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JEE • CHEMISTRY • d & f- BLOCK ELEMENTS

D & f- BLOCK ELEMENTS practice · Medium

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91 Medium
91 of 254
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}.
92 Medium
92 of 254
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.
93 Medium
93 of 254
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.
94 Medium
94 of 254
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.
95 Medium
95 of 254
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.
96 Medium
96 of 254
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'.
97 Medium
97 of 254
Understanding +4 | -1
In which one of the following four pairs of oxides are both oxides amphoteric?
View answer and explanation
Correct option: A
Cr2O3{Cr}_2 {O}_3 and V2O5{V}_2 {O}_5
Option A (Cr2O3{Cr}_2 {O}_3 and V2O5{V}_2 {O}_5) is correct because both chromium(III) oxide and vanadium(V) oxide can act as both acids and bases, depending on the reaction conditions.
98 Medium
98 of 254
Remembering +4 | -1
Which one of the following pairs of elements is called ‘chemical twins’ because of their very similar chemical properties?
View answer and explanation
Correct option: D
Hf and Zr
Option D (Hf and Zr) is correct because Hafnium (Hf) and Zirconium (Zr) have very similar chemical properties due to their similar ionic radii caused by lanthanide contraction, making them 'chemical twins'.
99 Medium
99 of 254
Understanding +4 | -1
Which of the following are correct order of the enthalpy of hydration of Ca+2,Mn+2\mathrm{Ca}^{+2}, \mathrm{Mn}^{+2} and Zn+2\mathrm{Zn}^+2
View answer and explanation
Correct option: D
Zn+2>Mn+2>Ca+2\mathrm{Zn}^{+2}>\mathrm{Mn}^{+2}>\mathrm{Ca}^{+2}
Option D is correct because the enthalpy of hydration is related to charge density, and Zn+2 has a higher charge density than Mn+2 and Ca+2, leading to greater hydration enthalpy.
100 Medium
100 of 254
Remembering +4 | -1
Pure conc. HNO3\mathrm{HN}{\mathrm{O}_3} makes iron passive as the surface is covered with protective layer of:
View answer and explanation
Correct option: B
Fe3O4\mathrm{F}{\mathrm{e}_3}{\mathrm{O}_4}
Option B (Fe3O4\mathrm{F}{\mathrm{e}_3}{\mathrm{O}_4}) is correct because concentrated nitric acid renders iron passive by forming a thin, protective layer of iron(II,III) oxide (Fe3O4\mathrm{F}{\mathrm{e}_3}{\mathrm{O}_4}) on the surface.
101 Medium
101 of 254
Application +4 | -1
The spin only magnetic moment of Fe2+\mathrm{F}{\mathrm{e}^{2 + }} ion (in BM) is approximately.
View answer and explanation
Correct option: B
7
