PRACTICAL CHEMISTRY MCQs for NEET UG CHEMISTRY — Medium

1 Medium

Analyzing +4 | -1

On heating colourless solid

A,

gas

B

(liquid at room temp.) and gas

C

are formed.

A

is decomposed by

\mathrm{NaOH}

on heating to form gas

D

giving white fumes with

\mathrm{HCl}.

Gas

{C}

occupies

800 \mathrm{~ml} / \mathrm{g}

at N.T.P. Hence

A

is

View answer and explanation

Correct option: D

\mathrm{NH}_4 \mathrm{NO}_2

Option D (

\mathrm{NH}_4 \mathrm{NO}_2

) is correct because ammonium nitrite decomposes upon heating to give nitrogen gas and water, and reacts with NaOH to produce ammonia gas.

Solution approach

Correct Answer + Core Reason: $\mathrm{NH}_4 \mathrm{NO}_2$ . It decomposes to give ${N}_2$ and ${H}_2{O}$ , and reacts with ${NaOH}$ to give ${NH}_3$ .

Step-by-Step Solution:

Step 1: Analyze the decomposition products of A.

Step 2: A decomposes to gas B (liquid at room temperature, likely water) and gas C.

Step 3: A is decomposed by ${NaOH}$ to form gas D, which gives white fumes with ${HCl}$ (likely ${NH}_3$ ).

Step 4: Gas C occupies $800 \mathrm{~ml} / \mathrm{g}$ at N.T.P. This suggests that gas C is ${N}_2$ (molar volume is $22.4 \mathrm{L} / \mathrm{mol}$ , molar mass is 28 g/mol, so volume per gram is $22400 / 28 = 800 \mathrm{ml} / \mathrm{g}$ ).

Step 5: Based on the products, A is likely ammonium nitrite ( $\mathrm{NH}_4 \mathrm{NO}_2$ ).
$\mathrm{NH}_4 \mathrm{NO}_2 \rightarrow {N}_2 + 2{H}_2{O}$
$\mathrm{NH}_4 \mathrm{NO}_2 + {NaOH} \rightarrow {NaNO}_2 + {NH}_3 + {H}_2{O}$

Why other options are wrong:
- Option A: $\mathrm{NH}_4 \mathrm{NO}_3$ is wrong because it decomposes to ${N}_2{O}$ and ${H}_2{O}$ , not ${N}_2$ and ${H}_2{O}$ .
- Option B: $\mathrm{NaNO}_2$ is wrong because it does not decompose to give a gas C and a liquid B.
- Option C: $\mathrm{NH}_4 \mathrm{Cl}$ is wrong because it does not decompose to give gas C and liquid B.

Key Concepts:
- Decomposition reactions
- Reactions of ammonium salts
- Qualitative analysis

Memory Aids:
- Remember the decomposition products of ammonium salts.

Quick Tips:
- Analyze the reaction products to identify the unknown compound.

2 Medium

2 of 50Question permalink

Application +4 | -1

Standard

HCl

solution is titrated to a pink phenolphthalein endpoint by adding a

\mathrm{NaOH}

solution while stirring. If a solution becomes pink throughout but loses its color upon standing for a short time, what should be done to restore the color?

View answer and explanation

Correct option: B

Add an additional drop of

\mathrm{NaOH}

solution

The correct answer is B because the pink color fading indicates the solution is becoming less basic, so adding NaOH will restore the basic conditions needed for phenolphthalein to be pink.

Solution approach

Correct Answer + Core Reason: Add an additional drop of $\mathrm{NaOH}$ solution. The pink color fades because the solution is losing its basicity.

Step-by-Step Solution:

Step 1: Understand the behavior of phenolphthalein indicator.

Step 2: Phenolphthalein is pink in basic solutions and colorless in acidic solutions.

Step 3: The pink color fading indicates the solution is becoming less basic.

Step 4: Adding $\mathrm{NaOH}$ will increase the basicity and restore the pink color.

Why other options are wrong:
- Option A: Adding more phenolphthalein will not change the pH of the solution.
- Option C: Adding $\mathrm{HCl}$ will make the solution more acidic and further diminish the pink color.
- Option D: Stirring will not change the pH of the solution.

Key Concepts:
- Acid-base indicators
- Phenolphthalein
- pH and basicity

Memory Aids:
- Phenolphthalein: Pink in base, colorless in acid

Quick Tips:
- Consider the pH range of the indicator and the effect of adding acid or base.

3 Medium

3 of 50Question permalink

Understanding +4 | -1

In V group,

\left(\mathrm{NH}_4\right)_2 \mathrm{CO}_3

is added to precipitate out the carbonates. We do not add

\mathrm{Na}_2 \mathrm{CO}_3

because

View answer and explanation

Correct option: C

\mathrm{MgCO}_3

will be precipitated out in

\mathrm{V}

group

Option C (

\mathrm{MgCO}_3

will be precipitated out in

\mathrm{V}

group) is correct because

\mathrm{Na}_2\mathrm{CO}_3

would increase the concentration of carbonate ions, leading to the precipitation of

\mathrm{MgCO}_3

along with the group V carbonates, interfering with the analysis.

Solution approach

Correct Answer + Core Reason: $\mathrm{MgCO}_3$ will be precipitated out in $\mathrm{V}$ group. $\mathrm{Na}_2\mathrm{CO}_3$ would precipitate $\mathrm{MgCO}_3$ along with Group V carbonates.

Step-by-Step Solution:

Step 1: In Group V analysis, the cations are precipitated as carbonates using ammonium carbonate (( $\mathrm{NH}_4)_2\mathrm{CO}_3$ ).

Step 2: Magnesium ions ( $Mg^{2+}$ ) are also present in the solution.

Step 3: If sodium carbonate ( $\mathrm{Na}_2\mathrm{CO}_3$ ) is used instead of ammonium carbonate, the higher concentration of carbonate ions will cause magnesium carbonate ( $MgCO_3$ ) to precipitate along with the Group V carbonates.

Step 4: This would interfere with the subsequent analysis of Group V cations.

Step 5: Ammonium carbonate is preferred because the ammonium ions suppress the ionization of magnesium hydroxide, preventing the precipitation of magnesium hydroxide or carbonate.

Why other options are wrong:
- Option A: $\mathrm{CaCO}_3$ is not soluble in $\mathrm{Na}_2\mathrm{CO}_3$ .
- Option B: $\mathrm{Na}_2\mathrm{CO}_3$ does not increase the solubility of Group V carbonates; it decreases it by the common ion effect.
- Option D: The correct answer is C.

Key Concepts:
- Qualitative analysis
- Group V cations
- Precipitation reactions
- Common ion effect

Memory Aids:
- Remember that magnesium ions can interfere with Group V analysis if sodium carbonate is used.

Quick Tips:
- Understand the role of each reagent in qualitative analysis.

4 Medium

4 of 50Question permalink

Remembering +4 | -1

\mathrm{As}_2 \mathrm{S}_3

dissolves in a solution of yellow ammonium sulphide to give:

View answer and explanation

Correct option: A

\left(\mathrm{NH}_4\right)_3 \mathrm{AsS}_4

Option A (

\left(\mathrm{NH}_4\right)_3 \mathrm{AsS}_4

) is correct because

\mathrm{As}_2\mathrm{S}_3

reacts with yellow ammonium sulfide to form ammonium thioarsenate.

Solution approach

Correct Answer + Core Reason: $\left(\mathrm{NH}_4\right)_3 \mathrm{AsS}_4$ . $\mathrm{As}_2\mathrm{S}_3$ reacts with yellow ammonium sulfide to form ammonium thioarsenate.

