Skip to main content
Back to d & f- BLOCK ELEMENTS

Your Progress

0 / 254 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • CHEMISTRY • d & f- BLOCK ELEMENTS

D & f- BLOCK ELEMENTS practice · Medium

Try a question, then open “View answer” to check it. Sign in to submit answers and save progress.

Practice format and chapter quiz

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.

Take a chapter quiz →
Showing 211-240 of 254 questions
211 Medium
211 of 254
Analyzing +4 | -1
Which of the following is not oxidized by O3{\mathrm{O}_3}?
View answer and explanation
Correct option: B
KMnO4\mathrm{KMn}{\mathrm{O}_4}
The correct answer is B (KMnO4\mathrm{KMn}{\mathrm{O}_4}) because manganese is already in its highest oxidation state (+7) in KMnO4\mathrm{KMn}{\mathrm{O}_4}, so it cannot be further oxidized by ozone.
212 Medium
212 of 254
Understanding +4 | -1
Addition of high proportions of manganese makes steel useful in making rails of rail roads because manganese ;
View answer and explanation
Correct option: A
Gives hardness to steel and can remove oxygen and sulphur
Option A (Gives hardness to steel and can remove oxygen and sulphur) is correct because manganese improves steel's hardness and acts as a deoxidizer and desulfurizer.
213 Medium
213 of 254
Remembering +4 | -1
Chemical name of corrosive sublimate is:
View answer and explanation
Correct option: C
Mercuric chloride
Option C (Mercuric chloride) is correct because the chemical name of corrosive sublimate is Mercuric chloride (HgCl2\mathrm{HgC}{\mathrm{l}_2}).
214 Medium
214 of 254
Remembering +4 | -1
Non-stoichiometric compounds are formed by:
View answer and explanation
Correct option: B
Transition elements
Option B (Transition elements) is correct because transition elements often exhibit variable oxidation states and crystal defects, leading to the formation of non-stoichiometric compounds.
215 Medium
215 of 254
Remembering +4 | -1
dd-block elements are arranged in ….of periodic table.
View answer and explanation
Correct option: D
Four series
Option D (Four series) is correct because the d-block elements are arranged in four series corresponding to the filling of the 3d, 4d, 5d, and 6d orbitals.
216 Medium
216 of 254
Analyzing +4 | -1
Which of the following are incorrect orders of property indicated against it?
View answer and explanation
Correct option: C
Sc+2>Ti+3>V+3>Cr+3{Sc}^{+2}>{Ti}^{+3}>{V}^{+3}>{Cr}^{+3} (order of polarizing Power)
The correct answer is C because the order of polarizing power is incorrect. Polarizing power increases with charge and decreases with size.
217 Medium
217 of 254
Analyzing +4 | -1
A mixture of inorganic substances on treatment with H2SO4{H}_2 {SO}_4 (dilute) gives out a gas which turns K2Cr2O7{K}_2 {Cr}_2 {O}_7 paper green and lead acetate paper black. The mixture contains
View answer and explanation
Correct option: D
Sulphide
Option D (Sulphide) is correct because the gas produced that turns K2Cr2O7 paper green and lead acetate paper black is hydrogen sulfide (H2S), which is produced from a sulfide.
218 Medium
218 of 254
Application +4 | -1
1 mole of Fe(NO2)2\mathrm{Fe}\left(\mathrm{NO}_2\right)_2 is oxidised into Fe3+\mathrm{Fe}^{3+} and NO3−\mathrm{NO}_3^{-} by x\mathrm{x} mole of Cr2O72−\mathrm{Cr}_2 \mathrm{O}_7^{2-} in acidic medium, x\mathrm{x} is
View answer and explanation
Correct option: D
0.830.83
Option D (0.83) is correct because it is calculated by equating the gram equivalents of Fe(NO2)2\mathrm{Fe(NO_2)_2} and Cr2O72−\mathrm{Cr_2O_7^{2-}}.
219 Medium
219 of 254
Remembering +4 | -1
The process of nitriding used in the treatment of steel is:
View answer and explanation
Correct option: A
Heating steel in an atmosphere of ammonia
Option A (Heating steel in an atmosphere of ammonia) is correct because nitriding involves heating steel in an atmosphere of ammonia to introduce nitrogen into the surface.
220 Medium
220 of 254
Understanding +4 | -1
KI and CuSO4\mathrm{CuS}{\mathrm{O}_4} solutions on mixing produce
View answer and explanation
Correct option: B
Cu2I2+I2+K2SO4\mathrm{C}{\mathrm{u}_2}{\mathrm{I}_2} + {\mathrm{I}_2} + {\mathrm{K}_2}\mathrm{S}{\mathrm{O}_4}
Option B is correct. KI reacts with CuSO4 to produce Cu2I2, I2, and K2SO4.
221 Medium
221 of 254
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 oxidation number of Mn in K2MnO4, the product of alkaline oxidative fusion of MnO2, is +6.
222 Medium
222 of 254
Understanding +4 | -1
K2Cr2O7{\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7} on strong heating gives:
View answer and explanation
Correct option: D
All of these
Option D is correct because strong heating of K2Cr2O7{\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7} yields K2CrO4{\mathrm{K}_2}\mathrm{Cr}{\mathrm{O}_4}, Cr2O3\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_3}, and O2{\mathrm{O}_2}.
223 Medium
223 of 254
Remembering +4 | -1
Which of the following compounds is metallic and ferromagnetic?
View answer and explanation
Correct option: A
CrO2Cr{O_2}
Option A (CrO2Cr{O_2}) is correct because Chromium(IV) oxide is a metallic and ferromagnetic compound used in magnetic tapes.
224 Medium
224 of 254
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.
225 Medium
225 of 254
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.
226 Medium
226 of 254
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.
227 Medium
227 of 254
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.
228 Medium
228 of 254
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.
229 Medium
229 of 254
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.
230 Medium
230 of 254
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.
231 Medium
231 of 254
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.
232 Medium
232 of 254
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.
233 Medium
233 of 254
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.
234 Medium
234 of 254
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.
235 Medium
235 of 254
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.
236 Medium
236 of 254
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.
237 Medium
237 of 254
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.
238 Medium
238 of 254
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.
239 Medium
239 of 254
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.
240 Medium
240 of 254
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.
Revision notes and key takeaways

Review d & f- BLOCK ELEMENTS in your textbook or class notes, then use this set to check your understanding.

Showing 211-240 of 254 questions
0 /254
0✓ 0✗
0% Accuracy
Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.