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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 1-30 of 410 questions
1 Easy
1 of 410
Remembering +4 | -1
Mercury is transported in metal containers made of:
View answer and explanation
Correct option: C
Iron
Option C (Iron) is correct because mercury does not form an amalgam with iron, making it a suitable material for transporting mercury.
2 Easy
2 of 410
Remembering +4 | -1
Cadmipone is a mixture of:
View answer and explanation
Correct option: A
CdS and BaSO4\mathrm{BaS}{\mathrm{O}_4}
Option A (CdS and BaSO4\mathrm{BaS}{\mathrm{O}_4}) is correct because Cadmipone is a mixture of cadmium sulfide (CdS) and barium sulfate (BaSO4\mathrm{BaS}{\mathrm{O}_4}).
3 Medium
3 of 410
Remembering +4 | -1
The number of element from the following that do not belong to lanthanoids is :
Eu, Cm, Er, Tb, Yb and Lu
View answer and explanation
Correct option: C
1
Option C (1) is correct because Cm (Curium) is an actinide, not a lanthanoid, while the others (Eu, Er, Tb, Yb, Lu) are lanthanoids.
4 Medium
4 of 410
Analyzing +4 | -1
Statement-I: Sm3+{Sm}^{3+}, Dy3+{Dy}^{3+} have same colour (yellow).

Statement-II: Both ions are having same number of unpaired electrons.
View answer and explanation
Correct option: B
Statement I is true, Statement II is true, Statement II is not the correct explanation of statement I
Option B is correct because Statement I is true (both have yellow color), Statement II is true (same number of unpaired f-electrons), but Statement II is not the correct explanation of Statement I. Color arises from similar electronic transitions, not just the number of unpaired electrons.
5 Medium
5 of 410
Understanding +4 | -1
The most common oxidation state of Lanthanoid
elements is +3. Which of the following is likely to
also form other oxidation state?
View answer and explanation
Correct option: A
Ce (At. No. 58)
Option A is correct because Cerium (Ce) is known to exhibit +3 and +4 oxidation states due to the stability of the Ce4+ ion with a noble gas configuration.
6 Medium
6 of 410
Remembering +4 | -1
A mixture of TiO2\mathrm{Ti}{\mathrm{O}_2} and BaSO4\mathrm{BaS}{\mathrm{O}_4} is called
View answer and explanation
Correct option: A
Titanox
Option A (Titanox) is correct because a mixture of titanium dioxide (TiO₂) and barium sulfate (BaSO₄) is commercially known as Titanox.
7 Medium
7 of 410
Understanding +4 | -1
Smallest ionic radii among the following is of
View answer and explanation
Correct option: A
Yb3+\mathrm{Yb}^{3+}
Option A (Yb3+\mathrm{Yb}^{3+}) is correct because, due to lanthanide contraction, ionic radii decrease with increasing atomic number across the lanthanide series.
8 Medium
8 of 410
Remembering +4 | -1
Density, malleability and ductility in coinage metals increase in the order:
View answer and explanation
Correct option: A
Cu, Ag, Au
Option A (Cu, Ag, Au) is correct because the density, malleability, and ductility of coinage metals increase in the order of copper (Cu), silver (Ag), and gold (Au).
9 Medium
9 of 410
Analyzing +4 | -1
Choose the correct statement
View answer and explanation
Correct option: D
High negative E∘\mathrm{E}^{\circ} value for Zn\mathrm{Zn} owing to completely filled 3d3 \mathrm{d} subshell in Zn2+\mathrm{Zn}^{2+}
Option D (High negative E∘\mathrm{E}^{\circ} value for Zn\mathrm{Zn} owing to completely filled 3d3 \mathrm{d} subshell in Zn2+\mathrm{Zn}^{2+}) is correct because the high stability of the completely filled d-orbital in Zn2+\mathrm{Zn}^{2+} contributes to its high negative reduction potential.
10 Easy
10 of 410
Application +4 | -1
Which of the following electronic arrangement will give highest value of magnetic moment?
View answer and explanation
Correct option: C
d4\mathrm{d}^4, weak field
Option C (d4, weak field) is correct because it results in the highest number of unpaired electrons, leading to the highest magnetic moment.
11 Medium
11 of 410
Remembering +4 | -1
Which is the salt of an organic acid?
View answer and explanation
Correct option: A
Rochelle salt
Option A (Rochelle salt) is correct because Rochelle salt is potassium sodium tartrate, which is a salt of tartaric acid, an organic acid.
12 Easy
12 of 410
Remembering +4 | -1
Which of the following shows maximum number of oxidation states
View answer and explanation
Correct option: C
Mn
Option C (Mn) is correct because Manganese (Mn) exhibits the maximum number of oxidation states, ranging from +2 to +7.
13 Medium
13 of 410
Understanding +4 | -1
A catalyst does not
View answer and explanation
Correct option: D
get consumed by itself in the reaction
Option D (get consumed by itself in the reaction) is correct because a catalyst participates in the reaction mechanism but is regenerated at the end, so it is not consumed.
14 Easy
14 of 410
Application +4 | -1
Which of the following has the maximum number of unpaired electrons?
View answer and explanation
Correct option: D
Fe2+{Fe}^{2+}
Option D (Fe2+{Fe}^{2+}) is correct because it has the electronic configuration (Ar) 3d⁶, resulting in four unpaired electrons.
15 Medium
15 of 410
Remembering +4 | -1
When hydrogen peroxide is added to acidified potassium dichromate, a blue colour is produced due to formation of
View answer and explanation
Correct option: C
CrO5\mathrm{Cr}{\mathrm{O}_5}
Option C (CrO5\mathrm{Cr}{\mathrm{O}_5}) is correct because hydrogen peroxide oxidizes acidified potassium dichromate to form unstable blue chromium peroxide (CrO5\mathrm{Cr}{\mathrm{O}_5}).

