\mathrm{Pb(NO_3)_2}$) is soluble in water and does not form a precipitate under normal conditions. Thus, its formation is not used as a confirmatory test for $Pb^{2+}$ ions.

So, the correct answer is Option D: lead nitrate.

In qualitative inorganic analysis, cations are divided into different groups based on their precipitation reactions. Group IV cations typically include those that form sulfides or hydroxides that are insoluble in water and dilute acids, but soluble in excess ammonia or sodium hydroxide.

Co$^{2+}$ (Cobalt)

Ni$^{2+}$ (Nickel)

Zn$^{2+}$ (Zinc)

These cations are generally placed in Group IV because they form precipitates with specific reagents used in the separation process.

Fe$^{3+}$ (Iron), on the other hand, is not typically included in Group IV. It belongs to Group III because it forms insoluble hydroxides that precipitate in the presence of ammonia.

Thus, the transition element cation that doesn't belong to Group IV is:

Option D

Fe$^{3+}$

	Element	$\begin{aligned}  &  \text { Colour imparted to } \\ &  \text { the flame }\end{aligned}$
(A)	$\mathrm{K}$	Violet
(B)	$\mathrm{Ca}$	Brick red
(C)	$\mathrm{Sr}$	Crimson red
(D)	$\mathrm{Ba}$	Apple green

Cations	Group No.	Group reagent
$\mathrm{Pb}^{+2}, \mathrm{Cu}^{+2}$	II	$\mathrm{H}_2 \mathrm{S}(\mathrm{g}) \text { in presence of } \text { dil HCl }$
$\mathrm{Al}^{+3}, \mathrm{Fe}^{+3}$	III	$\begin{aligned}  &  \mathrm{NH}_4 \mathrm{OH} \text { in presence of } \\ &  \mathrm{NH}_4 \mathrm{Cl}\end{aligned}$
$\mathrm{CO}^{+2}, \mathrm{Ni}^{+2}$	$\mathrm{IV}$	$\begin{aligned}  &  \mathrm{H}_2 \mathrm{~S} \text { in presence of } \\ &  \mathrm{NH}_4 \mathrm{OH}\end{aligned}$
$\mathrm{Ba}^{+2}, \mathrm{Ca}^{+2}$	$\mathrm{~V}$	$\begin{aligned}  &  \left(\mathrm{NH}_4\right)_2 \mathrm{CO}_3 \text { in presence } \\ &  \text { of } \mathrm{NH}_4 \mathrm{OH}\end{aligned}$

Anion is $\mathrm{SO}_{3}^{-2}$

$ \mathrm{BaSO}_{3} \stackrel{\text { dil } \mathrm{HCl}}{\longrightarrow} \underset{x(\text { gas })}{\mathrm{SO}_{2} \uparrow} $

Gas is released with smell of burning sulphur.

$\mathrm{CO}_{3}^{2-}$ : On action of dil sulphuric acid, $\mathrm{CO}_{2}$ gas is released which turns lime water milky.

$\mathrm{S}^{2-} $ : On action of dil sulphuric acid, $\mathrm{H}_{2} \mathrm{~S}$ gas is released which turns lead acetate paper black.

$\mathrm{SO}_{3}^{2-}$ : On action of dil $\mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{SO}_{2}$ gas is evolved which turns acidified potassium dichromate solution green.

$\mathrm{NO}_{2}^{-}$: On action of dil $\mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{NO}_{2}$ gas is evolved which turns $\mathrm{KI}$ solution containg starch blue.

During test for halogen, if nitrogen or sulphur is also present in the compound, then sodium fusion extract is first boiled with concentrated nitric acid to decompose cyanide or sulphide of sodium formed during Lassaigne's test.

Cupric salts give green flame with blue centre. The colour of other salts are

$\begin{array}{ll}\mathrm{Sr}^{2+} & \text { Crimson red } \\ \mathrm{Ca}^{2+} & \text { Brick red } \\ \mathrm{Ba}^{2+} & \text { Green }\end{array}$

On addition of dimethylglyoxime to alkaline solution of $\mathrm{Ni}^{+2}$, a bright red ppt. is obtained.

