D & f- BLOCK ELEMENTS practice · Mixed difficulty

Option A ($\mathrm{Cr}{\mathrm{O}_5} + {\mathrm{H}_2}\mathrm{O}$) is the correct product because chromic acid reacts with hydrogen peroxide to form chromium pentoxide ($\mathrm{Cr}{\mathrm{O}_5}$), which has a butterfly structure.

The nature of both the ions determines the color of an electrolyte solution because the electronic transitions within both cations and anions contribute to light absorption.

Option A (Have variable valency) is correct because actinides exhibit multiple oxidation states due to the small energy difference between the 5f, 6d, and 7s orbitals.

Option A (Convex upwards) is correct because mercury has a stronger cohesive force than adhesive force with glass, resulting in a convex meniscus.

Option C (To form complexes which are water soluble) is correct because gold is extracted by forming a water-soluble complex with cyanide ions in the hydrometallurgical process.

Option C (Ni) is correct because Nickel has the highest density among the given options.

Option D (82%) is correct because it is the calculated percentage of MnO2 in the pyrolusite ore based on the given reactions and titration data.

Option B ($\mathrm{HgC}{\mathrm{l}_2}$) is correct because mercuric chloride is known to sublime upon heating, meaning it transitions directly from solid to gas phase.

Option D (Cu) shows a jump in the second ionization potential because removing the second electron disrupts a stable, fully-filled d-orbital ($d^{10}$).

Option C (Assertion is correct, but reason is wrong statement) is correct because manganate ion ($MnO_4^{2-}$) has one unpaired electron and is paramagnetic, while permanganate ion ($MnO_4^{-}$) has no unpaired electrons and is diamagnetic. The reason states that manganate ion has two unpaired electrons, which is incorrect.

Option B (Case hardening) is correct because case hardening involves heating steel in a carbon-rich environment to create a hard outer layer.

Option B ($\mathrm{Hg}{\mathrm{I}_2}$) is correct because it matches the description of a red solid that dissolves in KI solution, and decomposes upon heating to release violet iodine fumes and metallic mercury.

The electronic configuration is $3 d^5 4 s^1$ because a magnetic moment of 6.92 BM corresponds to approximately 6 unpaired electrons, which is achieved with this configuration.

Titanium dioxide ($Ti{O_2}sol$) is a positive sol because metal oxides generally form positive sols.

Option D ($\mathrm{[Xe]4}{\mathrm{f}^\mathrm{7}}\mathrm{6}{\mathrm{s}^\mathrm{2}},\,\mathrm{[Xe]4}{\mathrm{f}^\mathrm{7}}\mathrm{5}{\mathrm{d}^\mathrm{1}}\mathrm{6}{\mathrm{s}^\mathrm{2}}\,\mathrm{and}\,\mathrm{[Xe]4}{\mathrm{f}^9}\mathrm{6}{\mathrm{s}^\mathrm{2}}$) is correct because it represents the correct electronic configurations of Eu, Gd, and Tb, considering Hund's rule and stability of half-filled orbitals.

Option B ($\mathrm{KMn}{\mathrm{O}_4}$) is the correct answer because Mn is already in its highest oxidation state (+7) and cannot be further oxidized by ${\mathrm{O}_3}$.

Option A (Formation of $Fe{\left( {CO} \right)_5}$ from $Fe$) is correct because the oxidation state of iron remains 0 in this process.

Option C is correct because copper sulphate does not react with KCl to form Cu2Cl2. The reaction is not thermodynamically favorable.

Option D (They are placed from third period to sixth period in modern periodic table) is wrong because d-block elements are placed from fourth period to seventh period in the modern periodic table.

Option C (Carnallite) is correct because Carnallite is a hydrated double chloride of potassium and magnesium, and does not contain iron.

Option B (White) is correct because pure zinc sulfide (ZnS) is white in color.

Option D (All of these) is correct because ferrous sulphate ($FeSO_4$) decomposes upon heating to produce $Fe_2O_3$, $SO_2$, and $SO_3$.

Option A (3, 5, 4 and 1) is correct because the number of electrons transferred corresponds to the change in oxidation state of Mn in each reduction.

Option B ($Z{n^{2 + }}$) is correct because $Z{n^{2 + }}$ has a $d^{10}$ configuration, meaning all d-orbitals are fully filled, and thus no d-d transitions are possible, resulting in a colorless solution.

Option A (Annealing) is correct because annealing involves heating steel to a high temperature and then slowly cooling it to relieve internal stresses and soften the metal.

Option B (Atomic as well as ${M^{3 + }}$ radii) is correct because lanthanoid contraction affects both the atomic and ionic radii (specifically ${M^{3 + }}$) of the lanthanides.

Option D (Partly iron carbide and partly as graphite) is correct because wrought iron contains a very small amount of carbon, which exists both as iron carbide ($Fe_3C$) and as free graphite.

Option C (Red) is incorrect. Option D (Brown) is correct because the reaction between $\mathrm{C}{\mathrm{u}^{2 + }}$ and ${\mathrm{K}_4}\mathrm{Fe}{\left( {\mathrm{CN}} \right)_6}$ produces a brown precipitate of copper ferrocyanide.

Option B is correct because SbF5 acts as a fluoride ion acceptor, oxidizing Mn(IV) in K2MnF6 to Mn(V) and releasing F2.

Option B ($\frac{{2(x-1)}}{{2-x}}$) is correct because it represents the ratio of M+1 to M+2 ions based on the given formula MxS and charge balance.