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JEE • CHEMISTRY • d & f- BLOCK ELEMENTS

D & f- BLOCK ELEMENTS practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 361-390 of 410 questions

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361 Easy
361 of 410
Understanding +4 | -1
Which product is obtained during reaction of MnO4−{MnO}_4^{-}with I−{I}^{-}in faintly alkaline condition?
View answer and explanation
Correct option: B
IO3−{IO}_3^{-}
Option B (IO3−{IO}_3^{-}) is correct because in a faintly alkaline condition, permanganate oxidizes iodide to iodate.
362 Medium
362 of 410
Analyzing +4 | -1
In the following salts the lowest value of magnetic moment is observed in
View answer and explanation
Correct option: A
MnSO4⋅4H2O{MnSO}_4 \cdot 4 {H}_2 {O}
The correct answer is ZnSO4 · 7 H2O because Zn2+ has a d10 configuration with no unpaired electrons, resulting in the lowest magnetic moment.
363 Medium
363 of 410
Understanding +4 | -1
In which of the following reactions, reaction of silver with the given acids lead to the formation of
compound & silver, along with liberation of gas ?
View answer and explanation
Correct option: D
All of these
The correct answer is D (All of these) because silver reacts with dilute HNO3HNO_3, concentrated HNO3HNO_3, and concentrated H2SO4H_2SO_4 to form silver compounds and liberate gases.
364 Medium
364 of 410
Remembering +4 | -1
V2O5\mathrm{V}_2 \mathrm{O}_5 reacts with alkalies to give
View answer and explanation
Correct option: A
VO4−3\mathrm{VO}_4^{-3}
Option A is correct because V2O5 reacts with alkalies to form vanadate ions, which are represented as VO4-3.
365 Easy
365 of 410
Application +4 | -1
In chromite ore, the oxidation number of iron and chromium are respectively.
View answer and explanation
Correct option: D
+2,+3+ 2, + 3
Option D (+2, +3) is correct because in chromite ore (FeCr2O4), iron exists as FeO (oxidation state +2) and chromium as Cr2O3 (oxidation state +3).
366 Medium
366 of 410
Analyzing +4 | -1
Which of the following statements are correct?

(i) Chromium has the highest melting point among the series 1 metals.

(ii) Number of unpaired electrons is greater in Cr than other elements of series 1 .

(iii) In any row the melting point of transition metal increases as the atomic number increases.
View answer and explanation
Correct option: B
(i) and (ii)
The correct answer is B ((i) and (ii)) because Chromium has the highest melting point in the first transition series, and it also has the maximum number of unpaired electrons in that series.
367 Easy
367 of 410
Application +4 | -1
Among the following the compound that is both paramagnetic and coloured is
View answer and explanation
Correct option: C
VOSO4\mathrm{VOS}{\mathrm{O}_4}
Option C (VOSO4\mathrm{VOS}{\mathrm{O}_4}) is correct because it contains vanadium in a +4 oxidation state, which has an unpaired electron, making it both paramagnetic and colored.
368 Medium
368 of 410
Remembering +4 | -1
Steel resistant to acid is:
View answer and explanation
Correct option: B
Molybdenum steel
Option B (Molybdenum steel) is correct because Molybdenum steel is known for its resistance to acid corrosion.
369 Hard
369 of 410
Analyzing +4 | -1
A transition metal M\mathrm{M} can exist in two oxidation states +1 and +2 . It forms a sulphide whose experimental formula is given by MxS\mathrm{M}_{\mathrm{x}} \mathrm{S} where 2>x>12 > x > 1. Then the ratio of metal ion in +1 state to those in +2 state in sulphide is given by
View answer and explanation
Correct option: B
2(x−1)2−x\frac{2(x-1)}{2-x}
Option B is correct. The ratio of metal ions in +1 to +2 oxidation states is derived using charge neutrality principles in the compound M_xS.
370 Easy
370 of 410
Analyzing +4 | -1
Which of the following statements are correct?

(A) The M3+/M2+\mathrm{M}^{3+} / \mathrm{M}^{2+} reduction potential for iron is greater than manganese.

