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JEE • CHEMISTRY • d & f- BLOCK ELEMENTS

D & f- BLOCK ELEMENTS practice · Medium

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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 254 questions
1 Medium
1 of 254
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.
2 Medium
2 of 254
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.
3 Medium
3 of 254
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.
4 Medium
4 of 254
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.
5 Medium
5 of 254
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.
6 Medium
6 of 254
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).
7 Medium
7 of 254
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.
8 Medium
8 of 254
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.
9 Medium
9 of 254
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.
10 Medium
10 of 254
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}).
11 Medium
11 of 254
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.
12 Medium
12 of 254
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.
13 Medium
13 of 254
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.
14 Medium
14 of 254
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.
15 Medium
15 of 254
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.

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16 Medium
16 of 254
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.
17 Medium
17 of 254
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.
18 Medium
18 of 254
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.
19 Medium
19 of 254
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.
20 Medium
20 of 254
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.
21 Medium
21 of 254
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.
22 Medium
22 of 254
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.
23 Medium
23 of 254
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.
24 Medium
24 of 254
Understanding +4 | -1
Interstitial compounds are formed when small atoms are trapped inside the crystal lattice of metals. Which of the following are the characteristic properties of interstitial compounds?

I. They have high melting points in comparison to pure metals.

II. They are very hard.

III. They retain metallic conductivity.

IV. They are chemically very reactive.

Choose the correct option.
View answer and explanation
Correct option: A
I, II and III
Option A (I, II and III) is correct because interstitial compounds have high melting points, are very hard, and retain metallic conductivity, but they are chemically inert, not reactive.
25 Medium
25 of 254
Remembering +4 | -1
The composition of duralumin is :
View answer and explanation
Correct option: D
Al(95%),Cu(4%),Mn(0.5%),Mg(0.5%){Al}(95 \%), {Cu}(4 \%), {Mn}(0.5 \%), {Mg}(0.5 \%).
Option D (Al(95%),Cu(4%),Mn(0.5%),Mg(0.5%){Al}(95 \%), {Cu}(4 \%), {Mn}(0.5 \%), {Mg}(0.5 \%)) is correct because this is the standard composition of duralumin, a lightweight and strong aluminum alloy.
26 Medium
26 of 254
Analyzing +4 | -1
A compound of iron exists as a dimer in vapour state. It is hygroscopic in nature and dissolves in water
giving brown acidic solution. The compound is :
View answer and explanation
Correct option: C
FeCl3\mathrm{FeCl}_3
Option C (FeCl3\mathrm{FeCl}_3) is correct because it forms a dimer in the vapor state, is hygroscopic, and dissolves in water to give a brown acidic solution due to the formation of Fe(OH)3\mathrm{Fe(OH)}_3 and HCl\mathrm{HCl}.
27 Medium
27 of 254
Remembering +4 | -1
Most common oxidation states of Ce (cerium) are
View answer and explanation
Correct option: A
+3,+4+ 3,\mathrm{ } + 4
Option A (+3, +4) is correct because Cerium (Ce) exhibits stable oxidation states of +3 and +4 due to its electronic configuration.
28 Medium
28 of 254
Analyzing +4 | -1
A metal which is ‘not’ affected by conc. H2SO4,  HNO3{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4},\mathrm{\;HN}{\mathrm{O}_3} or alkalies forms a compound XX. This compound XX can be used to give a complex which finds its application for toning in photography. The metal is:
View answer and explanation
Correct option: A
Au
Option A (Au) is correct because gold (Au) is inert to concentrated H₂SO₄, HNO₃, and alkalies, and its chloride (AuCl₃) forms a complex used in photography toning.
29 Medium
29 of 254
Remembering +4 | -1
The element californium belongs to the family of:
View answer and explanation
Correct option: A
Actinoid series
The correct answer is A (Actinoid series) because californium (Cf) with atomic number 98 belongs to the actinoid series.
30 Medium
30 of 254
Remembering +4 | -1
Which is used as substitute for platinum in jewellery?
View answer and explanation
Correct option: A
Rolled gold
Option A (Rolled gold) is correct because rolled gold is often used as a cheaper alternative to platinum in jewelry.
Revision notes and key takeaways

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Key Takeaways

  • ✓ This set contains 254 single-correct-answer MCQs on d & f- BLOCK ELEMENTS.
  • ✓ Try each question before opening its answer. Read the explanation where one is available.
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