A) \[T{{i}^{3+}}\]
B) \[S{{c}^{3+}}\]
C) \[M{{n}^{2+}}\]
D) \[Z{{n}^{2+}}\]
Correct Answer: C
Solution :
\[T{{i}^{3+}}\,\to \,\,3{{d}^{1}}4{{s}^{0}}\]; \[S{{c}^{3+}}\,\to \,\,3{{d}^{0}}\] \[M{{n}^{2+}}\,\to \,\,3{{d}^{5}}4{{s}^{0}}\]; \[Z{{n}^{2+}}\,\to \,\,3{{d}^{10}}4{{s}^{0}}\] In \[M{{n}^{2+}}\] number of unpaired d \[{{e}^{-}}=5\]. So it has maximum magnetic moment according to the formula. \[\mu =\sqrt{n(n+2)}\]
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