${\rm{FeS}}{{\rm{O}}_4}.7{{\rm{H}}_2}{\rm{O}}$

${\rm{ZnS}}{{\rm{O}}_4}.7{{\rm{H}}_2}{\rm{O}}$

${\rm{CuS}}{{\rm{O}}_4}.5{{\rm{H}}_2}{\rm{O}}$

${\rm{CaS}}{{\rm{O}}_4}.\frac{1}{2}{\rm{\;}}{{\rm{H}}_2}{\rm{O}}$

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JEE • CHEMISTRY • d & f- BLOCK ELEMENTS

D & f- BLOCK ELEMENTS practice · Mixed difficulty

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Practice format and chapter quiz

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 391-410 of 410 questions

391 Medium

391 of 410

Remembering +4 | -1

Stainless steel contains:

View answer and explanation

Correct option: C

14%Cr

Option C (14%Cr) is correct because stainless steel typically contains around 11-15% chromium for corrosion resistance.

392 Hard

392 of 410

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

d

-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

Option D (1 and 4) is correct because Cobalt(III) is more stable in octahedral complexes and Osmium shows +8 oxidation state.

393 Medium

393 of 410

Understanding +4 | -1

Effective magnetic moment of

\mathrm{S}{\mathrm{c}^{3 + }}

ion is

View answer and explanation

Correct option: B

Option D (1) is correct because each carbon atom in graphite is sp2 hybridized, leaving one free electron.

394 Hard

394 of 410

Remembering +4 | -1

Which of the following compounds is used as the starting material for the preparation of potassium dichromate?

View answer and explanation

Correct option: C

\mathrm{FeC}{\mathrm{r}_2}{\mathrm{O}_4}

(Chromite)

Option C (

\mathrm{FeC}{\mathrm{r}_2}{\mathrm{O}_4}

(Chromite)) is correct because chromite ore (

\mathrm{FeC}{\mathrm{r}_2}{\mathrm{O}_4}

) is the primary starting material for the industrial preparation of potassium dichromate.

395 Medium

395 of 410

Understanding +4 | -1

Transition elements exhibit variable oxidation states in their compounds due to

View answer and explanation

Correct option: B

Very small energy difference between

(n-1) d

and

ns

orbitals

Option B is correct because the small energy difference between (n-1)d and ns orbitals allows for variable oxidation states as both can participate in bonding.

396 Easy

396 of 410

Understanding +4 | -1

Copper sulphate is commercially made from copper scrap by:

View answer and explanation

Correct option: B

Action of dilute sulphuric acid and air

Option B (Action of dilute sulphuric acid and air) is correct because copper scrap reacts with dilute sulfuric acid in the presence of air (oxygen) to form copper sulfate.

397 Easy

397 of 410

Understanding +4 | -1

Which oxide of Mn is acidic in nature?

View answer and explanation

Correct option: B

\mathrm{M}{\mathrm{n}_2}{\mathrm{O}_7}

Option B (

\mathrm{M}{\mathrm{n}_2}{\mathrm{O}_7}

) is correct because Mn2O7 forms HMnO4 (permanganic acid) when dissolved in water, indicating its acidic nature.

398 Medium

398 of 410

Analyzing +4 | -1

Among the given options, identify the option representing incorrect statement?

View answer and explanation

Correct option: C

Solid potassium dichromate and concentrated

\mathrm{H}_2 \mathrm{SO}_4

liberates chlorine gas with solid sodium chloride which is used to confirm presence of

\mathrm{Cl}^{\ominus}

ion

The correct answer is C because solid potassium dichromate and concentrated H2SO4 do not liberate chlorine gas with solid sodium chloride; instead, they produce chromyl chloride (CrO2Cl2).

399 Medium

399 of 410

Analyzing +4 | -1

In chromyl chloride, the number of d-electrons present on chromium is same as in (Given at. no. of

\text{Ti:} 22, \, \text{V:} 23, \, \text{Cr:}24, \, \text{Mn:} 25, \, \text{Fe:} 26

)

View answer and explanation

Correct option: B

\mathrm{Mn}

(VII)

Option B is correct because Cr in chromyl chloride (CrO2Cl2) has a +6 oxidation state, resulting in a d0 configuration, which is the same as Mn(VII).

400 Easy

400 of 410

Remembering +4 | -1

Gold exhibits the variable oxidation states of:

View answer and explanation

Correct option: B

+ 1,\mathrm{ } + 3

Option B (

+ 1,\mathrm{ } + 3

) is correct because gold commonly exhibits oxidation states of +1 and +3.

