Which of the following is used to oxidise $\mathrm{Fe}^{+2}$ to $\mathrm{Fe}^{+3}$ in III group.

conc. $\mathrm{H}_2 \mathrm{SO}_4$

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JEE • CHEMISTRY • PRACTICAL CHEMISTRY

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Showing 1-30 of 84 questions

1 Medium

1 of 84

Remembering +4 | -1

Mercurous chloride turns black on treatment with ammonium hydroxide. This is due to the formation of

View answer and explanation

Correct option: C

\mathrm{Hg} \text{ and }\mathrm{HgNH}_2 \mathrm{Cl}

Option C (

\mathrm{Hg} \text{ and }\mathrm{HgNH}_2 \mathrm{Cl}

) is correct because mercurous chloride reacts with ammonium hydroxide to form a mixture of mercury and mercury(II) amido-chloride, which appears black.

2 Medium

2 of 84

Understanding +4 | -1

While preparing crystals of Mohr's salt, dil.

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

is added to a mixture of ferrous sulphate and ammonium sulphate, before dissolving this mixture in water, dil.

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

is added here to:

View answer and explanation

Correct option: A

Prevent the hydrolysis of ferrous sulphate

Option B (

\Delta {H} > {T} \Delta {S}

) is correct because for a spontaneous reaction, the change in Gibbs free energy must be negative, and in this case, the magnitude of

\Delta H

is greater than

T \Delta S

3 Medium

3 of 84

Understanding +4 | -1

Na_2CO_3

cannot be used to identify

View answer and explanation

Correct option: A

CO_3^{ 2 - }

Option A is correct.

Na_2CO_3

cannot be used to identify

CO_3^{2-}

because it is the same ion.

4 Medium

4 of 84

Analyzing +4 | -1

A substance on treatment with dil.

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

liberates a colourless gas which produces (i) turbidity with baryta water and (ii) turns acidified dichromate solution green. The reaction indicates the presence of

View answer and explanation

Correct option: C

\mathrm{SO}_3^{2-}

The correct answer is C (

\mathrm{SO}_3^{2-}

) because the colorless gas that produces turbidity with baryta water (

\mathrm{Ba(OH)_2}

) and turns acidified dichromate green is sulfur dioxide (

\mathrm{SO_2}

), which is released from sulfite ions (

\mathrm{SO}_3^{2-}

5 Medium

5 of 84

Remembering +4 | -1

Which of the following situation exists at the equivalence point of a titration.

View answer and explanation

Correct option: D

\frac{\left[H^{\oplus}\right]}{[\stackrel{\ominus}{O} H]}=10^{-14}

Option B ([H+]=[OH-]) is correct because at the equivalence point in a titration, the concentrations of hydrogen and hydroxide ions are equal, representing a neutral solution.

6 Medium

6 of 84

Remembering +4 | -1

\mathrm{PbCrO}_4

is yellow in colour due to

View answer and explanation

Correct option: B

\mathrm{Cr} \mathrm{O}_4^{-2}

Option B (

\mathrm{Cr} \mathrm{O}_4^{-2}

) is correct because the chromate ion (

\mathrm{Cr} \mathrm{O}_4^{-2}

) is responsible for the yellow color of

\mathrm{PbCrO}_4

7 Medium

7 of 84

Remembering +4 | -1

Identify the one which does not come under the chemical methods of quantitative analysis?

View answer and explanation

Correct option: D

Magnetic suceptibility

Option D (Magnetic suceptibility) is correct because magnetic susceptibility is a physical method of quantitative analysis, not a chemical method.

8 Medium

8 of 84

Remembering +4 | -1

\mathrm{Ca}, \mathrm{Ba}

and

\mathrm{Sr}

ions are precipitated in fifth group as their

View answer and explanation

Correct option: C

Carbonates

Option C (Carbonates) is correct because Ca, Ba, and Sr ions are precipitated as their carbonates in the fifth group of qualitative analysis.

9 Medium

9 of 84

Remembering +4 | -1

The cations are grouped on the basis of

View answer and explanation

Correct option: B

Solubility product of radicals

Option B (Solubility product of radicals) is correct because cations are grouped based on the solubility products of their radicals during qualitative analysis.

10 Medium

10 of 84

Remembering +4 | -1

Mark the compound which turns black with

\mathrm{NH}_4 \mathrm{OH}:

View answer and explanation

Correct option: B

Mercurous chloride

Option B (Mercurous chloride) is correct because

\mathrm{Hg}_2 \mathrm{Cl}_2

reacts with

\mathrm{NH}_4 \mathrm{OH}

to form mercury and

\mathrm{Hg(NH_2)Cl}

, which appears black.