Option C (5) is incorrect, the correct answer is closer to 4.9, which is approximately 5, but the closest answer is A (4) or D (6). Option B (7) is not correct because the formula is applied incorrectly. The correct answer is closest to A (4).
102 Medium
102 of 254
Analyzing +4 | -1
Consider the reactions
(A)    H2O+2HI→I2+2H2O\mathrm{(A)}\,\,\,\,{H_2}O + 2HI \to {I_2} + 2{H_2}O
(B)    HOCl+H2O2→H3O++Cl−+O2\mathrm{(B)}\,\,\,\,HOCl + {H_2}{O_2} \to {H_3}{O^ + } + C{l^ - } + {O_2}
Which of the following statement is correct about H2O2{H_2}{O_2} with reference to these reactions? Hydrogen peroxide is ____
View answer and explanation
Correct option: B
An oxidising agent in (A) and reducing agent in (B)
Option C (A reducing agent in (A) and oxidising agent in (B)) is correct because in reaction A, H2O2 oxidizes HI to I2, acting as an oxidizing agent, and in reaction B, H2O2 reduces HOCl to Cl-, acting as a reducing agent.
103 Medium
103 of 254
Analyzing +4 | -1
The amphoteric oxide among V2O3,\mathrm{V}_2 \mathrm{O}_3, V2O4\mathrm{V}_2 \mathrm{O}_4 and V2O5\mathrm{V}_2 \mathrm{O}_5 upon reaction with alkali leads to formation of an oxide anion. The oxidation state of V in the oxide anion is:
View answer and explanation
Correct option: C
5
Option C (5) is correct because the amphoteric oxide V2O5\mathrm{V}_2\mathrm{O}_5 reacts with alkali to form VO43−\mathrm{VO}_4^{3-}, where the oxidation state of V is +5.
104 Medium
104 of 254
Application +4 | -1
Magnetic moment 2.84 BM is given by
View answer and explanation
Correct option: C
Ni2+N{i^{2 + }}
Option C (Ni2+N{i^{2 + }}) is correct because it has two unpaired electrons, resulting in a magnetic moment of approximately 2.84 BM.
105 Medium
105 of 254
Analyzing +4 | -1
In the reaction Cr2O72−+SnCl2+H+→SnCl4+CrCl3+H2O{Cr}_2{O}_7{ }^{2-}+{SnCl}_2+{H}^{+} \rightarrow {SnCl}_4+{CrCl}_3+{H}_2{O}
View answer and explanation
Correct option: C
Cr2O72−{Cr}_2 {O}_7{ }^{2-} is reduced and SnCl2{SnCl}_2 is oxidised
Option C (Cr2O72−{Cr}_2 {O}_7{ }^{2-} is reduced and SnCl2{SnCl}_2 is oxidised) is correct because chromium is reduced (oxidation state decreases) and tin is oxidized (oxidation state increases).
106 Medium
106 of 254
Remembering +4 | -1
Solder is an alloy of
View answer and explanation
Correct option: B
30%30 \% lead, 70%70 \% tin
Option B (30%30 \% lead, 70%70 \% tin) is correct because solder is an alloy primarily composed of tin and lead, with tin typically being the larger component.
107 Medium
107 of 254
Understanding +4 | -1
Assertion: Highest oxidation states of dd-block metals are usually seen in oxides and oxoanions.

