One of the notable features of the transition element is the great variety of oxidation states it may show in its compounds. The elements which give the greatest number of oxidation states occur in or near the middle of the series. Manganese, for example exhibits all the oxidation states from +2 to +7. The lesser number of oxidation states at the extreme ends is either due to the loss or sharing of a few electrons (Sc, Ti) or too many d electrons of higher valence. Early in the first series scandium (II) is virtually unknown and titanium (IV) which is more stable than Ti (III) or Ti (II). On the other end, the only oxidation state of zinc is +2.

Oxidation state of scandium is

+3

The correct answer is (3) +3.

The oxidation state of an element in a compound is a measure of the number of electrons that it has gained or lost in order to form that compound. Oxidation states are often represented by Roman numerals, and they can be positive or negative.
Scandium (Sc) is a transition metal, and its most common oxidation state is +3. The reason for this is related to its electron configuration and the stability that is achieved when it loses three electrons.
The electron configuration of scandium is [Ar] 3d¹ 4s². When scandium forms compounds, it typically loses the three outer electrons (two from the 4s orbital and one from the 3d orbital) to achieve a stable electron configuration similar to that of the noble gas argon ([Ar] 3d⁰ 4s⁰). By losing these three electrons, scandium attains a +3 oxidation state.
It's important to note that while +3 is the most common oxidation state for scandium, there are instances where it can exhibit other oxidation states, such as +2 in certain compounds. However, +3 is the more prevalent and stable oxidation state for scandium.