d and f Block Elements

Substances that are strongly attracted in an applied magnetic field are known as ferromagnetic. Ferromagnetism is, in fact, considered as an extreme form of paramagnetism. Therefore, Statement I is correct.

For the ion $\mathrm{Cr}^{2+}$, the electronic configuration is $3d^4 \, 4s^0$, resulting in 4 unpaired electrons.

For the ion $\mathrm{Nd}^{3+}$, the electronic configuration is $4f^3 \, 6s^0$, resulting in 3 unpaired electrons.

Hence, Statement II is incorrect.

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 (KMnO₄).

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 (MnO₂).

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.

Magnetic moment $\mu=\sqrt{n(n+2)} B M$

$\mathrm{n} \Rightarrow$ Number of unpaired electrons

Hence (Cr$^{2+}$, Fe$^{2+}$) and (Ti$^{3+}$, Cu$^{2+}$) are the pair of same magnetic moment.

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.

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

Let's match each element in List II with the appropriate block or group in List I:

(A) Lanthanoid - The Lanthanides or Lanthanoids are a series of 15 metallic elements from Lanthanum (La) to Lutetium (Lu). Cerium (Ce) is part of this series.

(B) d-block element - The d-block elements are also known as transition metals. Manganese (Mn) belongs to this group.

(C) p-block element - The p-block contains elements from groups 13 to 18 of the periodic table; Arsenic (As) belongs to this block.

(D) s-block element - The s-block elements include the first two groups (alkali metals and alkaline earth metals). Cesium (Cs) is one of the alkali metals, hence it is part of the s-block.

So, the correct matching is:

• (A) Lanthanoid - (I) Ce
• (B) d-block element - (IV) Mn
• (C) p-block element - (II) As
• (D) s-block element - (III) Cs

Thus, the correct option is Option C:

A-I, B-IV, C-II, D-III

$\bullet$ $\mathrm{PEt}_3$ 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.

Spin only magnetic moment is given by $\sqrt{n(n+2)} \mathrm{BM}$

$\therefore \mathrm{Cr}^{2+}$ and $\mathrm{Fe}^{2+}$ will have same spin only magnetic moment.

$ \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.

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.

Reason is the correct explanation of Assertion.

Reason in correct explanation the above Assertion.

$\mathrm{Cu}^{+2}$ is more stable than $\mathrm{Cu}^{+1}$ because released hydration energy is more in case of $\mathrm{Cu}^{+2}$ than $\mathrm{Cu}^{+1}$.

$\mathrm{D} \rightarrow \mathrm{V}_{2} \mathrm{O}_{5}$ dissolve in acid to give $\mathrm{VO}_{4}^{-3}$ salts. This doesn't shown by $\mathrm{V}_{2} \mathrm{O}_{4}$

Statement I : Cr²⁺ is reducing as its configuration changes from d⁴ to d³, the latter having a half-filled t_2g level. On the other hand, the change from Mn³⁺ to Mn²⁺ results in the half-filled (d⁵) configuration which has extra stability.

Statement II : Sc³⁺ has zero unpaired electron, so magnetic moment is also zero. Hence, Sc³⁺ will repelled by the applied magnetic field.