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NEET • CHEMISTRY • d & f- BLOCK ELEMENTS

D & f- BLOCK ELEMENTS practice · Hard

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31 Hard
31 of 46
Analyzing +4 | -1
Metallic bonds do not play a role in:
View answer and explanation
Correct option: A
Brass
Option C (Germanium) is correct because Germanium is a metalloid and forms covalent bonds, not metallic bonds.
32 Hard
32 of 46
Analyzing +4 | -1
A solution when diluted with H2O{\mathrm{H}_2}\mathrm{O} and boiled gives a white ppt. On addition of excess NH4Cl/NH4OH\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}/\mathrm{N}{\mathrm{H}_4}\mathrm{OH}, the volume of precipitate decreases due to dissolution leaving behind a white gelatinous precipitate. The precipitate which dissolves in NH4OH/NH4Cl\mathrm{N}{\mathrm{H}_4}\mathrm{OH}/\mathrm{N}{\mathrm{H}_4}\mathrm{Cl} is:
View answer and explanation
Correct option: A
Zn(OH)2\mathrm{Zn}{\left( {\mathrm{OH}} \right)_2}
Option A (Zn(OH)2\mathrm{Zn}{\left( {\mathrm{OH}} \right)_2}) is correct because zinc hydroxide is amphoteric and dissolves in excess ammonium hydroxide due to the formation of a complex, while aluminum hydroxide forms a gelatinous precipitate.
33 Hard
33 of 46
Remembering +4 | -1
Which of the following compounds is used as the starting material for the preparation of potassium dichromate?
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Correct option: C
FeCr2O4\mathrm{FeC}{\mathrm{r}_2}{\mathrm{O}_4} (Chromite)
Option C (FeCr2O4\mathrm{FeC}{\mathrm{r}_2}{\mathrm{O}_4} (Chromite)) is correct because chromite ore is the primary source from which potassium dichromate is produced.
34 Hard
34 of 46
Understanding +4 | -1
A blue colouration is not obtained when:
View answer and explanation
Correct option: B
Copper sulphate solution reacts with K4[Fe(CN)6{\mathrm{K}_4}[\mathrm{Fe}{\left( {\mathrm{CN}} \right)_6}]
A blue coloration is not obtained when copper sulphate solution reacts with K4[Fe(CN)6]{\mathrm{K}_4}[\mathrm{Fe}{\left( {\mathrm{CN}} \right)_6}] because it forms a chocolate brown precipitate of copper ferrocyanide.
35 Hard
35 of 46
Analyzing +4 | -1
Which of the following reagents does not convert copper sulphate to cuprous state?
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Correct option: D
K4[Fe(CN)6]\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]
Option D (K4[Fe(CN)6]\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]) is correct because it does not have the reducing power to convert Cu2+ to Cu+.
36 Hard
36 of 46
Understanding +4 | -1
In the equation,