Step-by-Step Solution:

Step 1: Write the reaction between $\mathrm{As}_2\mathrm{S}_3$ and yellow ammonium sulfide, which is primarily $(\mathrm{NH}_4)_2\mathrm{S}_x$ , where x > 1.

Step 2: The reaction is: $\mathrm{As}_2\mathrm{S}_3 + 3(\mathrm{NH}_4)_2\mathrm{S}_x \rightarrow 2(\mathrm{NH}_4)_3\mathrm{AsS}_3 + (3x-3)S$ \nHowever, since the ammonium sulfide is yellow, it contains excess sulfur, which further reacts with $(\mathrm{NH}_4)_3\mathrm{AsS}_3$ to form $(\mathrm{NH}_4)_3\mathrm{AsS}_4$ .

Step 3: The overall reaction is: $\mathrm{As}_2\mathrm{S}_3 + 3(\mathrm{NH}_4)_2\mathrm{S}_2 \rightarrow 2(\mathrm{NH}_4)_3\mathrm{AsS}_4$

Step 4: The product formed is ammonium tetrathioarsenate, $(\mathrm{NH}_4)_3\mathrm{AsS}_4$ .

Why other options are wrong:
- Option B: $\left(\mathrm{NH}_4\right)_3 \mathrm{AsS}_5$ is not the correct formula for the product.
- Option C: $\left(\mathrm{NH}_4\right)_2 \mathrm{AsO}_2$ contains oxygen and is not formed in this reaction.
- Option D: $\left(\mathrm{NH}_4\right)_3 \mathrm{AsS}_3$ is an intermediate but not the final product in yellow ammonium sulfide.

Key Concepts:
- Qualitative analysis
- Reactions of sulfides

Memory Aids:
- Remember that yellow ammonium sulfide contains excess sulfur.

Quick Tips:
- Balance the chemical equation to determine the correct product.

5 Medium

5 of 50Question permalink

Understanding +4 | -1

Which is the safest method for performing a
flame test?

View answer and explanation

Correct option: C

Soak a wooden splint in an aqueous solution
of the metal salt, then burn the splint in the
flame from a Bunsen burner

Option C (Soak a wooden splint in an aqueous solution of the metal salt, then burn the splint in the flame from a Bunsen burner) is the safest method because it avoids the use of flammable solvents like methanol and provides a controlled way to introduce the metal salt into the flame.

Solution approach

Correct Answer + Core Reason: Soak a wooden splint in an aqueous solution of the metal salt, then burn the splint in the flame from a Bunsen burner. This method is the safest and most controlled.

Step-by-Step Solution:

Step 1: Flame tests are used to identify metal ions based on the color of the flame they produce when heated.

Step 2: The safest method involves using a wooden splint soaked in an aqueous solution of the metal salt.

Step 3: The splint is then carefully introduced into the flame of a Bunsen burner.

Step 4: This method avoids the use of flammable organic solvents and provides a controlled way to introduce the metal salt into the flame.

Why other options are wrong:
- Option A: Dissolving the metal salt in methanol and squirting it into a lit Bunsen burner is dangerous due to the flammability of methanol and the risk of uncontrolled burning.
- Option B: Dissolving the metal salt in methanol and pouring it into a crystallizing dish is also dangerous due to the large amount of flammable methanol being used.
- Option D: Heating the splint on top of a ceramic hotplate will not produce a flame and will not allow for the observation of the characteristic flame color.

Key Concepts:
- Flame tests
- Qualitative analysis

Memory Aids:
- Use aqueous solutions and wooden splints for safer flame tests.

Quick Tips:
- Always prioritize safety when performing chemical experiments.

6 Medium

6 of 50Question permalink

Remembering +4 | -1

When bismuth chloride is poured into a large volume of water the white precipitate produced is

View answer and explanation

Correct option: C

\mathrm{BiOCl}

Option C (

\mathrm{BiOCl}

) is correct because bismuth chloride reacts with water to form bismuth oxychloride as a white precipitate.

Solution approach

Correct Answer + Core Reason: $\mathrm{BiOCl}$ . Bismuth chloride undergoes hydrolysis to form bismuth oxychloride.

Step-by-Step Solution:
Step 1: Bismuth chloride ( $BiCl_3$ ) reacts with water ( $H_2O$ ).
Step 2: The reaction forms bismuth oxychloride ( $BiOCl$ ) and hydrochloric acid ( $HCl$ ). The balanced equation is: $BiCl_3 + H_2O \rightleftharpoons BiOCl + 2HCl$
Step 3: Bismuth oxychloride ( $BiOCl$ ) precipitates out as a white solid.

Why other options are wrong:
- Option A: $\mathrm{Bi(OH)}_3$ is not the primary product of the reaction between bismuth chloride and water. While bismuth hydroxide can form, it's less stable and readily converts to BiOCl.
- Option B: $\mathrm{Bi}_2 \mathrm{O}_3$ is bismuth oxide, which is not directly formed in this reaction.
- Option D: $\mathrm{Bi}_2 \mathrm{OCl}_3$ is not a known stable compound formed in this reaction.

Key Concepts:
- Hydrolysis of metal halides
- Precipitation reactions
- Formation of complex ions

Memory Aids:
- Remember that bismuth halides tend to form oxyhalides in water.

Quick Tips:
- Consider the relative stability of different bismuth compounds in aqueous solutions.

7 Medium

7 of 50Question permalink

Analyzing +4 | -1

In which of the following solvents,

\mathrm{AgBr}

will have the highest solubility

View answer and explanation

Correct option: B

10^{-3} \mathrm{~M} \mathrm{~NH}_4 \mathrm{OH}

Option B (

10^{-3} \mathrm{~M} \mathrm{~NH}_4 \mathrm{OH}

) is correct because

\mathrm{AgBr}

forms a complex with ammonia, increasing its solubility.

Solution approach

Correct Answer + Core Reason: $10^{-3} \mathrm{~M} \mathrm{~NH}_4 \mathrm{OH}$ . $\mathrm{AgBr}$ forms a complex with ammonia, increasing its solubility.

Step-by-Step Solution:

Step 1: Consider the solubility of $\mathrm{AgBr}$ in different solvents.

Step 2: $\mathrm{AgBr}$ is sparingly soluble in pure water.

Step 3: In $10^{-3} \mathrm{~M} \mathrm{~NaBr}$ and $10^{-3} \mathrm{~M} \mathrm{~HBr}$ , the solubility of $\mathrm{AgBr}$ is suppressed due to the common ion effect.

Step 4: In $10^{-3} \mathrm{~M} \mathrm{~NH}_4 \mathrm{OH}$ , $\mathrm{AgBr}$ forms a complex ion, $[\mathrm{Ag}(\mathrm{NH}_3)_2]^+$ , which increases its solubility.
$\mathrm{AgBr}(s) + 2\mathrm{NH}_3(aq) \rightleftharpoons [\mathrm{Ag}(\mathrm{NH}_3)_2]^+(aq) + \mathrm{Br}^-(aq)$

Why other options are wrong:
- Option A: $10^{-3} \mathrm{~M} \mathrm{~NaBr}$ is wrong because the common ion effect of $\mathrm{Br}^-$ decreases the solubility of $\mathrm{AgBr}$ .
- Option C: Pure water is wrong because $\mathrm{AgBr}$ is sparingly soluble in pure water, and the solubility is less than in ammonia solution.
- Option D: $10^{-3} \mathrm{~M} \mathrm{~HBr}$ is wrong because the common ion effect of $\mathrm{Br}^-$ decreases the solubility of $\mathrm{AgBr}$ .

