D & f- BLOCK ELEMENTS practice · Medium

The correct answer is From a mixed precipitate of AgCl and AgI, ammonia solution dissolves only AgCl because AgCl forms a soluble complex with ammonia, while AgI is less soluble and does not dissolve.

Option C (Their ability to adopt variable oxidation states) is correct because transition metals' catalytic activity stems from their ability to change oxidation states.

Option A is correct because back bonding from the metal to CO ligands weakens the C-O bond, making it longer.

Option C (Both (1) & (2)) is correct because the reaction between $\mathrm{KMnO}_4$ and $\mathrm{HCl}$ produces $\mathrm{KCl}$, $\mathrm{MnCl}_2$, $\mathrm{H}_2\mathrm{O}$, and $\mathrm{Cl}_2$.

Option A (Electronic configuration) is correct because lanthanoids and actinoids share similarities in their electronic configurations, particularly in the filling of f-orbitals.

Option B ($\mathrm{Fe}{\mathrm{S}_2}$) is correct because Fool's gold is the common name for iron pyrite, which has the formula $FeS_2$.

Option B is correct. Actinoids can exhibit +4 oxidation states more readily than lanthanoids, but the reason provided, while true, is not the correct explanation.

Option D (1) is correct because each carbon atom in graphite is sp2 hybridized, leaving one free electron.

Option A ($\mathrm{A}=\mathrm{Na}_2 \mathrm{CrO}_4, \mathrm{~B}=\mathrm{CrO}_5$) is correct because the reaction of chromyl chloride with NaOH forms sodium chromate, and the subsequent reaction with HCl and H2O2 forms chromium pentoxide.

Option A is correct because the deep blue complex formed is cuprammonium sulphate, which has the formula [Cu(NH3)4]SO4.

Option B is correct because the $3d^5$ configuration has the maximum number of unpaired electrons (5), leading to the highest magnetic moment.

Option B (If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion) is correct because both the assertion about Europium(II) stability and the reason about Cerium salts being catalysts are true, but unrelated.

Option D (At high temperature and low pressure) is correct because at high temperature and low pressure, the intermolecular forces and volume occupied by gas molecules become negligible, thus behaving like an ideal gas.

Option D (Only ${Zn}^{2+}$ is colourless and ${Ni}^{2+}$ and ${Cr}^{3+}$ are coloured) is correct because ${Zn}^{2+}$ has a completely filled d-orbital configuration ($d^{10}$), making it colorless, while ${Ni}^{2+}$ ($d^8$) and ${Cr}^{3+}$ ($d^3$) have unpaired d-electrons and exhibit d-d transitions, resulting in color.

Option A (${\mathrm{C}_2}\mathrm{H}_5^ -$) is the strongest base because it is the conjugate base of the weakest acid (ethane).

Option B (Is not blackened by the action of ${\mathrm{H}_2}\mathrm{S}$) is correct because zinc white does not react with hydrogen sulfide to form a black sulfide, unlike lead white.

Option A ($\mathrm{V}_2 \mathrm{O}_5, \mathrm{Cr}_2 \mathrm{O}_3$) is correct because both vanadium(V) oxide and chromium(III) oxide can react with both acids and bases.

Option B is false because the ionic radii of trivalent lanthanides steadily decrease with increasing atomic number due to lanthanide contraction.

Option D (Lu) is correct because Lutetium (Lu) is a lanthanoid element, belonging to the series from atomic number 58 to 71.

Option A (Both A and R are true and R is the correct explanation of A) is correct because the degree of complex formation depends on the charge and size of the ion, and actinides form complexes with pi-bonding ligands.

Option D (A white precipitate of $\mathrm{Cu}_2 \mathrm{I}_2$ in $\mathrm{I}$, an orange precipitate dissolving to $\mathrm{HgI}_4^{2-}$ in $\mathrm{II}$ and a yellow precipitate of $\mathrm{PbI}_2$ in III) is correct because it accurately describes the reactions and precipitates formed when excess KI is added to CuSO4, HgCl2, and Pb(NO3)2 solutions.

Option C (34.08 g) is correct because it is calculated by determining the limiting reagent and then using stoichiometry to find the mass of chlorine produced.

Option A is correct. Misch metal is an alloy of lanthanoid metal (~95%) and iron (~5%) with traces of S, C, Ca, and Al. It's used in Mg-based alloys for various applications, including making lighter flints, but not aeroplane bodies.

Option A is correct because La(OH)3 is indeed the least basic among lanthanide hydroxides, and Ce4+ can act as an oxidizing agent.

Option A is correct because both statements are true, and the lanthanoid contraction directly explains the similar radii of the 2nd and 3rd transition series.

Option A (3) is correct because the number of unpaired electrons (n) is calculated using the formula $\mathrm{\mu } = \sqrt {n(n + 2)}$ and solving for n when $\mathrm{\mu } = \sqrt{15}$.

Option D ($\mathrm{CuSO}_4.5 \mathrm{H}_2 \mathrm{O}$) is correct because the blue color of copper sulfate pentahydrate arises from d-d transitions in the copper(II) ion due to the ligand field splitting caused by water molecules.

Option B ($\mu=g \sqrt{{J}({J}+1)}$) is correct because the magnetic moment of lanthanide ions is determined using this formula, which accounts for both spin and orbital angular momentum contributions.

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.

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.