D & f- BLOCK ELEMENTS practice · Mixed difficulty

Option A (1) is correct because Xe in XeOF4 has sp3d2 hybridization with one lone pair of electrons.

Option A (${Cr}_2 {O}_3$ and ${V}_2 {O}_5$) is correct because both chromium(III) oxide and vanadium(V) oxide can act as both acids and bases, depending on the reaction conditions.

Option D is correct. Cr3+ is the most stable due to its d3 configuration, which leads to a stable t2g configuration in an octahedral field.

Option D (Hf and Zr) is correct because Hafnium (Hf) and Zirconium (Zr) have very similar chemical properties due to their similar ionic radii caused by lanthanide contraction, making them 'chemical twins'.

Option D is correct because the enthalpy of hydration is related to charge density, and Zn+2 has a higher charge density than Mn+2 and Ca+2, leading to greater hydration enthalpy.

Option B ($\mathrm{F}{\mathrm{e}_3}{\mathrm{O}_4}$) is correct because concentrated nitric acid renders iron passive by forming a thin, protective layer of iron(II,III) oxide ($\mathrm{F}{\mathrm{e}_3}{\mathrm{O}_4}$) on the surface.

Option C (5) is incorrect, the correct answer is closer to 4.9, which is approximately 5, but the closest answer is A (4) or D (6). Option B (7) is not correct because the formula is applied incorrectly. The correct answer is closest to A (4).

Option C (A reducing agent in (A) and oxidising agent in (B)) is correct because in reaction A, H2O2 oxidizes HI to I2, acting as an oxidizing agent, and in reaction B, H2O2 reduces HOCl to Cl-, acting as a reducing agent.

Option D (Mercury) is correct because mercury is the essential component of an amalgam, which is an alloy of mercury with another metal.

Option B (Mn) is correct because Mn has the highest 3rd ionization enthalpy due to its stable d5 configuration after the removal of two electrons.

Option C (5) is correct because the amphoteric oxide $\mathrm{V}_2\mathrm{O}_5$ reacts with alkali to form $\mathrm{VO}_4^{3-}$, where the oxidation state of V is +5.

Option B (Distillation in vacuum) is correct because mercury has a relatively low boiling point, and vacuum distillation is an effective method for purifying it.

Option C ($N{i^{2 + }}$) is correct because it has two unpaired electrons, resulting in a magnetic moment of approximately 2.84 BM.

Option C (${Cr}_2 {O}_7{ }^{2-}$ is reduced and ${SnCl}_2$ is oxidised) is correct because chromium is reduced (oxidation state decreases) and tin is oxidized (oxidation state increases).

Option B ($30 \%$ lead, $70 \%$ tin) is correct because solder is an alloy primarily composed of tin and lead, with tin typically being the larger component.

Option B (Paramagnetic) is correct because transition metals generally have unpaired electrons in their d-orbitals, which causes them to exhibit paramagnetism.

Option D (Lanthanoid contraction) is correct because the lanthanoid contraction causes the atomic radii of Zr and Hf to be very similar due to the poor shielding of the 4f electrons.

Option A (Both assertion and reason are correct statements, and reason is the correct explanation of the assertion) is correct because oxygen's ability to form $p\pi-d\pi$ bonds stabilizes high oxidation states of d-block metals in oxides and oxoanions.

Option D ($\mathrm{K}_2 \mathrm{MnO}_4, \mathrm{MnO}_2, \mathrm{O}_2$) is correct because it represents the products formed during the thermal decomposition of $\mathrm{KMnO}_4$.

Option B (Copper containing 2% impurity) is correct because blister copper is approximately 98% pure copper with about 2% impurities.

Option D (Potash alum) is correct because double salts are formed by the combination of two different salts, and potash alum (KAl(SO₄)₂·12H₂O) fits this description.

Option B (Black) is correct because copper sulfide (CuS) is a black solid.

Option D ($\mathrm{H}{\mathrm{g}_2}\mathrm{C}{\mathrm{l}_2}$) is correct because in this compound, mercury exists as Hg2²⁺, giving it an oxidation state of +1.

Option D (All of these) is correct because ferrous sulfate decomposes upon heating to produce ferric oxide, sulfur dioxide, and sulfur trioxide.

Option D (All of these) is correct because $\mathrm{MnO}_2$, $\mathrm{Zn}$, and $\mathrm{Ni} / \mathrm{Cd}$ are all used in different types of batteries.

Option A ($(n-1) d^{10} n s^2$) is correct because Zinc (Zn), Cadmium (Cd), and Mercury (Hg) are group 12 elements and have a completely filled d-orbital and two electrons in the s-orbital.

Option C is correct because the second ionization enthalpy of Cr is highest due to the stable half-filled d5 configuration after the first ionization, followed by Mn, V, and Ti.

Option C (Red oxide of copper) is correct because $\mathrm{C}{\mathrm{u}_2}\mathrm{O}$ is commonly known as cuprous oxide or red oxide of copper.

Option A (Zn) is correct because Zinc is more reactive than Gold and can displace it from the aurocyanide complex.

Option D ($95 \%$) is correct because mischmetall contains approximately 95% lanthanoids.