Key Concepts:
- Solubility product
- Common ion effect
- Complex formation

Memory Aids:
- Remember that silver halides are soluble in ammonia due to complex formation.

Quick Tips:
- Consider the common ion effect and complex formation when determining solubility.

8 Medium

8 of 50Question permalink

Analyzing +4 | -1

Which one among the following pairs of ions cannot be separated by

{H}_2 {S}

in dilute

{HCl}?

View answer and explanation

Correct option: A

\mathrm{Bi}^{3+}, \mathrm{Sn}^{4+}

Option A (

\mathrm{Bi}^{3+}, \mathrm{Sn}^{4+}

) is correct because both

\mathrm{Bi}^{3+}

and

\mathrm{Sn}^{4+}

form sulfide precipitates in dilute

{HCl}

, so they cannot be separated using

{H}_2{S}

in dilute

{HCl}

.

Solution approach

Correct Answer + Core Reason: $\mathrm{Bi}^{3+}, \mathrm{Sn}^{4+}$ . Both form sulfide precipitates under acidic conditions.

Step-by-Step Solution:

Step 1: Consider the group II cations which precipitate as sulfides in dilute ${HCl}$ .

Step 2: $\mathrm{Bi}^{3+}$ and $\mathrm{Sn}^{4+}$ both belong to group II and precipitate as ${Bi}_2{S}_3$ and ${SnS}_2$ respectively in dilute ${HCl}$ .

Step 3: $\mathrm{Hg}^{2+}$ precipitates as ${HgS}$ in acidic medium.

Step 4: $\mathrm{Cu}^{2+}$ precipitates as ${CuS}$ in acidic medium.

Step 5: $\mathrm{Al}^{3+}$ and $\mathrm{Zn}^{2+}$ do not precipitate as sulfides in dilute ${HCl}$ .

Step 6: $\mathrm{Ni}^{2+}$ precipitates as ${NiS}$ at higher pH.

Why other options are wrong:
- Option B: $\mathrm{Al}^{3+}, \mathrm{Hg}^{2+}$ is wrong because $\mathrm{Hg}^{2+}$ precipitates as ${HgS}$ while $\mathrm{Al}^{3+}$ does not precipitate in dilute ${HCl}$ .
- Option C: $\mathrm{Zn}^{2+}, \mathrm{Cu}^{2+}$ is wrong because $\mathrm{Cu}^{2+}$ precipitates as ${CuS}$ while $\mathrm{Zn}^{2+}$ does not precipitate in dilute ${HCl}$ .
- Option D: $\mathrm{Ni}^{2+}, \mathrm{Cu}^{2+}$ is wrong because $\mathrm{Cu}^{2+}$ precipitates as ${CuS}$ in dilute ${HCl}$ while $\mathrm{Ni}^{2+}$ does not.

Key Concepts:
- Group II cations
- Sulfide precipitation
- Qualitative analysis

Memory Aids:
- Remember the group II cations and their sulfide precipitates.

Quick Tips:
- Consider the pH dependence of sulfide precipitation.

9 Medium

9 of 50Question permalink

Understanding +4 | -1

In the precipitation of the iron group (III) in qualitative analysis, ammonium chloride is added before adding ammonium hydroxide to:

View answer and explanation

Correct option: B

Decrease concentration of

{ }^{-} \mathrm{OH}

ions

Option B (Decrease concentration of

{ }^{-} \mathrm{OH}

ions) is correct because ammonium chloride decreases the hydroxide ion concentration through the common ion effect, preventing the precipitation of group III hydroxides before group IV sulfides.

Solution approach

Correct Answer + Core Reason: Decrease concentration of ${ }^{-} \mathrm{OH}$ ions. Ammonium chloride suppresses the ionization of ammonium hydroxide, decreasing the hydroxide ion concentration.

Step-by-Step Solution:
Step 1: Ammonium hydroxide ( $NH_4OH$ ) is a weak base and dissociates in water as follows: $NH_4OH \rightleftharpoons NH_4^+ + OH^-$
Step 2: Ammonium chloride ( $NH_4Cl$ ) is a strong electrolyte and dissociates completely in water: $NH_4Cl \rightarrow NH_4^+ + Cl^-$
Step 3: Adding $NH_4Cl$ increases the concentration of $NH_4^+$ ions in the solution. This shifts the equilibrium of $NH_4OH$ dissociation to the left, according to Le Chatelier's principle.
Step 4: The shift to the left decreases the concentration of $OH^-$ ions in the solution, preventing the precipitation of group III hydroxides (e.g., $Fe(OH)_3$ , $Al(OH)_3$ ) before the group IV cations can be precipitated as sulfides.

Why other options are wrong:
- Option A: Prevent interference by phosphate ions: While phosphate ions can interfere, the primary reason for adding $NH_4Cl$ is to control the hydroxide ion concentration.
- Option C: Increase concentration of $\mathrm{Cl}^{-}$ ions: While the chloride ion concentration does increase, the primary purpose is to decrease the hydroxide ion concentration.
- Option D: Increase concentration of $\mathrm{NH}_4^{+}$ ions: While the ammonium ion concentration does increase, the primary purpose is to decrease the hydroxide ion concentration.

Key Concepts:
- Common ion effect
- Le Chatelier's principle
- Solubility product
- Qualitative analysis

Memory Aids:
- Remember that adding a common ion suppresses the ionization of a weak electrolyte.

Quick Tips:
- Understand the common ion effect and its application in qualitative analysis.

10 Medium

10 of 50Question permalink

Remembering +4 | -1

A solution of a salt of which metal produces
bright red colour in a flame test?

View answer and explanation

Correct option: A

Lithium

The correct answer is A (Lithium) because lithium salts produce a bright red color in a flame test.

Solution approach

Correct Answer + Core Reason: Lithium. Lithium salts produce a bright red color in a flame test.

Step-by-Step Solution:

Step 1: Flame tests are used to identify the presence of certain metal ions based on the characteristic color they impart to a flame.

Step 2: Lithium salts produce a bright red color.

Step 3: Sodium salts produce a yellow-orange color.

Step 4: Potassium salts produce a lilac (purple) color.

Step 5: Copper salts produce a green or blue-green color.

Why other options are wrong:
- Option B (Sodium): Sodium produces a yellow-orange color, not bright red.
- Option C (Potassium): Potassium produces a lilac (purple) color, not bright red.
- Option D (Copper): Copper produces a green or blue-green color, not bright red.

Key Concepts:
- Flame test
- Metal ion identification
- Characteristic colors

Memory Aids:
- Lithium = Red (think Li = Red)

Quick Tips:
- Memorize the characteristic colors produced by common metal ions in a flame test.

11 Medium

11 of 50Question permalink

Understanding +4 | -1

A gas is obtained by addition of dil.

\mathrm{H}_2 \mathrm{SO}_4

to a mixture which turns lead acetate paper black. It is

View answer and explanation

Correct option: A

\mathrm{H}_2 \mathrm{S}

Option A (

\mathrm{H}_2 \mathrm{S}

) is correct because hydrogen sulfide gas reacts with lead acetate to form lead sulfide, which is a black precipitate.

Solution approach

Correct Answer + Core Reason: $\mathrm{H}_2 \mathrm{S}$ . $\mathrm{H}_2 \mathrm{S}$ reacts with lead acetate to form a black precipitate of lead sulfide.

Step-by-Step Solution:

Step 1: The gas turns lead acetate paper black, indicating the presence of sulfide ions ( $S^{2-}$ ).

