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NEET • CHEMISTRY • d & f- BLOCK ELEMENTS

D & f- BLOCK ELEMENTS practice · Mixed difficulty

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Showing 181-210 of 213 questions

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181 Easy
181 of 213
Understanding +4 | -1
Most basic compound among the following is
View answer and explanation
Correct option: A
La(OH)3La{\left( {OH} \right)_3}
Option A (La(OH)3La{\left( {OH} \right)_3}) is the most basic because the basicity of lanthanide hydroxides decreases as the ionic size decreases from La to Lu.
182 Hard
182 of 213
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.
183 Easy
183 of 213
Application +4 | -1
The +3 ion of which one of the following has half-filled 4f4f sunshell?
View answer and explanation
Correct option: C
Gd
Option C is correct because Gd3+ has 7 electrons in its 4f subshell, resulting in a half-filled configuration (4f7).
184 Easy
184 of 213
Remembering +4 | -1
Most common oxidation state of vanadium is
View answer and explanation
Correct option: D
+5+ 5
Option D (+5+ 5) is correct because the most stable and common oxidation state of Vanadium is +5, found in compounds like V2O5V_2O_5.
185 Easy
185 of 213
Analyzing +4 | -1
Which of the following compounds has colour but no unpaired electrons?
View answer and explanation
Correct option: A
KMnO4\mathrm{KMnO}_4
Option A (KMnO4KMnO_4) is correct because it exhibits color due to charge transfer transitions, even though it has no unpaired d-electrons in the traditional sense.
186 Easy
186 of 213
Remembering +4 | -1
The magnetic moment μ\mathrm{\mu }, of transition metals is related to the number of unpaired elelctrnos nn as
View answer and explanation
Correct option: D
μ=n(n+2)\mathrm{\mu } = \sqrt {n\left( {n + 2} \right)}
Option D is correct because the magnetic moment is calculated using the spin-only formula involving the number of unpaired electrons.
187 Easy
187 of 213
Understanding +4 | -1
Which will reduce acidified potassium dichromate solution?
View answer and explanation
Correct option: B
Mohr’s salt
Option B (Mohr’s salt) is correct because Mohr's salt (ferrous ammonium sulfate) contains Fe2+Fe^{2+} ions, which can be oxidized to Fe3+Fe^{3+} by potassium dichromate, thus reducing the dichromate.
188 Medium
188 of 213
Application +4 | -1
Ferrous ion changes to XX ion, on reacting with acidified hydrogen peroxide. The number of dd-electrons present in XX and its magnetic moment (in BM) are, respectively
View answer and explanation
Correct option: B
5 and 5.92
Option B (5 and 5.92) is correct because ferrous ion (Fe2+Fe^{2+}) is oxidized to ferric ion (Fe3+Fe^{3+}), which has 5 d-electrons and a magnetic moment of approximately 5.92 BM.
189 Easy
189 of 213
Remembering +4 | -1
Brass is an alloy of:
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Correct option: A
Zn and Cu
Option A (Zn and Cu) is correct because brass is an alloy primarily composed of zinc and copper.
190 Hard
190 of 213
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.
191 Medium
191 of 213
Analyzing +4 | -1
Identify the incorrect statement among the following
View answer and explanation
Correct option: D
4f4f and 5f5f-orbitals are equally shielded.
Option D is correct because 4f orbitals are shielded more effectively than 5f orbitals due to their spatial distribution.
192 Easy
192 of 213
Application +4 | -1
Which of the following in its oxidation state shows the paramagnetism?
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Correct option: A
Tb(IV)\mathrm{Tb}(\mathrm{IV})
Option A (Tb(IV)\mathrm{Tb}(\mathrm{IV})) is correct because Tb(IV)\mathrm{Tb}(\mathrm{IV}) has unpaired electrons, making it paramagnetic.
193 Easy
193 of 213
Analyzing +4 | -1
The correct order of spin only magnetic moments for the following complex ions is
View answer and explanation
Correct option: C
[Fe(CN)6]3−<[Mn(CN)6]3−<[CoF6]3−<[MnBr4]2−\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-} < \left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}< \left[\mathrm{CoF}_{6}\right]^{3-} < \left[\mathrm{MnBr}_{4}\right]^{2-}
Option C ([Fe(CN)6]3−<[Mn(CN)6]3−<[CoF6]3−<[MnBr4]2−[Fe(CN)_6]^{3-} < [Mn(CN)_6]^{3-} < [CoF_6]^{3-} < [MnBr_4]^{2-}) is correct because it arranges the complexes in increasing order of their spin-only magnetic moments, which depends on the number of unpaired electrons.
194 Medium
194 of 213
Application +4 | -1
KMnO4KMn{O_4} reacts with oxalic acid according to the equation 2MnO4−+5C2O42−+16H+→2Mn2++10CO2+8H2O2MnO_4^ - + 5{C_2}O_4^{2 - } + 16{H^ + } \to 2M{n^{2 + }} + 10C{O_2} + 8{H_2}O
Here 20 ml of 0.1 M KMnO4KMn{O_4} will react with
View answer and explanation
Correct option: A
50 ml of 0.1 M H2C2O4{H_2}{C_2}{O_4}
The correct answer is 50 ml of 0.1 M H2C2O4{H_2}{C_2}{O_4} because the stoichiometry of the reaction dictates that 2 moles of KMnO4KMnO_4 react with 5 moles of oxalic acid.
195 Medium
195 of 213
Understanding +4 | -1
(NH4)2Cr2O7{\left( {\mathrm{N}{\mathrm{H}_4}} \right)_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7} on heating gives a gas which is also given by
View answer and explanation
