A) \[H{{e}^{+}}\]
B) \[L{{i}^{2+}}\]
C) \[B{{e}^{3+}}\]
D) all cases
Correct Answer: D
Solution :
\[H{{e}^{+}},\,\,L{{i}^{2+}}\]and\[B{{e}^{3+}}\]each has one electron, hence no screening.
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