Step 2: When dilute sulfuric acid is added to a mixture containing a sulfide salt, hydrogen sulfide gas ( $H_2S$ ) is produced.

Step 3: $H_2S$ reacts with lead acetate ( $(\mathrm{CH}_3\mathrm{COO})_2\mathrm{Pb}$ ) to form lead sulfide ( $PbS$ ), which is a black precipitate.

Step 4: The reaction is: $(\mathrm{CH}_3\mathrm{COO})_2\mathrm{Pb} + H_2S \rightarrow PbS + 2\mathrm{CH}_3\mathrm{COOH}$

Why other options are wrong:
- Option B: $\mathrm{NO}_2$ is a brown gas and does not turn lead acetate paper black.
- Option C: $\mathrm{CO}_2$ is a colorless gas and does not react with lead acetate.
- Option D: $\mathrm{SO}_2$ is a colorless gas with a pungent odor and does not turn lead acetate paper black directly, although it can be oxidized to sulfate which might indirectly react.

Key Concepts:
- Qualitative analysis
- Reactions of sulfides

Memory Aids:
- Hydrogen sulfide turns lead acetate paper black.

Quick Tips:
- Know the characteristic reactions of common gases.

12 Medium

12 of 50Question permalink

Understanding +4 | -1

The acidic solution of a salt produced a deep blue colour with starch iodine solution. The salt may be

View answer and explanation

Correct option: B

Nitrite

Option B (Nitrite) is correct because nitrites can oxidize iodide ions in the presence of acid to form iodine, which gives a blue color with starch.

Solution approach

Correct Answer + Core Reason: Nitrite. Nitrites oxidize iodide to iodine, which forms a blue complex with starch.

Step-by-Step Solution:
Step 1: The acidic solution of the salt contains an oxidizing agent that can oxidize iodide ions to iodine.
Step 2: Nitrite ions ( $NO_2^−$ ) in acidic solution can oxidize iodide ions ( $I^−$ ) to iodine ( $I_2$ ).
Step 3: The reaction is: $2NO_2^− + 2I^− + 4H^+ → 2NO + I_2 + 2H_2O$
Step 4: Iodine ( $I_2$ ) forms a deep blue complex with starch.

Why other options are wrong:
- Option A: Chloride: Chloride ions are not strong enough oxidizing agents to oxidize iodide ions to iodine under these conditions.
- Option C: Acetate: Acetate ions are not oxidizing agents and will not cause the formation of iodine.
- Option D: Bromide: While bromide can be an oxidizing agent, it's less likely to be the primary cause of the blue color with starch-iodine solution in this context compared to nitrite.

Key Concepts:
- Redox reactions
- Oxidation of iodide ions
- Starch-iodine complex
- Qualitative analysis

Memory Aids:
- Remember that nitrites can oxidize iodide to iodine, which gives a blue color with starch.

Quick Tips:
- Know the common oxidizing and reducing agents and their reactions.

13 Medium

13 of 50Question permalink

Remembering +4 | -1

Mark the correct statement:

View answer and explanation

Correct option: D

All the above statement are correct

Option D (All the above statement are correct) is correct because all the statements accurately describe the precipitation of basic radicals in different groups.

Solution approach

Correct Answer + Core Reason: All the above statement are correct. Each statement accurately describes the precipitation of basic radicals in the respective groups.

Step-by-Step Solution:

Step 1: Review the group analysis of basic radicals.

Step 2: Group I basic radicals (e.g., $\mathrm{Ag^+}, \mathrm{Pb^{2+}}, \mathrm{Hg_2^{2+}}$ ) precipitate as chlorides.

Step 3: Group IV basic radicals (e.g., $\mathrm{Zn^{2+}}, \mathrm{Mn^{2+}}, \mathrm{Co^{2+}}, \mathrm{Ni^{2+}}$ ) precipitate as sulfides in a neutral or alkaline medium.

Step 4: Group V basic radicals (e.g., $\mathrm{Ba^{2+}}, \mathrm{Sr^{2+}}, \mathrm{Ca^{2+}}$ ) precipitate as carbonates.

Why other options are wrong:
- Option A: I group basic radicals precipitate as chlorides - This is correct, but not the complete answer.
- Option B: IV group basic radicals precipitates as sulphides - This is correct, but not the complete answer.
- Option C: V group basic radicals precipitates as carbonates - This is correct, but not the complete answer.

Key Concepts:
- Group analysis of basic radicals
- Precipitation reactions

Memory Aids:
- Group I: Chlorides, Group IV: Sulfides, Group V: Carbonates.

Quick Tips:
- Understand the basis of group separation in qualitative analysis.

14 Medium

14 of 50Question permalink

Remembering +4 | -1

Which of the following give white precipitate when

\mathrm{HCl}

is added to its aqueous solution

View answer and explanation

Correct option: A

\mathrm{Hg}^{+}

The correct answer is A (

\mathrm{Hg}^{+}

) because mercurous ions (

\mathrm{Hg}^{+}

) react with

\mathrm{HCl}

to form insoluble mercurous chloride (

\mathrm{Hg_2Cl_2}

), which precipitates as a white solid.

Solution approach

Correct Answer + Core Reason: $\mathrm{Hg}^{+}$ . Mercurous ions form a white precipitate with HCl.

Step-by-Step Solution:

Step 1: When hydrochloric acid (HCl) is added to an aqueous solution containing mercurous ions ( $\mathrm{Hg}^{+}$ ), a white precipitate of mercurous chloride ( $\mathrm{Hg_2Cl_2}$ ) is formed.

Step 2: The reaction is: $\mathrm{Hg_2^{2+}(aq) + 2HCl(aq) \rightarrow Hg_2Cl_2(s) + 2H^+(aq)}$

Step 3: Magnesium, zinc, and cadmium ions do not form precipitates with HCl under normal conditions.

Why other options are wrong:
- **Option B ( $M g^{+}+$ ): Magnesium ions do not form a precipitate with HCl.
- Option C ( $Z n^{+}+$ ): Zinc ions do not form a precipitate with HCl.
- Option D ( $\mathrm{Cd}^{+}+$ ): Cadmium ions do not form a precipitate with HCl.

Key Concepts:
- Precipitation reactions
- Mercurous chloride
- Solubility rules

Memory Aids:
- Hg+ + HCl = White Precipitate

Quick Tips:**
- Remember that mercurous ions form a white precipitate with HCl.

15 Medium

15 of 50Question permalink

Remembering +4 | -1

Soda-lime test is used to detect _______ element of organic compound.

View answer and explanation

Correct option: C

N

The correct answer is C (N) because the soda-lime test is used to detect the presence of nitrogen in an organic compound.

Solution approach

Correct Answer + Core Reason: N. Soda-lime test is used to detect nitrogen.

Step-by-Step Solution:

Step 1: The soda-lime test involves heating an organic compound with soda-lime (a mixture of sodium hydroxide and calcium oxide).

Step 2: If nitrogen is present, it is converted to ammonia gas ( $NH_3$ ).

Step 3: The ammonia gas can be detected by its characteristic odor or by its ability to turn red litmus paper blue.

Why other options are wrong:
- Option A (C): Carbon is detected through other methods, such as combustion analysis.
- Option B (H): Hydrogen is detected through other methods, such as combustion analysis.
- Option D (S): Sulfur is detected using the Lassaigne's test or Carius method.

Key Concepts:
- Soda-lime test
- Nitrogen detection
- Ammonia gas

Memory Aids:
- Soda-lime = Nitrogen (think SLN)

Quick Tips:
- Remember that the soda-lime test is specific for nitrogen detection.

