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Question of 132
Q.

Match the following given in List I with List II.

List I (Elements)List II (Their maximum oxidation states)
i)Thorium (Th)a)+7
ii)Protactinium (Pa)b)+6
iii)Lawrencium (Lr)c)+3
iv)Uranium (U)d)+5
v)Plutonium (Pu)e)+4

Choose the correct option :

  1. i – e, ii – d, iii – a, iv – b, v – c
  2. i – e, ii – d, iii – c, iv – b, v – a
  3. i – b, ii – a, iii – c, iv – e, v – d
  4. i – b, ii – e, iii – c, iv – d, v – a
Karnataka PUCKarnataka II PUC Board 2026MCQ· 1mImportance★★★★★
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Matching each actinoid to its maximum oxidation state gives Th–+4+4, Pa–+5+5, Lr–+3+3, U–+6+6, Pu–+7+7, i.e. option (b).

Early actinoids show a rise in maximum oxidation state as the number of available 5f/6d5f/6d electrons increases, peaking around U–Np–Pu, then falling again toward the heavier members (which behave much like lanthanoids with a stable +3+3 state).

  • i) Thorium (Th): maximum oxidation state =+4= +4 → (e)
  • ii) Protactinium (Pa): maximum oxidation state =+5= +5 → (d) …

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