A) zero
B) 1.73
C) 2.84
D) 3.87
Correct Answer: A
Solution :
\[Z{{n}^{2+}}(Z=30):[Ar]\,3{{d}^{10}}4{{s}^{0}};\]zero unpaired\[{{e}^{-}}\]. Hence, its magnetic moment is zero. \[\mu =\sqrt{n(n+2)}=\sqrt{0(0+2)}\] \[\mu =0\]
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