16 Medium

16 of 50Question permalink

Remembering +4 | -1

Nitrogen in the organic compound can be
determined by

View answer and explanation

Correct option: B

Kjeldhal method

The correct answer is B (Kjeldhal method) because the Kjeldahl method is specifically designed for the quantitative determination of nitrogen in organic compounds.

Solution approach

Correct Answer + Core Reason: Kjeldhal method. The Kjeldahl method is used for nitrogen determination.

Step-by-Step Solution:

Step 1: The Kjeldahl method involves digesting the organic compound with concentrated sulfuric acid, which converts the nitrogen to ammonium sulfate.

Step 2: The ammonium sulfate is then treated with excess alkali (NaOH), liberating ammonia gas.

Step 3: The ammonia gas is distilled into a known volume of standard acid, and the excess acid is back-titrated to determine the amount of ammonia, and hence the amount of nitrogen in the original sample.

Why other options are wrong:
- Option A (Lucas method): The Lucas method is used to distinguish between primary, secondary, and tertiary alcohols.
- Option C (Victor Meyers method): The Victor Meyer's method is used to determine the molecular weight of volatile liquids.
- Option D (Beckmann method): The Beckmann method is used to determine the molecular weight of a substance by measuring the elevation in boiling point or depression in freezing point.

Key Concepts:
- Kjeldahl method
- Nitrogen determination
- Quantitative analysis

Memory Aids:
- Kjeldahl = Nitrogen (think K for Kjeldahl and N for Nitrogen)

Quick Tips:
- Remember that the Kjeldahl method is specifically for nitrogen determination.

17 Medium

17 of 50Question permalink

Remembering +4 | -1

Which of the following does not give

{NH}_3

on heating with

NaOH

solution?

View answer and explanation

Correct option: B

Aniline

The correct answer is B (Aniline) because aniline does not have an amide or imide group that can be hydrolyzed by NaOH to release ammonia.

Solution approach

Correct Answer + Core Reason: Aniline. Aniline does not give $NH_3$ on heating with $NaOH$ solution.

Step-by-Step Solution:

Step 1: Urea contains an amide group that can be hydrolyzed by NaOH to release ammonia.

Step 2: Acetanilide and benzanilide also contain amide groups that can be hydrolyzed by NaOH to release ammonia.

Step 3: Aniline is an amine and does not contain an amide group that can be hydrolyzed by NaOH to release ammonia.

Why other options are wrong:
- Option A (Urea): Urea contains an amide group and will release ammonia upon heating with NaOH.
- Option C (Acetanilide): Acetanilide contains an amide group and will release ammonia upon heating with NaOH.
- Option D (Benzanide): Benzanide contains an amide group and will release ammonia upon heating with NaOH.

Key Concepts:
- Hydrolysis of amides
- Reaction with NaOH
- Functional groups

Memory Aids:
- Amides + NaOH = Ammonia

Quick Tips:
- Identify the functional groups present in the compounds to determine if they will react with NaOH to release ammonia.

18 Medium

18 of 50Question permalink

Remembering +4 | -1

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

View answer and explanation

Correct option: A

Dimethyl glyoxime

Option A (Dimethyl glyoxime) is correct because it forms a bright red precipitate with Nickel(II) ions in a basic medium.

Solution approach

Correct Answer + Core Reason: Dimethyl glyoxime. It forms a bright red precipitate with Nickel(II) ions in a basic medium.

Step-by-Step Solution:

Step 1: Understand Nickel(II) Ion Reactions: Nickel(II) ions ( $Ni^{2+}$ ) react with specific reagents to form colored precipitates, which are used for their detection and quantitative estimation.

Step 2: Reaction with Dimethylglyoxime: Dimethylglyoxime (DMG) reacts with $Ni^{2+}$ in a basic medium to form a bright red precipitate of nickel dimethylglyoximate, $[Ni(DMG)_2]$ .

Step 3: Analyze the other reagents:
- Meta-dinitrobenzene: Used as an indicator, not for precipitation of nickel.
- Neutral ${FeCl}_3$ : Used to detect phenols, not nickel.
- Sodium nitroprusside: Used to detect sulfide ions, not nickel.

Step 4: Conclusion: Dimethyl glyoxime is the reagent that gives a brilliant red precipitate with Nickel ions in a basic medium.

Why other options are wrong:
- Option B: Meta-dinitrobenzene is not used for the detection of nickel ions.
- Option C: Neutral ${FeCl}_3$ is not used for the detection of nickel ions.
- Option D: Sodium nitroprusside is not used for the detection of nickel ions.

Key Concepts:
- Qualitative Analysis
- Nickel(II) Ion Reactions
- Dimethylglyoxime

Memory Aids:
- Nickel (Ni) and DMG give a red precipitate.

Quick Tips:
- Remember the specific reagents used for the detection of common metal ions.

19 Medium

19 of 50Question permalink

Understanding +4 | -1

Given below are two statements:

Statement (I): Kjeldahl method is applicable to
estimate nitrogen in pyridine.

Statement (II): The nitrogen present in pyridine
can easily be converted into ammonium sulphate in
Kjeldahl method.

In the light of the above statements, choose the
correct answer from the options given below.

View answer and explanation

Correct option: A

Both Statement I and Statement II is false

The correct answer is A (Both Statement I and Statement II is false) because Kjeldahl method is not applicable to estimate nitrogen in pyridine, and the nitrogen present in pyridine cannot be easily converted into ammonium sulphate in Kjeldahl method.

Solution approach

Correct Answer + Core Reason: Both Statement I and Statement II is false. Kjeldahl method is not suitable for nitrogen in pyridine, and the nitrogen in pyridine is not easily converted to ammonium sulfate.

Step-by-Step Solution:

Step 1: Statement I is false: The Kjeldahl method is not suitable for compounds where the nitrogen is part of a heterocyclic ring (like pyridine) or when nitrogen is directly bonded to oxygen or nitrogen atoms. This is because the nitrogen in these compounds is not easily converted into ammonium sulfate during digestion with sulfuric acid.

Step 2: Statement II is false: The nitrogen in pyridine is difficult to convert into ammonium sulfate because of the stability of the heterocyclic ring. The nitrogen is not readily protonated and released as ammonia.

Why other options are wrong:
- Option B (Statement I is false but Statement II is true): Statement II is also false.
- Option C (Both Statement I and Statement II is true): Both statements are false.
- Option D (Statement I is true but Statement II is false): Statement I is also false.

Key Concepts:
- Kjeldahl method limitations
- Heterocyclic nitrogen compounds
- Pyridine

Memory Aids:
- Kjeldahl doesn't work well with rings

Quick Tips:
- Remember the limitations of the Kjeldahl method, especially with heterocyclic compounds.

20 Medium

20 of 50Question permalink

Application +4 | -1

A hydrated salt of

\mathrm{Na}_2 \mathrm{SO}_3

loses

22.22 \%

of its mass on strong heating. The hydrated salt is:

View answer and explanation

Correct option: D

\mathrm{Na}_2 \mathrm{SO}_3 \cdot 2 \mathrm{H}_2 \mathrm{O}

Option D is correct because the mass loss corresponds to two water molecules being removed from the hydrated salt.

Solution approach

Correct Answer + Core Reason: $\mathrm{Na}_2 \mathrm{SO}_3 \cdot 2 \mathrm{H}_2 \mathrm{O}$ . The mass loss corresponds to 2 water molecules.

