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JEE • CHEMISTRY • d & f- BLOCK ELEMENTS

D & f- BLOCK ELEMENTS practice · Mixed difficulty

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Showing 61-90 of 410 questions

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61 Medium
61 of 410
Understanding +4 | -1
For the ions Zn2+,Ni2+{Zn}^{2+}, {Ni}^{2+} and Cr3+{Cr}^{3+} which among the following statements is correct?

(atomic number of Zn=30,Ni=28{Zn}=30, {Ni}=28 and Cr=24{Cr}=24 )
View answer and explanation
Correct option: D
Only Zn2+{Zn}^{2+} is colourless and Ni2+{Ni}^{2+} and Cr3+{Cr}^{3+} are coloured
Option D (Only Zn2+{Zn}^{2+} is colourless and Ni2+{Ni}^{2+} and Cr3+{Cr}^{3+} are coloured) is correct because Zn2+{Zn}^{2+} has a completely filled d-orbital configuration (d10d^{10}), making it colorless, while Ni2+{Ni}^{2+} (d8d^8) and Cr3+{Cr}^{3+} (d3d^3) have unpaired d-electrons and exhibit d-d transitions, resulting in color.
62 Easy
62 of 410
Understanding +4 | -1
Percentage of gold in 21.6 carat gold is:
View answer and explanation
Correct option: B
90
Option A (1) is correct because Xe in XeOF4 has sp3d2 hybridization with one lone pair of electrons.
63 Medium
63 of 410
Analyzing +4 | -1
Assertion: Fe3O4\mathrm{Fe}_3 \mathrm{O}_4 contains iron atoms in two different oxidation numbers.

Reason: Fe2+\mathrm{Fe}^{2+} ions decolourize KMnO4\mathrm{KMnO}_4 solution.
View answer and explanation
Correct option: B
Both assertion and reason are true but reason is not the correct explanation of assertion.
Option A (C2H5−{\mathrm{C}_2}\mathrm{H}_5^ -) is the strongest base because it is the conjugate base of the weakest acid (ethane).
64 Hard
64 of 410
Understanding +4 | -1
Assertion : KMnO4{\text{KMn}}{{\text{O}}_{\text{4}}} is dark pink coloured compound.

Reason : In the KMnO4{\text{KMn}}{{\text{O}}_{\text{4}}} charge transfer spectrum occurs.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is correct explanation of the Assertion.
The correct answer is A (If both Assertion & Reason are True & the Reason is correct explanation of the Assertion) because KMnO4's color arises from a charge transfer spectrum.
65 Medium
65 of 410
Remembering +4 | -1
Zinc white is a better white pigment than lead white because it:
View answer and explanation
Correct option: B
Is not blackened by the action of H2S{\mathrm{H}_2}\mathrm{S}
Option B (Is not blackened by the action of H2S{\mathrm{H}_2}\mathrm{S}) is correct because zinc white does not react with hydrogen sulfide to form a black sulfide, unlike lead white.
66 Easy
66 of 410
Analyzing +4 | -1
Which shows a jump in second ionization potential?
View answer and explanation
Correct option: D
Cu
Option D (Cu) is correct because copper has an electronic configuration of [Ar]3d¹⁰4s¹, and removing the second electron requires breaking the stable, fully filled 3d¹⁰ configuration, resulting in a large jump in ionization energy.
67 Easy
67 of 410
Remembering +4 | -1
Useful lanthanoid member is:
View answer and explanation
Correct option: A
Cerium
Option A (Cerium) is correct because Cerium is used in gas mantles, glass polishing, and lighter flints.
68 Medium
68 of 410
Understanding +4 | -1
Which of the following is amphoteric oxide?

