A) \[S{{c}^{3+}}\]
B) \[F{{e}^{2+}}\]
C) \[T{{i}^{3+}}\]
D) \[M{{n}^{2+}}\] (Atomic number Sc = 21, Fe = 26, Ti = 22, Mn = 25)
Correct Answer: A
Solution :
\[_{21}Sc\to 1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{1}}\] \[S{{c}^{+3}}\to 1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}\] Due to presence of paired electron it will be colourless.
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