Step-by-Step Solution:

Step 1: Calculate the molar mass of $\mathrm{Na}_2 \mathrm{SO}_3$ : $2(23) + 32 + 3(16) = 46 + 32 + 48 = 126 \text{{ g/mol}}$

Step 2: Let the hydrated salt be $\mathrm{Na}_2 \mathrm{SO}_3 \cdot x \mathrm{H}_2 \mathrm{O}$ .

Step 3: The molar mass of $x \mathrm{H}_2 \mathrm{O}$ is $18x \text{{ g/mol}}$ .

Step 4: The total molar mass of the hydrated salt is $(126 + 18x) \text{{ g/mol}}$ .

Step 5: The mass loss is $22.22 \%$ , so $0.2222 = \frac{{18x}}{{126 + 18x}}$

Step 6: Solve for $x$ : $0.2222(126 + 18x) = 18x \Rightarrow 28.00 + 4.00x = 18x \Rightarrow 28 = 14x \Rightarrow x = 2$

Step 7: The hydrated salt is $\mathrm{Na}_2 \mathrm{SO}_3 \cdot 2 \mathrm{H}_2 \mathrm{O}$ .

Why other options are wrong:
- Option A: $\mathrm{Na}_2 \mathrm{SO}_3 \cdot 4 \mathrm{H}_2 \mathrm{O}$ would have a different percentage mass loss.
- Option B: $\mathrm{Na}_2 \mathrm{SO}_3 \cdot 6 \mathrm{H}_2 \mathrm{O}$ would have a different percentage mass loss.
- Option C: $\mathrm{Na}_2 \mathrm{SO}_3 \cdot \mathrm{H}_2 \mathrm{O}$ would have a different percentage mass loss.

Key Concepts:
- Hydrated salts
- Molar mass
- Percentage mass loss

Memory Aids:
- Remember to calculate molar masses accurately.

Quick Tips:
- Set up an equation based on the percentage mass loss to find the number of water molecules.

21 Medium

21 of 50Question permalink

Remembering +4 | -1

Which of the following can be used to precipitate lead salt.

View answer and explanation

Correct option: D

All

Option D (All) is correct because

\mathrm{K}_2\mathrm{CrO}_4

,

\text{KI}

, and dil.

\mathrm{H}_2\mathrm{SO}_4

can all form insoluble precipitates with lead ions.

Solution approach

Correct Answer + Core Reason: All. All the given reagents form precipitates with lead ions.

Step-by-Step Solution:

Step 1: Consider the reaction of lead ions ( $\mathrm{Pb}^{2+}$ ) with each reagent.

Step 2: $\mathrm{K}_2\mathrm{CrO}_4$ reacts with $\mathrm{Pb}^{2+}$ to form lead chromate ( $\mathrm{PbCrO}_4$ ), which is a yellow precipitate.

Step 3: $\text{KI}$ reacts with $\mathrm{Pb}^{2+}$ to form lead iodide ( $\mathrm{PbI}_2$ ), which is a yellow precipitate.

Step 4: Dilute $\mathrm{H}_2\mathrm{SO}_4$ reacts with $\mathrm{Pb}^{2+}$ to form lead sulfate ( $\mathrm{PbSO}_4$ ), which is a white precipitate.

Step 5: Since all three reagents form precipitates, the answer is All.

Why other options are wrong:
- Option A: $\mathrm{K}_2 \mathrm{CrO}_4$ is a valid precipitating agent, but it's not the only one.
- Option B: $\text{KI}$ is a valid precipitating agent, but it's not the only one.
- Option C: dil. $\mathrm{H}_2 \mathrm{SO}_4$ is a valid precipitating agent, but it's not the only one.
- Option D: All is correct because A, B, and C are all correct.

Key Concepts:
- Precipitation reactions
- Solubility rules

Memory Aids:
- Lead salts are generally insoluble except for lead nitrate and lead acetate.

Quick Tips:
- Remember common precipitation reactions for qualitative analysis.

22 Medium

22 of 50Question permalink

Application +4 | -1

If

\mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \rightarrow \mathrm{H}_2 \mathrm{O}({l}) ;

\Delta H=-X \mathrm{~kJ},

then heat liberated for complete neutralisation of

1 \mathrm{~gm}

molecule of

\mathrm{H}_2 \mathrm{SO}_4

in aqueous solution wil be

View answer and explanation

Correct option: B

-2 X \mathrm{~kJ}

Option B (

-2X \text{{ kJ}}

) is correct because the neutralization of 1 mole of H

_2

SO

_4

involves 2 moles of H

^+

ions reacting with 2 moles of OH

^-

ions, thus releasing twice the given heat.

Solution approach

Correct Answer + Core Reason: $-2X \text{{ kJ}}$ . The complete neutralization of one mole of H $_2$ SO $_4$ requires two moles of OH $^-$ .

Step-by-Step Solution:

Step 1: Identify the reaction. The given reaction is $\mathrm{H}^{+}(aq) + \mathrm{OH}^{-}(aq) \rightarrow \mathrm{H}_2\mathrm{O}(l); \Delta H = -X \text{{ kJ}}$

Step 2: Consider the neutralization of H $_2$ SO $_4$ . The dissociation of H $_2$ SO $_4$ in water is: $\mathrm{H}_2\mathrm{SO}_4(aq) \rightarrow 2\mathrm{H}^{+}(aq) + \mathrm{SO}_4^{2-}(aq)$

Step 3: Determine the heat released. For each mole of H $_2$ SO $_4$ that is neutralized, two moles of H $^+$ ions react with two moles of OH $^-$ ions. Therefore, the heat released will be twice the heat released for one mole of H $^+$ ions.

Step 4: Calculate the total heat released. Since the heat released for one mole of H $^+$ is -X kJ, the heat released for two moles of H $^+$ is -2X kJ.

Final Answer: $-2X \text{{ kJ}}$

Why other options are wrong:
- **Option A ( $-57.0 \text{{ kJ}}$ ): This is the approximate heat of neutralization for a strong acid and strong base, but it doesn't account for the given value of -X kJ.
- Option C ( $-X \text{{ kJ}}$ ):** This is the heat released for one mole of H $^+$ ions, but H $_2$ SO $_4$ produces two moles of H $^+$ ions per mole of acid.
- **Option D ( $-X/2 \text{{ kJ}}$ ): This is incorrect because the heat released should be twice the given value, not half.

Key Concepts:
- Heat of neutralization
- Stoichiometry of acid-base reactions
- Enthalpy change

Memory Aids:**
- Remember that H $_2$ SO $_4$ is a diprotic acid, meaning it donates two protons.

Quick Tips:
- Always consider the stoichiometry of the reaction when calculating heat changes.
- Pay attention to whether the acid is monoprotic or polyprotic.

23 Medium

23 of 50Question permalink

Understanding +4 | -1

During the preparation of Mohr's salt solution (Ferrous ammonium sulphate), which of the following acid is added to prevent hydrolysis of

\mathrm{Fe}^{2+}

ion?

View answer and explanation

Correct option: C

Dilute sulphuric acid

Option C (Dilute sulphuric acid) is correct because it prevents the hydrolysis of

\mathrm{Fe}^{2+}

ions by providing

\mathrm{H}^+

ions, shifting the equilibrium away from hydroxide formation.

Solution approach

Correct Answer + Core Reason: Dilute sulphuric acid. Dilute sulfuric acid prevents the hydrolysis of $Fe^{2+}$ ions by maintaining a low pH.

