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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 121-150 of 410 questions

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121 Easy
121 of 410
Remembering +4 | -1
Monel metal is an alloy of?
View answer and explanation
Correct option: A
Cu, Ni, Fe, Mn
Option A (Cu, Ni, Fe, Mn) is correct because Monel metal is primarily an alloy of copper and nickel, with smaller amounts of iron and manganese.
122 Medium
122 of 410
Analyzing +4 | -1
Wrong match of the following is
View answer and explanation
Correct option: C
K2Cr2O7{K}_2{Cr}_2{O}_7 can disproportionate in acidic medium
Option C (K2Cr2O7{K}_2{Cr}_2{O}_7 can disproportionate in acidic medium) is the wrong match because K2Cr2O7{K}_2{Cr}_2{O}_7 cannot disproportionate as chromium is already in its highest oxidation state (+6).
123 Easy
123 of 410
Analyzing +4 | -1
Assertion: d5d^5 configuration is more stable than d4d^4

Reason: d5d^5 has more exchange energy as compared to d4d^4 because 10 & 6 exchanges are possible in d5d^5 & d4d^4 respectively.
View answer and explanation
Correct option: A
If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
Option A (If both Assertion and Reason are correct and Reason is the correct explanation of Assertion) is correct because d5 configuration has higher exchange energy due to maximum possible parallel spins.
124 Medium
124 of 410
Understanding +4 | -1
Which is obtained when SO2\mathrm{S}{\mathrm{O}_2} is bubbled through a solution of CuCl2\mathrm{CuC}{\mathrm{l}_2}?
View answer and explanation
Correct option: B
Cu2Cl2\mathrm{C}{\mathrm{u}_2}\mathrm{C}{\mathrm{l}_2}
Option B (Cu2Cl2\mathrm{C}{\mathrm{u}_2}\mathrm{C}{\mathrm{l}_2}) is correct because sulfur dioxide reduces cupric chloride to cuprous chloride in the presence of water.
125 Easy
125 of 410
Remembering +4 | -1
Chemical name of vermilion is:
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Correct option: A
Mercuric sulphide
Option A (Mercuric sulphide) is correct because the chemical name of vermilion is mercuric sulfide (HgS).
126 Easy
126 of 410
Remembering +4 | -1
Duraluminium is an alloy contains:
View answer and explanation
Correct option: B
Mg + Cu + Al + Mn + Si
Option B (Mg + Cu + Al + Mn + Si) is correct because duralumin is an alloy of aluminum with copper, magnesium, manganese, and silicon.
127 Easy
127 of 410
Remembering +4 | -1
Thermite is a mixture of iron oxide and:
View answer and explanation
Correct option: D
Al powder
Option D (Al powder) is correct because thermite is a mixture of iron oxide and aluminum powder, used in welding.
128 Hard
128 of 410
Application +4 | -1
A compound in which a metal ion Mx+{\mathrm{M}^{x + }}(Z=25) has a spin only magnetic moment of 24\sqrt {24}BM. The number of unpaired electrons in the compound and the oxidation state of the metal ion are respectively.
View answer and explanation
Correct option: D
4 and 3
Option D (4 and 3) is correct because the spin-only magnetic moment calculation yields 4 unpaired electrons, and the electronic configuration confirms a +3 oxidation state.
129 Medium
129 of 410
Understanding +4 | -1
Most basic lanthanoids hydroxide among the following is
View answer and explanation
Correct option: A
Eu(OH)3\mathrm{Eu}(\mathrm{OH})_3
Option A (Eu(OH)3\mathrm{Eu}(\mathrm{OH})_3) is the most basic lanthanoid hydroxide because basicity decreases with increasing atomic number across the lanthanoid series.
130 Medium
130 of 410
Remembering +4 | -1
Yellow mercury (II) oxide is obtained when
View answer and explanation
Correct option: B
HgCl2\mathrm{HgC}{\mathrm{l}_2} is treated with NaOH solution
Option B (HgCl2\mathrm{HgC}{\mathrm{l}_2} is treated with NaOH solution ) is correct because the reaction of mercury(II) chloride with sodium hydroxide produces yellow mercury(II) oxide.
