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JEE • CHEMISTRY • d & f- BLOCK ELEMENTS

D & f- BLOCK ELEMENTS practice · Medium

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Showing 151-180 of 254 questions

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151 Medium
151 of 254
Analyzing +4 | -1
Element of most positive value of EM3+/M2+∘{E}_{{M}^{3+} / {M}^{2+}}^{\circ} among the following is
View answer and explanation
Correct option: D
CoCo
Option D (CoCo) is correct because Cobalt has the most positive EM3+/M2+∘{E}_{{M}^{3+} / {M}^{2+}}^{\circ} value among the given elements.
152 Medium
152 of 254
Analyzing +4 | -1
Among Sc(III), Ti(IV), Pd(II) and Cu(II) ions
View answer and explanation
Correct option: D
Sc (III), Ti (IV) are diamagnetic and Pd(II), Cu(II) are paramagnetic
Option D (Sc (III), Ti (IV) are diamagnetic and Pd(II), Cu(II) are paramagnetic) is correct because Sc(III) and Ti(IV) have no d-electrons, while Pd(II) and Cu(II) have unpaired d-electrons.
153 Medium
153 of 254
Remembering +4 | -1
The composition of bell metal is
View answer and explanation
Correct option: A
Cu + Sn
Option A (Cu + Sn) is correct because bell metal is an alloy primarily composed of copper and tin.
154 Medium
154 of 254
Application +4 | -1
Cr2O72−+xH++ye−→2Cr3++AH2O\mathrm{Cr}_2 \mathrm{O}_7^{2-}+\mathrm{xH}^{+}+\mathrm{ye}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+\mathrm{AH}_2 \mathrm{O}
Balance the above reaction and find x,y\mathrm{x, y} and A\mathrm{A}.
View answer and explanation
Correct option: B
x=14,y=6,A=7\mathrm{x=14, y=6, A=7}
The correct answer is B (x=14, y=6, A=7) because balancing the redox reaction requires 14 H⁺ ions, 6 electrons, and produces 7 water molecules.
155 Medium
155 of 254
Analyzing +4 | -1
Which of the following conversions can be carried out by both acidified K2Cr2O4\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_4 and acidified KMnO4\mathrm{KMnO}_4 ?

(i) Fe2+→Fe3++e−\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-}

(ii) I−→IO3−\mathrm{I}^{-} \rightarrow \mathrm{IO}_3^{-}

(iii) I−→I2\mathrm{I}^{-} \rightarrow \mathrm{I}_2

(iv) H2 S→S\mathrm{H}_2 \mathrm{~S} \rightarrow \mathrm{S}
View answer and explanation
Correct option: C
(i), (iii) and (iv)
Option C is correct because acidified K2Cr2O4 and acidified KMnO4 can both oxidize Fe2+ to Fe3+, I- to I2, and H2S to S.
156 Medium
156 of 254
Remembering +4 | -1
Metal are generally hard, opaque, shiny and good conductor of heat and electricity. Which of the following elements has highest density?
View answer and explanation
Correct option: B
IrIr
Option B (IrIr) is correct because Iridium (Ir) has the highest density among the given elements.
157 Medium
157 of 254
Application +4 | -1
In the equation,

4M  +  8CN−  +  2H2O  +  O2    →  4[M(CN)2]−  +  4OH−4M\mathrm{\;} + \mathrm{\;}8\mathrm{C}{\mathrm{N}^ - }\mathrm{\;} + \mathrm{\;}2{\mathrm{H}_2}\mathrm{O\;} + \mathrm{\;}{\mathrm{O}_2}\mathrm{\;\;} \to \mathrm{\;}4{\left[ {M{{\left( {\mathrm{CN}} \right)}_2}} \right]^ - }\mathrm{\;} + \mathrm{\;}4\mathrm{O}{\mathrm{H}^ - }