$\mathrm{Ni}^{+2}+2 \mathrm{dmg} \rightarrow\left[\mathrm{Ni}(\mathrm{dmg})_{2}\right]^{+2}$ (Bright red ppt)

Mn²⁺ $\to$ IV group

As³⁺ $\to$ II B group

Cu²⁺ $\to$ II A group

Al³⁺ $\to$ III group

Note :

IV^th group cations – Zn²⁺ ,Mn²⁺ ,Ni²⁺ , Co²⁺

II B group cations – As³⁺ , Sb³⁺ , Sn²⁺

II A group cations – Hg²⁺ , Pb²⁺ , Bi³⁺ , Cu²⁺ , Cd²⁺

III^rd group cations – Al³⁺ , Cr³⁺ , Fe³⁺

Al³⁺ and Fe³⁺ sulphides hydrolyse in water.

Ni²⁺ and Zn²⁺ require basic medium with H₂S to form ppt

Ca²⁺ and Ba²⁺ sulphides are soluble

hence, we will receive only CuS ppt.

When a copper(II) salt (containing Cu²⁺) is treated with potassium iodide (KI), the cupric ions (Cu²⁺) are reduced to copper(I) (Cu⁺), and iodine (I₂) is released. The solid product formed in the reaction is copper(I) iodide, CuI.

The simplified reaction is:

$ \text{2 Cu}^{2+} + 4 \text{I}^{-} \;\longrightarrow\; 2 \,\text{CuI} \!\downarrow + \text{I}_{2} $

So, the correct answer is:

Option C: CuI.

Statement I is true. Dimethyl glyoxime (DMG) is a chelating reagent commonly used for the identification of Ni²⁺ ions in aqueous solution. In the presence of ammonium hydroxide (NH₄OH), Ni²⁺ reacts with dimethyl glyoxime to form a deep red-colored complex, which can be used to confirm the presence of Ni²⁺ ions.

Statement II is also true. Dimethyl glyoxime is a neutral bidentate ligand, which means it has two donor atoms that can coordinate with a metal ion, and its overall charge is neutral.

Lucas reagent – Conc. HCl/ZnCl₂

Dumas method – CuO/CO₂

Kjeldahl’s method – H₂SO₄

Hinsberg test – C₆H₅SO₂Cl/aq. KOH

m-chlorobenzoic acid dissolves in aq NaHCO₃ while m-chlorophenol dissolves in aq NaOH.

Solution containing Zn²⁺ ions forms white precipitate on reaction with H₂S gas.

Sodium nitroprusside test is used for detection of sulphur in organic compound.

$\stackrel{+}{\mathrm{NO}}$ ligand is converted into $\mathrm{NOS}^{-}$ligand.

