PRACTICAL CHEMISTRY MCQs for NEET UG CHEMISTRY — Easy

1 Easy

Remembering +4 | -1

Lassaigne’s test is used in qualitative analysis to detect

View answer and explanation

Correct option: D

All

The correct answer is D (All) because Lassaigne's test is used to detect nitrogen, sulfur, and halogens in organic compounds.

Solution approach

Correct Answer + Core Reason: All. Lassaigne's test is used to detect nitrogen, sulfur, and halogens.

Step-by-Step Solution:

Step 1: Lassaigne's test involves fusing the organic compound with sodium metal.

Step 2: This converts the elements present into ionic forms: nitrogen to sodium cyanide ( $NaCN$ ), sulfur to sodium sulfide ( $Na_2S$ ), and halogens to sodium halides ( $NaX$ , where X = Cl, Br, I).

Step 3: These ionic compounds can then be detected by simple chemical tests.

Why other options are wrong:
- **Option A ( $N$ ): While Lassaigne's test does detect nitrogen, it also detects other elements.
- Option B ( $S$ ): While Lassaigne's test does detect sulfur, it also detects other elements.
- Option C ( $Cl$ ): While Lassaigne's test does detect chlorine, it also detects other elements.

Key Concepts:
- Lassaigne's test
- Detection of N, S, and halogens
- Qualitative analysis

Memory Aids:
- Lassaigne's = N, S, Halogens (think of the acronym NHS)

Quick Tips:**
- Remember that Lassaigne's test is a general test for multiple elements, not just one.

2 Easy

2 of 18Question permalink

Analyzing +4 | -1

[x]+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow[y]

a colourless gas with irritating smell,

[{y}]+\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow

green solution.

[{x}]

and

[y]

are

View answer and explanation

Correct option: A

\mathrm{SO}_3^{2-}, \mathrm{SO}_2

Option A (

\mathrm{SO}_3^{2-}, \mathrm{SO}_2

) is correct because sulfite ions react with sulfuric acid to produce sulfur dioxide, which has an irritating smell and can reduce dichromate ions to a green solution of chromium(III) ions.

Solution approach

Correct Answer + Core Reason: $\mathrm{SO}_3^{2-}, \mathrm{SO}_2$ . Sulfite ions react with sulfuric acid to produce sulfur dioxide, which has an irritating smell and can reduce dichromate ions to a green solution of chromium(III) ions.

Step-by-Step Solution:\nStep 1: Consider the reactions of each anion with sulfuric acid.\nStep 2: Identify the gas produced in each case.\nStep 3: Determine which gas has an irritating smell and can reduce dichromate ions.

$\mathrm{SO}_3^{2-} + H_2SO_4 \rightarrow SO_2 + H_2O + SO_4^{2-}$
$SO_2$ is a colorless gas with an irritating smell.
$K_2Cr_2O_7 + 3SO_2 + H_2SO_4 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + H_2O$
$Cr_2(SO_4)_3$ is a green solution.

Why other options are wrong:
- Option B: $\mathrm{Cl}^{-}, \mathrm{HCl}$ : HCl is a colorless gas, but it doesn't readily reduce dichromate ions.
- Option C: $\mathrm{S}^{-2}, \mathrm{H}_2 \mathrm{S}$ : H2S is a colorless gas with a rotten egg smell, not an irritating smell.
- Option D: $\mathrm{CO}_3^{2-}, \mathrm{CO}_2$ : CO2 is a colorless and odorless gas.

Key Concepts:
- Reactions of anions with acids
- Redox reactions

Memory Aids:
- Remember the characteristic properties of common gases.

Quick Tips:
- Know the reactions of common anions with acids.

3 Easy

3 of 18Question permalink

Remembering +4 | -1

A salt gives violet vapours when treated with conc.

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

it contains

View answer and explanation

Correct option: B

\mathrm{I}^{-}

Option B (

\mathrm{I}^{-}

) is correct because iodide ions produce violet vapors when treated with concentrated

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

.

