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Magnetic moment 2.84 B.M. is given by: (At. nos. Ni = 28, Ti = 22, Cr = 24, Co = 27)
A
$Ni^{2+}$
B
$Ti^{3+}$
C
$Cr^{2+}$
D
$Co^{2+}$
Detailed Solution
$\mu = \sqrt{n(n+2)}$ B.M.
$2.84 = \sqrt{n(n+2)} \Rightarrow n^2 + 2n - 8.07 = 0 \Rightarrow n \approx 2$
$Ni^{2+}$: $[Ar]3d^84s^0$ has 2 unpaired electrons.
($Ti^{3+}$ has 1, $Cr^{2+}$ has 4 and $Co^{2+}$ has 3 unpaired electrons.)
$2.84 = \sqrt{n(n+2)} \Rightarrow n^2 + 2n - 8.07 = 0 \Rightarrow n \approx 2$
$Ni^{2+}$: $[Ar]3d^84s^0$ has 2 unpaired electrons.
($Ti^{3+}$ has 1, $Cr^{2+}$ has 4 and $Co^{2+}$ has 3 unpaired electrons.)