Q.Why do the transition elements exhibit higher enthalpies of atomisation?

Concept and Intuition

Interstitial compounds (e.g., TiC, TiN, Fe$_3$C, VH$_{0.56}$) form when small atoms such as H, C, N, or B fit into the interstitial (empty) spaces of a metal's crystal lattice without drastically disrupting the metallic bonding framework. Because the delocalised electron sea of the metal lattice is largely preserved, these compounds keep several metal-like characteristics: high melting point, hardness, and — crucially — metallic electrical conductivity.

Step-by-Step Solution

1. (A) High melting points: interstitial compounds are known for even higher melting points than the parent metal (interstitial atoms strengthen the lattice). TRUE.
2. (B) Loses electrical conductivity: since the metallic bonding/electron sea is retained, these compounds actually conduct electricity like the parent metal — they do NOT lose conductivity. FALSE — this is the incorrect statement.
3. (C) Chemically inert: interstitial compounds are indeed chemically quite inert/unreactive. TRUE. …

Concept and Intuition

Melting points of the d-block transition metals rise toward the middle of each series (peaking around Group 6) because of strong metallic bonding reinforced by (n−1)d electron participation, then fall off toward both ends. Among all transition metals, tungsten holds the record for the highest melting point.

Step-by-Step Solution

1. Compare typical high melting points: W ≈ 3422°C, Re ≈ 3186°C, Mo ≈ 2623°C, Cr ≈ 1907°C.
2. Tungsten's melting point is the highest among these (and the highest of any metal), due to very strong metallic/covalent-like bonding involving its d-electrons.

Common Mistakes …

Concept and Intuition

This is a fact-recall matching question about industrially important compounds/alloys and their real-world uses.

Step-by-Step Solution

1. i. TiO$_2$ – pigment industry: True. Titanium dioxide is the most widely used white pigment (titanium white) in paints, plastics, and paper.
2. ii. MnO$_2$ – dry battery cells: True. In the Leclanché dry cell, MnO$_2$ acts as a depolarizer, oxidizing the hydrogen gas produced at the cathode. …

Concept and Intuition

Enthalpy of atomization reflects the strength of metallic bonding, which comes largely from the overlap of unpaired d-orbital electrons between neighbouring metal atoms (in addition to the delocalized s-electrons). Metals with more unpaired d-electrons form stronger, more extensive metallic bonds and so have higher atomization enthalpies. Zinc has the electronic configuration $[\mathrm{Ar}]3d^{10}4s^2$ — its d-subshell is completely filled, leaving no unpaired d-electrons to participate in interatomic bonding, so its metallic bonding is comparatively weak.

Step-by-Step Solution

1. Write electron configurations: V = $[\mathrm{Ar}]3d^3 4s^2$ (3 unpaired d-electrons), Cr = $[\mathrm{Ar}]3d^5 4s^1$ (6 unpaired electrons total incl. 4s, exceptionally high atomization enthalpy), Fe = $[\mathrm{Ar}]3d^6 4s^2$ (4 unpaired d-electrons), Zn = $[\mathrm{Ar}]3d^{10}4s^2$ (0 unpaired d-electrons). …

Concept and Intuition

• Statement (i): A metal ion liberates $\mathrm{H_2}$ from a dilute acid when its reduction potential is more negative than that of the $\mathrm{H^+/H_2}$ couple (taken as $0\,V$). For vanadium, $E^\circ(\mathrm{V^{3+}/V^{2+}}) = -0.26\,V$. Since this is negative, the reverse reaction ($\mathrm{V^{2+}} \to \mathrm{V^{3+}} + e^-$) coupled with $2\mathrm{H^+} + 2e^- \to \mathrm{H_2}$ is spontaneous — so $\mathrm{V^{2+}}$ is a strong enough reducing agent to liberate hydrogen gas from dilute acid.
• Statement (ii): Lanthanoid contraction means ionic radii shrink steadily across the series. The early members (La, Ce, Pr…) have relatively large $\mathrm{Ln^{3+}}$ radii, close in size to $\mathrm{Ca^{2+}}$ — this is exactly why rare-earth minerals substitute for calcium in nature. They do not behave like aluminium (aluminium chemistry — small, highly charge-dense $\mathrm{Al^{3+}}$ — is a different comparison used elsewhere, e.g. for beryllium/diagonal relationships). So (ii) is false as stated.
• Statement (iii): Historically 'silver' coins in the UK were sterling silver, but since 1947 they have been struck in cupro-nickel (75% Cu, 25% Ni) — a genuine transition-metal alloy fact.
• Statement (iv): Actinoids show a wider range of oxidation states than lanthanoids because 5f, 6d and 7s levels are close in energy. Np, Pu, and Am can all reach $+7$ (e.g. as $\mathrm{NpO_5^{3-}}$) under strongly oxidising alkaline conditions, though $+5$/$+6$ are more common. So Np's maximum oxidation state of $+7$ is correct.