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16 Hard
16 of 410
Analyzing +4 | -1
Sodium salt of an organic acid ′X′'X' produces effervescence with conc. H2SO4{H_2}S{O_4}. ′X′'X' reacts with the acidified aqueous CaCl2CaC{l_2} solution to give a white precipitate which decolourises acidic solution of KMnO4KMn{O_4}. ′X′'X' is
View answer and explanation
Correct option: D
Na2C2O4N{a_2}{C_2}{O_4}
Option D is correct because sodium oxalate (Na₂C₂O₄) produces effervescence with conc. H₂SO₄, forms a white precipitate with CaCl₂, and decolorizes acidic KMnO₄.
17 Medium
17 of 410
Remembering +4 | -1
Which is the common oxidation state of the first transition series of elements?
View answer and explanation
Correct option: A
+2+ 2
Option A (+2+2) is correct because the most common oxidation state for the first transition series is +2 due to the loss of ns² electrons.
18 Medium
18 of 410
Analyzing +4 | -1
S-I: Many copper (I) compounds are undergo disproportion.

S-II: Cu2+{Cu}^{2+} is more stable than Cu+{Cu}^{+} because of much more negative Δhyd H∘\Delta_{\text {hyd }}{H}^{\circ} of Cu2+{Cu}^{2+} than Cu+{Cu}^{+}, which more than compensates for second ionisation enthalpy.
View answer and explanation
Correct option: A
Statement-I & statement-II are correct
Option A is correct because both statements are true: Copper(I) compounds disproportionate, and Cu2+ is more stable than Cu+ due to its higher hydration enthalpy.
19 Medium
19 of 410
Remembering +4 | -1
Which of the following oxides of chromium is amphoteric in nature?
View answer and explanation
Correct option: B
Cr2O3\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_3}
Option B (Cr2O3\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_3}) is correct because chromium(III) oxide is known to exhibit amphoteric behavior, reacting with both acids and bases.
20 Medium
20 of 410
Analyzing +4 | -1
Which of the following statements are correct, when a mixture of NaCl\mathrm{NaCl} and K2Cr2O7\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 is gently warmed with conc. H2SO4\mathrm{H}_2 \mathrm{SO}_4 ?

I. A deep red vapour is evolved.

II. The vapour when passed into NaOH\mathrm{NaOH} solution gives a yellow solution of Na2CrO4\mathrm{Na}_2 \mathrm{CrO}_4.

