A) \[F{{e}^{2+}}\]
B) \[C{{o}^{2+}}\]
C) \[C{{r}^{3+}}\]
D) \[N{{i}^{2+}}\]
Correct Answer: A
Solution :
: \[_{26}{{F}^{2+}}:\,\,3{{d}^{6}}\,4{{s}^{0}}\] contains 4 unpaired electrons. \[_{27}C{{o}^{2+}}:\,\,3{{d}^{7}}\,4{{s}^{0}}\]contains 3 unpaired electrons. \[_{27}C{{r}^{3+}}:\,\,3{{d}^{3}}\,4{{s}^{0}}\] contains 3 unpaired electrons. \[_{28}N{{i}^{2+}}:\,\,3{{d}^{8}}\,4{{s}^{0}}\] contains 2 unpaired electrons. Hence, \[F{{e}^{2+}}(n=4)\] has highest magnetic moment.
You need to login to perform this action.
You will be redirected in
3 sec
