NEET The d- and f-Block Elements Previous Year Questions with Solutions

Transition elements have exceptionally high melting and boiling points compared to s-block elements in the same period. For example, Vanadium melts at 1910 °C and Chromium at 1907 °C.

• • Assertion A is factually correct.

Metallic bonding strength is directly proportional to the number of valence electrons per atom available for delocalization throughout the crystal lattice. In 3d transition metals, the energies of 3d and 4s subshells are very close, allowing both 4s and (n-1)d electrons (especially unpaired d-electrons) to participate in covalent interatomic bonding and metallic lattice cohesion.

• • Reason R is factually correct.

Because the interatomic bonding is reinforced by the dual participation of (n-1)d and ns electrons, immense thermal energy is required to break the lattice bonds. Thus, Reason R is the precise, direct scientific explanation of Assertion A.

• • Matches Option (1).

Xenon (Xe) has 54 electrons. The ground-state configurations of the neutral atoms are: - Ce (58): [Xe] 4f¹ 5d¹ 6s² - Nd (60): [Xe] 4f⁴ 6s² - Tb (65): [Xe] 4f⁹ 6s² - Ho (67): [Xe] 4f¹¹ 6s²

• • Trivalent ionization removes outer 6s and 5d electrons first, then 4f.

For all lanthanoids, forming the stable Ln³⁺ cation removes 3 valence electrons (the two 6s electrons and one 5d or 4f electron): - Ce³⁺: 58 - 3 - 54 = 1 electron in 4f ⇒ [Xe] 4f¹ - Nd³⁺: 60 - 3 - 54 = 3 electrons in 4f ⇒ [Xe] 4f³ - Tb³⁺: 65 - 3 - 54 = 8 electrons in 4f ⇒ [Xe] 4f⁸ - Ho³⁺: 67 - 3 - 54 = 10 electrons in 4f ⇒ [Xe] 4f¹⁰

• • Total electrons in 4f subshell established.

The 4f subshell has 7 spatial orbitals (maximum 14 electrons): 1. Ce³⁺ (4f¹): 1 unpaired electron. 2. Nd³⁺ (4f³): 3 unpaired electrons. 3. Tb³⁺ (4f⁸): 7 single + 1 paired = 6 unpaired electrons. 4. Ho³⁺ (4f¹⁰): 7 orbitals with 10 electrons means 3 orbitals are doubly occupied and 4 orbitals are singly occupied. Number of unpaired electrons = 7 - 3 = 4.

• • Ho³⁺ has exactly 4 unpaired electrons.
• • Matches Option (4).

Potassium permanganate decomposes upon heating above 513 K (240 °C) in a solid-state disproportionation/decomposition reaction:

• • Products are potassium manganate (K₂MnO₄), black manganese dioxide (MnO₂), and oxygen gas (O₂).

1. Oxidation state: In K₂MnO₄, potassium is +1 and oxygen is -2. 2(+1) + x + 4(-2) = 0 ⇒ x = +6. 2. Electronic configuration: Mn (Z = 25) is [Ar] 3d⁵ 4s². Mn(VI) is [Ar] 3d¹. 3. Magnetism: The single unpaired electron in the 3d orbital makes the manganate ion [MnO₄]²⁻ PARAMAGNETIC (μ ≈ 1.73 BM). 4. Color: The tetraoxo anion [MnO₄]²⁻ absorbs light via ligand-to-metal charge transfer (LMCT) and d-d transitions, appearing intensely DARK GREEN.

• • Color: Green.
• • Magnetism: Paramagnetic.
• • Formula: K₂MnO₄.

In contrast, KMnO₄ has manganese in the +7 oxidation state (3d⁰), which has no unpaired d-electrons and is diamagnetic (dark purple due to LMCT).

• • Matches Option (1).

Analyze the physical and chemical definition of ferromagnetism from the NCERT text:

• • Paramagnetic substances are weakly attracted by a magnetic field due to unpaired electrons.
• • When exchange coupling forces magnetic dipoles of adjacent atoms to align spontaneously in large macroscopic domains, the substance exhibits ferromagnetism (e.g., Fe, Co, Ni).
• • NCERT d- and f-Block Elements explicitly states: "Ferromagnetism is an extreme form of paramagnetism."
• • Statement I is unequivocally TRUE.

