The magnetic moment for two complexes of empi | vedclass

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Question

(C) The empirical formula is $Ni(NH_3)_4(NO_3)_2 \cdot 2H_2O$. For the first complex,the magnetic moment is $0 \ BM$,which implies it is diamagnetic.

Vedclass · Accepted Answer

$[Ni(NH_3)_4](NO_3)_2 \cdot 2H_2O = 	ext{Square planar}$

Vedclass · Answer

(C) The empirical formula is $Ni(NH_3)_4(NO_3)_2 \cdot 2H_2O$. For the first complex,the magnetic moment is $0 \ BM$,which implies it is diamagnetic. $Ni^{2+}$ has a $d^8$ configuration. In a square planar geometry,$d^8$ ions form low-spin complexes where electrons pair up,resulting in $0$ unpaired electrons and a magnetic moment of $0 \ BM$. Thus,the first complex is $[Ni(NH_3)_4](NO_3)_2 \cdot 2H_2O$ with a square planar geometry. For the second complex,the magnetic moment is $2.84 \ BM$,which corresponds to $n=2$ unpaired electrons $(\mu = \sqrt{n(n+2)} = \sqrt{2(4)} = \sqrt{8} \approx 2.83 \ BM)$. This corresponds to an octahedral geometry where $Ni^{2+}$ is in a high-spin state,such as $[Ni(NH_3)_4(H_2O)_2](NO_3)_2$.

The magnetic moment for two complexes of empirical formula $Ni(NH_3)_4(NO_3)_2 \cdot 2H_2O$ is $0$ and $2.84 \ BM$ respectively. The second complex is not a neutral complex. The correct formula and geometry of the first complex is:

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