Step-by-Step Solution:
Step 1: Ferrous ions ( $Fe^{2+}$ ) in solution can undergo hydrolysis, reacting with water to form iron hydroxide and hydrogen ions: $Fe^{2+}(aq) + 2H_2O(l) \rightleftharpoons Fe(OH)_2(s) + 2H^+(aq)$
Step 2: Adding dilute sulfuric acid ( $H_2SO_4$ ) increases the concentration of $H^+$ ions in the solution.
Step 3: According to Le Chatelier's principle, this shifts the equilibrium to the left, suppressing the hydrolysis of $Fe^{2+}$ ions and preventing the formation of iron hydroxide precipitate.

Why other options are wrong:
- Option A: Concentrated sulphuric acid: Using concentrated sulfuric acid is not necessary and can lead to unwanted side reactions or decomposition.
- Option B: Dilute nitric acid: Nitric acid is an oxidizing agent and can oxidize $Fe^{2+}$ to $Fe^{3+}$ , which is undesirable.
- Option D: Dilute hydrochloric acid: While hydrochloric acid can provide $H^+$ ions, it also introduces chloride ions, which can form complexes with iron ions and potentially interfere with the crystallization of Mohr's salt. Sulfuric acid is preferred because the sulfate ion is less likely to form strong complexes.

Key Concepts:
- Hydrolysis of metal ions
- Le Chatelier's principle
- Common ion effect
- Preparation of Mohr's salt

Memory Aids:
- Remember that acids suppress the hydrolysis of metal ions.

Quick Tips:
- Understand the factors that affect the hydrolysis of metal ions.

24 Medium

24 of 50Question permalink

Understanding +4 | -1

The best indicator for the detection of the end point in the titration of a weak acid and a strong base is

View answer and explanation

Correct option: D

Phenolphthalein (pH range 8 to 10)

Option D (Phenolphthalein (pH range 8 to 10)) is correct because the titration of a weak acid with a strong base has an equivalence point in the basic range (pH > 7), and phenolphthalein's color change occurs within this range.

Solution approach

Correct Answer + Core Reason: Phenolphthalein (pH range 8 to 10). The equivalence point of a weak acid-strong base titration is basic.

Step-by-Step Solution:

Step 1: In the titration of a weak acid with a strong base, the salt formed at the equivalence point undergoes hydrolysis, resulting in a basic solution (pH > 7).

Step 2: The ideal indicator should change color near the equivalence point.

Step 3: Phenolphthalein has a pH range of 8 to 10, which is suitable for detecting the endpoint in a weak acid-strong base titration.

Step 4: Methyl orange and methyl red have pH ranges in the acidic region and are suitable for strong acid-weak base titrations.

Step 5: Thymol blue has two pH ranges, one in the acidic region and one in the basic region, but its basic range is too broad for precise endpoint detection in this case.

Why other options are wrong:
- Option A: Methyl orange (pH range 3 to 4) is suitable for strong acid-weak base titrations.
- Option B: Methyl red (pH range 4 to 6) is suitable for strong acid-weak base titrations.
- Option C: Thymol blue (pH range 8 to 3) has a wide range and is less precise.

Key Concepts:
- Acid-base titrations
- Indicators
- Equivalence point

Memory Aids:
- Remember the pH ranges of common indicators and the pH at the equivalence point for different types of titrations.

Quick Tips:
- Choose an indicator whose pH range includes the pH at the equivalence point.

25 Medium

25 of 50Question permalink

Understanding +4 | -1

Appearance of blood red colour, on treatment of the sodium fusion extract of an organic compound with

{FeSO}_4

in presence of concentrated

{H}_2 {SO}_4

indicates the presence of element/s

View answer and explanation

Correct option: A

{N}

and

{S}

The correct answer is A (

{N}

and

{S}

) because the appearance of a blood red color indicates the formation of iron thiocyanate, which requires both nitrogen and sulfur to be present in the organic compound.

Solution approach

Correct Answer + Core Reason: ${N}$ and ${S}$ . The blood red color indicates the presence of both nitrogen and sulfur.

Step-by-Step Solution:

Step 1: In the sodium fusion extract, if both nitrogen and sulfur are present, they combine to form thiocyanate ions ( $CNS^-$ ).

Step 2: These thiocyanate ions react with ferrous sulfate ( $FeSO_4$ ) in the presence of concentrated sulfuric acid ( $H_2SO_4$ ) to form iron(III) thiocyanate, which has a blood-red color.

Step 3: The reaction is: $Fe^{3+} + CNS^- \rightarrow [Fe(SCN)]^{2+}$ (blood red color).

Why other options are wrong:
- **Option B ( ${N}$ ): If only nitrogen were present, the solution would turn Prussian blue due to the formation of ferric ferrocyanide.
- Option C ( ${Br}$ ): Bromine is detected by a different test, typically involving silver nitrate.
- Option D ( ${S}$ ): If only sulfur were present, a violet color might be observed due to the formation of ferric sulfide, but not a blood-red color.

Key Concepts:
- Lassaigne's test
- Detection of nitrogen and sulfur
- Iron thiocyanate complex

Memory Aids:
- Blood red = N and S (think Blood = Nitrogen and Sulfur)

Quick Tips:**
- Remember that the blood-red color specifically indicates the presence of both nitrogen and sulfur together.

26 Medium

26 of 50Question permalink

Application +4 | -1

\mathrm{Fe}(\mathrm{OH})_3+\mathrm{HCl} \rightarrow \mathrm{A} \xrightarrow{\mathrm{K}_4\left[\mathrm{Fe} (\mathrm{CN})_6\right]} \mathrm{C};

What is (C)?

View answer and explanation

Correct option: C

\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3

Option C is correct because the reaction of FeCl3 with K4[Fe(CN)6] produces ferric ferrocyanide, which has the formula Fe4[Fe(CN)6]3.

Solution approach

Correct Answer + Core Reason: $\mathrm{Fe}_4[\mathrm{Fe}(\mathrm{CN})_6]_3$ . This is the formula for ferric ferrocyanide (Prussian blue).

Step-by-Step Solution:

Step 1: $\mathrm{Fe}(\mathrm{OH})_3 + 3\mathrm{HCl} \rightarrow \mathrm{FeCl}_3 + 3\mathrm{H}_2\mathrm{O}$ . So, A is $\mathrm{FeCl}_3$ .

Step 2: $4\mathrm{FeCl}_3 + 3\mathrm{K}_4[\mathrm{Fe}(\mathrm{CN})_6] \rightarrow \mathrm{Fe}_4[\mathrm{Fe}(\mathrm{CN})_6]_3 + 12\mathrm{KCl}$ .

Step 3: Therefore, C is $\mathrm{Fe}_4[\mathrm{Fe}(\mathrm{CN})_6]_3$ .

Why other options are wrong:
- Option A: $\mathrm{Fe}_2[\mathrm{Fe}(\mathrm{CN})_6]$ is not the correct formula for the precipitate formed.
- Option B: $\mathrm{Fe}_3[\mathrm{Fe}(\mathrm{CN})_6]_2$ is not the correct formula for the precipitate formed.
- Option D: $\mathrm{K}_4[\mathrm{Fe}(\mathrm{CN})_6]$ is a reactant, not the product.

Key Concepts:
- Precipitation reactions
- Complex ion formation
- Ferric ferrocyanide (Prussian blue)

Memory Aids:
- Remember the formula for Prussian blue.

Quick Tips:
- Balance the chemical equation to determine the correct stoichiometry.

27 Medium

27 of 50Question permalink

Understanding +4 | -1

When water is added to a salt solution containing chloroform, chloroform layer turns violet. Salt contains

View answer and explanation

Correct option: B

I^-

Option B (

I^-

) is correct because iodine dissolves in chloroform to give a violet color.

