A) \[T{{i}^{3+}}\]
B) \[T{{i}^{4+}}\]
C) \[F{{e}^{2+}}\]
D) \[F{{e}^{3+}}\] (The at. no. of \[Ti\] and \[Fe\] are 22 and 26 respectively)
Correct Answer: B
Solution :
\[T{{i}^{+4}}\,\to \,3{{d}^{0}}4{{s}^{0}}\] \ no unpaired \[{{e}^{-}}\].
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