Mn2O7,CrO3,Cr2O3,CrO,V2O5 and V2O4\mathrm{Mn}_2 \mathrm{O}_7, \mathrm{CrO}_3, \mathrm{Cr}_2 \mathrm{O}_3, \mathrm{CrO}, \mathrm{V}_2 \mathrm{O}_5 \text { and } \mathrm{V}_2 \mathrm{O}_4
View answer and explanation
Correct option: A
V2O5,Cr2O3\mathrm{V}_2 \mathrm{O}_5, \mathrm{Cr}_2 \mathrm{O}_3
Option A (V2O5,Cr2O3\mathrm{V}_2 \mathrm{O}_5, \mathrm{Cr}_2 \mathrm{O}_3) is correct because both vanadium(V) oxide and chromium(III) oxide can react with both acids and bases.
69 Medium
69 of 410
Understanding +4 | -1
Which one of the following statements concerning lanthanide elements is false?
View answer and explanation
Correct option: B
The ionic radii of trivalent lanthanides steadily increase with increase in atomic number.
Option B is false because the ionic radii of trivalent lanthanides steadily decrease with increasing atomic number due to lanthanide contraction.
70 Easy
70 of 410
Application +4 | -1
Electronic configuration of a transition element x in +3 oxidation state is [Ar]3d5\left[ {Ar} \right]3{d^5}. It’s atomic number is
View answer and explanation
Correct option: B
26
Option B (26) is correct because the atomic number is calculated by adding the number of electrons in Argon, the 3d orbital, and the oxidation state.
71 Easy
71 of 410
Remembering +4 | -1
Which of the following is the correct sequence of atomic weights of given elements?
View answer and explanation
Correct option: A
Co>Ni>Fe\mathrm{Co} > Ni > Fe
Option A (Co>Ni>Fe\mathrm{Co} > Ni > Fe) is correct because the atomic weights are approximately Co (58.9), Ni (58.7), and Fe (55.8), thus Co > Ni > Fe.
72 Medium
72 of 410
Remembering +4 | -1
Which of the following is a lanthanoid?
View answer and explanation
Correct option: D
Lu
Option D (Lu) is correct because Lutetium (Lu) is a lanthanoid element, belonging to the series from atomic number 58 to 71.
73 Easy
73 of 410
Remembering +4 | -1
Which ion in aqueous medium has orange colour?
View answer and explanation
Correct option: A
Cr2O72−\mathrm{C}{\mathrm{r}_2}\mathrm{O}_7^{2 - }
Option A (Cr2O72−\mathrm{C}{\mathrm{r}_2}\mathrm{O}_7^{2 - }) is correct because dichromate ions (Cr2O72−\mathrm{C}{\mathrm{r}_2}\mathrm{O}_7^{2 - }) are known to have an orange color in aqueous solutions.
74 Medium
74 of 410
Analyzing +4 | -1
Assertion (A): The degree of complex formation in actinoides decreases in the order

M4+>MO22+>M3+>MO2+{M}_4^{+}>{MO}_2^{2+}>{M}^{3+}>{MO}_2^{+}

Reason (R): Actinoides form complexes with π\pi-bonding ligands such as alkyl phosphines and thioethers.
View answer and explanation
Correct option: A
Both A and R are true and R is the correct explanation of A
Option A (Both A and R are true and R is the correct explanation of A) is correct because the degree of complex formation depends on the charge and size of the ion, and actinides form complexes with pi-bonding ligands.
75 Medium
75 of 410
Understanding +4 | -1
When KI\mathrm{KI} (excess) is added to the following solutions

