The transition elements exhibit variable oxidation states because (n−1)d electrons can also participate in bonding, as the energy difference between (n−1)d and ns orbitals is very small. The maximum oxidation state that can be exhibited by a transition element is equal to the total number of electrons present in both ns and (n−1)d orbitals. The elements in the middle exhibit more oxidation states, e.g., manganese exhibits all oxidation states from +2 to +7. In the second series, a stable maximum oxidation state is exhibited by technetium, and an unstable maximum oxidation state of +8 is exhibited by ruthenium. In the third series, a stable maximum oxidation state of +8 is exhibited by osmium. Transition metal ions having a completely filled or an exactly half-filled d subshell are stable. The Fe3+ ion is more stable than the Fe2+ ion because of the stable half-filled 3d5 electronic configuration of Fe3+.
The maximum oxidation state shown by a transition element is equal to the total number of its
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → ns and (n−1)d electrons
The passage states that the maximum oxidation state a transition element can show is equal to the total number of electrons in its ns and (n−1)d orbitals.
Because these two sets of orbitals are very close in energy, the (n−1)d electrons can take part in bonding along with the ns electrons.
For example, Mn (3d54s2) has 5 + 2 = 7 such electrons and shows a maximum oxidation state of +7.
Why the other options are incorrect
Option (1) ns electrons — Incorrect. The number of ns electrons gives the minimum oxidation state, not the maximum.
Option (2) (n−1)d electrons — Incorrect. This leaves out the ns electrons, which are lost first when the ions form.
Option (3) ns and np electrons — Incorrect. Transition elements have the outer configuration (n−1)d1–10ns1–2; their np orbitals are empty.