11 Medium

11 of 84

Analyzing +4 | -1

A metal

X

on heating with nitrogen gas gives

Y.

Y

on treatment with

\mathrm{H}_2 \mathrm{O}

gives a colourless gas which on passing through

\mathrm{CuSO}_4

gives blue solution then

Y

View answer and explanation

Correct option: B

\mathrm{Mg}_3 \mathrm{N}_2

Option B (

\mathrm{Mg}_3 \mathrm{N}_2

) is correct because magnesium reacts with nitrogen to form magnesium nitride, which then reacts with water to produce ammonia, which gives a blue solution with

\mathrm{CuSO}_4

12 Medium

12 of 84

Remembering +4 | -1

Which of the following is insoluble in yellow ammonium polysulphide.

View answer and explanation

Correct option: A

\mathrm{CuS}

Option A (

\mathrm{CuS}

) is correct because copper sulfide is insoluble in yellow ammonium polysulfide, while the sulfides of arsenic, antimony, and tin dissolve due to the formation of thio salts.

13 Medium

13 of 84

Remembering +4 | -1

On addition of aqueous

\mathrm{NaOH}

to salt solution, a white gelatinous precipitate is formed which dissolves in excess of alkali. The salt solution contains

View answer and explanation

Correct option: A

Aluminium ions

Option A (Aluminium ions) is correct because aluminium ions react with NaOH to form a white gelatinous precipitate of aluminium hydroxide, which dissolves in excess NaOH to form sodium aluminate.

14 Medium

14 of 84

Remembering +4 | -1

Which group anions don't give any characteristic gas with dil and con.

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

View answer and explanation

Correct option: C

III Group

Option III Group is correct because the anions in Group III (

\mathrm{SO}_4^{-2}, \mathrm{PO}_4^{-3}, \mathrm{BO}_3^{-2}, \mathrm{F}^{-}

) do not produce characteristic gases with dilute or concentrated

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

15 Medium

15 of 84

Remembering +4 | -1

\mathrm{CuSO}_4

when reacts with

\mathrm{KCN}

forms

\mathrm{CuCN}

which is insoluble in water, it is soluble in excess of

\text{KCN},

due to formation of the following complex

View answer and explanation

Correct option: B

\text{K}_3\left[\text{Cu}(\text{CN})_4\right]

Option B (

\text{K}_3\left[\text{Cu}(\text{CN})_4\right]

) is correct because

\text{CuCN}

dissolves in excess

\text{KCN}

to form the complex ion

\left[\text{Cu}(\text{CN})_4\right]^{3-}

, which is balanced by three potassium ions.

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16 Medium

16 of 84

Remembering +4 | -1

Which colour confirms the presence of
phosphorus in organic compound:

View answer and explanation

Correct option: C

yellow

Option C is correct because the presence of phosphorus in an organic compound is confirmed by a yellow precipitate of ammonium phosphomolybdate when reacted with ammonium molybdate.

17 Medium

17 of 84

Remembering +4 | -1

Which of the following statement is incorrect:

View answer and explanation

Correct option: A

Colour of lead sulphide (PbS) precipitate is yellow

Option A (Colour of lead sulphide (PbS) precipitate is yellow) is correct because lead sulphide (PbS) precipitate is black, not yellow.

18 Medium

18 of 84

Analyzing +4 | -1

Given below are two statements one is labelled as Assertion (A) and the other is labeilled as Reason.

Assertion (A): In the titration of HCl against NaOH, phenolphthalein is used as suitable indicator.

Reason (R): Phenolphthalein is pink coloured in basic medium.

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

View answer and explanation

Correct option: D

If both (A) and (R) are correct, but (R) is not the
correct explanation of (A)

Option D is correct. Both the assertion that phenolphthalein is used as an indicator in the titration of HCl against NaOH and the reason that phenolphthalein is pink in a basic medium are correct, but the reason is not the correct explanation for the assertion. Phenolphthalein is chosen because its color change occurs near the equivalence point of the titration.

19 Medium

19 of 84

Remembering +4 | -1

Nitrogen in the organic compound can be
determined by

View answer and explanation

Correct option: B

Kjeldhal method

Option B (Kjeldhal method) is correct because the Kjeldahl method is a quantitative method used to determine the nitrogen content in organic and inorganic substances.

20 Medium

20 of 84

Remembering +4 | -1

In organic analysis, Brady’s reagent is used for
the detection of which of the following functional
groups:

View answer and explanation

Correct option: C

Aledhyde/ketones

Option C (Aledhyde/ketones) is correct because Brady's reagent (2,4-dinitrophenylhydrazine) reacts with aldehydes and ketones to form hydrazones, which are typically colored precipitates.

