d and f Block Elements Questions — NEET Chemistry

Q1

Although +3 oxidation state is most common in lanthanoids, cerium still shows +4 oxidation state because:

A After losing one more electron, it acquires

4 f^{14}

electronic configuration.

B Its nearest inert gas is Radon.

C Its atomic number is 61 .

D After losing one more electron, it acquires

4 f^0

electronic configuration.

Correct Answer

Option D

Solution

Although +3 oxidation state is most common in lanthanoids, cerium still shows +4 oxidation state because after losing one more electron it acquires $4 f^{\circ}$ electronic configuration.

Q2

The calculated 'spin-only' magnetic moment

\mathrm{Ti}^{2+}\left(3 \mathrm{~d}^2\right)

is :

A 5.92 BM

B 3.87 BM

C 2.84 BM

D 4.90 BM

Correct Answer

Option C

Solution

The formula for the spin-only magnetic moment is given by:

\mu = \sqrt{n(n+2)} \ \text{B.M.}

Here, $n$ = number of unpaired electrons. For $\mathrm{Ti}^{2+}$ , the electronic configuration is:

\mathrm{Ti}^{2+} : [\mathrm{Ar}] \ 4s^0 \ 3d^2

In the $3d^2$ configuration, there are 2 unpaired electrons, so $n = 2$ . Substituting in the formula:

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

\mu = \sqrt{8} = 2.84 \ \mathrm{B.M.}

Therefore, the spin-only magnetic moment of $\mathrm{Ti}^{2+}$ is $2.84 \ \mathrm{B.M.}$ .

Q3

Baeyer's reagent is :

A Acidic potassium permanganate solution

B Acidic potassium dichromate solution

C Cold, dilute, aqueous solution of potassium permanganate

D Hot, concentrated solution of potassium permanganate

Correct Answer

Option C

Solution

The correct answer is Option C: Cold, dilute, aqueous solution of potassium permanganate .

Here's why: Baeyer's reagent is a chemical test used to detect the presence of unsaturation in organic compounds, particularly the presence of carbon-carbon double bonds (C=C).

It relies on the oxidizing power of potassium permanganate (KMnO 4 ).

Here's a breakdown of why the other options are incorrect: Option A: Acidic potassium permanganate solution - While acidic potassium permanganate is a strong oxidizing agent, it is too reactive and would oxidize other functional groups present in the molecule, making it unsuitable for specifically detecting unsaturation.

Option B: Acidic potassium dichromate solution - Acidic potassium dichromate is another strong oxidizing agent, but it is not typically used for detecting unsaturation.

It is commonly used for oxidizing alcohols to aldehydes or ketones.

Option D: Hot, concentrated solution of potassium permanganate - A hot, concentrated solution of potassium permanganate would be too harsh and could lead to over-oxidation, making it unreliable for detecting unsaturation.

Why Cold, Dilute, Aqueous Potassium Permanganate Works In Baeyer's test, the cold, dilute, aqueous solution of potassium permanganate acts as a mild oxidant.

The purple color of the permanganate solution disappears as it reacts with the double bond, forming a brown precipitate of manganese dioxide (MnO 2 ).

The reaction can be represented as follows:

3R-CH=CH-R' + 2KMnO_4 + 4H_2O \longrightarrow 3R-CH(OH)-CH(OH)-R' + 2MnO_2 + 2KOH

The disappearance of the purple color and the formation of the brown precipitate indicate the presence of an unsaturated compound.

This reaction is highly specific to alkenes and alkynes because they are readily oxidized by the permanganate ion.

Q4

Which of the following set of ions act as oxidising agents?

A

\mathrm{Ce}^{4+}

and

\mathrm{Tb}^{4+}

B

\mathrm{La}^{3+}

and

\mathrm{Lu}^{3+}

C

\mathrm{Eu}^{2+}

and

\mathrm{Yb}^{2+}

D

\mathrm{Eu}^{2+}

and

\mathrm{Tb}^{4+}

Correct Answer

Option A

Solution

Most stable oxidation state of lanthanoids is +3

\mathrm{Ce}^{4+}

and

\mathrm{Tb}^{4+}

will get reduced easily and will be good oxidising agents.

Q5

The UV-visible absorption bands in the spectra of lanthanoid ions are 'X', probably because of the excitation of electrons involving 'Y'. The 'X' and 'Y', respectively, are : Narrow and d and f orbitals

A Broad and f orbitals

B Narrow and f orbitals

C Broad and d and f orbitals

D Narrow and d and f orbitals

Correct Answer

Option B

Solution

The UV-visible absorption bands in the spectra of lanthanoid ions are primarily influenced by the electronic transitions that occur within the f orbitals.

Lanthanoid ions have partially filled 4f orbitals, and the transitions within these orbitals are generally quite narrow.

