Salt Analysis

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 :

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.

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

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

ions This option is correct. Adding ammonium chloride (

) introduces a common ion effect, where the concentration of

ions is reduced due to the formation of

. This reaction can be represented as:

By reducing the concentration of

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

ions While it is true that ammonium chloride increases the concentration of

ions, this is not the primary reason for its addition in this context. The focus is more on regulating

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

ion concentration rather than dealing directly with phosphate ions. Option D: increase concentration of

ions While ammonium chloride does increase the concentration of

ions, which in turn helps in maintaining the

concentration at a lower level, this is again a secondary consideration to the primary goal of controlling the

ion concentration to ensure selective precipitation.

Therefore, the correct answer is: Option A: decrease concentration of

ions

The

ion present in the salt gives pale yellow ppt. of

with

, insoluble in dil

but partially soluble in aq.

.

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

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.

Conclusion Statement (I) is True.

Statement (II) is False.

Therefore, the correct choice is: (B) Statement I is true but Statement II is false.

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.

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

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

$$2 \mathrm{CuSO}_4+\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \xrightarrow{\mathrm{CH}_3 \mathrm{COOH}}\begin{aligned} &\mathrm{Cu}_2\left[\mathrm{Fe}(\mathrm{CN})_6\right]+2 \mathrm{~K}_2 \mathrm{SO}_4\\ &\text{ (Chocolate brown ppt.) } \end{aligned}$$ $$4 \mathrm{FeCl}_3+3 \mathrm{~K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \longrightarrow \begin{aligned} &\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3+12 \mathrm{KCl}\\ &\text{ (Prussian Blue ppt.) } \end{aligned}$$ $3 \mathrm{ZnCl}_4+2 \mathrm{~K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \xrightarrow{\mathrm{NH}_4 \mathrm{OH}}\underset{\text{ (White or bluish white ppt.) }}{\mathrm{K}_2 \mathrm{Zn}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2+6 \mathrm{KCl}}$