(C) The atomic number of scandium $(Sc)$ is $21$. Its electronic configuration is $[Ar] 3d^1 4s^2$. To achieve a stable noble gas configuration $([Ar]

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(C) The atomic number of scandium $(Sc)$ is $21$. Its electronic configuration is $[Ar] 3d^1 4s^2$. To achieve a stable noble gas configuration $([Ar])$,scandium loses three electrons (two from $4s$ and one from $3d$). Therefore,the most common and stable oxidation state of scandium is $+ 3$.