Write the electronic configuration of chromium and its divalent cation:

• • According to Hund’s rule of maximum multiplicity, the 4 electrons occupy 4 degenerate d-orbitals singly.
• • Number of unpaired electrons in Cr²⁺: n = 4.

Write the electronic configuration of neodymium and its trivalent lanthanoid cation:

• • The 3 electrons occupy 3 of the seven degenerate 4f-orbitals singly.
• • Number of unpaired electrons in Nd³⁺: n = 3.
• • Since 4 ≠ 3, Statement II is FALSE.
• • Statement I is true but Statement II is false, matching Option (3).

For transition metal ions of the first series, orbital angular momentum is largely quenched by the crystal/ligand field. The magnetic moment depends exclusively on the number of unpaired electrons n:

Determining the electronic configurations of the 3d transition metal cations: - Ti3+ (Z = 22): [Ar] 3d1 -> n = 1 unpaired electron (μ = √3 = 1.73 BM) - Cr2+ (Z = 24): [Ar] 3d4 -> n = 4 unpaired electrons (μ = √24 = 4.90 BM) - Mn2+ (Z = 25): [Ar] 3d5 -> n = 5 unpaired electrons (μ = √35 = 5.92 BM) - Fe2+ (Z = 26): [Ar] 3d6 -> 5 + 1 paired -> n = 4 unpaired electrons (μ = √24 = 4.90 BM) - Sc3+ (Z = 21): [Ar] 3d0 -> n = 0 unpaired electrons (μ = 0 BM, diamagnetic)

Manganese has atomic number Z = 25 with neutral ground-state configuration [Ar] 3d5 4s2. Removing valence electrons gives: - Mn3+: [Ar] 3d4 (partially filled, relatively less stable) - Mn2+: [Ar] 3d5 (symmetrically half-filled, possessing maximal exchange energy and high stabilization)

Standard reduction potential E° is related to the standard Gibbs free energy change by ΔG° = -nFE°. Because the conversion of d4 into the half-filled d5 configuration is energetically extremely favorable, ΔG° is large and negative, rendering E° unusually high and positive (+1.57 V):

For chromium: Cr3+ is [Ar] 3d3 (stable half-filled t2g³ level in octahedral fields), so reducing it to Cr2+ (3d4) is unfavorable (E° = -0.41 V). For iron: Fe3+ is already 3d5, so reducing it to Fe2+ (3d6) loses the d5 stability (E° = +0.77 V). Only in the manganese couple does reduction produce the stable d5 configuration from d4.

Lanthanoids have valence electrons filling the 4f subshell, with neutral ground states: - Cerium (Ce, Z = 58): [Xe] 4f¹ 5d¹ 6s² - Ytterbium (Yb, Z = 70): [Xe] 4f¹⁴ 6s² - Europium (Eu, Z = 63): [Xe] 4f⁷ 6s² - Gadolinium (Gd, Z = 64): [Xe] 4f⁷ 5d¹ 6s²

Removing electrons from the outermost orbitals: 1. Ce⁴⁺: Loses 6s², 5d¹, and 4f¹ ⟹ [Xe] 4f⁰. Possesses an empty 4f subshell (n = 0 unpaired electrons) ⟹ Diamagnetic. 2. Yb²⁺: Loses 6s² ⟹ [Xe] 4f¹⁴. Possesses a completely filled 4f subshell (n = 0 unpaired electrons) ⟹ Diamagnetic. 3. Gd³⁺: Loses 6s² and 5d¹ ⟹ [Xe] 4f⁷ (n = 7 unpaired electrons) ⟹ Strongly paramagnetic. 4. Eu³⁺: Loses 6s² and one 4f ⟹ [Xe] 4f⁶ (n = 6 unpaired electrons) ⟹ Paramagnetic. 5. Eu²⁺: Loses 6s² ⟹ [Xe] 4f⁷ (n = 7 unpaired electrons) ⟹ Paramagnetic.

