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The "spin-only" magnetic moment [in units of Bohr magneton,$(\mu_B)$] of $Ni^{2+}$ in aqueous solution would be (Atomic no Ni=28)

$(a)\;1.73\qquad(b)\;2.84\qquad(c)\;4.90\qquad(d)\;0$

1 Answer

$Ni^{2+}$ has two unpaired electrons.
$\therefore \mu=\sqrt{n(n+2)}=\sqrt 8=2.84BM$
Hence (b) is the correct answer.
answered Jan 13, 2014 by sreemathi.v
 

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