(B) The higher oxidation states of first row dd-block elements get stabilized by oxide ion

(C) Aqueous solution of Cr2+\mathrm{Cr}^{2+} can liberate hydrogen from dilute acid

(D) Magnetic moment of V2+\mathrm{V}^{2+} is observed between 4.4−5.2BM4.4-5.2 \mathrm{BM}

Choose the correct answer from the options given below:
View answer and explanation
Correct option: A
(B), (C) only
Option A ((B), (C) only) is correct because the higher oxidation states of first row d-block elements are stabilized by oxide ions, and aqueous solutions of Cr2+{}^{2+} can liberate hydrogen from dilute acid.
371 Medium
371 of 410
Analyzing +4 | -1
The metal ion whose electronic configuration is
not affected by the nature of the ligand and which
gives a violet colour in non-luminous flame under
hot condition in borax bead test is
View answer and explanation
Correct option: B
Ni2+\mathrm{Ni}^{2+}
The correct answer is B (Ni²⁺) because Ni²⁺ has a d⁸ configuration that is less affected by ligand field strength, and it gives a violet color in the borax bead test under hot conditions.
372 Medium
372 of 410
Understanding +4 | -1
The highest magnetic moment is shown by the transition metal ion with the configuration
View answer and explanation
Correct option: B
3d53{d^5}
Option B (3d53{d^5}) is correct because it has the maximum number of unpaired electrons (5), leading to the highest magnetic moment.
373 Easy
373 of 410
Remembering +4 | -1
The formula of azurite is
View answer and explanation
Correct option: B
2CuCO3.Cu(OH)22\mathrm{CuC}{\mathrm{O}_3}.\mathrm{Cu}{\left( {\mathrm{OH}} \right)_2}
Option B (2CuCO3.Cu(OH)22\mathrm{CuC}{\mathrm{O}_3}.\mathrm{Cu}{\left( {\mathrm{OH}} \right)_2}) is correct because it is the chemical formula for azurite, a copper carbonate hydroxide mineral.
374 Medium
374 of 410
Analyzing +4 | -1
Statement I : Generally transition elements exhibit higher enthalpies of atomisation.

Statement II : Cr2+\mathrm{Cr}^{2+} is reducing while Mn3+\mathrm{Mn}^{3+} is oxidizing.

In the light of above statements, choose the correct option among the following.
View answer and explanation
Correct option: B
Both statement I and statement II are correct
Option B is correct because transition elements have strong interatomic interactions leading to high enthalpies of atomisation, and the redox properties of Cr2+ and Mn3+ are as stated.
375 Medium
375 of 410
Understanding +4 | -1
Assertion(A) Mn-atom loses nsn s electrons first during ionisation as compared to (n−1)d(n-1) d electrons.

