The spin-only magnetic moment of a divalent i | vedclass

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(A) The atomic number of the element is $Z = 29$,which corresponds to Copper $(Cu)$.The electronic configuration of $Cu$ is $[Ar] 3d^{10} 4s^1$.For th

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$2$

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(A) The atomic number of the element is $Z = 29$,which corresponds to Copper $(Cu)$.The electronic configuration of $Cu$ is $[Ar] 3d^{10} 4s^1$.For the divalent ion $Cu^{2+}$,the electronic configuration is $[Ar] 3d^9$.In the $3d$ orbital,there are $9$ electrons,which means there is $1$ unpaired electron.The spin-only magnetic moment formula is $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.Substituting $n = 1$: $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.Rounding $1.73$ to the nearest integer gives $2 \ BM$.

$A$. Tetrahedral $NiCl_2Br_2^{2-}$	$B$. Square planar $Pt(NH_3)_2Cl_2$
$C$. Octahedral $Co(NH_3)_3Cl_3$	$D$. Square planar $Pd(NH_3)_3Br^{+}$
$E$. Octahedral $Co(en)_3^{3+}$

The spin-only magnetic moment of a divalent ion in aqueous solution (atomic number $29$) is ...... $BM.$

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Which of the following molecules/ions can exhibit isomerism:
$A$. Tetrahedral $NiCl_2Br_2^{2-}$ $B$. Square planar $Pt(NH_3)_2Cl_2$
$C$. Octahedral $Co(NH_3)_3Cl_3$ $D$. Square planar $Pd(NH_3)_3Br^{+}$
$E$. Octahedral $Co(en)_3^{3+}$

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