d and f Block Elements

JEE Chemistry · 278 questions · Page 1 of 28 · Click an option or "Show Solution" to reveal answer

The electrolytes usually used in the electroplating of gold and silver, respectively are :

A [Au(CN)2]– and [Ag(CN)2]–

B [Au(CN)2]– and [Ag(Cl)2]–

C [Au(OH)4]– and [Ag(OH)2]–

D [Au(NH3)2]– and [Ag(CN)2]–

Correct Answer

Option A

Solution

[Au(CN)2]– and [Ag(CN)2]– both are soluble complexes.

[Au(CN)2]– ⇌ Au+ + 2CN- [Ag(CN)2]– ⇌ Ag+ + 2CN- The concentration of produced Au+ ion and Ag+ ion due to ionisation is very less which ensure slow deposition and this slow deposition will generate uniform layer on the substance to be electroplated.

Ellingham diagram is a graphical representation of :

\Delta

G vs P

\Delta

G vs T

C (

\Delta

-

\Delta

S) vs T

\Delta

H vs T

Correct Answer

Option B

Solution

Ellingham diagram is a graphical representation of

\Delta

G vs T when metal heated with oxygen to form metal oxide.

The method used for the purification of Indium is :

A zone refining

B vapour phase refining

C liquation

D van Arkel method

Correct Answer

Option A

Solution

Zone refining is used for the purification of indium.

The statement is incorrect about Ellingham diagram is

A provides idea about the reaction rate.

B provides idea about free energy change.

C provides idea about changes in the phases during the reaction.

D provides idea about reduction of metal oxide.

Correct Answer

Option A

Solution

Ellingham diagram is a plot between

\Delta

^\circ

and T and does not give any information regarding rate of reaction.

Which one of the following set of elements can be detected using sodium fusion extract?

A Sulfur, Nitrogen, Phosphorous, Halogens

B Phosphorous, Oxygen, Nitrogen, Halogens

C Nitrogen, Phosphorous, Carbon, Sulfur

D Halogens, Nitrogen, Oxygen, Sulfur

Correct Answer

Option A

Solution

By sodium fusion extract we can detect Sulphur, Nitrogen, Phosphorous and Halogens, because they are converted in to their ionic form with sodium metal.

The addition of silica during the extraction of copper from its sulphide ore :-

A converts copper sulphide into copper silicate

B converts iron oxide into iron silicate

C reduces copper sulphide into metallic copper

D reduces the melting point of the reaction mixture

Correct Answer

Option B

Solution

Silica is used to remove FeO impurity from the ore of copper

FeO + Si{O_2} \to \mathop {FeSi{O_3}}\limits_{iron\,silicate\,(Slag)}

Given below are two statements: one is labelled as "Assertion A" and the other is labelled as "Reason R" Assertion A : In the complex

\mathrm{Ni}(\mathrm{CO})_{4}

and

\mathrm{Fe}(\mathrm{CO})_{5}

, the metals have zero oxidation state. Reason R : Low oxidation states are found when a complex has ligands capable of

\pi

-donor character in addition to the

\sigma

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

A A is not correct but R is correct

B A is correct but R is not correct

C Both A and R are correct but R is NOT the correct explanation of A

D Both A and R are correct and R is the correct explanation of A

Correct Answer

Option B

Solution

Option B: "A is correct but R is incorrect" is the right answer.

Here is the reasoning: Assertion A is correct.

In the complex

\mathrm{Ni(CO)_4}

, if we assume the oxidation state of Ni to be x, we can set up the equation: $x + (0 \times 4) = 0$ From which we find that (x = 0).

Similarly, in

\mathrm{Fe(CO)_5}

, if we let the oxidation state of Fe be y, the equation becomes: $y + (0 \times 5) = 0$ Solving this gives (y = 0).

Therefore, the oxidation states of both Ni and Fe are zero in their respective complexes, verifying that Assertion A is true.

Regarding Reason R, it is incorrect.

The low oxidation states in these complexes are stabilized through synergic bonding, which involves the ligand being a $\pi$ -acceptor, not a $\pi$ -donor.

The CO ligand in these complexes accepts electron density from the metal into its antibonding molecular orbitals, characterizing a $\pi$ -acid ligand, and not a $\pi$ -donor ligand as stated.

Consequently, Reason R does not hold true.

Which of the following ore is concentrated using group 1 cyanide salt ?

A Sphalerite

B Malachite

C Siderite

D Calamine

Correct Answer

Option A

Solution

The chemical formulas of given ore in options are as follows Sphalerite ore : ZnS Calamine ore : ZnCO3 Siderite ore : FeCO3 Malachite ore : Cu(OH)2.

