A) \[T{{i}^{+4}}\]
B) \[C{{u}^{+}}\]
C) \[Z{{n}^{+2}}\]
D) \[C{{r}^{+3}}\]
Correct Answer: D
Solution :
[d] \[T{{i}^{+4}}=[Ar]4{{s}^{0}}3{{d}^{0}}(0\,unpaired\,\,{{e}^{-}})\] \[C{{u}^{+}}=\left[ Ar \right]4{{s}^{0}}3{{d}^{10}}\left( 0\text{ }unpaired\text{ }{{e}^{-}} \right)\] \[Z{{n}^{+2}}=[Ar]4{{s}^{0}}3{{d}^{10}}(0\,unpaired\,{{e}^{-}})\] \[C{{r}^{+3}}=\left[ Ar \right]4{{s}^{0}}3{{d}^{3}}\left( 3\text{ }unpaired\text{ }{{e}^{-}} \right)\]
You need to login to perform this action.
You will be redirected in
3 sec