Solution approach

Correct Answer + Core Reason: $\mathrm{I}^{-}$ . Iodide ions produce violet vapours when treated with concentrated $\mathrm{H}_2\mathrm{SO}_4$ .

Step-by-Step Solution:

Step 1: Reaction with Conc. $\mathrm{H}_2\mathrm{SO}_4$ : Concentrated sulfuric acid is a strong oxidizing agent. When it reacts with halide ions, it oxidizes them to their corresponding halogens.

Step 2: Analyze the given anions:
- A. $\mathrm{Cl}^{-}$ : Chloride ions produce colorless HCl gas.
- B. $\mathrm{I}^{-}$ : Iodide ions produce violet vapors of iodine ( $I_2$ ).
- C. $\mathrm{Br}^{-}$ : Bromide ions produce reddish-brown vapors of bromine ( $Br_2$ ).
- D. $\mathrm{NO}_3^{-}$ : Nitrate ions produce reddish-brown $NO_2$ gas.

Step 3: Conclusion: Iodide ions ( $\mathrm{I}^{-}$ ) produce violet vapors when treated with concentrated $\mathrm{H}_2\mathrm{SO}_4$ due to the formation of iodine ( $I_2$ ).

Why other options are wrong:
- Option A: $\mathrm{Cl}^{-}$ produces colorless HCl gas, not violet vapors.
- Option C: $\mathrm{Br}^{-}$ produces reddish-brown vapors of bromine, not violet vapors.
- Option D: $\mathrm{NO}_3^{-}$ produces reddish-brown $NO_2$ gas, not violet vapors.

Key Concepts:
- Reactions of Halides with Conc. $\mathrm{H}_2\mathrm{SO}_4$
- Oxidation of Halide Ions

Memory Aids:
- Iodine ( $I_2$ ) is violet.
- Bromine ( $Br_2$ ) is reddish-brown.

Quick Tips:
- Remember the colors of the halogens to quickly identify the halide ion in this type of reaction.

4 Easy

4 of 18Question permalink

Understanding +4 | -1

Phosphorous can be estimated as

View answer and explanation

Correct option: A

M{g_2}{P_2}{O_7}

Option A (

M{g_2}{P_2}{O_7}

) is correct because phosphorus is quantitatively converted to magnesium ammonium phosphate, which is then ignited to magnesium pyrophosphate for gravimetric estimation.

Solution approach

Correct Answer + Core Reason: $M{g_2}{P_2}{O_7}$ . Phosphorous is estimated gravimetrically as magnesium pyrophosphate.

Step-by-Step Solution:

Step 1: Phosphorus in a sample is converted to phosphate ions ( $PO_4^{3-}$ ).

Step 2: Phosphate ions are then precipitated as magnesium ammonium phosphate ( $MgNH_4PO_4 \cdot 6H_2O$ ) by adding magnesium ions ( $Mg^{2+}$ ) and ammonium ions ( $NH_4^+$ ).

Step 3: The precipitate is then ignited to form magnesium pyrophosphate ( $Mg_2P_2O_7$ ).

Step 4: The mass of $Mg_2P_2O_7$ is used to calculate the amount of phosphorus in the original sample.

Why other options are wrong:
- Option B: $N{a_3}P{O_4}$ is not a suitable form for gravimetric estimation.
- Option C: ${P_2}{O_3}$ is not a stable form for estimation.
- Option D: ${P_2}{O_5}$ is not a suitable form for gravimetric estimation.

Key Concepts:
- Gravimetric analysis
- Precipitation reactions

Memory Aids:
- Remember the steps involved in the gravimetric estimation of phosphorus.

Quick Tips:
- Know the different methods for estimating elements quantitatively.

5 Easy

5 of 18Question permalink

Remembering +4 | -1

Which of the following cations is detected by the flame test?

View answer and explanation

Correct option: B

\mathrm{K}^{+}

Option B (

\mathrm{K}^{+}

) is correct because potassium ions impart a characteristic lilac or violet color to the flame.

Solution approach

Correct Answer + Core Reason: $\mathrm{K}^{+}$ . Potassium ions impart a characteristic lilac or violet color to the flame.