131 Medium
131 of 410
Remembering +4 | -1
A developer used in photography is:
View answer and explanation
Correct option: C
A mild reducing agent
Option C (A mild reducing agent) is correct because photographic developers reduce silver halide to metallic silver.
132 Easy
132 of 410
Application +4 | -1
The octahedral diamagnetic low spin complex among the following is
View answer and explanation
Correct option: A
[Co(NH3)6]3+\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3 +}
Option A (Co(NH3)63+\mathrm{Co}(\mathrm{NH}_3)_6^{3 +}) is correct because it forms a low-spin, diamagnetic octahedral complex due to the strong field ligand NH3.
133 Medium
133 of 410
Remembering +4 | -1
The stainless steel developed in India contains the following special components:
View answer and explanation
Correct option: C
Manganese and chromium
Option C (Manganese and chromium) is correct because stainless steel typically contains chromium for corrosion resistance and manganese for increased strength and toughness.
134 Medium
134 of 410
Understanding +4 | -1
In context of the lanthanoids, which of the following statements is not correct?
View answer and explanation
Correct option: D
A vailability of 4ff-electrons results in the formation of compounds in +4 state for all members of the series.
Option D (A vailability of 4ff-electrons results in the formation of compounds in +4 state for all members of the series) is not correct because not all lanthanoids exhibit a stable +4 oxidation state.
135 Medium
135 of 410
Understanding +4 | -1
which of the following does not exist
View answer and explanation
Correct option: A
[CCl6]2{\left[ {CC{l_6}} \right]^2}
Option A ([CCl6]2−{\left[ {CC{l_6}} \right]^2}^ -) is correct because carbon cannot expand its octet due to the absence of available d-orbitals.
136 Medium
136 of 410
Understanding +4 | -1
Increasing order of oxidising power is:
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Correct option: D
All of these
The correct answer is D (All of these) because all the given options represent correct increasing orders of oxidizing power for the specified species.
137 Medium
137 of 410
Remembering +4 | -1
Anhydrous ferric chloride is prepared by
View answer and explanation
Correct option: D
Passing dry Cl2\mathrm{C}{\mathrm{l}_2} gas over heated iron scrap.
Option D (Passing dry Cl2\mathrm{C}{\mathrm{l}_2} gas over heated iron scrap) is correct because it directly reacts iron with chlorine gas to form anhydrous ferric chloride.
138 Medium
138 of 410
Analyzing +4 | -1
Which of the following arrangements does not represent the correct order of the property stated against it?
View answer and explanation
Correct option: A
V2+<Cr2+<Mn2+<Fe2+:{V^{2 + }} < C{r^{2 + }} < M{n^{2 + }} < F{e^{2 + }}: paramagnetic behaviour
Option A is correct. The paramagnetic behavior order is incorrect because Fe²⁺ has the same number of unpaired electrons as Cr²⁺, making their paramagnetic behavior similar.
139 Medium
139 of 410
Understanding +4 | -1
Which of the following ions will exhibit colour in aqueous solutions?
View answer and explanation
Correct option: C
Ti3+(Z=22){Ti}^{3+}({Z}=22)
Option C (Ti3+(Z=22){Ti}^{3+}({Z}=22)) is correct because it has one unpaired d-electron, allowing for d-d transitions and thus exhibiting color.
140 Medium
140 of 410
Understanding +4 | -1
The number of ions formed on dissolving one molecule of FeSO4(NH4)2SO4.6H2O\mathrm{FeS}{\mathrm{O}_4}{\left( {\mathrm{N}{\mathrm{H}_4}} \right)_2}\mathrm{S}{\mathrm{O}_4}.6{\mathrm{H}_2}\mathrm{O} is:
View answer and explanation
Correct option: B
5
Option D (H3PO3) is correct because PCl3 reacts with water to form phosphorous acid (H3PO3) and HCl.
141 Medium
141 of 410
Remembering +4 | -1
One of the important uses of ferrous sulphate is in the:
View answer and explanation
Correct option: A
Manufacture of blue-black ink
Option A (Manufacture of blue-black ink) is correct because ferrous sulfate is used in the production of iron gall ink, which is a type of blue-black ink.
