Atomic numbers of Cr,Fe,and Co are 24,26,and | vedclass

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(D) To determine which complex is paramagnetic,we analyze the electronic configuration of the central metal ion in each complex:$1$. $[Co(NH_3)_6]^{3+

Vedclass · Accepted Answer

$[Cr(CN)_6]^{3-}$

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(D) To determine which complex is paramagnetic,we analyze the electronic configuration of the central metal ion in each complex:$1$. $[Co(NH_3)_6]^{3+}$: $Co^{3+}$ is $3d^6$. $NH_3$ is a strong field ligand,causing pairing. All electrons are paired $(t_{2g}^6 e_g^0)$,so it is diamagnetic.$2$. $[Co(CN)_6]^{3-}$: $Co^{3+}$ is $3d^6$. $CN^-$ is a strong field ligand,causing pairing. All electrons are paired $(t_{2g}^6 e_g^0)$,so it is diamagnetic.$3$. $[Fe(CN)_6]^{4-}$: $Fe^{2+}$ is $3d^6$. $CN^-$ is a strong field ligand,causing pairing. All electrons are paired $(t_{2g}^6 e_g^0)$,so it is diamagnetic.$4$. $[Cr(CN)_6]^{3-}$: $Cr^{3+}$ is $3d^3$. The configuration is $t_{2g}^3 e_g^0$. There are $3$ unpaired electrons in the $3d$ orbitals. Thus,it is paramagnetic.Therefore,the correct option is $D$.

List-$I$ (Complexes)	List-$II$ (Types)
$a$. $[Co(NH_3)_5NO_2]Cl_2$ and $[Co(NH_3)_5ONO]Cl_2$	$i$. Linkage isomerism
$b$. $[Cr(NH_3)_6][Co(CN)_6]$ and $[Cr(CN)_6][Co(NH_3)_6]$	$ii$. Coordination isomerism
$c$. $[Co(NH_3)_5(SO_4)]Br$ and $[Co(NH_3)_5Br]SO_4$	$iii$. Ionisation isomerism
$d$. $[Cr(H_2O)_6]Cl_3$ and $[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O$	$iv$. Solvate isomerism

Atomic numbers of $Cr$,$Fe$,and $Co$ are $24$,$26$,and $27$ respectively. Which of the following inner orbital octahedral complexes are paramagnetic?

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