Transition Metals: Variable Oxidation State, Colour and Catalysis · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

What makes a transition metal

Science · Chemistry · ages 17-18
Name ______________________   Date ____________
  1. Why does zinc not qualify as a transition metal?

    • Zinc sits outside the d block of the periodic table
    • Its Zn2+ ion has an empty d sub-shell, like Sc3+
    • Its Zn2+ ion has a full d sub-shell
  2. What makes an element a transition metal?

    • At least one stable ion with a partly filled d sub-shell
    • A full d sub-shell in every one of its common ions
    • Atoms that lose only their two 4s electrons in every compound
  3. Manganese is +7 in KMnO4 and +2 in MnSO4, so it shows more than one oxidation state.

    Circle one:   True   False

  4. Why does scandium not qualify as a transition metal?

    • Its only ion Sc3+ has an empty d sub-shell
    • Scandium has no d electrons even as an uncombined atom
    • Scandium forms coloured ions with several oxidation states
  5. Why are copper(II) compounds usually coloured while zinc compounds are white?

    • Copper atoms are heavier, so their compounds reflect less light
    • In Cu2+ a d electron can jump to a higher, split d level
    • Zinc ions have empty d orbitals, so they cannot hold ligands
  6. Mn2+ ions catalyse the reaction between manganate(VII) and ethanedioate ions in solution. Mn2+ is oxidised to Mn3+ in the first step. What must happen next for Mn2+ to be a catalyst?

    • Mn3+ is oxidised again to Mn7+ and stays there until the end
    • Mn3+ is reduced back to Mn2+ by the ethanedioate ions
    • Mn3+ leaves the solution as a solid, so it cannot react again
  7. Which pair gives one heterogeneous and one homogeneous transition metal catalyst?

    • Solid iron in the Haber process, and Fe2+ ions in solution
    • Solid iron in the Haber process, and solid V2O5 in the Contact process
    • Fe2+ ions in solution, and Fe3+ ions in the same solution
  8. In excess ammonia, four of the six water ligands around Cu2+ are replaced by ammonia, and the colour of the solution changes. Why?

    • Ammonia fills the copper d sub-shell, so no light is absorbed any more
    • The new ligands split the d levels by a different amount
    • Ammonia changes copper into a different element with its own colour
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Answer key

For grown-ups. Fold this page away before handing over the rest.

What makes a transition metal W1-mt_UIVC6d7wak-s1

  1. Its Zn2+ ion has a full d sub-shell · Zn2+ holds ten d electrons, a full d sub-shell, so it fails the partly filled test.
  2. At least one stable ion with a partly filled d sub-shell · The test is a stable ion with a partly filled d sub-shell. Scandium and zinc fail because their only ions are empty (Sc3+) or full (Zn2+), even though the scandium atom is 3d1.
  3. True · In KMnO4, K is +1 and four O give -8, so Mn is +7. In MnSO4 the sulfate is 2-, so Mn is +2. One metal shows two states, so the sentence is true.
  4. Its only ion Sc3+ has an empty d sub-shell · Sc3+ has lost all its d electrons, so the sub-shell is empty, not partly filled.
  5. In Cu2+ a d electron can jump to a higher, split d level · Ligands split the d levels. Cu2+ is 3d9, so a d electron can absorb visible light and jump up. Zn2+ is 3d10, so every d orbital is full and no jump is possible.
  6. Mn3+ is reduced back to Mn2+ by the ethanedioate ions · A catalyst must be remade. Mn3+ oxidises ethanedioate ions and is reduced back to Mn2+, so the manganese changes state and changes back, as Fe2+ does in the lesson.
  7. Solid iron in the Haber process, and Fe2+ ions in solution · Solid iron is in a different phase from the nitrogen and hydrogen gases, so it is heterogeneous. Fe2+ ions are in the same solution as the reacting ions, so they are homogeneous.
  8. The new ligands split the d levels by a different amount · Ligands split the d levels, and different ligands split them by different amounts. A different split absorbs a different part of the spectrum, so the colour changes. The copper is still copper(II).
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