MCQ
$3{d^{10}}4{s^0}$ electronic configuration exhibits by
- ✓$Z{n^{ + + }}$
- B$C{u^{ + + }}$
- C$C{d^{ + + }}$
- D$H{g^{ + + }}$
$Z{n^{ + + }} = \,[Ar]\,3{d^{10}}\,4{s^0}$
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( $Ba = 137,\,\,Cl = 35.5,\,\,S = 32,\,\,H = 1$ and $O = 16$ )