Ce⁴⁺ (4f⁰) and Yb²⁺ (4f¹⁴) both possess zero unpaired f-electrons. Consequently, both are diamagnetic, matching Option (3).

In the isolated gaseous state, copper exhibits the following configurations:

The stability of an oxidation state in aqueous solution is dictated by the total standard electrode potential (E°), which is governed by the Born-Haber energy cycle involving sublimation enthalpy, ionization enthalpy, and hydration enthalpy:

Although removing a second electron from copper requires a very large amount of energy (high second ionization enthalpy IE₂ = 1958 kJ mol⁻¹ to break the stable 3d¹⁰ shell), the much smaller ionic radius and higher charge of Cu²⁺ gives it a massive negative hydration enthalpy (-2121 kJ mol⁻¹ compared to only -593 kJ mol⁻¹ for Cu⁺):

Because the hydration enthalpy heavily tilts the free energy balance toward Cu²⁺, Cu⁺ ions are unstable in water and undergo rapid disproportionation:

Statement A: Transition metals (except scandium, which only forms Sc₂O₃) form ionic MO oxides (TiO to NiO) — verbatim NCERT (TRUE). Statement B: Highest oxidation states matching group number occur from Group 3 (Sc₂O₃, +3) to Group 7 (Mn₂O₇, +7) — verbatim NCERT (TRUE):

As the oxidation state of a transition metal increases, covalent character increases and basic character decreases: V₂O₃ (+3) is basic, V₂O₄ (+4) is amphoteric, and V₂O₅ (+5) is amphoteric/acidic. Statement C claims basic character increases, which is directly FALSE:

V₂O₄ has vanadium in oxidation state +4. It dissolves in non-oxidizing acids to give vanadyl salts containing VO²⁺ (+4). VO₄³⁻ contains vanadium in oxidation state +5 (orthovanadate). Statement D is FALSE:

CrO (Cr²⁺) is basic, Cr₂O₃ (Cr³⁺) is amphoteric, and CrO₃ (Cr⁶⁺) is acidic. Statement E correctly identifies CrO as basic and Cr₂O₃ as amphoteric (TRUE):

The steady decrease in atomic and ionic radii along the actinoid series (actinoid contraction) is larger from element to element than lanthanoid contraction. This arises because $5f$ electrons have more diffuse radial wavefunctions and exert even poorer electrostatic shielding of nuclear charge than $4f$ electrons. Statement (1) is scientifically correct.

According to NCERT Chapter on d- and f-Block Elements: *"Many trivalent lanthanoid ions are coloured both in the solid state and in aqueous solutions. Colour of these ions may be attributed to the presence of partially filled f electrons ($f$-$f$ transitions)."* Only ions with an empty subshell ($La^{3+}: 4f^0$) or a completely filled subshell ($Lu^{3+}: 4f^{14}$) are colorless. The vast majority ($Pr^{3+}, Nd^{3+}, Sm^{3+}, Eu^{3+}, Gd^{3+}, Tb^{3+}, Dy^{3+}, Ho^{3+}, Er^{3+}, Tm^{3+}$) possess characteristic vivid colors (e.g., $Nd^{3+}$ is pink, $Pr^{3+}$ is green). Therefore, Statement (2) is demonstrably incorrect.

- Statement 3: Lanthanoids are silvery-white metals and good conductors of heat and electricity (NCERT fact, correct). - Statement 4: Actinoids are highly electropositive and reactive metals, readily attacked by boiling water and oxygen, especially when in finely divided state (NCERT fact, correct).

Zirconium ($Zr$, atomic number 40) is a member of the second ($4d$) transition series in Group 4. Hafnium ($Hf$, atomic number 72) is located directly beneath $Zr$ in the third ($5d$) transition series. In moving from $Zr$ to $Hf$, 32 protons are added to the nucleus, including the 14 elements of the lanthanoid series ($_{58}\text{Ce}$ to $_{71}\text{Lu}$).

