Question

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. $TiCl_4$

Answer 1

Answer: (B)

$[ Cu (NH_3)_4]^{2+}$

Answer 2

Magnetic moment 1.73 BM
$\mu = \sqrt{n (n +2) } B.M$
n = no. of unpaired e-
$\mu = 1.73$
$1.73 = \sqrt{n ( n+2)} B.M$
$n = 1$
$* [CoCl_6]^{4-} \to Co^{+2} ; d^7$
Cl- (weak field ligand) $t_2g^5eg^2$ unpaired $e^-= 3$
$* [Cu(NH_3)_4]^{2+} Cu^{+2} - d^9$
$NH_3$ Strong field ligand, hybridisation $dsp^2$
$*$ one $e^-$ of $3d$
$* [Ni(CN)_4]^{2-} \to Ni^{+2} - d^8$ unpaired $e^- =0$
$CN^-$ - Strong field ligand $dsp^2$
$* TiCl_4 \to Ti^{+4} d^{\circ}$ unpaired e- = zero.