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

Elements are categorized into specific blocks and series within the periodic table according to the subshell that receives their final valence electron. Cerium represents an inner-transition metal belonging to the lanthanoid series, aligning A with I. Manganese features its differentiating electron in the d-subshell, classifying it as a d-block element and connecting B with IV. Arsenic accommodates its outermost electrons in the p-orbital, making it a p-block element that pairs C with II. Cesium finishes filling the s-orbital as an alkali metal, establishing it as an s-block element and linking D with III.

The most stable and characteristic oxidation state across the lanthanoid series is $+3$, which heavily dictates the redox behavior of these elements. Ions like $\mathrm{Ce}^{4+}$ and $\mathrm{Tb}^{4+}$ have a strong thermodynamic tendency to gain an electron to attain this stable configuration. Because they undergo reduction easily themselves, they function effectively as strong oxidizing agents in chemical reactions.

A lanthanoid ion exhibits diamagnetic behavior only when all of its f-orbitals are either entirely vacant or completely filled, leaving no unpaired electrons to contribute to a net magnetic moment. Cerium in the $+4$ oxidation state adopts a noble gas core configuration with a completely empty $4f^0$ shell, while ytterbium in the $+2$ oxidation state features a fully saturated $4f^{14}$ configuration. Because both species completely lack unpaired electrons, they are unable to respond to magnetic fields and remain strictly diamagnetic.

The spin-only magnetic moment of a transition metal ion is determined by the total number of unpaired electrons housed within its d-orbitals, quantified by the expression $\mu = \sqrt{n(n+2)}$. Evaluating the electronic configurations reveals that $\mathrm{Cr}^{2+}$ has four unpaired electrons in its $3d^4$ shell, while $\mathrm{Fe}^{2+}$ also possesses four unpaired electrons in its $3d^6$ configuration. Applying the magnetic moment formula yields an identical value for both ions since they share the same number of unpaired electrons. In contrast, $\mathrm{Ti}^{3+}$ contains one, $\mathrm{Mn}^{2+}$ has five, and $\mathrm{Sc}^{3+}$ holds zero unpaired electrons, resulting in distinct magnetic moments for each of them.

Small size and high electronegativity of nitrogen atoms cause intense inter-electronic repulsion between their non-bonding lone pairs when forced into a single bond. This strong repulsion drastically weakens the $\mathrm{N}-\mathrm{N}$ single bond compared to the significantly larger phosphorus atoms in a $\mathrm{P}-\mathrm{P}$ single bond. Meanwhile, tertiary phosphines and arsines possess vacant d-orbitals that readily overlap with filled metal d-orbitals to establish robust multiple bonding characteristics, while nitrogen lacks accessible d-orbitals for such bonding.

Moving horizontally across a transition series results in a progressive rise in nuclear charge because each added proton significantly pulls the electron cloud closer to the nucleus. Although electrons enter the inner d-subshells simultaneously, their mutual shielding is relatively poor, meaning the effective nuclear charge experienced by the outermost valence electrons steadily climbs. This continuous enhancement in attractive pull requires progressively greater energy to remove an electron, explaining the overall upward trend in ionization enthalpy alongside minor fluctuations.

The overall stability of an ion in an aqueous medium is determined by a delicate balance between the energy required to form the gaseous ion and the energy released when the gaseous ion gets hydrated by water molecules. Although the second ionisation enthalpy of copper is quite high, the hydration enthalpy of the doubly charged copper ion is exceptionally negative, releasing a vast amount of energy. This massive release of hydration energy more than compensates for the high energy cost of removing the second electron, rendering $\mathrm{Cu}^{2+}$ salts significantly more stable than $\mathrm{Cu}^{+}$salts in water.

Metal oxide acidity and basicity are intrinsically linked to the oxidation state of the metal, where higher oxidation states lead to increased covalent character and enhanced acidity. As the oxidation state of vanadium increases from $\mathrm{V}_2\mathrm{O}_3$ to $\mathrm{V}_2\mathrm{O}_5$, the acidic character increases rather than the basic character, rendering statement C incorrect. Furthermore, vanadium(IV) oxide dissolves in acidic solutions to yield vanadyl ions ($\mathrm{VO}^{2+}$) rather than $\mathrm{VO}_4^{3-}$ salts, making statement D incorrect as well. Since statements A, B, and E represent accurate chemical behaviors of transition metal oxides, statements C and D are the only incorrect ones.

Ionisation enthalpy depends on how tightly the outermost electrons are held by the nuclear charge, which is heavily influenced by the shielding effect of inner orbitals. In actinoids, electrons enter the $5f$ subshell, which is more diffuse and exhibits a greater shielding effect compared to the contracted $4f$ orbitals of lanthanoids. Consequently, the effective nuclear charge experienced by the valence electrons in early actinoids is lower than in early lanthanoids, making it easier to remove an electron. Thus, both the assertion regarding lower ionisation enthalpies and the reason concerning $5f$ orbital shielding are correct, with the shielding directly explaining the reduced ionisation energy.

