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JEE • CHEMISTRY • d & f- BLOCK ELEMENTS

D & f- BLOCK ELEMENTS practice · Hard

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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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1 Hard

1 of 17

Analyzing +4 | -1

Sodium salt of an organic acid

'X'

produces effervescence with conc.

{H_2}S{O_4}

'X'

reacts with the acidified aqueous

CaC{l_2}

solution to give a white precipitate which decolourises acidic solution of

KMn{O_4}

'X'

View answer and explanation

Correct option: D

N{a_2}{C_2}{O_4}

Option D is correct because sodium oxalate (Na₂C₂O₄) produces effervescence with conc. H₂SO₄, forms a white precipitate with CaCl₂, and decolorizes acidic KMnO₄.

2 Hard

2 of 17

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.

The correct answer is A (If both Assertion & Reason are True & the Reason is correct explanation of the Assertion) because KMnO4's color arises from a charge transfer spectrum.

3 Hard

3 of 17

Analyzing +4 | -1

Assertion : Actinides show much higher range of oxidation states

Reason : Energy difference between 5f and 6d orbitals is large

View answer and explanation

Correct option: C

If Assertion is True but the Reason is False.

Option C is correct. Actinides exhibit a wider range of oxidation states due to smaller energy differences between 5f, 6d, and 7s orbitals, not a large difference as stated in the reason.

4 Hard

4 of 17

Analyzing +4 | -1

Consider the following statements

I.

Sc

has a low value of

E_{{M^{ + 3}}/{M^{ + 2}}}^0

II.

Zn

has the highest value of

E_{{M^{ + 3}}/{M^{ + 2}}}^0

III.

Mn

and

Fe

have comparatively high and low values of

E_{{M^{ + 3}}/{M^{ + 2}}}^0

respectively

The correct set of statements is (Choose the low values of

E_{{M^{ + 3}}/{M^{ + 2}}}^0

respectively).

The correct set of statements is (Choose the appropriate option)

View answer and explanation

Correct option: B

I, II and III

Option B (I, II and III) is correct because all three statements accurately describe the trends in

E_{{M^{ + 3}}/{M^{ + 2}}}^0

values for Sc, Zn, Mn, and Fe.

5 Hard

5 of 17

Analyzing +4 | -1

{I}_2+{Na}_2 {CO}_3

sol

^{\mathrm{n}} . \stackrel{\Delta}{\longrightarrow} {X}+{Y}

If '

{X}

' gives coloured ppt. with

{Pb}\left({CH}_3 {COO}\right)_2

solution, then '

{Y}

' will not be respond to which of the following

View answer and explanation

Correct option: B

{Y}+{Cr}_2 {O}_7{ }^{2-}\text{(aq)} +{OH}^{-}(\mathrm{aq})

The reaction produces NaI (X) and NaIO3 (Y). NaI gives a colored precipitate with lead acetate. NaIO3 is an oxidizing agent and will not oxidize Cr2O7^2- in a basic medium. Therefore, the correct answer is B.

6 Hard

6 of 17

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 the spin-only magnetic moment calculation yields 4 unpaired electrons, and the electronic configuration confirms a +3 oxidation state.

7 Hard

7 of 17

Understanding +4 | -1

A blue colouration is not obtained when:

View answer and explanation

Correct option: B

Copper sulphate solution reacts with

{\mathrm{K}_4}[\mathrm{Fe}{\left( {\mathrm{CN}} \right)_6}

]

Option B (Copper sulphate solution reacts with

{\mathrm{K}_4}[\mathrm{Fe}{\left( {\mathrm{CN}} \right)_6}

]) is correct because the reaction produces a brown precipitate, not a blue coloration.

8 Hard

8 of 17

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.

The correct answer is C (If Assertion is True but the Reason is False) because while Eu²⁺ and Yb²⁺ are reducing agents, it's due to their tendency to achieve stable f⁷ and f¹⁴ configurations, not just a general half-filled configuration.

9 Hard

9 of 17

Understanding +4 | -1

In chromyl chloride test for confirmation of

{Cl}^{-}

ion, a yellow solution is obtained. Acidification of the solution and addition of amyl alcohol and

10 \% {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

Option A (6) is correct because the oxidation state of chromium in

CrO_5

is +6, even though it has a peroxo linkage.

10 Hard

10 of 17

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 particles to scatter, is known as spitting of silver.

11 Hard

11 of 17

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}}

is indeed an oxidizing agent, but +4 is not the common oxidation state for all lanthanides.

12 Hard

12 of 17

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, potassium permanganate (

\mathrm{KMnO}_4

) oxidizes hydrogen peroxide (

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

) and itself gets reduced to

\mathrm{Mn}^{2+}

13 Hard

13 of 17

Analyzing +4 | -1

\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

\mathrm{x}, \mathrm{y}

and

\mathrm{z}

are respectively

View answer and explanation

Correct option: A

\mathrm{I}_2, \mathrm{IO}_3^{-}, \mathrm{MnO}_2

Option A (

\mathrm{I}_2, \mathrm{IO}_3^{-}, \mathrm{MnO}_2

) is correct because it correctly identifies the products of the redox reactions under different pH conditions.

14 Hard

14 of 17

Analyzing +4 | -1

A transition metal

\mathrm{M}

can exist in two oxidation states +1 and +2 . It forms a sulphide whose experimental formula is given by

\mathrm{M}_{\mathrm{x}} \mathrm{S}

where

2 > x > 1

. Then the ratio of metal ion in +1 state to those in +2 state in sulphide is given by

View answer and explanation

Correct option: B

\frac{2(x-1)}{2-x}

Option B is correct. The ratio of metal ions in +1 to +2 oxidation states is derived using charge neutrality principles in the compound M_xS.

15 Hard

15 of 17

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.

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16 Hard

16 of 17

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.

17 Hard

17 of 17

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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Key Takeaways

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