Step-by-Step Solution:

Step 1: Understand Flame Test: The flame test is an analytical procedure used to detect the presence of certain elements, primarily metal ions, based on their characteristic emission spectra when heated in a flame.

Step 2: Analyze the given cations:
- A. $\mathrm{NH}_4^{+}$ : Ammonium ions do not give a characteristic flame color.
- B. $\mathrm{K}^{+}$ : Potassium ions give a lilac or violet color to the flame.
- C. $\mathrm{Mg}^{2+}$ : Magnesium ions do not give a characteristic flame color.
- D. $\mathrm{Al}^{3+}$ : Aluminum ions do not give a characteristic flame color.

Step 3: Conclusion: Potassium ions ( $\mathrm{K}^{+}$ ) are detected by the flame test due to their characteristic lilac or violet flame color.

Why other options are wrong:
- Option A: $\mathrm{NH}_4^{+}$ ions do not produce a characteristic color in the flame test.
- Option C: $\mathrm{Mg}^{2+}$ ions do not produce a characteristic color in the flame test.
- Option D: $\mathrm{Al}^{3+}$ ions do not produce a characteristic color in the flame test.

Key Concepts:
- Flame Test
- Emission Spectra
- Detection of Metal Ions

Memory Aids:
- Potassium (K) gives a lilac/violet flame.

Quick Tips:
- Remember the characteristic flame colors of common metal ions to quickly identify them in a flame test.

6 Easy

6 of 18Question permalink

Remembering +4 | -1

In detection of

N

and

S

both in the compound by Lassaigne’s fusion test, blood red colour is due to formation of

View answer and explanation

Correct option: D

{\left[ {Fe{{(CNS)}_2}} \right]^{2 + }}

The correct answer is D (

{\left[ {Fe{{(CNS)}_2}} \right]^{2 + }}

) because the blood red color in Lassaigne's test indicates the formation of iron(III) thiocyanate complex, specifically

{\left[ {Fe{{(CNS)}_2}} \right]^{2 + }}

.

Solution approach

Correct Answer + Core Reason: ${\left[ {Fe{{(CNS)}_2}} \right]^{2 + }}$ . The blood red color is due to the formation of the iron(III) thiocyanate complex.

Step-by-Step Solution:

Step 1: In Lassaigne's test, if both nitrogen and sulfur are present, they combine to form thiocyanate ions ( $CNS^-$ ).

Step 2: These thiocyanate ions react with ferric ions ( $Fe^{3+}$ ) to form a blood-red colored complex, which is identified as ${\left[ {Fe{{(CNS)}_n}} \right]^{3-n}}$ , where n is typically 1 or 2.

Step 3: The most stable and commonly formed complex is ${\left[ {Fe{{(CNS)}_2}} \right]^{2 + }}$ .

Why other options are wrong:
- **Option A ( ${\left[ {Fe(CN)} \right]^ + }$ ): This complex would form if only nitrogen were present, not both nitrogen and sulfur.
- Option B ( ${\left[ {Fe{{(CNS)}_2}} \right]^ + }$ ): The charge on the complex is incorrect. It should be 2+.
- Option C ( ${\left[ {Fe{{(CNS)}_3}} \right]^ - }$ ):** While this complex can form, it's less stable and less likely to be the primary cause of the blood-red color compared to ${\left[ {Fe{{(CNS)}_2}} \right]^{2 + }}$ .

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

Memory Aids:
- Blood red = Iron Thiocyanate (FeSCN)

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

7 Easy

7 of 18Question permalink

Remembering +4 | -1

Mercurous ion is represented as:

View answer and explanation

Correct option: A

\mathrm{Hg}_2^{2+}

Option A (

\mathrm{Hg}_2^{2+}

) is correct because mercurous ion exists as a diatomic species with a +2 charge overall.

Solution approach

Correct Answer + Core Reason: $\mathrm{Hg}_2^{2+}$ . Mercurous ion exists as a diatomic species with a +2 charge overall.