The radius of an orbit in a hydrogen-like species follows the proportionality $r_n \propto \frac{n^2}{Z}$. Setting up the comparative ratio for the third orbit of the lithium ion and the second orbit of the helium ion relates their respective principal quantum numbers and atomic numbers. Substituting $n = 3$ and $Z = 3$ for lithium along with $n = 2$ and $Z = 2$ for helium into this expression yields $105.8 \times \frac{3^2}{3} \times \frac{2}{2^2}$. Evaluating this calculation results in a final radius of $158.7$ pm.

The spontaneity of a redox reaction is determined by the standard cell potential, calculated as the difference between the reduction potential of the cathode and the anode, where a positive value indicates a spontaneous process ($\Delta G^\circ < 0$). Reversing the given reduction potential for the manganese half-reaction yields an oxidation potential of +1.51 V for the permanganate reduction half-cell acting as the cathode. Subtracting the standard reduction potential of the oxygen half-reaction from the permanganate half-reaction gives a net cell potential of +0.287 V. Since the resulting standard cell potential is positive, the reaction is thermodynamically spontaneous, meaning permanganate successfully liberates oxygen from water in acidic conditions.

Inner transition elements experience a progressive contraction across a period due to the imperfect shielding of nuclear charge by interior electrons. The orbitals involved in actinoids are more spatially extended and diffuse than those in lanthanoids, leading to a significantly reduced ability to screen outer electrons from the nucleus. Consequently, this exceptionally poor shielding by the 5f subshell causes the effective nuclear charge to pull the valence shell inward much more strongly, making the actinoid contraction greater and more gradual than the lanthanoid contraction.

Oxidation states and magnetic behavior in transition metal ions depend heavily on electronic configurations and exchange energy stability. Chromium(II) possesses a $3d^4$ configuration and readily loses an electron to attain the stable half-filled $t_{2g}^3$ configuration of chromium(III), making it a strong reducing agent rather than an oxidizing one. Conversely, manganese(III) tends to gain an electron to form the exceptionally stable half-filled $3d^5$ configuration of manganese(II), rendering it strongly oxidizing. Meanwhile, scandium(III) has a completely empty d-subshell with a noble gas configuration of argon, meaning it contains no unpaired electrons and is therefore diamagnetic, causing its compounds to be repelled by a magnetic field.

The stability of a half-filled subshell is exceptionally high due to the symmetrical distribution of electrons and maximum exchange energy, which lowers the overall energy state of the system. Gadolinium has an atomic configuration of $[\mathrm{Xe}] 4f^7 5d^1 6s^2$, meaning the removal of the third electron involves extracting a single $5d$ electron and leaving behind a perfectly half-filled $4f^7$ core. This resulting half-filled configuration is exceptionally stable owing to high exchange enthalpy, requiring less energy than expected for the third ionisation step.

A metal displacement reaction occurs when a metal in a compound is replaced by another metal in its uncombined elemental state based on their relative reactivities. Examining the given choices, options involving thermal decomposition of salts or reactions releasing hydrogen gas do not fit this specific criterion. Specifically, the reaction featuring chromium oxide and aluminum involves a more reactive metal displacing a less reactive metal from its oxide, successfully producing elemental chromium.

Inner transition series elements exhibit distinct periodic trends and physical characteristics dictated by their electron configurations. Trivalent lanthanoid ions typically contain unpaired electrons that undergo f-f electronic transitions when interacting with light, resulting in characteristic coloration in both solid and aqueous states rather than being colorless. Furthermore, lanthanoids possess typical metallic bonding making them effective conductors of heat and electricity, while actinoids display high chemical reactivity, particularly when divided into fine powders. The actinoid contraction is also more pronounced element-by-element than the lanthanoid contraction due to the even poorer screening ability of 5f orbitals.

The filling of 4f orbitals prior to the third transition series introduces poor shielding of nuclear charge by inner electrons, leading to a steady contraction in atomic sizes known as the lanthanoid contraction. This cumulative pulling effect counteracts the expected regular increase in atomic and ionic radii down a group. As a result, zirconium and hafnium end up possessing nearly identical dimensions despite belonging to different periods.

The oxidation state of a transition metal in an oxoanion is determined by balancing its charge against the known negative charge of each coordinated oxygen atom. For both chromate ($\mathrm{CrO}_4^{2-}$) and dichromate ($\mathrm{Cr}_2\mathrm{O}_7^{2-}$), setting up the algebraic sum yields an oxidation state of +6 for chromium in both species. Therefore, claiming that their oxidation states differ is factually incorrect, making this statement the right choice among the options.

Vanadium pentoxide ($\mathrm{V}_2\mathrm{O}_5$) acts as the catalyst for oxidizing sulfur dioxide to sulfur trioxide during the industrial manufacture of sulfuric acid via the Contact process. The mixture of titanium tetrachloride and trimethylaluminum constitutes a Ziegler-Natta catalyst, which effectively drives the polymerization of ethylene. Additionally, palladium chloride ($\mathrm{PdCl}_2$) facilitates the Wacker process for converting ethyne to ethanal, and nickel complexes are employed in the polymerization of alkynes.

Permanganate acts as an oxidizing agent whose reduction product and oxidation capacity depend strongly on the pH of canyons and reaction mediums. In neutral or mildly alkaline solutions, potassium permanganate oxidizes iodide ions up to the iodate state rather than stopping at free iodine. Balancing this redox process yields manganese dioxide alongside potassium iodate, identifying the conversion product definitively.