The 'spin-only' magnetic moment [in units of Bohr magneton,\[({{\mu }_{\beta }})]\]of\[N{{i}^{2+}}\]in aqueous solution would be (Atomic number of\[Ni=28\])
AIEEE Solved Paper-2006
A)2.84
B) 4.90
C) 0
D) 1.73
Correct Answer:
A
Solution :
\[N{{i}^{2+}}=[Ar]3{{d}^{8}}\] Number of unpaired electrons = 2 Hence, magnetic moment\[=\sqrt{n(n+2)}=\sqrt{8}=2.84\]