Step-by-Step Solution:

Step 1: Understand Mercurous and Mercuric Ions: Mercury can exist in two common oxidation states: +1 (mercurous) and +2 (mercuric).

Step 2: Mercurous Ion: The mercurous ion is a diatomic species with the formula $\mathrm{Hg}_2^{2+}$ . It consists of two mercury atoms covalently bonded together, with a total charge of +2.

Step 3: Mercuric Ion: The mercuric ion is a monatomic species with the formula $\mathrm{Hg}^{2+}$ .

Step 4: Conclusion: Mercurous ion is represented as $\mathrm{Hg}_2^{2+}$ .

Why other options are wrong:
- Option B: $\mathrm{Hg}^{2+}$ represents the mercuric ion, not the mercurous ion.
- Option C: $\mathrm{Hg}+\mathrm{Hg}^{2+}$ is not a standard representation of the mercurous ion.
- Option D: $\mathrm{Hg}_2^{+}$ is not a stable or commonly recognized form of mercury ion.

Key Concepts:
- Mercury Ions
- Mercurous Ion ( $\mathrm{Hg}_2^{2+}$ )
- Mercuric Ion ( $\mathrm{Hg}^{2+}$ )

Memory Aids:
- MercurOUS is di-atomic (Hg22+), MercurIC is single (Hg2+).

Quick Tips:
- Remember the difference between mercurous and mercuric ions and their respective formulas.

8 Easy

8 of 18Question permalink

Remembering +4 | -1

The group reagent for group

\mathrm{V}

radicals is

View answer and explanation

Correct option: A

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

Option A ((

\mathrm{NH}_4)_2 \mathrm{CO}_3

) is correct because ammonium carbonate is the group reagent for group V cations in qualitative analysis.

Solution approach

Correct Answer + Core Reason: $(\mathrm{NH}_4)_2 \mathrm{CO}_3$ . Ammonium carbonate is the group reagent for group V radicals.

Step-by-Step Solution:

Step 1: Understand Group V Radicals: In qualitative analysis, cations are divided into groups based on their behavior with certain reagents. Group V cations are precipitated as carbonates.

Step 2: Group V Cations: The group V cations are $Ba^{2+}, Sr^{2+}, Ca^{2+}$ .

Step 3: Group Reagent: The group reagent for group V cations is ammonium carbonate, $(\mathrm{NH}_4)_2 \mathrm{CO}_3$ , in the presence of ammonium chloride ( $\mathrm{NH}_4\mathrm{Cl}$ ) to control the carbonate ion concentration.

Step 4: Conclusion: The group reagent for group V radicals is $(\mathrm{NH}_4)_2 \mathrm{CO}_3$ .

Why other options are wrong:
- Option B: $(\mathrm{NH}_4)_2 \mathrm{SO}_4$ is not the group reagent for group V radicals.
- Option C: $\mathrm{NH}_4 \mathrm{Cl}$ is used to control the carbonate ion concentration but is not the group reagent itself.
- Option D: $(\mathrm{NH}_4)_2 \mathrm{C}_2 \mathrm{O}_4$ is not the group reagent for group V radicals.

Key Concepts:
- Qualitative Analysis
- Group Reagents
- Group V Cations

Memory Aids:
- Group V: Barium, Strontium, Calcium - precipitated by Carbonate.

Quick Tips:
- Remember the group reagents for each group of cations in qualitative analysis.

9 Easy

9 of 18Question permalink

Understanding +4 | -1

Which of the following is the best scientific method to test the presence of water in a liquid

View answer and explanation

Correct option: A

Use of anhydrous -

CuS{O_4}

Option A (Use of anhydrous -

CuS{O_4}

) is correct because anhydrous

CuSO_4

turns blue in the presence of water, indicating its presence.

Solution approach

Correct Answer + Core Reason: Use of anhydrous - $CuS{O_4}$ . Anhydrous $CuSO_4$ turns blue in the presence of water.

Step-by-Step Solution:

Step 1: Anhydrous copper sulfate ( $CuSO_4$ ) is white in color.

