Correct Answer – D
Scandium always exists in the oxidation state (+3), and occurs as `Sc^(3+)` ion. The formation of `Sc^(3+)` requires the removal of the two 4s and one d-d specturm is impossible. Consequently the ion and its compounds are coplorless and dimagnetic. Of the simple ions of the 3d-serios, only `Sc^(3+)` is known to have `d^(0)` configuration. Zn (+2) is the only state known for Zn. However the answer of this question is (4), because Zn is a d-block element but not a transition element. The reduced tendency of higher oxidation states towards the end of the series is due to steady increase in the effective nuclear charge along the series. This pulls the (n-1) d orbitals into the electron core and hence they are not readily available for bonding. For instance, the only oxidation state for Zn is Zn(+2) where no (n-1) d electron is involved. On the other hand, early in the series, it is difficult to form species that do not utillise the (n-1) d electron, e.g., Sc(+2) is virually unknown while Ti(+4) is more stable than Ti(+2).