D & f- BLOCK ELEMENTS practice · Mixed difficulty

Option A (Ag) is correct because Silver (Ag) belongs to the second transition series (4d series), while Fe, Cu, and V belong to the first transition series (3d series).

Option B (Decomposed in sunlight) is correct because silver nitrate is photosensitive and decomposes when exposed to sunlight.

The correct answer is C (${\mathrm{N}_2}\left( {1\,\mathrm{g}\,\mathrm{atom}} \right) \to {\mathrm{N}_2}\left( {8\,\mathrm{g}\,\mathrm{atom}} \right)$) because compressing a gas reduces its entropy, leading to a negative change in entropy (ΔS).

Option A (Complete $d$-subshell) is correct because Zinc (Zn) has a completely filled d-orbital, which prevents it from exhibiting variable valency.

Option B ($Ag$) is correct because Silver (Ag) has the highest melting point among the given options.

Option B (Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion) is correct because while both statements are true, the larger size of actinoids is not the direct cause of their more varied oxidation states. The availability of 5f, 6d, and 7s electrons for bonding is the primary reason.

Option A (1,3,1) is correct because it correctly balances the given chemical reaction, ensuring the conservation of mass and charge.

Option B is correct. Zinc and mercury have completely filled d-orbitals, preventing them from easily losing or sharing d-electrons to exhibit variable valency.

Option B (92.5 % Ag and 7.5% Cu) is correct because sterling silver, commonly used in case hardening, consists of 92.5% silver and 7.5% copper.

Option D (2) is correct because Tl3+ and Pb4+ are strong oxidizing agents due to the inert pair effect, making them more stable in their lower oxidation states.

Option A (Rare earths) is correct because the 4f subshell is filled in the lanthanide series, also known as rare earth elements.

Option D ($({n}-1) {d}$ ns orbitals) is correct because transition elements lose electrons from both the (n-1)d and ns orbitals during ionization.

Option D (Fe) is correct because mercury does not form amalgams with iron.

The correct answer is A (Blue colour) because the addition of $Na_2O_2$ to an acidified dichromate solution results in the formation of a blue-colored perchromate.

Option A (Ferric chloride) is correct because ferric chloride ($\mathrm{FeC}{\mathrm{l}_3}$) is used as a styptic agent to stop bleeding due to its ability to coagulate blood.

Option B ($\mathrm{Ni}^{2+}$) is correct because, for isoelectronic species, ionic radii decrease with increasing nuclear charge (atomic number).

Option B (${Mn}$) is correct because manganese has an anomalous electronic configuration that results in a lower melting point compared to its neighbors in the periodic table.

Option A (LnC) is correct because lanthanides form carbides of the types Ln3C, Ln2C3, and LnC2, but not LnC.

Option D ($+5$) is correct because vanadium's most stable and common oxidation state is +5, corresponding to the loss of all its valence electrons.

Option D (Transition elements) is correct because transition elements exhibit similar properties both vertically (within a group) and horizontally (across a period) due to their electronic configurations and metallic bonding.

Option A (Mn) is correct because Mn has the highest 3rd ionization energy due to its stable d5 configuration after losing two electrons.

Option A ($\mathrm{Mn}^{2+}$) is correct because in acidic medium, potassium permanganate ($\mathrm{KMnO}_4$) oxidizes hydrogen peroxide ($\mathrm{H}_2 \mathrm{O}_2$) and itself gets reduced to $\mathrm{Mn}^{2+}$.

The correct answer is C (The +4 oxidation state of cerium is not known in solutions) because Ce(IV) is a strong oxidizing agent and exists in solutions.

Option D (Zinc, nickel and copper) is correct because German silver is an alloy primarily composed of zinc, nickel, and copper.

Option A (${Cr}_2\left({SO}_4\right)_3$) is correct because ${K}_2 {Cr}_2 {O}_7$ is reduced to ${Cr}^{3+}$ which forms ${Cr}_2\left({SO}_4\right)_3$ in the presence of sulfate ions.

Option D is correct because $\mathrm{Cu}^{2+}$ is unstable with $\mathrm{I}^{-}$ due to the formation of $\mathrm{CuI}$ and $\mathrm{I}_2$.

Option C (KI) is correct because KI reacts with copper sulfate to form a yellow-brown precipitate that turns white with excess Na2S2O3, and it forms a yellow curdy precipitate with Ag+ that is insoluble in NH4OH.

Option C (2) is correct because the outermost orbit of an Fe atom ($4s^2$) contains 2 electrons.

Option D (Green) is correct because $KMnO_4$ in a basic medium ($KOH$) is reduced to $MnO_4^{2-}$, which is green in color.

Option A (Mixture of potassium ferricyanide and ammonium ferric citrate or ferric oxalate) is correct because blueprint paper is coated with a mixture of these compounds which react upon exposure to UV light to form Prussian blue.