Step 2: When water is added to anhydrous copper sulfate, it forms hydrated copper sulfate ( $CuSO_4 \cdot 5H_2O$ ), which is blue in color.

Step 3: This color change is a specific test for the presence of water.

Why other options are wrong:
- Option B: Use of litmus paper indicates acidity or basicity, not the presence of water.
- Option C: Taste is not a scientific method and can be dangerous.
- Option D: Smell is not a reliable indicator of the presence of water.

Key Concepts:
- Qualitative analysis
- Hydration reactions

Memory Aids:
- Anhydrous copper sulfate turns blue with water.

Quick Tips:
- Use specific chemical tests for accurate identification of substances.

10 Easy

10 of 18Question permalink

Understanding +4 | -1

Phenolphthalein does not act as an indicator for the titration between

View answer and explanation

Correct option: A

\mathrm{HCl} \,\, \& \,\,\mathrm{NH}_4 \mathrm{OH}

Option A (

\mathrm{HCl} \, \& \,\mathrm{NH}_4 \mathrm{OH}

) is correct because the titration of a strong acid (HCl) with a weak base (

\mathrm{NH}_4\mathrm{OH}

) results in a weakly acidic solution at the equivalence point, and phenolphthalein's color change occurs in a more basic range (pH 8-10).

Solution approach

Correct Answer + Core Reason: $\mathrm{HCl} \, \& \,\mathrm{NH}_4 \mathrm{OH}$ . Phenolphthalein is not suitable for titrations where the equivalence point is acidic.

Step-by-Step Solution:

Step 1: Phenolphthalein is a weak organic acid that is colorless in acidic solutions and pink in basic solutions. Its pH range for color change is 8.3 - 10.0.

Step 2: In the titration of a strong acid (HCl) with a weak base ( $\mathrm{NH}_4\mathrm{OH}$ ), the equivalence point is acidic (pH < 7) due to the hydrolysis of the ammonium ion.

Step 3: Since the pH at the equivalence point is less than 8.3, phenolphthalein will remain colorless, and the endpoint will not be easily detected.

Step 4: In the other options, the equivalence point is either neutral or basic, making phenolphthalein a suitable indicator.

Why other options are wrong:
- Option B: $\mathrm{Ca}(\mathrm{OH})_2 \,\&\, \mathrm{HCl}$ : Strong base - strong acid titration, equivalence point is near 7, phenolphthalein can be used.
- Option C: $\mathrm{NaOH} \,\&\, \mathrm{H}_2 \mathrm{SO}_4$ : Strong base - strong acid titration, equivalence point is near 7, phenolphthalein can be used.
- Option D: $\mathrm{KOH} \,\&\, \mathrm{CH}_3 \mathrm{COOH}$ : Strong base - weak acid titration, equivalence point is basic, phenolphthalein can be used.

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

Memory Aids:
- Remember the pH range of phenolphthalein 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.

11 Easy

11 of 18Question permalink

Remembering +4 | -1

Methyl orange gives red colour in

View answer and explanation

Correct option: B

\mathrm{HCl}

solution

Option B (

\mathrm{HCl}

solution) is correct because methyl orange is red in acidic solutions.

Solution approach

Correct Answer + Core Reason: $\mathrm{HCl}$ solution. Methyl orange is an indicator that turns red in acidic solutions.

Step-by-Step Solution:

Step 1: Recall the properties of methyl orange as an acid-base indicator.

Step 2: Methyl orange is red in solutions with a pH less than 3.1, yellow in solutions with a pH greater than 4.4, and orange in between.

Step 3: $\mathrm{HCl}$ is a strong acid, so its solution will have a low pH.

Why other options are wrong:
- Option A: $\mathrm{KOH}$ is a strong base, so its solution will be alkaline, and methyl orange will be yellow.
- Option C: $\mathrm{Na}_2\mathrm{CO}_3$ is a basic salt, so its solution will be alkaline, and methyl orange will be yellow.
- Option D: $\mathrm{NaCl}$ is a neutral salt, so its solution will be neutral, and methyl orange will be orange or yellow depending on the exact pH.