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

Identify the metal (M)
View answer and explanation
Correct option: A
Au\mathrm{Au}
Option A (Au\mathrm{Au}) is correct because gold is extracted by leaching with cyanide, forming a complex ion.
37 Hard
37 of 46
Application +4 | -1
In order to oxidise a mixture of one mole of each of FeC2O4,  Fe2(C2O4)3,  FeSO4Fe{C_2}{O_4},\,\,F{e_2}{({C_2}{O_4})_3},\,\,FeS{O_4} and Fe2(SO4)3F{e_2}{(S{O_4})_3} in acidic medium, the number of moles of KMnO4KMn{O_4} required is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the total number of moles of KMnO4KMnO_4 required to oxidize the mixture is calculated based on the stoichiometry of the redox reactions involved.
38 Hard
38 of 46
Understanding +4 | -1
The actinoids exhibit more number of oxidation states in general than the lanthanoids. This is because
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Correct option: D
The 5f5f-orbitals extend further from the nucleus than the 4f4f -orbitals.
Option D (The 5f5f-orbitals extend further from the nucleus than the 4f4f -orbitals) is correct because the greater extension of 5f orbitals in actinoids allows for more effective shielding and participation in bonding, leading to variable oxidation states.
39 Hard
39 of 46
Understanding +4 | -1
Which of the following statements regarding copper salts is not true?
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Correct option: C
Copper(I) oxide is red powder
Option A (Copper(I) Disproportionates into Cu and Cu(II) in aqueous solution) is not true because Copper(I) disproportionates into Cu and Cu(II) in aqueous solution is a true statement. Copper(I) oxide is actually a red powder.
40 Hard
40 of 46
Analyzing +4 | -1
The colour of 62Sm3+_{62}\mathrm{S}{\mathrm{m}^{3 + }} is yellow. The expected colour of   66Dy3+{\mathrm{\;}_{66}}\mathrm{D}{\mathrm{y}^{3 + }} is
View answer and explanation
Correct option: A
Yellow
Option A (Yellow) is correct because Dy3+ and Sm3+ are complementary and exhibit similar colors due to similar electronic transitions.
41 Hard
41 of 46
Application +4 | -1
KMnO4\mathrm{KMnO}_4 reacts with KI\mathrm{KI}, in weakly basic medium to form I2\mathrm{I}_2 and MnO2\mathrm{MnO}_2. When 250 ml250 \mathrm{~ml} of 0.1M KI0.1 \mathrm{M} \mathrm{~KI} solution reacts with 250 mL250 \mathrm{~mL} of 0.02 M KMnO40.02 \mathrm{~M} \mathrm{~KMnO}_4 in basic medium, what is the number of moles of I2I_2 formed?
View answer and explanation
Correct option: B
0.00750.0075
Option B (0.00750.0075) is correct because it represents the number of moles of I2I_2 formed based on the stoichiometry of the reaction between KMnO4KMnO_4 and KIKI in a weakly basic medium.
42 Hard
42 of 46
Understanding +4 | -1
In chromyl chloride test for confirmation of Cl−{Cl}^{-}ion, a yellow solution is obtained. Acidification of the solution and addition of amyl alcohol and 10%H2O210 \% {H}_2 {O}_2 turns organic layer blue indicating formation of chromium pentoxide. The oxidation state of chromium in that is
View answer and explanation
Correct option: A
6
The oxidation state of chromium in chromium pentoxide is 6 because chromium pentoxide has a butterfly structure with two peroxide linkages.
43 Hard
43 of 46
Analyzing +4 | -1
I(aq)Θ+MnO4−(aq)→H+x+Mn(aq)+2I(aq)Θ+MnO4−(aq)→ weakly OHΘ neutral ( or )y+MnO2MnO4(aq)−+Mn(aq)+2→ZnSO4z+4H+\begin{aligned} & \mathrm{I}_{(\mathrm{a q})}^{\Theta}+\mathrm{MnO}_4^{-}(\mathrm{a q}) \xrightarrow{\mathrm{H}^{+}} \mathrm{x}+\mathrm{Mn}_{(\mathrm{a q})}^{+2} \\ & \mathrm{I}_{(\mathrm{a q})}^{\Theta}+\mathrm{MnO}_4^{-}(\mathrm{a q}) \xrightarrow[\text { weakly } \mathrm{OH}^{\Theta}]{\text { neutral }(\text { or })} \mathrm{y}+\mathrm{MnO}_2 \\ & \mathrm{MnO}_{4(\mathrm{a q})}^{-}+\mathrm{Mn}_{(\mathrm{a q})}^{+2} \xrightarrow{\mathrm{ZnSO}_4} \mathrm{z}+4 \mathrm{H}^{+} \end{aligned}
products x,y\mathrm{x}, \mathrm{y} and z\mathrm{z} are respectively
View answer and explanation
Correct option: A
I2,IO3−,MnO2\mathrm{I}_2, \mathrm{IO}_3^{-}, \mathrm{MnO}_2
Option A (I2,IO3−,MnO2\mathrm{I}_2, \mathrm{IO}_3^{-}, \mathrm{MnO}_2) is the correct answer because iodide is oxidized to iodine in acidic medium, to iodate in neutral or weakly basic medium, and MnO4−MnO_4^- is reduced to Mn2+Mn^{2+} and MnO2MnO_2 respectively. MnO4−MnO_4^- and Mn2+Mn^{2+} react to form MnO2MnO_2.
44 Hard
44 of 46
Understanding +4 | -1
Arrange Ce3+,  La3+,  Pm3+\mathrm{C}{\mathrm{e}^{3 + }},\mathrm{\;L}{\mathrm{a}^{3 + }},\mathrm{\;P}{\mathrm{m}^{3 + }} and Yb3+\mathrm{Y}{\mathrm{b}^{3 + }} in increasing order of their ionic radii
View answer and explanation
Correct option: A
Yb3+<Pm3+<Ce3+<La3+\mathrm{Y}{\mathrm{b}^{3 + }} < P{\mathrm{m}^{3 + }} < C{\mathrm{e}^{3 + }} < L{\mathrm{a}^{3 + }}
Option A (Yb3+<Pm3+<Ce3+<La3+\mathrm{Y}{\mathrm{b}^{3 + }} < P{\mathrm{m}^{3 + }} < C{\mathrm{e}^{3 + }} < L{\mathrm{a}^{3 + }}) is correct because ionic radii decrease with increasing atomic number across the lanthanide series due to lanthanide contraction.
45 Hard
45 of 46
Analyzing +4 | -1
Assertion : FeCl3{\text{FeC}}{{\text{l}}_3} does not affect iodometric titration of CuSO4{\text{CuS}}{{\text{O}}_{\text{4}}} Solution

Reason : FeCl3{\text{FeC}}{{\text{l}}_3} is formed.
View answer and explanation
Correct option: D
If both Assertion & Reason are false.
Option D (If both Assertion & Reason are false) is correct because FeCl3{\text{FeC}}{{\text{l}}_3} *does* affect the iodometric titration of CuSO4{\text{CuS}}{{\text{O}}_{\text{4}}} solution, and the reason is incomplete. FeCl3{\text{FeC}}{{\text{l}}_3} can oxidize I−I^- to I2I_2, interfering with the titration.
46 Hard
46 of 46
Understanding +4 | -1
Pick out the correct statements from the following.

I. Cobalt (III) is more stable in octahedral complexes.

II. Zinc forms coloured ions or complexes.

III. Most of the dd-block elements and their compounds are ferromagnetic.

IV. Osmium shows (VIII) oxidation state.

V. Cobalt (II) is more stable in octahedral complexes.
View answer and explanation
Correct option: D
1 and 4
Cobalt (III) is more stable in octahedral complexes due to crystal field stabilization energy, and Osmium shows (VIII) oxidation state, making statements I and IV correct.
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