The spin-only magnetic moment value of the sq | vedclass

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(D) The complex is $[Pt(NH_3)_2 Cl(NH_2 CH_3)] Cl$.In this complex,the oxidation state of $Pt$ is $+2$.$Pt$ $(Z=78)$ has an electronic configuration o

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$0$

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(D) The complex is $[Pt(NH_3)_2 Cl(NH_2 CH_3)] Cl$.In this complex,the oxidation state of $Pt$ is $+2$.$Pt$ $(Z=78)$ has an electronic configuration of $[Xe] 4f^{14} 5d^9 6s^1$. Thus,$Pt^{2+}$ has a $5d^8$ configuration.Since the complex is square planar,it undergoes $dsp^2$ hybridization.In a $5d^8$ square planar complex,the electrons pair up in the $5d$ orbitals,leaving no unpaired electrons.Therefore,the number of unpaired electrons $(n)$ is $0$.The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} \ B.M.$Substituting $n=0$,we get $\mu = \sqrt{0(0+2)} = 0 \ B.M.$

List-$I$ (Complex)	List-$II$ (Spin only magnetic moment in $B$.$M$.)
$A$. $[Cr(NH_3)_6]^{3+}$	$I$. $4.90$
$B$. $[NiCl_4]^{2-}$	$II$. $3.87$
$C$. $[CoF_6]^{3-}$	$III$. $0.0$
$D$. $[Ni(CN)_4]^{2-}$	$IV$. $2.83$

The spin-only magnetic moment value of the square planar complex $[Pt(NH_3)_2 Cl(NH_2 CH_3)] Cl$ is $B.M.$ (Nearest integer). (Given atomic number for $Pt = 78$)

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