Key Concepts:
- Acid-base indicators
- pH scale

Memory Aids:
- Methyl Orange: Red in Acid, Yellow in Base.

Quick Tips:
- Know the color changes of common acid-base indicators.

12 Easy

12 of 18Question permalink

Remembering +4 | -1

In the borax bead test of

\mathrm{Co}^{2+},

the blue colour of bead is due to the formation of:

View answer and explanation

Correct option: C

\mathrm{Co}\left(\mathrm{BO}_2\right)_2

Option C (

\mathrm{Co}(\mathrm{BO}_2)_2

) is correct because cobalt metaborate is formed, which is blue in color.

Solution approach

Correct Answer + Core Reason: $\mathrm{Co}(\mathrm{BO}_2)_2$ . The blue color is due to the formation of cobalt metaborate.

Step-by-Step Solution:

Step 1: Understand Borax Bead Test: The borax bead test is a qualitative analytical method used to detect the presence of certain metal ions. Borax ( $Na_2B_4O_7 \cdot 10H_2O$ ) is heated to form a glassy bead of boric anhydride ( $B_2O_3$ ) and sodium metaborate ( $NaBO_2$ ).

Step 2: Reaction with $\mathrm{Co}^{2+}$ : When the borax bead is brought into contact with a cobalt salt and heated, cobalt oxide (CoO) is formed, which then reacts with boric anhydride to form cobalt metaborate:
$CoO + B_2O_3 \rightarrow Co(BO_2)_2$

Step 3: Color of Cobalt Metaborate: Cobalt metaborate ( $Co(BO_2)_2$ ) is blue in color.

Step 4: Conclusion: The blue color in the borax bead test of $\mathrm{Co}^{2+}$ is due to the formation of cobalt metaborate, $\mathrm{Co}(\mathrm{BO}_2)_2$ .

Why other options are wrong:
- Option A: $\mathrm{B}_2\mathrm{O}_3$ is boric anhydride, which is colorless.
- Option B: $\mathrm{Co}_3\mathrm{B}_2$ is not the compound formed in the borax bead test.
- Option D: $\mathrm{CoO}$ is cobalt oxide, which is black or brown, not blue.

Key Concepts:
- Borax Bead Test
- Formation of Metal Metaborates
- Characteristic Colors of Metal Ions

Memory Aids:
- Cobalt (Co) gives a blue bead.

Quick Tips:
- Remember the characteristic colors of metal metaborates in the borax bead test.

13 Easy

13 of 18Question permalink

Remembering +4 | -1

The aqueous solution of which salt is coloured

View answer and explanation

Correct option: C

\mathrm{Co}\left(\mathrm{NO}_3\right)_2

Option C (

\mathrm{Co}(\mathrm{NO}_3)_2

) is correct because cobalt(II) salts are typically colored in aqueous solution.

Solution approach

Correct Answer + Core Reason: $\mathrm{Co}(\mathrm{NO}_3)_2$ . Cobalt(II) salts are typically colored in aqueous solution.

Step-by-Step Solution:

Step 1: Understand Color in Aqueous Solutions: The color of an aqueous solution of a salt is due to the presence of transition metal ions that can absorb certain wavelengths of light in the visible region. This absorption is due to d-d transitions within the metal ion.

Step 2: Analyze the given salts:
- A. $\mathrm{Zn}(\mathrm{NO}_3)_2$ : Zinc is a d10 ion and does not exhibit d-d transitions, so its solutions are colorless.
- B. $\mathrm{LiNO}_3$ : Lithium is an alkali metal and does not form colored solutions.
- C. $\mathrm{Co}(\mathrm{NO}_3)_2$ : Cobalt(II) is a transition metal ion that exhibits d-d transitions, so its solutions are colored (typically pink or red).
- D. Potash alum: Potash alum, $KAl(SO_4)_2 \cdot 12H_2O$ , contains $Al^{3+}$ which is a colorless ion.

Step 3: Conclusion: The aqueous solution of $\mathrm{Co}(\mathrm{NO}_3)_2$ is colored due to the presence of the $Co^{2+}$ ion.