Reason(R) The effective nuclear charge experienced by (n−1)d(n-1) d electrons is greater than that by nsn s electrons.
View answer and explanation
Correct option: A
If both (A) and (R) are correct and (R) is the correct explanation of (A).
Option A is correct. Mn loses ns electrons before (n-1)d electrons because the effective nuclear charge experienced by (n-1)d electrons is greater than that by ns electrons, making it more difficult to remove them.
376 Easy
376 of 410
Remembering +4 | -1
Catalyst used in oxo-process is
View answer and explanation
Correct option: D
[COH(CO)4]\left[ {COH{{\left( {CO} \right)}_4}} \right]
Option D ([COH(CO)_4]) is the correct catalyst used in the oxo-process, as it is a known fact in organometallic chemistry.
377 Easy
377 of 410
Remembering +4 | -1
Highest oxidation state exhibited by an actinoid is
View answer and explanation
Correct option: D
7
Option D (7) is correct because Neptunium (Np) and Plutonium (Pu) can exhibit an oxidation state of +7, which is the highest among the actinoids.
378 Medium
378 of 410
Remembering +4 | -1
For making good quality mirrors, plates of float glasses are used. These are obtained by floating molten glass over a liquid metal which does not solidify before glass. The metal can be:
View answer and explanation
Correct option: C
Hg
Option C (Hg) is correct because mercury (Hg) is a liquid metal at room temperature and does not solidify before the molten glass.
379 Easy
379 of 410
Analyzing +4 | -1
Which of the following compounds has colour but no unpaired electrons?
View answer and explanation
Correct option: A
KMnO4\mathrm{KMnO}_4
Option A is correct because KMnO4 exhibits color due to charge transfer transitions even though it has no unpaired electrons.
380 Medium
380 of 410
Analyzing +4 | -1
Assertion: Out of CrO3 & Al2O3,CrO3CrO_3 \,\&\, Al_2O_3, CrO_3 having lower melting point than Al2O3Al_2O_3.
Reason: Oxidation state of CrCr in CrO3CrO_3 is high
View answer and explanation
Correct option: B
If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
Option B is correct because both the assertion and the reason are correct, but the reason does not explain the assertion. The high oxidation state of Cr makes it more covalent, leading to a lower melting point.
381 Medium
381 of 410
Understanding +4 | -1
Formation of coloured solution is possible when metal ion in the compound contains
View answer and explanation
Correct option: C
Unpaired electrons
Option C is correct because the presence of unpaired electrons in a metal ion allows for d-d transitions, which absorb light and result in a colored solution.
382 Medium
382 of 410
Understanding +4 | -1
The pair that has similar atomic radii is
View answer and explanation
Correct option: D
MoMo and WW
Option D (MoMo and WW) is correct because MoMo and WW have similar atomic radii due to lanthanoid contraction.
383 Medium
383 of 410
Analyzing +4 | -1
The incorrect statement(s) among (1)-(3) is (are)
1) W(VI) is more stable than Cr(VI).
2) In the presence of HCl, permanganate titrations provide satisfactory results.
3) Some lanthanoid oxides can be used as phosphors.
View answer and explanation
Correct option: B
2 only
The correct answer is B (2 only) because KMnO4 in the presence of HCl will oxidize HCl to Cl2, leading to inaccurate results in permanganate titrations.
384 Medium
384 of 410
Remembering +4 | -1
Which of the following is not a member of 3d3d-transition series?
View answer and explanation
Correct option: C
Au\mathrm{Au}
Option C (Au\mathrm{Au}) is correct because gold (Au) is a member of the 5d transition series, not the 3d transition series.
385 Medium
385 of 410
Analyzing +4 | -1
The correct statement are
View answer and explanation
Correct option: C
Both (1) and (2)
Option C (Both (1) and (2)) is correct because both KMnO4\mathrm{KMnO}_4 and K2Cr2O7\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 exhibit color due to charge transfer, and paramagnetic gases are evolved upon heating CuSO4\mathrm{CuSO}_4 and AgNO3\mathrm{AgNO}_3.
386 Easy
386 of 410
Remembering +4 | -1
Magnetite is:
View answer and explanation
Correct option: C
Fe3O4\mathrm{F}{\mathrm{e}_3}{\mathrm{O}_4}
Option C (Fe3O4\mathrm{F}{\mathrm{e}_3}{\mathrm{O}_4}) is correct because it is the chemical formula for Magnetite.
387 Easy
387 of 410
Remembering +4 | -1
Blue vitriol is:
View answer and explanation
Correct option: C
CuSO4.5H2O\mathrm{CuS}{\mathrm{O}_4}.5{\mathrm{H}_2}\mathrm{O}
Option C (CuSO4.5H2O\mathrm{CuS}{\mathrm{O}_4}.5{\mathrm{H}_2}\mathrm{O}) is correct because blue vitriol is the common name for copper(II) sulfate pentahydrate.
388 Medium
388 of 410
Application +4 | -1
The oxidation number of Mn in the product of alkaline oxidative fusion of MnO2\mathrm{Mn}{\mathrm{O}_2} is
View answer and explanation
Correct option: D
6
Option D (6) is correct because the alkaline oxidative fusion of MnO2 produces K2MnO4, where Mn has an oxidation number of +6.
389 Medium
389 of 410
Understanding +4 | -1
An acid radical (A) when treated with dil. H2SO4\mathrm{H}_2 \mathrm{SO}_4, a colourless gas (B) comes out which is suffocating with the smell of burning sulphur. Gas B when treated with K2Cr2O7\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 acidified with dil. H2SO4\mathrm{H}_2 \mathrm{SO}_4, green colour appeared. A\mathrm{A} and B\mathrm{B} are respectively
View answer and explanation
Correct option: D
SO32−\mathrm{SO}_3^{2-} and SO2\mathrm{SO}_2
Option D (SO32−\mathrm{SO}_3^{2-} and SO2\mathrm{SO}_2) is correct because sulfite ions react with acid to produce sulfur dioxide, which has a suffocating smell and turns acidified potassium dichromate green.
390 Medium
390 of 410
Remembering +4 | -1
The transition element that has lowest enthalpy of atomization is:
View answer and explanation
Correct option: B
CuCu
Option B (CuCu) is correct because copper has a completely filled d-orbital in its +1 oxidation state, leading to weaker metallic bonding and thus lower enthalpy of atomization compared to Fe, V, and Zn.
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