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NEET • CHEMISTRY • d & f- BLOCK ELEMENTS

D & f- BLOCK ELEMENTS practice · Mixed difficulty

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61 Easy
61 of 213
Application +4 | -1
The spin only magnetic moment [in units of Bohr magneton] of Ni2+{Ni}^{2+} in aqueous solution would be (At. No. 28) :
View answer and explanation
Correct option: A
2.84
Option A (2.84) is correct because Ni2+{Ni}^{2+} has two unpaired electrons, resulting in a spin-only magnetic moment of 2.84 BM.
62 Easy
62 of 213
Remembering +4 | -1
The compound used in preservation of wood is:
View answer and explanation
Correct option: C
ZnCl2\mathrm{ZnC}{\mathrm{l}_2}
Option C (ZnCl2\mathrm{ZnC}{\mathrm{l}_2}) is correct because zinc chloride is a common wood preservative that protects wood from decay and insect damage.
63 Easy
63 of 213
Remembering +4 | -1
On strongly heating AgNO3\mathrm{AgN}{\mathrm{O}_3} we get:
View answer and explanation
Correct option: C
Ag
Option C (Ag) is correct because upon strong heating, silver nitrate decomposes to form silver metal, nitrogen dioxide, and oxygen.
64 Easy
64 of 213
Analyzing +4 | -1
Statement I: Fe+2\mathrm{Fe}^{+2} is more stable than Fe+3\mathrm{Fe}^{+3} ion

Statement II: Fe+3\mathrm{Fe}^{+3} ion contain half-filled 3d3 \mathrm{d} subshell
View answer and explanation
Correct option: D
Statement I if false but statement II is correct
Option D is correct because Statement I is false, and Statement II is correct. Fe3+{Fe}^{3+} is more stable than Fe2+{Fe}^{2+} due to the half-filled d-orbital configuration.
65 Medium
65 of 213
Application +4 | -1
A solution of 10 mLM10FeSO410 \mathrm{~mL} \frac{{M}}{10} {FeSO}_4 is titrated with KMnO4{KMnO}_4 solution in acidic medium the amount of KMnO4{KMnO}_4 used will be
View answer and explanation
Correct option: D
10 mL10 \mathrm{~mL} of 0.02M0.02 \mathrm{M}
The correct answer is 10 mL10 \mathrm{~mL} of 0.02M0.02 \mathrm{M} because the stoichiometry of the reaction between FeSO4FeSO_4 and KMnO4KMnO_4 requires careful calculation of the molar ratios and volumes.
66 Easy
66 of 213
Understanding +4 | -1
Which of the following metal ions is not coloured?
View answer and explanation
Correct option: D
Zn2+\mathrm{Z}{\mathrm{n}^{2 + }}
Option D (Zn2+\mathrm{Z}{\mathrm{n}^{2 + }}) is correct because zinc(II) ion has a completely filled d-orbital configuration (d10d^{10}), preventing d-d transitions and thus resulting in no color.
67 Hard
67 of 213
Understanding +4 | -1
Which of the following is an acidic oxide?
View answer and explanation
Correct option: C
Mn2O7\mathrm{M}{\mathrm{n}_2}{\mathrm{O}_7}
Option C (Mn2O7\mathrm{M}{\mathrm{n}_2}{\mathrm{O}_7}) is correct because higher oxidation states of manganese tend to form acidic oxides, and Mn2O7Mn_2O_7 has Mn in the +7 oxidation state.
68 Medium
68 of 213
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 derived from balancing the oxidation and reduction half-reactions, showing that 1 mole of Fe(NO2)2Fe(NO_2)_2 requires 5/6 moles of Cr2O72−Cr_2O_7^{2-} for complete oxidation.
69 Easy
69 of 213
Application +4 | -1
Magnetic moment in B.M. of Ni2+\mathrm{Ni}^{2+} ion is
View answer and explanation
Correct option: B
8\sqrt{8}
The magnetic moment of Ni2+\mathrm{Ni}^{2+} ion is 8\sqrt{8} B.M. because Ni2+\mathrm{Ni}^{2+} has two unpaired electrons, and the spin-only formula is used to calculate the magnetic moment.
70 Medium
70 of 213
Remembering +4 | -1
The extraction of which of the following metals involves bessemerization?
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Correct option: D
Cu
Option D (Cu) is correct because bessemerization is a process used in the extraction of copper to remove iron impurities.
71 Hard
71 of 213
Understanding +4 | -1
Consider the following reaction:

