D & f- BLOCK ELEMENTS practice · Mixed difficulty

Option D ($S{r^{2 + }}$) is correct because among the given alkaline earth metal ions, $Sr^{2+}$ has the lowest charge density due to its larger size, leading to the highest ionic mobility in aqueous solution.

Option A ($d^5$) is correct because it has the maximum number of unpaired electrons (5), leading to the highest magnetic moment.

The correct answer is B (Both A and R are true but R is not the correct explanation of A) because while both statements are true, the difficulty in separating Zr and Hf is due to their similar ionic radii, not just their group placement.

Option A (25) is correct because the atomic number is calculated by adding the electrons in the [Ar] core and the d⁴ configuration in the +3 oxidation state, then adding back the 3 electrons lost during ionization.

Option D (Ag) is correct because Argentite is a silver sulfide mineral with the chemical formula Ag₂S, making it an ore of silver.

Option D ($3d^5 4s^2$) is correct because it corresponds to Manganese (Mn), which exhibits the largest number of oxidation states (up to +7).

Option D (C and D only) is correct because $Ce^{4+}$ and $Tb^{4+}$ are strong oxidizing agents as they tend to gain electrons to achieve the more stable +3 oxidation state.

Option C (Hg) is correct because mercury (Hg) has the lowest melting point among the given metals, at -38.9°C.

Option A (first transition series) is correct because Iron (Fe) is the most abundant d-block metal and it belongs to the first transition series.

Option B (Silver reacts with sulphur of egg yolk forming black $\mathrm{A}{\mathrm{g}_2}\mathrm{S}$) is correct because egg yolk and mustard contain sulfur compounds that react with silver to form silver sulfide, which is black.

Option A (${Mn}^{+2}$) is correct because it has the maximum number of unpaired electrons, leading to the highest magnetic moment.

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.

Option D (Natural radioactivity) is correct because natural radioactivity is a property of unstable nuclei, not a general characteristic of transition elements.

Option A (They retain metallic conductivity) is correct because interstitial compounds of transition metals retain metallic conductivity due to the presence of metallic bonding.

Option A ($\mathrm{M}{\mathrm{n}^{2 + }}$) is correct because it has 5 unpaired electrons, resulting in a magnetic moment of approximately 5.93 BM.

Option B (Colloidal gold solution) is correct because Purple of Cassius is the trade name for gold sol in water.

Option C (Carnallite) is correct because Carnallite ($\mathrm{KCl}.\mathrm{MgCl}_2.6{\mathrm{H}_2}\mathrm{O}$) does not contain iron.

Option A ($\left[\mathrm{MnO}_4\right]^{2-}$) is both paramagnetic and colored because it has one unpaired electron and undergoes d-d transition.

Option B (Increases) is correct because increasing the carbon content in iron increases its hardness due to the formation of iron carbide (cementite), which is a hard and brittle phase.

Option A (Reductant and oxidant) is correct because $C{r^{2 + }}$ tends to oxidize to $C{r^{3 + }}$ (reductant) and $M{n^{2 + }}$ can be oxidized further but also can be reduced (oxidant).

Option B (Casiterite) is correct because Casiterite is an ore of tin ($Sn$), not iron.

Option C (Interstitial hydrides) is correct because d-block elements form metallic hydrides where hydrogen atoms occupy the interstitial sites in the metal lattice.

Option A (C) is correct because carbon (in the form of coke) is used as the reducing agent in the extraction of iron from iron oxide.

Option A (Fe, Co, Ni) is correct because Iron (Fe), Cobalt (Co), and Nickel (Ni) are the metals that are known as ferrous metals.

Option D is correct. Anhydrous copper sulfate dissolves in water to form a blue solution due to the formation of hydrated copper ions.

Option A is correct because both statements are true: Cr²⁺ is a stronger reducing agent than Fe²⁺, and in aqueous medium, d³ is more stable than d⁵ configuration.

Option A (Fe) is correct because Iron (Fe) is the most abundant transition element in the Earth's crust.

Option D (A mixture of hydrochloric and nitric acids) is correct because platinum dissolves in aqua regia, which is a mixture of concentrated hydrochloric and nitric acids.

Option D (V) is correct because Vanadium (V) is not ferromagnetic, while Iron (Fe), Cobalt (Co), and Nickel (Ni) are well-known ferromagnetic materials.

Option D ($\mathrm{H}{\mathrm{g}_2}$) is correct because the grey color observed is due to the formation of metallic mercury ($\mathrm{Hg}$) via reduction of $\mathrm{HgC}{\mathrm{l}_2}$ by excess $\mathrm{SnC}{\mathrm{l}_2}$.