21 Medium

21 of 84

Remembering +4 | -1

An yellow coloured precipitate

\mathrm{PbCrO}_4

is insoluble in

View answer and explanation

Correct option: C

\mathrm{CH}_3 \mathrm{COOH}

Option C (

\mathrm{CH}_3 \mathrm{COOH}

) is correct because

\mathrm{PbCrO}_4

is insoluble in acetic acid.

22 Medium

22 of 84

Analyzing +4 | -1

In the separation of

\mathrm{Cu}^{2+}

and

\mathrm{Cd}^{2+}

2^{\text {nd }}

group qualitative analysis of cations tetrammine copper (II) sulphate and tetrammine cadmium (II) sulphate react with

\text{KCN}

to form the corresponding cyano complexes. Which one of the following pairs of the complexes and their relative stability enables the separation of

\mathrm{Cu}^{2+}

and

\mathrm{Cd}^{2+}?

View answer and explanation

Correct option: A

\mathrm{K}_3\left[\mathrm{Cu}(\mathrm{CN})_4\right]

more stable and

\mathrm{K}_2\left[\mathrm{Cd}(\mathrm{CN})_4\right]

less stable

Option A (K3[Cu(CN)4] more stable and K2[Cd(CN)4] less stable) is correct because the higher stability of the copper complex and lower stability of the cadmium complex allows for selective precipitation of CdS.

23 Medium

23 of 84

Remembering +4 | -1

\mathrm{MgSO}_4

on reaction with

\mathrm{NH}_4 \mathrm{OH}

and

\mathrm{Na}_2 \mathrm{HPO}_4

forms a white crystalline precipitate. What is its formula?

View answer and explanation

Correct option: A

\mathrm{Mg}\left(\mathrm{NH}_4\right) \mathrm{PO}_4

Option A (

\mathrm{Mg(NH_4)PO_4}

) is correct because the reaction of

\mathrm{MgSO_4}

with

\mathrm{NH_4OH}

and

\mathrm{Na_2HPO_4}

forms a white crystalline precipitate of magnesium ammonium phosphate.

24 Medium

24 of 84

Understanding +4 | -1

Acidic

\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7

reacts with

\mathrm{H}_2 \mathrm{S}

to produce

View answer and explanation

Correct option: C

\text{S}

Option C is correct because acidic K2Cr2O7 oxidizes H2S to elemental sulfur (S).

25 Medium

25 of 84

Understanding +4 | -1

In the preparation of p-nitro acetanilide from aniline, nitration is not done by nitrating mixture (a mixture of con.

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

and

\mathrm{HNO}_3)

because

View answer and explanation

Correct option: C

-\mathrm{NH}_2

group gets oxidized

The correct answer is that the

-\mathrm{NH}_2

group gets oxidized because aniline is highly reactive and prone to oxidation under strongly oxidizing conditions like a nitrating mixture.

26 Medium

26 of 84

Remembering +4 | -1

Which is most soluble in water?

View answer and explanation

Correct option: D

\mathrm{AgF}

Option D (AgF) is the most soluble in water because fluorides of silver are more soluble than other halides due to the higher hydration energy of fluoride ions.

27 Medium

27 of 84

Remembering +4 | -1

Test by which starch and cellulose can be distinguished from each other is

View answer and explanation

Correct option: C

Iodine test

Option C (Iodine test) is correct because starch gives a blue-black color with iodine, while cellulose does not.

28 Medium

28 of 84

Understanding +4 | -1

The ion which can be precipitated with

\mathrm{H}_2 \mathrm{S}

in presence of

\mathrm{NH}_4 \mathrm{OH}

View answer and explanation

Correct option: B

\mathrm{Ni}^{+2}

The correct answer is B.

Ni^{+2}

is a group IV cation that precipitates as a sulfide in the presence of

H_2S

and

NH_4OH

29 Medium

29 of 84

Remembering +4 | -1

Which of the following is used to oxidise

\mathrm{Fe}^{+2}

\mathrm{Fe}^{+3}

in III group.

View answer and explanation

Correct option: C

conc.

\mathrm{HNO}_3

Option C (conc.

\mathrm{HNO}_3

) is correct because concentrated nitric acid is a strong oxidizing agent that can oxidize

\mathrm{Fe}^{+2}

\mathrm{Fe}^{+3}

30 Medium

30 of 84

Remembering +4 | -1

The sulphides of which one of the following groups of elements are soluble in yellow ammonium sulphide?

View answer and explanation

Correct option: C

\mathrm{As}, \mathrm{Sb} \text{ and }\mathrm{Sn}

Option C (

\mathrm{As}, \mathrm{Sb} \text{ and }\mathrm{Sn}

) is correct because the sulfides of Arsenic (As), Antimony (Sb), and Tin (Sn) are soluble in yellow ammonium sulfide.

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