Reason: Oxygen is capable of forming pr−drpr-dr bonds with transition metals.
View answer and explanation
Correct option: A
Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.
Option A (Both assertion and reason are correct statements, and reason is the correct explanation of the assertion) is correct because oxygen's ability to form pπ−dπp\pi-d\pi bonds stabilizes high oxidation states of d-block metals in oxides and oxoanions.
108 Medium
108 of 254
Remembering +4 | -1
KMnO4→ΔX+Y+Z\mathrm{KMnO}_4 \xrightarrow{\Delta} \mathrm{X}+\mathrm{Y}+\mathrm{Z}

In this chemical reaction X, Y and Z are:
View answer and explanation
Correct option: D
K2MnO4,MnO2,O2\mathrm{K}_2 \mathrm{MnO}_4, \mathrm{MnO}_2, \mathrm{O}_2
Option D (K2MnO4,MnO2,O2\mathrm{K}_2 \mathrm{MnO}_4, \mathrm{MnO}_2, \mathrm{O}_2) is correct because it represents the products formed during the thermal decomposition of KMnO4\mathrm{KMnO}_4.
109 Medium
109 of 254
Remembering +4 | -1
Blister copper is:
View answer and explanation
Correct option: B
Copper containing 2% impurity
Option B (Copper containing 2% impurity) is correct because blister copper is approximately 98% pure copper with about 2% impurities.
110 Medium
110 of 254
Remembering +4 | -1
Compound(s) useful in the battery industries is/are
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because MnO2\mathrm{MnO}_2, Zn\mathrm{Zn}, and Ni/Cd\mathrm{Ni} / \mathrm{Cd} are all used in different types of batteries.
111 Medium
111 of 254
Analyzing +4 | -1
The correct order of decreasing Second ionisation enthalpy of Ti, V, CrTi,\,V,\,Cr and MnMn is
View answer and explanation
Correct option: C
Cr>Mn>V>TiCr > Mn > V > Ti
Option C is correct because the second ionization enthalpy of Cr is highest due to the stable half-filled d5 configuration after the first ionization, followed by Mn, V, and Ti.
112 Medium
112 of 254
Understanding +4 | -1
From gold aurocyanide Na[Au(CN)2]\mathrm{Na}\left[ {\mathrm{Au}{{\left( {\mathrm{CN}} \right)}_2}} \right], gold can be precipitated by adding powder of:
View answer and explanation
Correct option: A
Zn
Option A (Zn) is correct because Zinc is more reactive than Gold and can displace it from the aurocyanide complex.
113 Medium
113 of 254
Remembering +4 | -1
Approximate percentage of lanthanoids in mischmetall is
View answer and explanation
Correct option: D
95%95 \%
Option D (95%95 \%) is correct because mischmetall contains approximately 95% lanthanoids.
114 Medium
114 of 254
Remembering +4 | -1
The form of iron having the highest carbon content is
View answer and explanation
Correct option: A
Cast iron
Option A (Cast iron) is correct because cast iron has the highest carbon content among the given options, ranging from 2 to 4.5%.
115 Medium
115 of 254
Remembering +4 | -1
Coinage alloy has the composition of:
View answer and explanation
Correct option: A
Ag + Cu + Ni
Option A (Ag + Cu + Ni) is correct because coinage alloys are typically composed of silver, copper, and nickel.
116 Medium
116 of 254
Analyzing +4 | -1
Which of the following metal nitrate can show given change.

Aqueous solution of metal nitrate→NH4OHExcess(No change in colour and no precipitate in test tube) {\text{Aqueous solution of metal nitrate}}\xrightarrow[{{\text{N}}{{\text{H}}_{\text{4}}}{\text{OH}}}]{{{\text{Excess}}}}\left( {{\text{No change in colour and no precipitate in test tube}}} \right){\text{ }}
View answer and explanation
Correct option: C
AgNO3\mathrm{AgNO}_3
Option C (AgNO3\mathrm{AgNO}_3) is correct because silver nitrate forms a soluble complex with excess ammonium hydroxide, resulting in no precipitate or color change.
117 Medium
117 of 254
Remembering +4 | -1
The metal that shows highest and maximum number of oxidation state is:
View answer and explanation
Correct option: B
Mn\mathrm{Mn}
Option B (Mn\mathrm{Mn}) is correct because Manganese exhibits the highest number of oxidation states among the given options, ranging from +2 to +7.
118 Medium
118 of 254
Analyzing +4 | -1
For which of the following pairs, magnetic moment is same?
View answer and explanation
Correct option: B
CuCl2,TiCl3\mathrm{CuCl}_{2}, \mathrm{TiCl}_{3}
Option B (CuCl2,TiCl3\mathrm{CuCl}_{2}, \mathrm{TiCl}_{3}) is correct because both Cu2+Cu^{2+} and Ti3+Ti^{3+} have one unpaired electron, resulting in the same magnetic moment.
119 Medium
119 of 254
Analyzing +4 | -1
Which of the following ions exhibits d-d transition and para magnetism as well?
View answer and explanation
Correct option: C
[Fe(CN)6]3−{\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }}
The correct answer is C ([Fe(CN)6]3−{\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }}) because it exhibits both d-d transition (color) and paramagnetism due to unpaired electrons.
120 Medium
120 of 254
Application +4 | -1
A compound of vanadium chloride has spin only magnetic moment of 1.73 Bm1.73 \mathrm{~Bm}. Its formula is
View answer and explanation
Correct option: C
VCl4\mathrm{VCl}_4
Option C (VCl4\mathrm{VCl}_4) is correct because a magnetic moment of 1.73 Bm corresponds to one unpaired electron, which is consistent with VCl4.
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