Solution approach

Correct Answer + Core Reason: $I^-$ . Iodine dissolves in chloroform to give a violet color.

Step-by-Step Solution:

Step 1: When water is added to a salt solution containing chloroform, the chloroform layer extracts any non-polar species present.

Step 2: Iodine ( $I_2$ ) is a non-polar molecule that dissolves in non-polar solvents like chloroform.

Step 3: The resulting solution of iodine in chloroform is violet in color.

Step 4: If the salt contains iodide ions ( $I^-$ ), they can be oxidized to iodine by an oxidizing agent present in the solution or by atmospheric oxygen.

Step 5: The iodine then dissolves in the chloroform layer, turning it violet.

Why other options are wrong:
- Option A: $Cl^-$ does not produce a violet color in chloroform.
- Option C: $NO_3^-$ does not produce a violet color in chloroform.
- Option D: $SO_4^{2-}$ does not produce a violet color in chloroform.

Key Concepts:
- Qualitative analysis
- Solubility
- Redox reactions

Memory Aids:
- Iodine in chloroform is violet.

Quick Tips:
- Remember the colors of common elements and compounds in different solvents.

28 Medium

28 of 50Question permalink

Application +4 | -1

When

\mathrm{MnO}_2

and

\mathrm{H}_2 \mathrm{SO}_4

is added to a salt

(\mathrm{A})

, the greenish yellow gas liberated as salt

(\mathrm{A})

is:

View answer and explanation

Correct option: D

\mathrm{NH}_4 \mathrm{Cl}

The question is flawed. The correct answer should be a chloride salt such as NaCl or KCl. When

\mathrm{MnO}_2

and

\mathrm{H}_2\mathrm{SO}_4

are added to a chloride salt, chlorine gas is liberated.

Solution approach

Correct Answer + Core Reason: $\mathrm{NH}_4 \mathrm{Cl}$ (Incorrect). The question is flawed. The correct answer should be a chloride salt such as NaCl or KCl. The reaction should produce chlorine gas.

Step-by-Step Solution:\nStep 1: Understand the reaction between metal halides, $\mathrm{MnO}_2$ , and $\mathrm{H}_2\mathrm{SO}_4$ .\nStep 2: Identify the gas liberated in each case.\nStep 3: Determine which salt produces a greenish-yellow gas.

$\mathrm{MnO}_2 + 4\mathrm{HCl} \rightarrow \mathrm{MnCl}_2 + 2\mathrm{H}_2\mathrm{O} + \mathrm{Cl}_2$
$\mathrm{Cl}_2$ is a greenish-yellow gas.
$\mathrm{NaBr}$ : Would produce bromine, which is reddish-brown.
$\mathrm{CaI}_2$ : Would produce iodine, which is violet.
$\mathrm{KNO}_3$ : Would not produce a gas under these conditions.
$\mathrm{NH}_4\mathrm{Cl}$ : Would produce chlorine gas if the reaction proceeds with MnO2 and H2SO4.

Why other options are wrong:
- Option A: $\mathrm{NaBr}$ : Bromine is reddish-brown, not greenish-yellow.
- Option B: $\mathrm{CaI}_2$ : Iodine is violet, not greenish-yellow.
- Option C: $\mathrm{KNO}_3$ : No gas is produced under these conditions.

Key Concepts:
- Reactions of metal halides with oxidizing agents
- Properties of halogens

Memory Aids:
- Remember the colors of the halogens: fluorine (pale yellow), chlorine (greenish-yellow), bromine (reddish-brown), iodine (violet).

Quick Tips:
- Know the characteristic reactions of halides.

29 Medium

29 of 50Question permalink

Application +4 | -1

{Fe}({OH})_3

can be separated from

{Al}({OH})_3

by addition of

View answer and explanation

Correct option: C

{NaOH}

solution

Option C (

{NaOH}

solution) is correct because

Al(OH)_3

is amphoteric and dissolves in NaOH, while

Fe(OH)_3

is not.

Solution approach

Correct Answer + Core Reason: ${NaOH}$ solution. $Al(OH)_3$ is amphoteric and dissolves in NaOH, while $Fe(OH)_3$ is not.

Step-by-Step Solution:\nStep 1: Understand the properties of $Fe(OH)_3$ and $Al(OH)_3$ .\nStep 2: Consider the effect of each reagent on the hydroxides.\nStep 3: Determine which reagent will selectively dissolve one hydroxide.

$Fe(OH)_3$ is a basic hydroxide and is insoluble in water and dilute bases.
$Al(OH)_3$ is an amphoteric hydroxide and can react with both acids and bases. It dissolves in NaOH to form sodium aluminate.
$Al(OH)_3 + NaOH \rightarrow NaAlO_2 + 2H_2O$

Why other options are wrong:
- Option A: dil. ${HCl}$ : Both hydroxides will dissolve in dilute HCl.
- Option B: ${NaCl}$ solution: NaCl will not dissolve either hydroxide.
- Option D: ${NH}_4 {Cl}$ and ${NH}_4 {OH}$ : This is a buffer solution and will not selectively dissolve one hydroxide.

Key Concepts:
- Amphoteric hydroxides
- Solubility rules

Memory Aids:
- Remember that aluminum hydroxide is amphoteric.

Quick Tips:
- Know the properties of common metal hydroxides.

30 Medium

30 of 50Question permalink

Remembering +4 | -1

On passing a gas, '

X

', through Nessler's regent, a brown precipitate is obtained. The gas '

X

' is

View answer and explanation

Correct option: B

{NH}_3

Option B (

\mathrm{NH}_3

) is correct because ammonia gas reacts with Nessler's reagent to form a brown precipitate.

Solution approach

Correct Answer + Core Reason: $\mathrm{NH}_3$ . Ammonia gas reacts with Nessler's reagent to form a brown precipitate.

Step-by-Step Solution:

Step 1: Understand Nessler's Reagent: Nessler's reagent is an alkaline solution of potassium tetraiodomercurate(II), $K_2[HgI_4]$ . It is used to detect the presence of ammonia.

Step 2: Reaction with Ammonia: When ammonia gas ( $NH_3$ ) is passed through Nessler's reagent, it reacts to form a brown precipitate of iodide of Millon's base, $NH_2Hg_2OI \cdot H_2O$ .

Step 3: Analyze the other gases:
- Chlorine ( $Cl_2$ ): Does not form a brown precipitate with Nessler's reagent.
- Hydrogen sulfide ( $H_2S$ ): Forms a black precipitate with mercury ions, but not the specific brown precipitate of Nessler's test.
- Carbon dioxide ( $CO_2$ ): Does not react with Nessler's reagent.

Step 4: Conclusion: The gas ' $X$ ' that gives a brown precipitate with Nessler's reagent is ammonia ( $NH_3$ ).

Why other options are wrong:
- Option A: ${Cl}_2$ does not form a brown precipitate with Nessler's reagent.
- Option C: ${H}_2 {S}$ forms a black precipitate, not a brown precipitate, with mercury ions.
- Option D: ${CO}_2$ does not react with Nessler's reagent.

Key Concepts:
- Nessler's Reagent
- Detection of Ammonia
- Qualitative Analysis

Memory Aids:
- Nessler's reagent detects Ammonia (NH3).

Quick Tips:
- Remember that Nessler's reagent is specifically used for the detection of ammonia.

Your Progress

Practical chemistry practice · Medium

Explore Other Chapters in CHEMISTRY

Key Takeaways

Continue Your Practice

Your practice