NaCl+K2Cr2O7+H2SO4→A+K2SO4+Na2SO4+H2O\mathrm{NaCl}+\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathbf{A}+\mathrm{K}_2 \mathrm{SO}_4+\mathrm{Na}_2 \mathrm{SO}_4+\mathrm{H}_2 \mathrm{O} The formula and colour of the product ' AA ' respectively will be
View answer and explanation
Correct option: D
CrO2Cl2\mathrm{CrO}_2 \mathrm{Cl}_2, red
The correct answer is CrO2Cl2\mathrm{CrO}_2 \mathrm{Cl}_2, red because chromyl chloride (CrO2Cl2\mathrm{CrO}_2 \mathrm{Cl}_2) is formed in the reaction, which has a reddish-brown color.
72 Medium
72 of 213
Remembering +4 | -1
Each transition series contains:
View answer and explanation
Correct option: B
10 elements
Option B (10 elements) is correct because each transition series corresponds to the filling of a d-orbital set, which can hold 10 electrons, hence 10 elements.
73 Hard
73 of 213
Application +4 | -1
Magnetic moment of manganese in (NH4)2MnBr2{\left( {\mathrm{N}{\mathrm{H}_4}} \right)_2}\mathrm{MnB}{\mathrm{r}_2} is
View answer and explanation
Correct option: B
5.91 BM
Option B (5.91 BM) is correct because in (NH4)2MnBr2, Mn is in the +2 oxidation state with a d5 configuration, leading to 5 unpaired electrons and a magnetic moment of approximately 5.92 BM.
74 Medium
74 of 213
Application +4 | -1
Magnetic moment of Cr2+C{r^{2 + }} ion is
View answer and explanation
Correct option: A
4.9 BM
Option A (4.9 BM) is correct because Cr2+Cr^{2+} has 4 unpaired electrons, and the magnetic moment is calculated using the spin-only formula.
75 Easy
75 of 213
Analyzing +4 | -1
Which of the following statements are INCORRECT?
A. All the transition metals except scandium form MO\text{MO} oxides which are ionic.
B. The highest oxidation number corresponding to the group number in transition metal oxides is attained in Sc2O3\mathrm{Sc}_2 \mathrm{O}_3 to Mn2O7\mathrm{Mn}_2 \mathrm{O}_7.
C. Basic character increases from V2O3\mathrm{V}_2 \mathrm{O}_3 to V2O4\mathrm{V}_2 \mathrm{O}_4 to V2O5\mathrm{V}_2 \mathrm{O}_5.
D. V2O4\mathrm{V}_2 \mathrm{O}_4 dissolves in acids to give VO43−\mathrm{VO}_4^{3-} salts.
E. CrO\mathrm{CrO} is basic but Cr2O3\mathrm{Cr}_2 \mathrm{O}_3 is amphoteric.
Choose the correct answer from the options given below:
View answer and explanation
Correct option: A
C and D only
Option A (C and D only) is correct because statements C and D are incorrect regarding the properties of vanadium oxides.
76 Easy
76 of 213
Remembering +4 | -1
Essential constituent of an amalgam is:
View answer and explanation
Correct option: D
Mercury
Option D (Mercury) is correct because an amalgam is defined as an alloy of mercury with another metal.
77 Medium
77 of 213
Analyzing +4 | -1
A laboratory reagent imparts green colour to the flame. On heating with solid K2Cr2O7{K}_2 {Cr}_2 {O}_7 and conc. H2SO4{H}_2 {SO}_4 it evolves a red gas. Identify the reagent
View answer and explanation
Correct option: B
BaCl2{BaCl}_2
Option C (CuCl2{CuCl}_2) is correct because copper salts impart a green color to the flame, and upon heating with K2Cr2O7{K}_2 {Cr}_2 {O}_7 and conc. H2SO4{H}_2 {SO}_4, CuCl2{CuCl}_2 evolves a red gas (chromyl chloride).
78 Hard
78 of 213
Understanding +4 | -1
Which one of the following reactions will occur on heating AgNO3\mathrm{AgN}{\mathrm{O}_3} above its melting point?
View answer and explanation
Correct option: C
2AgNO3→2AgNO2+O22\mathrm{AgN}{\mathrm{O}_3} \to 2\mathrm{AgN}{\mathrm{O}_2} + {\mathrm{O}_2}
Option C (2AgNO3→2AgNO2+O22\mathrm{AgN}{\mathrm{O}_3} \to 2\mathrm{AgN}{\mathrm{O}_2} + {\mathrm{O}_2}) is correct because silver nitrate decomposes to silver nitrite and oxygen upon heating.
79 Hard
79 of 213
Analyzing +4 | -1
Assertion : UF6{\text{U}}{{\text{F}}_6} is more covalent than UF4{\text{U}}{{\text{F}}_4}.

