D And F Block Elements – NEET UG ChemistryPractice Questions & PYQs

When copper sulphate reacts with potassium cyanide, cupric ions are initially reduced by cyanide ions to form cuprous cyanide alongside the release of cyanogen gas. This freshly precipitated cuprous cyanide readily dissolves when exposed to an excess of cyanide ions due to the formation of a remarkably stable coordination complex. In this resulting soluble complex, copper exists in the plus one oxidation state coordinated by four cyanide ligands to yield the tetracyanocuprate ion.

In an alkaline environment, permanganate acts as an oxidizing agent and is reduced to manganese dioxide, involving a three-electron change per manganese atom as its oxidation state drops from plus seven to plus four. Concurrently, iodide is oxidized to iodate, which involves a six-electron change per iodine atom as its oxidation state increases from minus one to plus five. Equating the total electrons lost and gained requires two moles of permanganate for every single mole of iodide consumed in the reaction.

Colour in transition metal complexes arises from $d-d$ electronic transitions made possible by the presence of unpaired electrons in the $d$ orbitals. Scandium with an atomic number of $21$ has the neutral configuration $[\mathrm{Ar}] 3d^1 4s^2$, meaning its tripositive ion $\mathrm{Sc}^{3+}$ completely empties its outer shells to achieve a noble gas configuration of $[\mathrm{Ar}] 3d^0$. Lacking any unpaired electrons, this ion cannot absorb visible light via $d-d$ transitions, rendering its aqueous solution entirely colourless unlike the other transition metal ions listed.

Removing a third electron from a divalent transition metal ion forces the disruption of the prevailing electronic configuration, with half-filled and fully-filled subshells requiring exceptionally high energy. Manganese has an atomic number of $25$ with a neutral configuration of $[\mathrm{Ar}] 3d^5 4s^2$, resulting in its $\mathrm{Mn}^{2+}$ ion possessing an extremely stable, half-filled $3d^5$ arrangement. Because pulling an electron away from this symmetrical and stable half-filled shell requires overcoming vast exchange energy, the third ionisation enthalpy peaks sharply at manganese compared to its neighbours.

The inner-transition elements known as lanthanoids occupy a specific block bridging group three and group four within the sixth period of the periodic table. This series spans fourteen distinct elements ranging consecutively from atomic number $58$ to $71$ as electrons are sequentially added to the anti-penultimate $4f$ sublevel. Because of spatial constraints and periodic table layout conventions, these fourteen elements are traditionally displayed in a separate row beneath the main body.

Achieving a uniform $3d^2$ electron configuration requires balancing the atomic number of each transition element with the positive charge of its specific ion to leave exactly two electrons in the $d$ orbital. Titanium with atomic number $22$ forms a divalent ion by losing its outer electrons, vanadium with atomic number $23$ forms a trivalent ion, chromium with atomic number $24$ forms a tetravalent ion, and manganese with atomic number $25$ forms a pentavalent ion. Across this specific sequence, each successive loss of electrons exactly compensates for the increase in nuclear charge, preserving the identical $3d^2$ outer structure.

The basicity of metal monoxides depends directly on the metallic character and the degree of ionic bonding present within the crystal lattice, where lower covalent contributions enhance basic properties. As effective nuclear charge and electronegativity shift across the period, the electron-donating capacity of the metal atom modulates the ionic nature of the oxygen bond. Chromium monoxide exhibits the highest ionic character and electron density release among these specific oxides, leading to the decreasing basic trend of $\mathrm{CrO} > \mathrm{VO} > \mathrm{TiO} > \mathrm{FeO}$.

Catalytic efficiency in transition metals stems primarily from their capacity to adopt multiple oxidation states and provide vacant coordination sites for reactants. By shifting easily between different valence states, these metal ions can readily form unstable intermediate compounds that lower the activation energy of reaction pathways. Furthermore, their large surface area and ability to activate adsorbed molecules significantly accelerate reaction rates in various chemical processes.