III. Chlorine gas is evolved.

IV. Chromyl chloride is formed.

Choose the correct option.
View answer and explanation
Correct option: C
I, II and IV
Option C (I, II and IV) is correct because the reaction produces chromyl chloride (CrO2Cl2CrO_2Cl_2) which gives a yellow solution of Na2CrO4Na_2CrO_4 when passed into NaOH.
21 Medium
21 of 410
Remembering +4 | -1
The oxide that shows magnetic property is
View answer and explanation
Correct option: B
Mn3O4M{n_3}{O_4}
Option B (Mn3O4M{n_3}{O_4}) is correct because it exhibits magnetic properties due to the presence of unpaired electrons in Mn.
22 Easy
22 of 410
Remembering +4 | -1
An element which is highly toxic for plants and animals is:
View answer and explanation
Correct option: C
Hg
Option C (Hg) is correct because Mercury (Hg) and its compounds are highly toxic to both plants and animals.
23 Medium
23 of 410
Understanding +4 | -1
Hardness of transition elements is due to:
View answer and explanation
Correct option: C
Covalent bonds
Option C (Covalent bonds) is correct because the hardness of transition elements is attributed to the formation of strong covalent bonds due to the presence of unpaired d-electrons.
24 Medium
24 of 410
Application +4 | -1
Select the coloured and paramagnetic ions
View answer and explanation
Correct option: C
Cu2+,Cr+,Mn2+{Cu}^{2+}, {Cr}^{+}, {Mn}^{2+}
Option C (Cu2+,Cr+,Mn2+{Cu}^{2+}, {Cr}^{+}, {Mn}^{2+}) is correct because these ions have unpaired electrons in their d-orbitals, making them both colored and paramagnetic.
25 Medium
25 of 410
Application +4 | -1
The magnetic moment of Cu2+\mathrm{C}{\mathrm{u}^{2 + }} ion is
View answer and explanation
Correct option: D
1.73
Option D (1.73) is correct because Cu2+\mathrm{C}{\mathrm{u}^{2 + }} has one unpaired electron, and its magnetic moment is calculated as n(n+2)=1(1+2)=1.73\sqrt {n(n + 2)} = \sqrt {1(1 + 2)} = 1.73 BM.
26 Medium
26 of 410
Application +4 | -1
A first row transition metal in its +2+2 oxidation state has a spin-only magnetic moment value of 3.86 BM3.86 \mathrm{~BM}. The atomic number of the metal is
View answer and explanation
Correct option: D
23
Option D (23) is correct because V+2V^{+2} has 3 unpaired electrons, resulting in a spin-only magnetic moment of 3.86 BM.
27 Medium
27 of 410
Analyzing +4 | -1
During the detection of acidic radical present in a salt, a student gets a pale yellow precipitate soluble with difficulty in NH4OH\mathrm{NH}_4 \mathrm{OH} solution when sodium carbonate extract was first acidified with dil. HNO3\mathrm{HNO}_3 and then AgNO3\mathrm{AgNO}_3 solution was added. This indicates presence of:
View answer and explanation
Correct option: A
Br−\mathrm{Br}^{-}
Option A (Br-) is correct because AgBr is a pale yellow precipitate that is sparingly soluble in NH4OH.
28 Medium
28 of 410
Analyzing +4 | -1
Given below are two statements one is labelled as Assertion(A) and the other is labelled as Reason(R).

Assertion(A): Size of BK+3{BK}^{+3} is less than that of NP+3{NP}^{+3} ion.

Reason(R): The above is a consequence of lanthanoid contraction.
In the light of the above statements, choose the most appropriate answer from the options given below.
View answer and explanation
Correct option: B
A is true but R is false
Option B is correct because under identical conditions of P and T, equal volumes contain equal moles, and since the molar masses of N2 and CO are both 28 g/mol, and the adjusted molar mass of O2 is also effectively 28 g/mol, their weights are equal.
29 Medium
29 of 410
Analyzing +4 | -1
Among the following compounds, the one which is both paramagnetic and coloured is
View answer and explanation
Correct option: C
VOSO4\mathrm{VOSO}_4
Option C (VOSO4) is correct because vanadium in VOSO4 has a +4 oxidation state with a d1 electronic configuration, making it paramagnetic and colored.
30 Medium
30 of 410
Remembering +4 | -1
Transuranic elements begins with
View answer and explanation
Correct option: A
Np
Option A (Np) is correct because Neptunium (Np) is the first transuranic element, following Uranium (U) in the periodic table.
Revision notes and key takeaways

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  • ✓ This set contains 410 single-correct-answer MCQs on d & f- BLOCK ELEMENTS.
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