401 Medium

401 of 410

Remembering +4 | -1

Green vitriol is

View answer and explanation

Correct option: A

\mathrm{FeS}{\mathrm{O}_4}.7{\mathrm{H}_2}\mathrm{O}

Option A (

\mathrm{FeS}{\mathrm{O}_4}.7{\mathrm{H}_2}\mathrm{O}

) is correct because Green vitriol is iron(II) sulfate heptahydrate (

\mathrm{FeS}{\mathrm{O}_4}.7{\mathrm{H}_2}\mathrm{O}

402 Medium

402 of 410

Application +4 | -1

When

\mathrm{KMnO}_4

reacts with iodide ion

\left(\mathrm{I}^{-}\right)

in neutral medium then the product obtained will be:

View answer and explanation

Correct option: D

\mathrm{IO}_3^{-}

Option D (

\mathrm{IO}_3^{-}

) is correct because in a neutral medium, permanganate oxidizes iodide ions to iodate ions.

403 Medium

403 of 410

Understanding +4 | -1

Positive sol among the following is

View answer and explanation

Correct option: C

Ti{O_2}sol

Option C (

Ti{O_2}sol

) is correct because

Ti{O_2}

sol is a positively charged sol, while the others are negatively charged.

404 Medium

404 of 410

Remembering +4 | -1

Permanent magnets are generally made of alloys of

View answer and explanation

Correct option: B

Option B (Co) is correct because permanent magnets are often made of alloys containing cobalt, along with aluminum and nickel.

405 Easy

405 of 410

Remembering +4 | -1

In 3d-series atomic number

({Z})

varies from

View answer and explanation

Correct option: A

Z=21-30

Option A (

Z=21-30

) is correct because the 3d series starts with Scandium (Z=21) and ends with Zinc (Z=30).

Explore Other Chapters in CHEMISTRY

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P-BLOCK ELEMENTS 736 MCQs Equilibrium 696 MCQs STRUCTURE OF ATOM 634 MCQs THERMODYNAMICS 536 MCQs CHEMICAL BONDING & MOLECULAR STRUCTURE 514 MCQs SOLUTIONS 480 MCQs

406 Medium

406 of 410

Understanding +4 | -1

Which of the following is the correct electronic configuration of Gadolonium.

View answer and explanation

Correct option: A

\left[ {Xe} \right]4{f^7}5{d^1}6{s^2}

Option A (

{Xe} 4{f^7}5{d^1}6{s^2}

) is correct because Gadolinium (Gd) has an electronic configuration that favors a half-filled 4f subshell for stability.

407 Medium

407 of 410

Understanding +4 | -1

Reaction of potassium chromate and

\mathrm{CuSO}_4

in aqueous solution produces :

View answer and explanation

Correct option: D

both

\mathrm{CuCrO}_4

and

\mathrm{Cu}_2 \mathrm{Cr}_2 \mathrm{O}_7

Option D (both

\mathrm{CuCrO}_4

and

\mathrm{Cu}_2 \mathrm{Cr}_2 \mathrm{O}_7

) is correct because the reaction produces both copper chromate and copper dichromate due to the acidic nature of the copper sulfate solution.

408 Easy

408 of 410

Understanding +4 | -1

Which of the following metals do not react with

\mathrm{HCl}

View answer and explanation

Correct option: B

Silver and gold

Option B is correct. Silver (Ag) and gold (Au) are noble metals and are unreactive with HCl because they have higher reduction potentials than hydrogen.

409 Medium

409 of 410

Application +4 | -1

A compound of a metal ion

{M^{x + }}

(Z = 24) has a spin only magnetic moment of

\sqrt {15}

Bohr Magnetons. The number of unpaired electrons in the compound are:

View answer and explanation

Correct option: D

Option D (3) is correct because the number of unpaired electrons is found using the formula

\mathrm{\mu } = \sqrt {n(n + 2)}

and solving for n when

\mathrm{\mu } = \sqrt{15}

, which gives n=3.

410 Hard

410 of 410

Analyzing +4 | -1

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason(R)

Assertion(A):

{La}({OH})_3

is the most basic while

{Lu}({OH})_3

is the least basic.

Reason(R): As the size of lanthanoids decreases covalent character of hydroxides decreases and basic character increases.

In the light of the above statements, choose the most appropriate answer from the options given below.

View answer and explanation

Correct option: D

A is correct but R is not correct

Option C (

\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}

and

\mathrm{CH}_3 \mathrm{COCH}_3

) is correct because both ethanol and acetone give a positive iodoform test.

Revision notes and key takeaways

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