Step-by-Step Solution …

Concept and Intuition (Transition Element Definition)

Transition metals (like Fe, Ni, Pt, Pd) and their complexes are famous for their catalytic activity. This is not just because they lower activation energy — all catalysts do that. The special reason lies in their unique electronic structure: they have partially filled d-orbitals, which allow them to:

• adopt multiple oxidation states,
• form temporary bonds with reactants,
• provide a surface or coordination site where reactants can come together in the right orientation.

The Reason (R) simply states the universal property of any catalyst. It is true, but it does not explain why transition metals in particular are so effective. So (R) is not the correct explanation of (A).

Step-by-step reasoning:

1. Check Assertion (A):

   Transition metals and their complexes are indeed widely used as catalysts — e.g., iron in the Haber process, platinum in catalytic converters, nickel in hydrogenation. This is a well-known fact.

   → So (A) is correct.

2. Check Reason (R):

   A catalyst, by definition, lowers the activation energy of a reaction, thereby increasing the rate without being consumed. This is a fundamental principle of catalysis.

   → So (R) is also correct.

3. Determine if (R) explains (A): …

Concept and Intuition

IUPAC defines a transition element as one whose atom (in the ground state) or common ion has an incompletely filled d-subshell. Zinc, cadmium and mercury all have the configuration $(n-1)d^{10}ns^2$ and lose only the $ns^2$ electrons to form $M^{2+}$, which is still $d^{10}$ — no partially filled d-orbital ever appears, so they are excluded from the transition series even though they sit in the d-block.

Step-by-Step Solution

1. Write electron configurations: Zn = $[Ar]3d^{10}4s^2$; Cd = $[Kr]4d^{10}5s^2$; Hg = $[Xe]4f^{14}5d^{10}6s^2$.
2. In each case the d-subshell is completely filled ($d^{10}$), both in the atom and in the common $M^{2+}$ ion. …

Concept and Intuition

Enthalpy of atomization measures the energy needed to convert one mole of metal atoms in the solid state into gaseous atoms — essentially, the strength of the metallic bonding holding the solid lattice together. Transition metals show unusually high atomization enthalpies compared to their neighbouring s- and p-block metals. NCERT explains this by noting that in transition metals, in addition to the delocalized valence-electron ('electron sea') metallic bonding common to all metals, the partially filled (n-1)d orbitals allow additional localized, covalent-like overlap between neighbouring atoms' d-orbitals. The greater the number of unpaired d-electrons available for this extra overlap, the stronger the overall interatomic bonding — which is exactly why atomization enthalpies of transition metals peak somewhere in the middle of each series (where the number of unpaired d-electrons is often highest) and are generally much larger than for s-/p-block metals.

Step-by-Step Solution

1. Check Assertion (A): transition elements have higher enthalpies of atomization — this is a well-established, textbook-supported fact (compare, e.g., atomization enthalpies of 3d transition metals to those of Ca, K, or Ga/Ge). True.
2. Check Reason (R): a large number of unpaired electrons facilitate strong interatomic interaction and strong bonding between atoms — also a textbook-supported mechanistic explanation. True. …

Concept and Intuition

Enthalpy of atomisation measures the energy needed to convert one mole of metal atoms in the solid (metallic) state into gaseous atoms — essentially a measure of the strength of metallic bonding. In transition metals, metallic bonding strength depends on the number of unpaired $d$ electrons and how well the $d$-orbitals overlap between neighbouring atoms.

Step-by-Step Solution

1. Across a transition series, atomisation enthalpy is influenced by the number of unpaired electrons — it rises to a maximum near the middle of the series (where the number of unpaired electrons is highest) and falls off toward both ends.
2. Comparing the same group across the three transition series (3d, 4d, 5d), the outer $d$-orbitals become progressively larger and more diffuse — $5d$ orbitals overlap more effectively with neighbouring atoms' orbitals than $4d$, which in turn overlap better than $3d$. …