In the lanthanoids, electrons enter the inner $(n-2)f = 4f$ orbitals. Because $f$-orbitals have highly diffuse angular shapes, they exert exceptionally poor screening (shielding) of the nuclear charge on the outer $5d$ and $6s$ electrons. Consequently, as the nuclear charge increases by 14 units, the effective nuclear charge ($Z_{\text{eff}}$) experienced by outer electrons increases steadily, pulling the electron clouds inwards.

This contraction in size (lanthanoid contraction) almost exactly cancels the expected expansion due to the addition of an extra quantum shell ($n=5 \to n=6$): - Atomic radii: $Zr = 160\text{ pm}, \quad Hf = 159\text{ pm}$ - Ionic radii ($M^{4+}$): $Zr^{4+} = 79\text{ pm}, \quad Hf^{4+} = 78\text{ pm}$ Hence, $Zr$ and $Hf$ possess almost identical physical and chemical properties and occur together in nature.

Chromium has atomic number $$Z = 24$$. Its exceptional, half-filled stable ground state electronic configuration is: $$\text{Cr} = [\text{Ar}] 3d^5 4s^1$$

During ionization to form $$Cr^{2+}$$, electrons are removed first from the outermost $$4s$$ subshell, followed by the $$3d$$ subshell: $$\text{Cr}^{2+} = [\text{Ar}] 3d^4 4s^0$$ In the five degenerate $$3d$$ orbitals, according to Hund's rule of maximum multiplicity, all 4 electrons occupy separate orbitals with parallel spins: $$\uparrow, \uparrow, \uparrow, \uparrow, \_$$. Therefore, the number of unpaired electrons is $$n = 4$$.

The spin-only magnetic moment ($$\mu_s$$) is calculated using the formula: $$\mu_s = \sqrt{n(n + 2)} \text{ BM}$$ Substituting $$n = 4$$: $$\mu_s = \sqrt{4(4 + 2)} = \sqrt{4 \times 6} = \sqrt{24} \text{ BM}$$ $$\sqrt{24} = 4.8989... \approx 4.90 \text{ BM}$$

- $$n = 3 \implies \mu = \sqrt{15} = 3.87\text{ BM}$$ (e.g. $$Cr^{3+}$$) - $$n = 5 \implies \mu = \sqrt{35} = 5.92\text{ BM}$$ (e.g. $$Fe^{3+}, Mn^{2+}$$) - $$n = 2 \implies \mu = \sqrt{8} = 2.84\text{ BM}$$ (e.g. $$Ni^{2+}$$).

In the yellow chromate ion ($$\text{CrO}_4^{2-}$$), oxygen has an oxidation state of $$-2$$: $$x + 4(-2) = -2$$ $$x - 8 = -2 \implies x = +6$$ Thus, chromium has an oxidation state of $$+6$$.

In the orange dichromate ion ($$\text{Cr}_2\text{O}_7^{2-}$$), oxygen has an oxidation state of $$-2$$: $$2x + 7(-2) = -2$$ $$2x - 14 = -2 \implies 2x = +12 \implies x = +6$$ Thus, chromium also has an oxidation state of $$+6$$.

Chromate and dichromate ions interconvert depending upon pH without any change in oxidation state: $$2\text{CrO}_4^{2-} \text{(yellow)} + 2\text{H}^+ \rightleftharpoons \text{Cr}_2\text{O}_7^{2-} \text{(orange)} + \text{H}_2\text{O}$$ Because chromium remains in the $$+6$$ oxidation state in both species, Statement (4) is undeniably false.

- Statement (1): $$Cr^{2+}$$ ($$d^4$$) is reducing because $$d^4 \to d^3$$ ($$t_{2g}^3$$, half-filled stable configuration in aqueous medium), whereas $$Fe^{2+}$$ ($$d^6 \to d^5$$) has a positive standard potential ($$E^\circ_{\text{Cr}^{3+}/\text{Cr}^{2+}} = -0.41\text{ V}$$, $$E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = +0.77\text{ V}$$). True. - Statement (2): Transition metals are catalysts due to multiple oxidation states and complex formation. True. - Statement (3): Interstitial compounds form when small non-metal atoms (H, C, N) occupy interstitial voids. True.