This is because the 4f orbitals are well shielded by the outer 5s and 5p orbitals, which results in minimal interaction with the surrounding crystal field, leading to narrow absorption bands.

Therefore, the correct answer is: Option B: Narrow and f orbitals

Q6

Identify the incorrect statement.

A

\mathrm{PEt}_3

and

\mathrm{AsPh}_3

as ligands can form

\mathrm{d} \pi-\mathrm{d} \pi

bond with transition metals

B The

\mathrm{N}-\mathrm{N}

single bond is as strong as the

\mathrm{P}-\mathrm{P}

single bond

C Nitrogen has unique ability to form

p \pi-p \pi

multiple bonds with nitrogen, carbon and oxygen

D Nitrogen cannot form

\mathrm{d} \pi-\mathrm{p} \pi

bond as other heavier elements of its group

Correct Answer

Option B

Solution

$\bullet$

\mathrm{PEt}_3

and

\mathrm{AsPh}_3

as ligands can form

\mathrm{d} \pi-\mathrm{d} \pi

bond with transition metals.

$\bullet$ The N – N single bond is weaker than the single P – P bond because of high inter-electronic repulsion of the non-bonding electrons.

$\bullet$ Nitrogen has unique ability to form p $\pi$ -p $\pi$ multiple bonds with itself, carbon and oxygen.

$\bullet$ Nitrogen cannot form d $\pi$ -p $\pi$ bond as other heavier elements of its group.

Q7

The

\mathrm{E}^{\circ}

value for the

\mathrm{Mn}^{3+} / \mathrm{Mn}^{2+}

couple is more positive than that of

\mathrm{Cr}^{3+} / \mathrm{Cr}^{2+}

or

\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}

due to change of

A

d^5

to

d^4

configuration

B

d^5

to

d^2

configuration

C

d^4

to

d^5

configuration

D

d^3

to

d^5

configuration

Correct Answer

Option C

Solution

\mathrm{E}_{\mathrm{Mn}^{3+} / \mathrm{Mn}^{2+}}^0>\mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}^{2+}}^{\circ} \text{ or } \mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\circ}

Electronic configuration of

\mathrm{Mn}^{3+}=[\mathrm{Ar}] 3 d^4

Electronic configuration of

\mathrm{Mn}^{2+}=[\mathrm{Ar}] 3 d^5

Electronic configuration of

\mathrm{Cr}^{3+}=[\mathrm{Ar}] 3 d^3

Electronic configuration of

\mathrm{Cr}^{2+}=[\mathrm{Ar}] 3 d^4

As

\mathrm{Mn}^{3+}

from

d^4

configuration goes to more stable

d^5

configuration (Half filled), due to more exchange energy in

d^5

configuration.

Q8

The pair of lanthanoid ions which are diamagnetic is

A

\mathrm{Ce}^{4+}

and

\mathrm{Yb}^{2+}

B

\mathrm{Ce}^{3+}

and

\mathrm{Eu}^{2+}

C

\mathrm{Gd}^{3+}

and

\mathrm{Eu}^{3+}

D

\mathrm{Pm}^{3+}

and

\mathrm{Sm}^{3+}

Correct Answer

Option A

Solution

Magnetic moment

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

\mathrm{n} \rightarrow

number of unpaired electron Hence

\mathrm{Ce}^{4+}

and

\mathrm{Yb}^{2+}

are only diamagnetic.

Q9

The stability of

\mathrm{Cu}^{2+}

is more than

\mathrm{Cu}^{+}

salts in aqueous solution due to -

A enthalpy of atomization.

B hydration energy.

C second ionisation enthalpy.

D first ionisation enthalpy.

Correct Answer

Option B

Solution

\mathrm{Cu}^{+2}

is more stable than

\mathrm{Cu}^{+1}

because released hydration energy is more in case of

\mathrm{Cu}^{+2}

than

\mathrm{Cu}^{+1}

.

Q10

The incorrect statement among the following is :

A Actinoids are highly reactive metals, especially when finely divided.

B Actinoid contraction is greater for element to element than lanthanoid contraction.

C Most of the trivalent Lanthanoid ions are colorless in the solid state.

D Lanthanoids are good conductors of heat and electricity.

Correct Answer

Option C

Solution

$\bullet$ The surface area increases when actinoids are finely divided which results in exposure of more reactant molecules to react.

Hence, rate increases and so, actinoids are highly reactive metals when finely divided.

$\bullet$ The shielding effect of 5f-orbitals in actinoids is poor than the shielding effect of4f-orbitals.

So, the effective nuclear charge on valence electrons is more in actinoids.

Hence, actinoid contraction is greater than lanthanoid contraction.

$\bullet$ Trivalent lanthanoid ions are coloured in the solid state due to presence of f-electrons.

$\bullet$ Lanthanoids are inner transition metals.

So, they are good conductors of heat and electricity.