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

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

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

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

The reaction of ${{\rm{K}}_2}{\rm{C}}{{\rm{r}}_2}{{\rm{O}}_7}$ with NaCl and conc ${{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}$ gives

${\rm{Cr}}{{\rm{O}}_2}{\rm{C}}{{\rm{l}}_2}$

${\rm{C}}{{\rm{r}}_2}{{\rm{O}}_3}$

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NEET • CHEMISTRY • d & f- BLOCK ELEMENTS

D & f- BLOCK ELEMENTS practice · Mixed difficulty

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1 Medium

1 of 213

Application +4 | -1

A substance which is not paramagnetic is:

View answer and explanation

Correct option: B

\mathrm{KMn}{\mathrm{O}_4}

The correct answer is

\mathrm{KMn}{\mathrm{O}_4}

because it is diamagnetic, having no unpaired electrons due to the

Mn

being in +7 oxidation state with

d^0

configuration.

2 Medium

2 of 213

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 the common name for iron(II) sulfate heptahydrate.

3 Medium

3 of 213

Analyzing +4 | -1

Which is the correct order of ionic sizes?) At.no.∶Ce=58,Sn=50,Yb=70 and Lu=71)

View answer and explanation

Correct option: C

\mathrm{Sn} > Ce > Yb > Lu

Option C (

\mathrm{Sn} > Ce > Yb > Lu

) is correct because ionic size decreases across the lanthanide series due to lanthanide contraction, and Sn is not a lanthanide.

4 Easy

4 of 213

Remembering +4 | -1

In the first transition series, the differentiating electron enters:

View answer and explanation

Correct option: C

d

-orbital

Option C (3

d

-orbital) is correct because the first transition series involves filling the 3

d

orbitals.

5 Easy

5 of 213

Understanding +4 | -1

Philosopher’s wool on heating with BaO at 1100

^\circ

C produce

View answer and explanation

Correct option: C

\mathrm{BaZn}{\mathrm{O}_2}

Option C (

\mathrm{BaZn}{\mathrm{O}_2}

) is correct because philosopher's wool (ZnO) reacts with BaO at high temperatures to form barium zincate.

6 Hard

6 of 213

Application +4 | -1

A compound in which a metal ion

{\mathrm{M}^{x + }}

(Z=25) has a spin only magnetic moment of

\sqrt {24}

BM. The number of unpaired electrons in the compound and the oxidation state of the metal ion are respectively.

View answer and explanation

Correct option: D

4 and 3

Option D (4 and 3) is correct because a magnetic moment of

\sqrt{24}

BM corresponds to 4 unpaired electrons, and with Z=25, an oxidation state of +3 leaves 4 d-electrons unpaired.

7 Medium

7 of 213

Understanding +4 | -1

The reaction of

{\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}

with NaCl and conc

{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

gives

View answer and explanation

Correct option: A

\mathrm{Cr}{\mathrm{O}_2}\mathrm{C}{\mathrm{l}_2}

Option A (

\mathrm{Cr}{\mathrm{O}_2}\mathrm{C}{\mathrm{l}_2}

) is correct because the reaction of potassium dichromate with NaCl and concentrated sulfuric acid produces chromyl chloride.

8 Hard

8 of 213

Analyzing +4 | -1

Assertion :

{\text{C}}{{\text{e}}^{ + 4}}

acts as oxidising agent in aqueous medium

Reason :

+4

is common oxidation state of lanthanides

View answer and explanation

Correct option: C

If Assertion is True but the Reason is False.

Option C (If Assertion is True but the Reason is False) is correct because

{\text{C}}{{\text{e}}^{ + 4}}

acts as an oxidizing agent, but +4 is not the common oxidation state of lanthanides.

9 Easy

9 of 213

Remembering +4 | -1

{\mathrm{K}_3}\mathrm{Fe}{\left( {\mathrm{CN}} \right)_6}

is used as ……. Indicator for

\mathrm{FeS}{\mathrm{O}_4}vs.{\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}

titrations.

View answer and explanation

Correct option: B

External

Option B (External) is correct because

{\mathrm{K}_3}\mathrm{Fe}{\left( {\mathrm{CN}} \right)_6}

is used as an external indicator in

\mathrm{FeS}{\mathrm{O}_4}vs.{\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}

titrations, meaning it's used outside the titration flask.

10 Hard

10 of 213

Analyzing +4 | -1

Assertion :

{\text{E}}{{\text{u}}^{2 + }}\,\& \,{\text{Y}}{{\text{b}}^{{\text{2 + }}}}

are reducing agents for their ions.

Reason: Both ions have stable half - filled configuration.

View answer and explanation

Correct option: C

If Assertion is True but the Reason is False.

Option C is correct because while Eu2+ and Yb2+ are indeed reducing agents, the reason given about stable half-filled configuration is not entirely accurate or complete.

