d and f Block Elements

The metal atom whose oxidation potential is less than that of hydrogen can release $\mathrm{H}_{2}$ from mineral acids.

$\mathrm{E}_{\mathrm{Zn} / \mathrm{Zn}^{+2}}=0.76$ $\mathrm{E}_{\mathrm{Ni/Ni}^{+ 2}}=0.25$ $E_{M n / M n^{+2}}=1.18$ $\mathrm{E}_{\mathrm{Cu}^{\circ} / \mathrm{Cu}^{+2}}=-0.34$

In neutral or Faintly alkaline medium, thiosulphate is oxidised almost quantitatively to sulphate ion according to reaction given below, $8 \mathrm{MnO}_{4}^{-}+3 \mathrm{~S}_{2} \mathrm{O}_{3}^{2-}+\mathrm{H}_{2} \mathrm{O} \rightarrow 8 \mathrm{MnO}_{2}+6 \mathrm{SO}_{4}^{2-}+2 \mathrm{OH}^{-}$ Here the $\mathrm{Mn}$ changes from $\mathrm{Mn}^{+7}$ to $\mathrm{Mn}^{+4}$ Thus overall change in its oxidation number would be of 3 .

Cu2+ ion is more stable than Cu+ ion due to high charge density on Cu2+ ion, it has higher hydration energy when it bonds to water molecules thus it forms stronger bonds.

$\therefore$ (A) is correct but (R) is not correct.

The indicator 'X' is starch, and it forms a blue colored complex with iodine (compound 'A').

Therefore, the correct option is: Option D: Starch and iodine In this situation, starch is commonly used as an indicator to detect the presence of iodine because starch forms a deep blue complex with iodine.

This color change is visible and helps in monitoring the progress of the reaction involving iodide ions and hydrogen peroxide, which ultimately produces iodine.

The blue color indicates the presence of iodine in the solution.

The correct statements about

are: (A) Mn is tetrahedrally surrounded by oxygen atoms. (C) Contains Mn-O-Mn bridge. Manganese (Mn) in

exhibits its highest oxidation state of +7.

The oxidation state is the charge an atom would have if the electrons in a chemical bond were shared equally.

In

, the manganese atoms are surrounded by oxygen atoms in a tetrahedral arrangement.

This means that each manganese atom is bonded to four oxygen atoms arranged in a pyramid shape.

This tetrahedral arrangement of oxygen atoms results in the highest oxidation state for manganese in

.

also contains Mn-O-Mn bridges, which are bonds between two manganese atoms through a shared oxygen atom.

These bonds help to reinforce the structure of the molecule and contribute to its stability.

It is incorrect to state that Mn is octahedrally surrounded by oxygen atoms or that

contains a Mn-Mn bond.

An octahedral arrangement would consist of six oxygen atoms surrounding a central manganese atom, and a Mn-Mn bond would involve two manganese atoms sharing electrons without an oxygen atom in between.

Neither of these are present in

.

Nd(60) = [Xe] 4f4 5d0 6s2 Nd2+ = [Xe] 4f4 5d0 5s0

$\mathrm{V}_{2} \mathrm{O}_{3}>\mathrm{V}_{2} \mathrm{O}_{4}>\mathrm{V}_{2} \mathrm{O}_{5}$ As positive oxidation state increases acidic nature increases and basic nature decreases.

Nessler's reagent is

$\mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{AgNO}_{3}(\mathrm{aq}) \stackrel{\text{ dil } \mathrm{HNO}_{3}}{\longrightarrow} \underset{(\mathrm{A})}{\mathrm{AgCl}} \downarrow$

$\therefore(\mathrm{A})$ is $\mathrm{AgCl}$ and $(\mathrm{B})$ is $\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right] \mathrm{Cl}$

1.

Eu²⁺: [Xe] 4f⁷ 2.

Yb²⁺: [Xe] 4f¹⁴ Eu²⁺ has a half-filled 4f orbital, while Yb²⁺ has a fully-filled 4f orbital.

Both of these configurations lead to increased stability due to the symmetrical distribution of electrons in the 4f subshell, resulting in higher ionization energies for both Europium (Eu) and Ytterbium (Yb).

Thus, the correct answer is Eu (Europium) and Yb (Ytterbium), as both elements have high third ionization energies due to their half-filled and fully-filled 4f subshell configurations in their divalent cation forms.