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

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

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

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

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JEE • CHEMISTRY • d & f- BLOCK ELEMENTS

D & f- BLOCK ELEMENTS practice · Medium

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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 241-254 of 254 questions

241 Medium

241 of 254

Understanding +4 | -1

An acid radical (A) when treated with dil.

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

, a colourless gas (B) comes out which is suffocating with the smell of burning sulphur. Gas B when treated with

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

acidified with dil.

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

, green colour appeared.

\mathrm{A}

and

\mathrm{B}

are respectively

View answer and explanation

Correct option: D

\mathrm{SO}_3^{2-}

and

\mathrm{SO}_2

Option D (

\mathrm{SO}_3^{2-}

and

\mathrm{SO}_2

) is correct because sulfite ions react with acid to produce sulfur dioxide, which has a suffocating smell and turns acidified potassium dichromate green.

242 Medium

242 of 254

Remembering +4 | -1

The transition element that has lowest enthalpy of atomization is:

View answer and explanation

Correct option: B

Cu

Option B (

Cu

) is correct because copper has a completely filled d-orbital in its +1 oxidation state, leading to weaker metallic bonding and thus lower enthalpy of atomization compared to Fe, V, and Zn.

243 Medium

243 of 254

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.

244 Medium

244 of 254

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.

245 Medium

245 of 254

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.

246 Medium

246 of 254

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).

247 Medium

247 of 254

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).

248 Medium

248 of 254

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}

249 Medium

249 of 254

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.

250 Medium

250 of 254

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.

251 Medium

251 of 254

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.

252 Medium

252 of 254

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.

253 Medium

253 of 254

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.

254 Medium

254 of 254

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.

Revision notes and key takeaways

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