(i) CuSO4\mathrm{CuSO}_4

(ii) HgCl2\mathrm{HgCl}_2

(iii) Pb(NO3)2\mathrm{Pb}\left(\mathrm{NO}_3\right)_2

separately:
View answer and explanation
Correct option: D
A white precipitate of Cu2I2\mathrm{Cu}_2 \mathrm{I}_2 in I\mathrm{I}, an orange precipitate dissolving to HgI42−\mathrm{HgI}_4^{2-} in II\mathrm{II} and a yellow precipitate of PbI2\mathrm{PbI}_2 in III
Option D (A white precipitate of Cu2I2\mathrm{Cu}_2 \mathrm{I}_2 in I\mathrm{I}, an orange precipitate dissolving to HgI42−\mathrm{HgI}_4^{2-} in II\mathrm{II} and a yellow precipitate of PbI2\mathrm{PbI}_2 in III) is correct because it accurately describes the reactions and precipitates formed when excess KI is added to CuSO4, HgCl2, and Pb(NO3)2 solutions.
76 Medium
76 of 410
Application +4 | -1
The reaction below can be used as a laboratory method for preparing small quantities of Cl2( g)\mathrm{Cl}_2(\mathrm{~g}). If a 49 g49 \mathrm{~g} sample that is 96% K2Cr2O796 \% \mathrm{~K}_2 \mathrm{Cr}_2 \mathrm{O}_7 by mass is allowed to react with 325 mL325 \mathrm{~mL} of HCl(aq)\mathrm{HCl}_{(\mathrm{aq})} with a density of 1.15 g/ml1.15 \mathrm{~g} / \mathrm{ml} and 30.1%HCl30.1 \% \mathrm{HCl} by mass, how many grams of Cl2( g)\mathrm{Cl}_2(\mathrm{~g}) are produced?

K2Cr2O7+HCl→CrCl3+Cl2+H2O+KCl\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+\mathrm{HCl} \rightarrow \mathrm{CrCl}_3+\mathrm{Cl}_2+\mathrm{H}_2 \mathrm{O}+\mathrm{KCl}

(Atomic weights: K=39,Cr=52,Cl=35.5\mathrm{K}=39, \mathrm{Cr}=52, \mathrm{Cl}=35.5 )
View answer and explanation
Correct option: C
34.08 g34.08 \mathrm{~g}
Option C (34.08 g) is correct because it is calculated by determining the limiting reagent and then using stoichiometry to find the mass of chlorine produced.
77 Easy
77 of 410
Remembering +4 | -1
Lightest transition element is:
View answer and explanation
Correct option: B
Sc
Option B (Sc) is correct because Scandium (Sc) is the first element in the first transition series and has the lowest atomic mass among the given options.
78 Medium
78 of 410
Understanding +4 | -1
Choose the appropriate option about the misch metal.
View answer and explanation
Correct option: C
It finds application in making aeroplane body
Option A is correct. Misch metal is an alloy of lanthanoid metal (~95%) and iron (~5%) with traces of S, C, Ca, and Al. It's used in Mg-based alloys for various applications, including making lighter flints, but not aeroplane bodies.
79 Easy
79 of 410
Remembering +4 | -1
Which is less reactive?
View answer and explanation
Correct option: C
Pt
The correct answer is C (Pt) because Platinum (Pt) is a noble metal and is known for its inertness and resistance to chemical reactions, making it less reactive than the other options.
80 Medium
80 of 410
Understanding +4 | -1
Consider the following statements.

(I)La(OH)3\mathrm{La}{\left( {\mathrm{OH}} \right)_3} is the least basic among hydroxides of lanthanides

(II) Zr4+\mathrm{Z}{\mathrm{r}^{4 + }} and Hf4+\mathrm{H}{\mathrm{f}^{4 + }} possess almost the same ionic radii

(III) Ce4+\mathrm{C}{\mathrm{e}^{4 + }} can act as an oxidizing agent

Which of the above is/are true?
View answer and explanation
Correct option: D
(I) only
Option A is correct because La(OH)3 is indeed the least basic among lanthanide hydroxides, and Ce4+ can act as an oxidizing agent.
81 Hard
81 of 410
Analyzing +4 | -1
Assertion : Actinides show much higher range of oxidation states

Reason : Energy difference between 5f and 6d orbitals is large
View answer and explanation
Correct option: C
If Assertion is True but the Reason is False.
Option C is correct. Actinides exhibit a wider range of oxidation states due to smaller energy differences between 5f, 6d, and 7s orbitals, not a large difference as stated in the reason.
82 Medium
82 of 410
Understanding +4 | -1
Statement (I): 2nd and 3rd transition series exhibit similar radii.