11 Medium

11 of 213

Understanding +4 | -1

Which pair consists only acidic oxides?

View answer and explanation

Correct option: A

\mathrm{Cr}{\mathrm{O}_3},\mathrm{M}{\mathrm{n}_2}{\mathrm{O}_7}

Option A (

\mathrm{Cr}{\mathrm{O}_3},\mathrm{M}{\mathrm{n}_2}{\mathrm{O}_7}

) is correct because both chromium trioxide and manganese heptoxide are acidic oxides.

12 Medium

12 of 213

Application +4 | -1

Which of the following ions has a magnetic moment of 5.93 BM?

(At. no. V=23, Cr=24, Mn=25, Fe=26)

View answer and explanation

Correct option: A

\mathrm{M}{\mathrm{n}^{2 + }}

Option A (

\mathrm{M}{\mathrm{n}^{2 + }}

) is correct because

\mathrm{M}{\mathrm{n}^{2 + }}

has 5 unpaired electrons, resulting in a magnetic moment close to 5.93 BM.

13 Easy

13 of 213

Analyzing +4 | -1

Given below are two statements:

Statement I : Aqueous solution of

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

is preferred as a primary standard in volumetric analysis over

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

aqueous solution.

Statement II :

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

has a higher solubility in water than

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

.

In the light of the above statements, choose the correct answer from the options given below:

View answer and explanation

Correct option: C

Statement I is true but Statement II is false

Option C (Statement I is true but Statement II is false) is correct because K2Cr2O7 is preferred as a primary standard due to its lower hygroscopicity and higher molar mass, but it has lower solubility than Na2Cr2O7.

14 Easy

14 of 213

Analyzing +4 | -1

Which of the following statements is not correct about

{Cu}^{2+}

and

{Cu}^{+}:

View answer and explanation

Correct option: D

Both

{Cu}^{+}

and

{Cu}^{2+}

are stable with

{I}^{-}

Option D is correct because

{Cu}^{+}

is not stable with

{I}^{-}

due to the formation of

CuI

precipitate.

15 Medium

15 of 213

Remembering +4 | -1

Which does not possess allotropic forms?

View answer and explanation

Correct option: C

Option C (Fe) is correct because Iron (Fe) exhibits allotropy, but the question asks which does NOT possess allotropic forms. Carbon, Tin, and Phosphorus all exhibit allotropy.

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16 Easy

16 of 213

Application +4 | -1

Maximum oxidation state is presented by:

View answer and explanation

Correct option: A

\mathrm{Cr}{\mathrm{O}_2}\mathrm{C}{\mathrm{l}_2}\mathrm{\;and\;MnO}_4^ -

Option A is correct because CrO2Cl2 has Cr in +6 oxidation state and MnO4- has Mn in +7 oxidation state, which is the maximum possible for Mn.

17 Hard

17 of 213

Understanding +4 | -1

{NaCl}

reacts with conc.

{H}_2 {SO}_4

and

{K}_2 {Cr}_2 {O}_7

to give reddish fumes (B), which react with

{NaOH}

to give yellow solution (C). (B) and (C) respectively are:

View answer and explanation

Correct option: A

{CrO}_2 {Cl}_2, {Na}_2 {CrO}_4

The correct answer is

{CrO}_2 {Cl}_2, {Na}_2 {CrO}_4

because the reddish fumes (B) are chromyl chloride, and it reacts with NaOH to form sodium chromate (yellow solution).

18 Medium

18 of 213

Remembering +4 | -1

Which metal gives hydrogen gas on heating with hot concentrated alkali?

View answer and explanation

Correct option: C

Option C (Zn) is correct because Zinc reacts with hot concentrated alkali to produce hydrogen gas and a zincate salt.

19 Easy

19 of 213

Analyzing +4 | -1

Which one of the following ions exhibits d-d transition and paramagnetism as well ?

View answer and explanation

Correct option: A

\,\,\mathrm{Mn}\,\mathrm{O}_\mathrm{4}^{\mathrm{2 - }}

Option A (

MnO_4^{2-}

) is correct because it exhibits both d-d transition (due to the presence of d electrons) and paramagnetism (due to the presence of unpaired electrons).

20 Easy

20 of 213

Application +4 | -1

Which of the following d-block ions show the maximum value of magnetic moment?

View answer and explanation

Correct option: A

\mathrm{Mn}^{+2}

Option A (

\mathrm{Mn}^{+2}

) is correct because it has the maximum number of unpaired electrons, leading to the highest magnetic moment.

21 Medium

21 of 213

Application +4 | -1

1 mole of

{K}_2 {Cr}_2 {O}_7

cannot oxidise

View answer and explanation

Correct option: D

4 moles of

{Sn}^{2+}

{Sn}^{4+}

Option D (4 moles of Sn2+ to Sn4+) is correct because 1 mole of K2Cr2O7 can oxidize a maximum of 3 moles of Sn2+.