P) Passing $\mathrm{H}_2 \mathrm{~S}$ in the presence of $\mathrm{NH}_4 \mathrm{OH}$ It is fourth group reagent. The metal ions (cations) included in this group are $\mathrm{Mn}^{2+}, \mathrm{Zn}^{2+}, \mathrm{Ni}^{2+}$ and $\mathrm{Co}^{2 \mathrm{t}}$.This group reagent is used for precipitating the cations $\mathrm{Mn}^{2+}, \mathrm{Zn}^{2+}, \mathrm{Ni}^{2+}$ and $\mathrm{C}_0{ }^{2+}$. They are precipitated as sulfides.From the given cations, $\mathrm{Mn}^{2+}$ is precipitated by passing $\mathrm{H}_2 \mathrm{~S}$ in the presence of $\mathrm{NH}_4OH$ Precipitate of $\mathrm{Mn}^{2+}$ is MnS . $ P-3 $ Q) $\left(\mathrm{NH}_4\right)_2 \mathrm{CO}_3$ in the presence of $\mathrm{NH}_4 \mathrm{OH}$ It is fifth group reagent. The cations, included in group $\bar{V}$ are $\mathrm{Ba}^{2+}, \mathrm{Ca}^{2+}$ and $\mathrm{Sr}^{2+}$. These cations are precipitated using the reagent $\left(\mathrm{NH}_4\right)_2 \mathrm{CO}_3$ in the presence of. $\mathrm{NH}_4 \mathrm{OH}$. These cations are precipitated as carbonates. From the given cations, $\mathrm{Ba}^{2+}$ is precipitated by using $\left(\mathrm{NH}_4\right)_2 \mathrm{CO}_3$ in the presence of $\mathrm{NH}_4 \mathrm{OH}$. Precipitate of $\mathrm{Ba}^{2+}$ is $\mathrm{BaCO}_3$. $ Q-4 $R) $\mathrm{NH}_4 \mathrm{OH} \,\,{\text {in }}$ the presence of $\mathrm{NH}_4 \mathrm{Cl}$ It is group III reagent. Group III cations are $\mathrm{Al}^{3+}, \mathrm{Cr}^{3+}$ and $\mathrm{Fe}^{3+}$. These cations get precipitated by using $\mathrm{NH}_4 \mathrm{OH}$ with $\mathrm{NH}_4 \mathrm{Cl}$. The cations are precipitated as hydroxides. From the given cations, $\mathrm{Al}^{3+}$ is precipitated by using $\mathrm{NH}_4 \mathrm{OH}$ in the presence of $\mathrm{NH}_4 \mathrm{Cl}$. Precipitate of $\mathrm{Al}^{3+}$ is $\mathrm{Al}(\mathrm{OH})_3$ $ R-2 $ S) Passing $\mathrm{H}_2 \mathrm{~S}$ in the presence of dilute HCl It is group II reagent. Group II cations are $\mathrm{Cu}^{2+}, \mathrm{Cd}^{2+}, \mathrm{Hg}^{2+}, \mathrm{Bi}^{2+}, \mathrm{Pb}^{2+}, \mathrm{Sn}^{2+}, \mathrm{Sn}^{4+}, \mathrm{Sb}^{3+}, \mathrm{As}^{3+}$ and $A s^{5+}$. They are precipitated as sulfides that are insoluble in dilute HCl . From the given cations, $\mathrm{Cu}^{2+}$ is precipitated by passing $H_2 S$ in the presence of dilute HCl Precipitate of $\mathrm{Cu}^{2+}$ is Cus. $ S-1 $ $ \begin{aligned} &\text { Answer: Option (A) }\\ &P \rightarrow 3, Q \rightarrow 4, \quad R \rightarrow 2, \quad S \rightarrow 1 \end{aligned} $

Precipitate X is K₂Fe[Fe(CN)₆]

Brown ring is due to the formation of [Fe(NO)(H₂O)₅]²⁺

The reactions involved are

$N{i^{2 + }} + HCl \to \mathop {{{\left[ {NiC{l_4}} \right]}^2}}\limits_{Tetrahedral} $

$N{i^{2 + }} + KCN \to \mathop {{{\left[ {Ni{{(CN)}_4}} \right]}^{2 - }}}\limits_{Square\,planar} $

The reaction involved is

Zn²⁺ + OH^$-$ $\to$ Zn(OH)₂ $\buildrel {O{H^ - }} \over \longrightarrow $ [Zn(OH)₄]^2$-$ (soluble)

Pb salts form white precipitate of PbCl₂ with dil. HCl, which dissolves in hot water.

The reactions involved are as follows:

$\mathop {C{r^{3 + }}}\limits_{(Q)} + 3N{H_4}OH\buildrel {{H_2}S} \over \longrightarrow \mathop {Cr{{(OH)}_3}}\limits_{(R)\,(Green\,ppt.)} + 3NH_4^ + $

$\mathop {2Cr{{(OH)}_3}}\limits_{(R)} + 3{H_2}{O_2} + 4NaOH \to \mathop {2N{a_2}Cr{O_4} + 8{H_2}O}\limits_{(S)\,(Yellow\,solution)} $