Correct option: A
Heating NH4NO2\mathrm{N}{\mathrm{H}_4}\mathrm{N}{\mathrm{O}_2}
Option A (Heating NH4NO2\mathrm{N}{\mathrm{H}_4}\mathrm{N}{\mathrm{O}_2}) is correct because it also produces nitrogen gas upon heating, similar to the given reaction.
196 Medium
196 of 213
Analyzing +4 | -1
Na2O2{Na}_2 {O}_2
View answer and explanation
Correct option: D
All are correct properties of Na2O2{Na}_2 {O}_2
Option D (All are correct properties of Na2O2{Na}_2 {O}_2) is correct because Na2O2{Na}_2 {O}_2 is diamagnetic, it's a salt of dibasic acid H2O2{H}_2 {O}_2, and it oxidizes Cr3+{Cr}^{3+} to CrO42−{CrO}_4^{2-}.
197 Medium
197 of 213
Application +4 | -1
Which of the following ions has the highest magnetic moment?
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Correct option: C
Mn2+\mathrm{M}{\mathrm{n}^{2 + }}
Option C (Mn2+\mathrm{Mn}^{2+}) is correct because it has the highest number of unpaired electrons, leading to the highest magnetic moment.
198 Medium
198 of 213
Application +4 | -1
Which of the following electronic configuration represents the maximum magnetic moment?
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Correct option: D
d6{d^6}
The correct answer is d6{d^6} because it has the maximum number of unpaired electrons (4) when considering Hund's rule, leading to the highest magnetic moment.
199 Hard
199 of 213
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.
200 Hard
200 of 213
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.
201 Medium
201 of 213
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) has an atomic number of 93. After losing 3 electrons to form Np3+Np^{3+}, the resulting electronic configuration is [Rn]5f4[Rn]5f^4.
202 Medium
202 of 213
Analyzing +4 | -1
In which of the following cases, the stability of two oxidation states is correctly represented
View answer and explanation
Correct option: B
Mn2+>Mn3+M{n^{2 + }} > M{n^{3 + }}
Option B (Mn2+>Mn3+M{n^{2 + }} > M{n^{3 + }}) is correct because Mn2+Mn^{2+} has a stable half-filled d5d^5 configuration, making it more stable than Mn3+Mn^{3+}.
203 Easy
203 of 213
Remembering +4 | -1
The nitrate of which metal leaves metallic globule on heating strongly?
View answer and explanation
Correct option: B
AgNO3\mathrm{AgN}{\mathrm{O}_3}
Option B (AgNO3\mathrm{AgN}{\mathrm{O}_3}) is correct because silver nitrate decomposes upon heating to produce metallic silver, nitrogen dioxide, and oxygen.
204 Easy
204 of 213
Analyzing +4 | -1
Which of the following is most stable?
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Correct option: D
Cr3+\mathrm{C}{\mathrm{r}^{3 + }}
Option D (Cr3+\mathrm{C}{\mathrm{r}^{3 + }}) is the most stable because it has a half-filled t2gt_{2g} orbital configuration.
205 Medium
205 of 213
Understanding +4 | -1
When excess of SnCl2\mathrm{SnC}{\mathrm{l}_2} is added to a solution of HgCl2\mathrm{HgC}{\mathrm{l}_2}, a white ppt. turning to grey is obtained. This grey colour is due to the formation of:
View answer and explanation
Correct option: D
Hg2\mathrm{H}{\mathrm{g}_2}
The grey color is due to the formation of mercury (Hg2\mathrm{H}{\mathrm{g}_2}) because SnCl2\mathrm{SnC}{\mathrm{l}_2} reduces HgCl2\mathrm{HgC}{\mathrm{l}_2} to metallic mercury.
206 Medium
206 of 213
Remembering +4 | -1
A hard and resistant alloy generally used in tip of nib of pen is:
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Correct option: B
Pt, Cr
Option A (Os, Ir) is correct because an alloy of Osmium and Iridium is very hard and resistant, making it suitable for pen nibs.
207 Hard
207 of 213
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.
208 Easy
208 of 213
Understanding +4 | -1
4K2Cr2O7→Heat4K2CrO4+3O2+X4{\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}\xrightarrow{{\mathrm{Heat}}}4{\mathrm{K}_2}\mathrm{Cr}{\mathrm{O}_4} + 3{\mathrm{O}_2} + X In the above reaction, XX is
View answer and explanation
Correct option: C
Cr2O3\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_3}
Option C (Cr2O3\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_3}) is correct because when potassium dichromate is heated, it decomposes to form potassium chromate, oxygen, and chromium(III) oxide.
209 Easy
209 of 213
Analyzing +4 | -1
The set of correct statements is:

(i) Manganese exhibits +7 oxidation state in its oxide.

(ii) Ruthenium and Osmium exhibit +8 oxidation in their oxides.

(iii) Sc shows +4 oxidation state which is oxidizing in nature.

(iv) Cr shows oxidising nature in +6 oxidation state.
View answer and explanation
Correct option: B
(i), (ii) and (iv)
Option B ((i), (ii) and (iv)) is correct because Manganese exhibits +7 in its oxide, Ruthenium and Osmium exhibit +8 in their oxides, and Cr shows oxidizing nature in +6 oxidation state.
210 Medium
210 of 213
Understanding +4 | -1
Transition elements exhibit positive oxidation states only. This is because of:
View answer and explanation
Correct option: B
Their electropositive nature
Option B (Their electropositive nature) is correct because transition elements readily lose electrons due to their relatively low ionization energies, leading to positive oxidation states.
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