142 Easy
142 of 410
Understanding +4 | -1
Which one of the following is an example of non-typical transition elements?
View answer and explanation
Correct option: B
Zn,Cd,Hg{Zn}, {Cd}, {Hg}
Option B (Zn,Cd,Hg{Zn}, {Cd}, {Hg}) is correct because these elements have a completely filled d-orbital in their common oxidation states, and thus do not exhibit typical transition metal properties.
143 Medium
143 of 410
Understanding +4 | -1
The prussian blue colour obtained in the Lassaigne's test for nitrogen is due to the formation of
View answer and explanation
Correct option: B
Iron (III) hexacyanoferrate (II)
Option B (Iron (III) hexacyanoferrate (II)) is correct because the Prussian blue color in Lassaigne's test is due to the formation of ferric ferrocyanide.
144 Medium
144 of 410
Understanding +4 | -1
Statement I: In square planar complex dx2−y2\text{d}_{\text{x}^2-\text{y}^2} is higher in energy than dz2\mathrm{d}_{\mathrm{z}^2}

Statement II: Ligand approach along x\mathrm{x} and y\mathrm{y} axis in square planar complex.
View answer and explanation
Correct option: A
Statement I and Statement II both are correct
Option A is correct. Statement I and Statement II are both correct because in a square planar complex, the dx2−y2_{x^2-y^2} orbital experiences more repulsion from the ligands approaching along the x and y axes, thus having higher energy than dz2_{z^2}.
145 Medium
145 of 410
Remembering +4 | -1
Thermal decomposition of zinc nitrate gives:
View answer and explanation
Correct option: B
ZnO
Option B (ZnO) is correct because thermal decomposition of zinc nitrate yields zinc oxide, nitrogen dioxide, and oxygen.
146 Easy
146 of 410
Remembering +4 | -1
Ferric sulphate on heating gives:
View answer and explanation
Correct option: C
SO3\mathrm{S}{\mathrm{O}_3} only
Option C (SO3\mathrm{S}{\mathrm{O}_3} only) is correct because ferric sulfate decomposes to ferric oxide and sulfur trioxide upon heating.
147 Medium
147 of 410
Analyzing +4 | -1
Identify correct statements from below:
A. The chromate ion is square planar.
B. Dichromates are generally prepared from
chromates.
C. The green manganate ion is diamagnetic
D. Dark green coloured K2MnO4\mathrm{K}_2 \mathrm{MnO}_4 disproportionates
in a neutral or acidic medium to give
permanganate.
E. With increasing oxidation number of transition
metal, ionic character of the oxides decreases.
Choose the correct answer from the options given
below:
View answer and explanation
Correct option: A
B, D, E only
Option A is correct because statements B, D, and E are accurate descriptions of chromate/dichromate chemistry, manganate/permanganate chemistry, and the trend in ionic character of oxides with increasing oxidation state.
148 Medium
148 of 410
Analyzing +4 | -1
Which of the following statements is not true?
View answer and explanation
Correct option: C
The oxidation state of Cr\mathrm{Cr} is highest in CrCl4\mathrm{CrCl}_4
Option C is correct because the highest oxidation state of Cr is in CrF6CrF_6, not CrCl4CrCl_4.
149 Medium
149 of 410
Analyzing +4 | -1
Assertion (A): Ti3+\mathrm{Ti}^{3+} acts as an oxidizing agent.

Reason (R): Due to inert pair effect, Ti+\mathrm{Ti}^{+} is more stable than Ti3+.\mathrm{Ti}^{3+}.
View answer and explanation
Correct option: A
Both (A) & (R) are true and the (R) is the correct explanation of the (A)
Option D (Both (A) and (R) are false) is the correct answer because Ti3+Ti^{3+} acts as a reducing agent and the inert pair effect is not relevant for Titanium.
150 Medium
150 of 410
Remembering +4 | -1
‘Hydride gap’ is referred to which region of the Periodic Table?
View answer and explanation
Correct option: D
Groups 7, 8 and 9
Option D (Groups 7, 8 and 9) is correct because elements in these groups generally do not form stable hydrides, leading to the 'hydride gap'.
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