Why other options are wrong:
- Option A: $\mathrm{Zn}(\mathrm{NO}_3)_2$ is colorless because $Zn^{2+}$ does not exhibit d-d transitions.
- Option B: $\mathrm{LiNO}_3$ is colorless because $Li^{+}$ does not exhibit d-d transitions.
- Option D: Potash alum is colorless because $Al^{3+}$ does not exhibit d-d transitions.

Key Concepts:
- Color of Transition Metal Ions
- d-d Transitions

Memory Aids:
- Cobalt (Co) salts are often colored.

Quick Tips:
- Remember that transition metal ions with partially filled d-orbitals often form colored solutions.

14 Easy

14 of 18Question permalink

Remembering +4 | -1

In Carius method for the estimation of sulphur precipitate is formed due to the formation of

View answer and explanation

Correct option: A

BaS{O_4}

The correct answer is A (

BaS{O_4}

) because in the Carius method, sulfur is converted to sulfuric acid, which then reacts with barium chloride to form barium sulfate precipitate.

Solution approach

Correct Answer + Core Reason: $BaS{O_4}$ . The precipitate formed in the Carius method for sulfur estimation is barium sulfate.

Step-by-Step Solution:

Step 1: In the Carius method, the organic compound is heated with fuming nitric acid in the presence of silver nitrate.

Step 2: Sulfur present in the compound is oxidized to sulfuric acid ( $H_2SO_4$ ).

Step 3: The sulfuric acid then reacts with barium chloride ( $BaCl_2$ ) to form barium sulfate ( $BaSO_4$ ) precipitate: $H_2SO_4 + BaCl_2 \rightarrow BaSO_4 + 2HCl$ .

Step 4: The precipitate is then filtered, dried, and weighed to determine the percentage of sulfur in the original compound.

Why other options are wrong:
- **Option B ( $Ba{\left( {HS{O_4}} \right)_2}$ ): This is barium bisulfate, which is not the precipitate formed in the Carius method.
- Option C ( $BaS$ ): This is barium sulfide, which is not formed during the Carius method.
- Option D ( $A{g_2}S$ ): This is silver sulfide, which might form if silver is used but the main precipitate for sulfur estimation is barium sulfate.

Key Concepts:
- Carius method
- Sulfur estimation
- Barium sulfate precipitate

Memory Aids:
- Carius = Sulfate (think C for Carius and S for Sulfate)

Quick Tips:**
- Remember that the Carius method converts sulfur to sulfuric acid, which then forms barium sulfate precipitate.

15 Easy

15 of 18Question permalink

Remembering +4 | -1

Select the incorrect statement regarding analytical balance.

View answer and explanation

Correct option: D

It is not a sensitive instrument

Option D (It is not a sensitive instrument) is correct because analytical balances are highly sensitive instruments.

Solution approach

Correct Answer + Core Reason: It is not a sensitive instrument. Analytical balances are designed to be highly sensitive.

Step-by-Step Solution:

Step 1: Define what an analytical balance is.

Step 2: Analytical balances are used for precise quantitative chemical analysis.

Step 3: They are designed to measure mass with very high accuracy and sensitivity.

Why other options are wrong:
- Option A: It is the fundamental kit in quantitative analysis - This is correct; analytical balances are essential for quantitative analysis.
- Option B: It measures samples very accurately - This is correct; accuracy is a key feature of analytical balances.
- Option C: It could measure the difference in mass upto 0.1 mg - This is correct; analytical balances have high resolution.

Key Concepts:
- Analytical balance
- Quantitative analysis
- Accuracy and sensitivity

Memory Aids:
- Analytical = Accurate.

Quick Tips:
- Understand the purpose and characteristics of analytical balances.

16 Easy

16 of 18Question permalink

Remembering +4 | -1

Which is soluble in

\mathrm{NH}_4 \mathrm{OH}?

View answer and explanation

Correct option: B

\mathrm{AgCl}

Option B (

\mathrm{AgCl}

) is correct because it forms a soluble complex with ammonia.