Statement (II): It is the net result of lanthanoid contraction.
View answer and explanation
Correct option: A
Statement I and Statement II are correct and statement II is the correct explanation of I\mathrm{I}.
Option A is correct because both statements are true, and the lanthanoid contraction directly explains the similar radii of the 2nd and 3rd transition series.
83 Medium
83 of 410
Application +4 | -1
The magnetic moment of a transition metal ion is 15  \sqrt {15\;}BM. Therefore, the number of unpaired electrons present in it, is
View answer and explanation
Correct option: A
3
Option A (3) is correct because the number of unpaired electrons (n) is calculated using the formula μ=n(n+2)\mathrm{\mu } = \sqrt {n(n + 2)} and solving for n when μ=15\mathrm{\mu } = \sqrt{15}.
84 Easy
84 of 410
Analyzing +4 | -1
A solid (A) which has photographic effect reacts with the solution of a sodium salt (B) to give a pale yellow ppt. (C). Sodium salt on heating gives brown vapours.

Identify A, B and C .
View answer and explanation
Correct option: A
AgNO3,  NaBr,  AgBr\mathrm{AgN}{\mathrm{O}_3},\mathrm{\;NaBr},\mathrm{\;AgBr}
Option A (AgNO3,  NaBr,  AgBr\mathrm{AgN}{\mathrm{O}_3},\mathrm{\;NaBr},\mathrm{\;AgBr}) is correct because silver nitrate reacts with sodium bromide to form silver bromide, a pale yellow precipitate used in photography, and sodium bromide produces brown fumes upon heating.
85 Easy
85 of 410
Remembering +4 | -1
Aqueous solution of which ion is pink in colour
View answer and explanation
Correct option: C
Mn2+{Mn}^{2+}
Option C (Mn2+{Mn}^{2+}) is correct because aqueous solutions of manganese(II) ions are typically pale pink in color.
86 Hard
86 of 410
Analyzing +4 | -1
Consider the following statements

I. ScSc has a low value of EM+3/M+20E_{{M^{ + 3}}/{M^{ + 2}}}^0

II. ZnZn has the highest value of EM+3/M+20E_{{M^{ + 3}}/{M^{ + 2}}}^0

III. MnMn and FeFe have comparatively high and low values of EM+3/M+20E_{{M^{ + 3}}/{M^{ + 2}}}^0 respectively

The correct set of statements is (Choose the low values of EM+3/M+20E_{{M^{ + 3}}/{M^{ + 2}}}^0 respectively).

The correct set of statements is (Choose the appropriate option)
View answer and explanation
Correct option: B
I, II and III
Option B (I, II and III) is correct because all three statements accurately describe the trends in EM+3/M+20E_{{M^{ + 3}}/{M^{ + 2}}}^0 values for Sc, Zn, Mn, and Fe.
87 Easy
87 of 410
Remembering +4 | -1
Verdigris is
View answer and explanation
Correct option: B
Basic copper acetate
Option B (Basic copper acetate) is correct because verdigris is a green pigment consisting of basic copper acetate.
88 Medium
88 of 410
Understanding +4 | -1
Which of the following compound has colour due to d−dd-d transition?
View answer and explanation
Correct option: D
CuSO4.5H2O\mathrm{CuSO}_4.5 \mathrm{H}_2 \mathrm{O}
Option D (CuSO4.5H2O\mathrm{CuSO}_4.5 \mathrm{H}_2 \mathrm{O}) is correct because the blue color of copper sulfate pentahydrate arises from d-d transitions in the copper(II) ion due to the ligand field splitting caused by water molecules.
89 Medium
89 of 410
Remembering +4 | -1
The magnetic moment of lanthanide ions is determined from which one of the following relation?
View answer and explanation
Correct option: B
μ=gJ(J+1)\mu=g \sqrt{{J}({J}+1)}
Option B (μ=gJ(J+1)\mu=g \sqrt{{J}({J}+1)}) is correct because the magnetic moment of lanthanide ions is determined using this formula, which accounts for both spin and orbital angular momentum contributions.
90 Easy
90 of 410
Understanding +4 | -1
Which of the following electronic configurations belong to transition elements?
View answer and explanation
Correct option: A
KL 3s2p6d5,4s13{s^2}{p^6}{d^5},4{s^1}
Option A is correct because transition elements are characterized by having partially filled d-orbitals, which is the case in the given electronic configuration KL 3s²3p⁶3d⁵, 4s¹.
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