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A magnetic moment of 1.73 BM will be shown by one among the following:
A
$[CoCl_6]^{4-}$
B
$[Cu(NH_3)_4]^{2+}$
C
$[Ni(CN)_4]^{2-}$
D
$TiCl_4$
Detailed Solution
$\mu = \sqrt{n(n+2)}$ BM = 1.73 gives n = 1 unpaired electron.
$[CoCl_6]^{4-}$: $Co^{2+}$, $d^7$, $Cl^-$ is a weak field ligand, $t_{2g}^5e_g^2$ — 3 unpaired electrons.
$[Cu(NH_3)_4]^{2+}$: $Cu^{2+}$, $d^9$, $dsp^2$ hybridisation — 1 unpaired electron.
$[Ni(CN)_4]^{2-}$: $Ni^{2+}$, $d^8$, $CN^-$ is a strong field ligand, $dsp^2$ — 0 unpaired electrons.
$TiCl_4$: $Ti^{4+}$, $d^0$ — 0 unpaired electrons.
$[CoCl_6]^{4-}$: $Co^{2+}$, $d^7$, $Cl^-$ is a weak field ligand, $t_{2g}^5e_g^2$ — 3 unpaired electrons.
$[Cu(NH_3)_4]^{2+}$: $Cu^{2+}$, $d^9$, $dsp^2$ hybridisation — 1 unpaired electron.
$[Ni(CN)_4]^{2-}$: $Ni^{2+}$, $d^8$, $CN^-$ is a strong field ligand, $dsp^2$ — 0 unpaired electrons.
$TiCl_4$: $Ti^{4+}$, $d^0$ — 0 unpaired electrons.