Nitric acid acts as a powerful oxidizing agent that readily undergoes reduction itself rather than allowing hydrogen ions to be reduced to hydrogen gas. When a metal reacts with hydrochloric or sulfuric acid, the hydronium ions accept electrons to form hydrogen gas because no stronger oxidizing species is present. However, in the presence of the nitrate ion, this oxidizing agent is preferentially reduced to nitrogen oxides or other products instead of liberating hydrogen.

Ionic size across the lanthanide series progressively diminishes from lanthanum to lutetium due to the steady accumulation of nuclear charge and the poor shielding effect of $4f$ orbitals, a phenomenon known as lanthanide contraction. Consequently, the radius of ${\mathrm{Lu}}^{3+}$ drops below that of ${\mathrm{Eu}}^{3+}$ and ${\mathrm{La}}^{3+}$ because of this cumulative shrinkage. Furthermore, because yttrium resides in the fifth period directly above lanthanum, it possesses one fewer electron shell, making the ${\mathrm{Y}}^{3+}$ ion smaller than the ${\mathrm{La}}^{3+}$ ion. Arranging these trends together yields the exact order of ${\mathrm{Y}}^{3+} < {\mathrm{Lu}}^{3+} < {\mathrm{Eu}}^{3+} < {\mathrm{La}}^{3+}$.

The electronic structure of inner transition elements involves the gradual population of the ante-penultimate electron shell with $f$ electrons while outer shells maintain standard occupancy. For lanthanides, this corresponds to the general formula $((n-2)f^{1-14}(n-1)s^{2}p^{6}d^{0-1}ns^{2})$, where $n$ represents the outermost principal quantum level. The $f$ orbitals accommodate anywhere from one to fourteen electrons, and an occasional electron enters the penultimate $d$ subshell before the outer $s$ orbital is completely filled.

The placement of a transition element in the periodic table is determined by summing the electrons residing in the outermost $s$ subshell and the incomplete penultimate $d$ subshell. For the given configuration of $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{3}4s^{2}$, combining the two electrons in the $4s$ orbital with the three electrons in the $3d$ orbital yields a total of five valence electrons. This sum places the element directly into group five of the $d$-block series.

The maximum oxidation state displayed by a transition metal is governed by the total number of valence electrons available from both its outermost $s$ orbitals and its inner $d$ subshell, which typically corresponds to its group number. Among the given options, manganese occupies group seven and utilizes all seven of its valence electrons to attain a maximum oxidation state of plus seven, such as in $\mathrm{KMnO}_{4}$. This valence capacity surpasses the maximum oxidation states accessible to vanadium, chromium, or iron under standard chemical conditions.

Marine structures and ship hulls submerged in seawater are highly susceptible to corrosion and the accumulation of marine organisms like barnacles. Applying a layer of white tin plating to the iron surface provides a physical barrier that prevents direct contact with corrosive salts and moisture while simultaneously discouraging the settlement of marine growth. This protective coating effectively safeguards the underlying iron structure from structural degradation over extended periods at sea.

The basic strength of lanthanide hydroxides is directly related to the ionic radius of the metal ion, where a larger radius results in a more ionic bond and easier release of hydroxide ions. Moving across the lanthanide series from lanthanum to lutetium, the steady decrease in ionic radius-known as lanthanide contraction-increases the covalent character of the hydroxide bond. This reduction in ionic character causes the basicity of the hydroxides to decrease progressively, making lanthanum hydroxide more basic than lutetium hydroxide.

Transition metal ions typically exhibit color due to $d-d$ electronic transitions that occur when there are partially filled $d$ orbitals with unpaired electrons. Scandium in its neutral ground state has an outer electronic configuration of $3d^1 4s^2$, but the tripositive scandium ion loses all three valence electrons to attain a noble gas configuration with an empty $3d$ subshell. Without any $d$ electrons available to absorb visible light through transitions, the ion transmits all wavelengths and appears entirely colourless in solution.