Identify the metal MM
View answer and explanation
Correct option: C
Silver
Option C (Silver) is correct because silver forms a complex with cyanide ions in the presence of oxygen and water, as shown in the cyanide process.
158 Medium
158 of 254
Understanding +4 | -1
The maximum number of possible oxidation states of actinoides are shown by
View answer and explanation
Correct option: D
neptunium (Np) and plutonium (Pu)
Option D (neptunium (Np) and plutonium (Pu)) is correct because these elements exhibit a wider range of oxidation states (+3 to +6) compared to the other actinoids listed.
159 Medium
159 of 254
Remembering +4 | -1
What is the oxidation state of iron in Mohr’s salt?
View answer and explanation
Correct option: C
+2+ 2
Option C (+2+ 2) is correct because in Mohr's salt, the iron is present as Fe2+Fe^{2+} ion, thus having an oxidation state of +2.
160 Medium
160 of 254
Understanding +4 | -1
When mercury (I) chloride is heated and the vapour so evolved are cooled, the substance on sublimation thus collected consists of
View answer and explanation
Correct option: A
Mercury and mercury (II) chloride
The correct answer is Mercury and mercury (II) chloride because when mercury (I) chloride is heated, it decomposes into mercury and mercury (II) chloride.
161 Medium
161 of 254
Analyzing +4 | -1
Which is the correct order of ionic sizes?) At.no.∶Ce=58,Sn=50,Yb=70 and Lu=71)
View answer and explanation
Correct option: C
Sn>Ce>Yb>Lu\mathrm{Sn} > Ce > Yb > Lu
Option C (Sn>Ce>Yb>Lu\mathrm{Sn} > Ce > Yb > Lu) is correct because ionic radii decrease with increasing effective nuclear charge, and Sn has the lowest charge among the given ions.
162 Medium
162 of 254
Remembering +4 | -1
Which among the following metals does not dissolve in aqua regia?
View answer and explanation
Correct option: D
Ir
Option D (Ir) is correct because iridium (Ir) is highly resistant to acids, including aqua regia.
163 Medium
163 of 254
Analyzing +4 | -1
Which of the following does not involve d→dd \rightarrow d transition for causing colour ?
View answer and explanation
Correct option: C
KMnO4{KMnO}_4
The correct answer is KMnO4 because the color of KMnO4 arises from ligand-to-metal charge transfer (LMCT), not a d-d transition.
164 Medium
164 of 254
Understanding +4 | -1
Observe the following statements
1) Lanthanides actively participate in chemical reactions
2) The basic nature of hydroxides of lanthanides increases from La(OH)3La{\left( {OH} \right)_3} to Lu(OH)3Lu{\left( {OH} \right)_3}
3) Most stable oxidation number shown by Lanthanides is + 3
The correct statements are
View answer and explanation
Correct option: D
1 and 3
Option D (1 and 3) is correct because lanthanides actively participate in reactions and primarily exhibit a +3 oxidation state, while the basicity of their hydroxides decreases across the series.
165 Medium
165 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.
166 Medium
166 of 254
Application +4 | -1
Correct outer electronic configuration of trivalent cationic form of Np{Np} (Neptunium, atomic number of 93 ) is
View answer and explanation
Correct option: C
[Rn]5f4[R n] 5 f^4
Option C ([Rn]5f4[R n] 5 f^4) is correct because Neptunium (Np) with atomic number 93 loses 3 electrons to form Np3+, resulting in the electronic configuration [Rn] 5f4.
167 Medium
167 of 254
Remembering +4 | -1
Most common oxidation state of Mn\mathrm{Mn} (atomic No: 25)25) is
View answer and explanation
Correct option: B
+2+2
Option B (+2) is correct because the most common oxidation state of manganese (Mn) is +2 due to the stability of the d5d^5 configuration after losing two electrons.
168 Medium
168 of 254
Understanding +4 | -1
Which of the following statements is not true?
View answer and explanation
Correct option: B
Na2Cr2O7\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7 is preferred over K2Cr2O7\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 in volumetric analysis
Option B (Na2Cr2O7\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7 is preferred over K2Cr2O7\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 in volumetric analysis) is not true because sodium dichromate is hygroscopic and cannot be used in volumetric analysis.
169 Medium
169 of 254
Understanding +4 | -1
Assertion : Zr and Hf have about the same atomic radius.

