d and f Block Elements — Questions & Answers

Sample questions from d and f Block Elements

1. Q1. A metallurgist notes that TUNGSTEN (W, a third-row/5d transition metal) has an exceptionally HIGH melting point (among the highest of all metals), substantially higher than chromium (Cr, its first-row/3d congener). Which explanation best accounts for this trend when comparing metals WITHIN the same group down the transition series?

   • A.Heavier transition metals within a group generally have MORE d electrons available for stronger metallic/covalent-character interatomic bonding (with greater orbital overlap due to more diffuse, further-extending 5d orbitals compared to 3d orbitals), producing progressively stronger interatomic bonding and correspondingly higher melting points moving down a transition-metal group✓
   • B.Melting point within a transition-metal group depends only on atomic mass, with heavier atoms always melting at a higher temperature regardless of bonding
   • C.Tungsten's high melting point is unrelated to its d-electron configuration and results purely from an unusual crystal packing arrangement not found in chromium
   • D.Third-row transition metals have completely filled d subshells that eliminate metallic bonding, paradoxically increasing melting point through purely ionic interactions
   Solution

   Moving DOWN a transition-metal group (from a first-row/3d element like chromium to its heavier third-row/5d congener tungsten), the valence d orbitals become progressively MORE spatially DIFFUSE/extended (5d orbitals extend further from the nucleus and overlap more effectively with neighbouring atoms' orbitals than the more compact 3d orbitals) -- this greater orbital overlap allows for STRONGER interatomic bonding interactions (with greater covalent-type d-orbital-overlap character contributing to the overall metallic bonding, in addition to the standard delocalized ns-electron metallic bonding) -- this progressively stronger interatomic bonding, moving down a transition-metal group, is a major contributing factor to the generally INCREASING melting points observed moving down many transition-metal groups, which is precisely why tungsten (5d) has a substantially higher melting point than chromium (3d), despite both belonging to the same group and both nominally having a comparable number of valence d electrons available for bonding.

2. Q2. The maximum oxidation state shown by uranium and the most common one in its aqueous chemistry are respectively:

   • A.+4, as in UF4
   • B.+3, as in U2O3
   • C.+6, as in the uranyl ion UO2^2+✓
   • D.+7, as in UO4^-
   Solution

   Uranium shows +3 to +6; the highest is +6, encountered as the very stable uranyl ion UO2^2+ (e.g. in uranyl nitrate), which dominates its aqueous and extraction chemistry.

3. Q3. The couple E(Cr3+/Cr) = -0.74 V while E(Cr2O7^2-/Cr3+) = +1.33 V. A student concludes that chromium metal in acid should be oxidised only as far as Cr3+ and not to Cr(VI). The conclusion is:

   • A.Incorrect, because the two potentials add to give a spontaneous overall oxidation to Cr(VI)
   • B.Correct, but only because Cr3+ is insoluble in acid
   • C.Incorrect, because H+ is a stronger oxidant than dichromate
   • D.Correct, because further oxidation of Cr3+ to Cr(VI) would need an oxidant with a potential above +1.33 V, which H+ does not provide✓
   Solution

   H+ (E = 0.00 V) can oxidise Cr to Cr3+ since E(Cr3+/Cr) is negative. Taking Cr3+ up to dichromate requires reversing a couple with E = +1.33 V, which needs a very strong oxidant such as persulphate or H2O2 in alkali, so acid alone stops at Cr3+.

4. Q4. A 0.02 M KMnO4 solution is used to titrate 25.0 mL of 0.10 M FeSO4 in dilute H2SO4. The volume of permanganate required at the end point is:

   • A.12.5 mL
   • B.25.0 mL✓
   • C.50.0 mL
   • D.5.0 mL
   Solution

   Moles Fe2+ = 0.0250 x 0.10 = 2.5 x 10^-3. MnO4- needs 1 mol per 5 mol Fe2+, so 5.0 x 10^-4 mol KMnO4 is required; volume = 5.0 x 10^-4 / 0.02 = 0.025 L = 25.0 mL.

5. Q5. Cu+ disproportionates in aqueous solution: 2Cu+ -> Cu2+ + Cu. What thermodynamic quantity chiefly favours this disproportionation?

   • A.The higher hydration enthalpy of Cu2+ relative to Cu+ makes the process energetically favourable✓
   • B.Cu+ has higher lattice energy than Cu2+
   • C.The reaction is entropy-driven only, with no enthalpy contribution
   • D.Disproportionation of Cu+ never actually occurs in water
   Solution

   The greater hydration enthalpy of Cu2+ compensates for the second ionization energy, making disproportionation of Cu+ into Cu2+ and Cu thermodynamically favourable in water.

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