22 Medium

22 of 213

Analyzing +4 | -1

Which one of the following does not decolourise an acidified

\mathrm{KMn}{\mathrm{O}_4}

solution?

View answer and explanation

Correct option: B

\mathrm{FeC}{\mathrm{l}_3}

Option B (

\mathrm{FeC}{\mathrm{l}_3}

) is the correct answer because it is already in its highest oxidation state (+3) and cannot be further oxidized by acidified

\mathrm{KMn}{\mathrm{O}_4}

23 Medium

23 of 213

Analyzing +4 | -1

Assertion (A): Green solution of potassium manganate (VI),

\mathrm{K}_2 \mathrm{MnO}_4

turns purple and brown solid is precipitated when

\mathrm{CO}_2

is bubbled into it.

Reason (R):

\mathrm{K}_2 \mathrm{MnO}_4

changes to

\mathrm{K}_3 \mathrm{MnO}_8

\mathrm{CO}_2

View answer and explanation

Correct option: C

\mathbf{A}

is true but

\mathbf{R}

is false.

Option C (

\mathbf{A}

is true but

\mathbf{R}

is false) is correct because the green solution of potassium manganate(VI) turns purple due to disproportionation, not because it changes to

\mathrm{K}_3\mathrm{MnO}_8

24 Easy

24 of 213

Remembering +4 | -1

Mercury is purified by:

View answer and explanation

Correct option: B

Distillation in vacuum

Option B (Distillation in vacuum) is correct because it allows mercury to be purified at lower temperatures, preventing decomposition or reaction with air.

25 Medium

25 of 213

Application +4 | -1

Number of electrons in 3

d

-orbital of

{\mathrm{V}^{2 + }},\mathrm{\;C}{\mathrm{r}^{2 + }},\mathrm{\;M}{\mathrm{n}^{2 + }}

, and

\mathrm{F}{\mathrm{e}^{2 + }}

are 3, 4, 5 and 6 respectively. Which of the following ions will have largest value of magnetic moment

\left( \mathrm{\mu } \right)

View answer and explanation

Correct option: C

\mathrm{M}{\mathrm{n}^{2 + }}

Option C is correct because Mn2+ has 5 unpaired electrons in its 3d orbitals, resulting in the largest magnetic moment according to the formula μ = √n(n+2) BM, where n is the number of unpaired electrons.

26 Hard

26 of 213

Remembering +4 | -1

Oxygen is absorbed by molten Ag, which is evolved on cooling and the silver particles are scattered; the phenomenon is known as:

View answer and explanation

Correct option: B

Spitting of silver

Option B (Spitting of silver) is correct because the phenomenon of oxygen being absorbed by molten silver and then released upon cooling, causing the silver to scatter, is known as spitting of silver.

27 Easy

27 of 213

Understanding +4 | -1

Which of the following statements is not true for Mohr’s salt?

View answer and explanation

Correct option: C

Oxidation state of iron is +3

Option C (Oxidation state of iron is +3) is correct because Mohr's salt is

FeSO_4.(NH_4)_2SO_4.6H_2O

, where iron is in the +2 oxidation state, not +3.

28 Hard

28 of 213

Understanding +4 | -1

The reaction of

\mathrm{H}_2 \mathrm{O}_2

with potassium permanganate in acidic medium leads to the formation of mainly

View answer and explanation

Correct option: A

\mathrm{Mn}^{2+}

Option A (

\mathrm{Mn}^{2+}

) is correct because in acidic medium,

\mathrm{H}_2\mathrm{O}_2

reduces

\mathrm{KMnO}_4

\mathrm{Mn}^{2+}

29 Hard

29 of 213

Understanding +4 | -1

Assertion :

{\text{KMn}}{{\text{O}}_{\text{4}}}

is dark pink coloured compound.

Reason : In the

{\text{KMn}}{{\text{O}}_{\text{4}}}

charge transfer spectrum occurs.

View answer and explanation

Correct option: A

If both Assertion & Reason are True & the Reason is correct explanation of the Assertion.

Option A (If both Assertion & Reason are True & the Reason is correct explanation of the Assertion) is correct because

KMnO_4

is dark pink due to ligand to metal charge transfer.

30 Medium

30 of 213

Understanding +4 | -1

The radius of

\mathrm{L}{\mathrm{a}^{3 + }}

(Atomic number of La = 57) is 1.06 Å. Which one of the following given values will be closest to the radius of

\mathrm{L}{\mathrm{u}^{3 + }}

?
(Atomic number of Lu=71)

View answer and explanation

Correct option: D

0.85 Å

Option D (0.85 Å) is correct because, due to lanthanide contraction, the ionic radii decrease across the lanthanide series.

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Key Takeaways

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