Reason : Fluorine is smaller in size.
View answer and explanation
Correct option: B
If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion.
Option A (If both Assertion & Reason are True & the Reason is correct explanation of the Assertion) is correct because smaller fluorine size leads to higher polarization and increased covalent character, making UF6{\text{U}}{{\text{F}}_6} more covalent than UF4{\text{U}}{{\text{F}}_4}.
80 Medium
80 of 213
Application +4 | -1
A magnetic moment of 1.73 BM will be shown by one among the following compounds:
View answer and explanation
Correct option: A
[Cu(NH3)4]2+{\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2 + }}
The correct answer is [Cu(NH3)4]2+{\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2 + }} because Cu2+Cu^{2+} has a d9d^9 configuration with one unpaired electron, resulting in a magnetic moment close to 1.73 BM.
81 Medium
81 of 213
Remembering +4 | -1
When hydrogen peroxide is added to acidified potassium dichromate, a blue colour is produced due to formation of
View answer and explanation
Correct option: C
CrO5\mathrm{Cr}{\mathrm{O}_5}
Option C (CrO5\mathrm{Cr}{\mathrm{O}_5}) is correct because the reaction of hydrogen peroxide with acidified potassium dichromate forms chromium pentoxide, which has a blue color.
82 Medium
82 of 213
Remembering +4 | -1
Ruthenium carbonyl is:
View answer and explanation
Correct option: B
Ru(CO)5\mathrm{Ru}{\left( {\mathrm{CO}} \right)_5}
Option B (Ru(CO)5\mathrm{Ru}{\left( {\mathrm{CO}} \right)_5}) is correct because Ruthenium pentacarbonyl has the formula Ru(CO)5\mathrm{Ru}{\left( {\mathrm{CO}} \right)_5}.
83 Easy
83 of 213
Understanding +4 | -1
Zr(Z=40)andHf(Z=72)Zr\mathrm{ }(Z = 40)\mathrm{ and }Hf(Z = 72) have similar atomic and ionic radii because of:
View answer and explanation
Correct option: A
Lanthanoid contraction
Option A (Lanthanoid contraction) is correct because Lanthanoid contraction causes ZrZr and HfHf to have similar atomic and ionic radii.
84 Medium
84 of 213
Remembering +4 | -1
Lanthanides are
View answer and explanation
Correct option: C
14 elements in the sixth period (atomic no. = 58 to 71) that are filling 4f4f sub-level.
Option C (14 elements in the sixth period (atomic no. = 58 to 71) that are filling 4f4f sub-level) is correct because lanthanides are defined as the 14 elements following lanthanum in the sixth period, filling the 4f sub-level.
85 Hard
85 of 213
Understanding +4 | -1
Copper exhibits only +2 oxidation state in its stable compounds. Why?
View answer and explanation
Correct option: B
+2+ 2 state compounds of copper are formed by exothermic reactions.
Option B is correct because the stability of Cu in the +2 oxidation state is attributed to the exothermic nature of the reactions forming its compounds.
86 Hard
86 of 213
Understanding +4 | -1
Transition metal complex with highest value of crystal field splitting (Δ0)({\Delta _0}) will be
View answer and explanation
Correct option: D
[Os(H2O)6]3+{\left[ {Os{{({H_2}O)}_6}} \right]^{3 + }}
Option B (Ag+\mathrm{A}{\mathrm{g}^ + }) is correct because Ag+Ag^+ forms a linear complex [Ag(CN)2]−[Ag(CN)_2]^- with a coordination number of 2.
87 Easy
87 of 213
Analyzing +4 | -1
An aqueous solution of FeSO4,Al2(SO4)3\mathrm{FeS}{\mathrm{O}_4},\mathrm{A}{\mathrm{l}_2}{\left( {\mathrm{S}{\mathrm{O}_4}} \right)_3} and chrome alum is heated with excess of Na2O2\mathrm{N}{\mathrm{a}_2}{\mathrm{O}_2} and filtered. The materials obtained are:
View answer and explanation
Correct option: C
A yellow filtrate and a brown residue
Option C (A yellow filtrate and a brown residue) is correct because FeSO4\mathrm{FeS}{\mathrm{O}_4} forms iron(III) oxide (brown residue) and chrome alum forms sodium chromate (yellow filtrate) in the presence of excess Na2O2\mathrm{N}{\mathrm{a}_2}{\mathrm{O}_2}.
88 Medium
88 of 213
Understanding +4 | -1
The trivalent ion having largest size in lanthanide series is
View answer and explanation
Correct option: D
La
Option D (La) is correct because La (Lanthanum) is the first element in the lanthanide series and has the largest ionic size among the trivalent lanthanide ions.
89 Hard
89 of 213
Understanding +4 | -1
Impurities of Cu and Ag from gold are removed by
View answer and explanation
Correct option: D
Both (2) and (3)
Impurities of Cu and Ag from gold are removed by boiling impure gold with concentrated H2SO4{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} and electrolytically, as these methods effectively separate the metals based on their chemical properties.
90 Medium
90 of 213
Remembering +4 | -1
Which form of iron is least ductile?
View answer and explanation
Correct option: B
Cast iron
Option B (Cast iron) is correct because cast iron has a high carbon content, making it brittle and less ductile compared to other forms of iron.
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