Solution approach

Correct Answer + Core Reason: $\mathrm{AgCl}$ . $\mathrm{AgCl}$ forms a soluble complex with $\mathrm{NH}_4\mathrm{OH}$ .

Step-by-Step Solution:

Step 1: Identify the reaction of $\mathrm{AgCl}$ with $\mathrm{NH}_4\mathrm{OH}$ .

Step 2: $\mathrm{AgCl}$ reacts with $\mathrm{NH}_4\mathrm{OH}$ to form a soluble complex, diamminesilver(I) chloride: $\mathrm{AgCl}(s) + 2\mathrm{NH}_4\mathrm{OH}(aq) \rightleftharpoons [\mathrm{Ag(NH_3)_2}]Cl(aq) + 2\mathrm{H_2O}(l)$ .

Step 3: The other options do not form soluble complexes with $\mathrm{NH}_4\mathrm{OH}$ .

Why other options are wrong:
- Option A: $\mathrm{PbCl}_2$ is soluble in hot water, not in $\mathrm{NH}_4\mathrm{OH}$ .
- Option C: $\mathrm{PbSO}_4$ is generally insoluble.
- Option D: $\mathrm{CaCO}_3$ is insoluble in water and does not readily dissolve in $\mathrm{NH}_4\mathrm{OH}$ .

Key Concepts:
- Solubility of salts
- Complex formation

Memory Aids:
- Silver halides (especially $\mathrm{AgCl}$ ) dissolve in ammonia due to complex formation.

Quick Tips:
- Remember the solubility rules for common salts. Consider complex formation reactions.

17 Easy

17 of 18Question permalink

Remembering +4 | -1

Beilstein test used for

View answer and explanation

Correct option: A

Halogens

The correct answer is A (Halogens) because the Beilstein test is specifically used to detect the presence of halogens in a chemical compound.

Solution approach

Correct Answer + Core Reason: Halogens. The Beilstein test is a qualitative test for the presence of halogens.

Step-by-Step Solution:

Step 1: The Beilstein test involves heating a copper wire in a flame until it no longer imparts a green or blue-green color to the flame.

Step 2: The substance to be tested is then placed on the wire and heated again. If halogens are present, they will react with the copper oxide on the wire to form a copper halide, which is volatile and imparts a green or blue-green color to the flame.

Step 3: Conclusion: The Beilstein test is positive for halogens.

Why other options are wrong:
- Option B (Nitrogen): Nitrogen is detected using the Lassaigne's test.
- Option C (Sulphur): Sulphur is detected using the Lassaigne's test.
- Option D (Carbon): Carbon is a fundamental element in organic compounds and is not specifically detected by the Beilstein test.

Key Concepts:
- Beilstein test
- Halogen detection
- Qualitative analysis

Memory Aids:
- Beilstein = Halogens (think B for Bromine, a halogen)

Quick Tips:
- Remember that the Beilstein test is specific to halogens and not other elements.

18 Easy

18 of 18Question permalink

Understanding +4 | -1

In qualitative analysis

\mathrm{NH}_4 \mathrm{Cl}

is added before

\mathrm{NH}_4 \mathrm{OH}

View answer and explanation

Correct option: A

to decrease

\left[\mathrm{OH}^{-}\right]

conc.

Option A (to decrease

\left[\mathrm{OH}^{-}\right]

conc.) is correct because adding ammonium chloride before ammonium hydroxide decreases the hydroxide ion concentration due to the common ion effect, preventing the precipitation of group III hydroxides.

Solution approach

Correct Answer + Core Reason: to decrease $\left[\mathrm{OH}^{-}\right]$ conc.. 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 B: to increase $\left[\mathrm{OH}^{-}\right]$ conc.: This is incorrect because adding $NH_4Cl$ decreases the hydroxide ion concentration.
- Option C: for making HCl: This is incorrect. $NH_4Cl$ does not directly form HCl in this context.
- Option D: to increase $\left[\mathrm{Cl}^{-}\right]$ conc.: While the chloride 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.

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