CuCO3 Sphalerite ore containing ZnS is concentrated using group is cyanide salt (NaCN) as a depressant by froth flotation method.

Note : ( Remember all those ores names.

Any one of those can be asked in the exam.)

Oxides Ores : (1) ZnO - Zincite (2) Fe2O3 - Haematite (3) Fe3O4 - Magnetite (FeO + Fe2O3 mixture) (4) Fe2O3 .

3H2O - Limonite (5) MnO2 - Pyrolusite (6) Cu2O - Cuprite or Ruby Copper (7) TiO2 - Rutile (8) FeCr2O4 - Chromite (FeO + Cr2O3) (9) FeTiO3 - Illmenite (FeO + TiO2) (10) Na2B4O7 .

10H2O - Borax or Tincal (11) U3O8 - Pitch Blende (12) SnO2 - Tin Stone or Cassiterite (13) Ca2B6O11 .

5H2O - Colemanite (2 Cao + 3 B2O3) (14) Al2O3 .

2H2O - Bauxite (15) Al2O3 .

H2O - Diaspore (16) Al2O3 - Corundum Sulphides Ores : (1) ZnS - Zinc Blende or Sphalerite (2) PbS - Galena (3) Ag2S - Argentite or Silver Glance (4) HgS - Cinnabar (5) Cu2S - Chalcocite or Copper glance (6) CuFeS2 - Copper pyrites or Chalco pyrites (Cu2S + Fe2S3 mixture) (7) FeS2 - Iron pyrites or Fool's Gold (8) 3Ag2S .

Sb2S2 - Pyrargyrite or ruby silver Halides Ores : (1) NaCl - Rock Salt (2) KCl - Sylvine (3) Na3AlF6 - Cryolite [3NaF + AlF6] (4) CaF2 - Fluorspar (5) KCl .

MgCl2 .

6H2O - Carnalite (6) AgCl - Horn Silver Carbonates Ores : (1) CaCO3 - Limestone (2) MgCO3 - Magnesite (3) CaCO3 .

MgCO3 - Dolomite (4) ZnCO3 - Calamine (5) PbCO3 - Cerrusite (6) FeCO3 - Siderite (7) CuCO3 .

Cu(OH)2 or Cu2CO3(OH)2 - Malachite or Basic Copper Carbonates (8) 2 CuCO3 .

Cu(OH)2 - Azurite Sulphates Ores : (1) CuSO4 .

2H2O - Gypsum (2) MgSO4 .

7H2O - Epson Salt (3) Na2SO4 .

10 H2O - Glauber's Salt (4) PbSO4 - Anglesite (5) ZnSO4 .

7H2O - White Vitriol (6) FeSO4 .

7H2O - Green Vitriol (7) CuSO4 .

5H2O - Blue Vitriol or Chalcanthite Nitrate Ores : (1) KNO3 - Indian Saltpetre (2) NaNO3 - Chile Saltpetre Arsenides Ores : (1) NiAs - Kupfernickel (2) NiAsS - Nickel glance

Given below are two statements : Statement I : Sphalerite is a sulphide or of zinc and copper glance is a sulphide ore of copper. Statement II : It is possible to separate two sulphide ores by adjusting proportion of oil to water or by using 'depressants' in a forth flotation method. Choose the most appropriate answer from the options given below :

A Statement I is true but Statement II is false.

B Both Statement I and Statement II are true.

C Statement I is false but Statement II is true.

D Both Statement I and Statement II are false.

Correct Answer

Option B

Solution

sphalerite-ZnS, copper glance - Cu2S two sulphide ores can be separated by adjusting proportions of oil to water or by using 'Depressants'.

Q10

Match with .tg .tg

List - I		List - II
(A)	Concentration of gold ore	(I)	Aniline
(B)	Leaching of alumina	(II)	NaOH
(C)	Froth stabiliser	(III)	SO $_2$
(D)	Blister copper	(IV)	NaCN

A (A) - (IV), (B) - (III), (C) - (II), (D) - (I)

B (A) - (IV), (B) - (II), (C) - (I), (D) - (III)

C (A) - (III), (B) - (II), (C) - (I), (D) - (IV)

D (A) - (II), (B) - (IV), (C) - (III), (D) - (I)

Correct Answer

Option B

Solution

Gold is concentrated by cyanidation Leaching of alumina is done by NaOH Froth stabiliser is aniline Blister copper has condensed SO2 on the surface

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