Question

The spin-only magnetic moment value of an octahedral complex among $CoCl_3⋅4NH_3$, $NiCl_2⋅6H_2O$ and $PtCl_4⋅2HCl$, which upon reaction with excess of $AgNO_3$ gives 2 moles of $AgCl$ is _____ B.M. (Nearest Integer)

Answer 1

$CoCl_3⋅4NH_3 \xrightarrow{\text {excess} \ AgNO_3 } [Co(NH_3)_4⋅Cl_2]+AgCl$
$NiCl_2⋅6H_2O\xrightarrow{\text {excess} \ AgNO_3 }[Ni(H_2O)_6]_2+ +2AgCl$
$PtCl_4⋅2HCl→[PtCl_6]^{4−} + \text {No AgCl ppt}$
$[Ni^{2+}(H_2O)_6]=d^8=t_2g^6 e^2_g=2$ unpaired~electrons
Magnetic moment,
$=\sqrt {2(2+2)}$
$=\sqrt 8$
$≈3$

So, the answer is $3$.