Reason: Zr and Hf lies in the same group.
View answer and explanation
Correct option: B
If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion.
Option B is correct because both the assertion and reason are true, but the reason is not the correct explanation. The lanthanide contraction is the actual reason.
170 Medium
170 of 254
Application +4 | -1
In order to prepare one litre 1 N1 \mathrm{~N} solution of KMnO4{KMnO}_4, how many grams of KMnO4{KMnO}_4 are required, if the solution to be used in acid medium for oxidation?
View answer and explanation
Correct option: C
31.60 g31.60 \mathrm{~g}
Option C (31.60 g) is correct because the equivalent mass of KMnO4 in acidic medium is its molecular mass divided by 5, which equals 31.60 g.
171 Medium
171 of 254
Understanding +4 | -1
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: Permanganate titrations are not performed in presence of hydrochloric acid.
Reason R: Chlorine is formed as a consequence of oxidation of hydrochloric acid.
In the light of the above statements, choose the correct answer from the options given below.
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 HCl can be oxidized by KMnO4, leading to inaccurate titration results.
172 Medium
172 of 254
Remembering +4 | -1
Silver nitrate is usually supplied in coloured bottles because it is:
View answer and explanation
Correct option: B
Decomposed in sunlight
Option B (Decomposed in sunlight) is correct because silver nitrate is photosensitive and decomposes when exposed to sunlight.
173 Medium
173 of 254
Understanding +4 | -1
Zn does not show variable valency because of
View answer and explanation
Correct option: A
Complete dd-subshell
Option A (Complete dd-subshell) is correct because Zinc (Zn) has a completely filled d-orbital, which prevents it from exhibiting variable valency.
174 Medium
174 of 254
Remembering +4 | -1
Metals generally melt at very high temperature. Amongst the following, the metal with the highest melting point will be
View answer and explanation
Correct option: B
AgAg
Option B (AgAg) is correct because Silver (Ag) has the highest melting point among the given options.
175 Medium
175 of 254
Analyzing +4 | -1
Assertion: Actinoids show more variety in oxidation state than lanthanoids.

Reason: Actinoids have comparatively larger size than lanthanoids.
View answer and explanation
Correct option: B
Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion.
Option B (Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion) is correct because while both statements are true, the larger size of actinoids is not the direct cause of their more varied oxidation states. The availability of 5f, 6d, and 7s electrons for bonding is the primary reason.
176 Medium
176 of 254
Application +4 | -1
In the chemical reaction,

K2Cr2O7+XH2SO4+YSO2→K2SO4+Cr2(SO4)3+ZH2O   X,Y{K}_2 {Cr}_2 {O}_7+{XH}_2 {SO}_4+{YSO}_2 \rightarrow {K}_2 {SO}_4+{Cr}_2\left({SO}_4\right)_3+{ZH}_2 {O}\,\,\, {X}, {Y} and Z{Z} are
View answer and explanation
Correct option: A
1,3,11,3,1
Option A (1,3,1) is correct because it correctly balances the given chemical reaction, ensuring the conservation of mass and charge.
177 Medium
177 of 254
Remembering +4 | -1
Case hardening
View answer and explanation
Correct option: B
92.5 % Ag and 7.5% Cu
Option B (92.5 % Ag and 7.5% Cu) is correct because sterling silver, commonly used in case hardening, consists of 92.5% silver and 7.5% copper.
178 Medium
178 of 254
Analyzing +4 | -1
The number of ions from the following that are expected to behave as oxidising agent is:

Sn4+,Sn2+,Pb2+,Tl3+,Pb4+,Tl+\mathrm{Sn}^{4+}, \mathrm{Sn}^{2+}, \mathrm{Pb}^{2+}, \mathrm{Tl}^{3+}, \mathrm{Pb}^{4+}, \mathrm{Tl}^{+}
View answer and explanation
Correct option: D
2
Option D (2) is correct because Tl3+ and Pb4+ are strong oxidizing agents due to the inert pair effect, making them more stable in their lower oxidation states.
179 Medium
179 of 254
Understanding +4 | -1
To an acidified dichromate solution, a pinch of Na2O2{Na}_2 {O}_2 is added and shaken. Which of the following is observed?
View answer and explanation
Correct option: A
Blue colour
The correct answer is A (Blue colour) because the addition of Na2O2Na_2O_2 to an acidified dichromate solution results in the formation of a blue-colored perchromate.
180 Medium
180 of 254
Remembering +4 | -1
Which is used for stopping bleeding?
View answer and explanation
Correct option: A
Ferric chloride
Option A (Ferric chloride) is correct because ferric chloride (FeCl3\mathrm{FeC}{\mathrm{l}_3}) is used as a styptic agent to stop bleeding due to its ability to coagulate blood.
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