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(A) In $[FeCl_4]^-$,$Fe^{3+}$ has the electronic configuration $[Ar] 3d^5$. Since $Cl^-$ is a weak field ligand,it forms a high-spin tetrahedral compl

Vedclass · Accepted Answer

$A, C$

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(A) In $[FeCl_4]^-$,$Fe^{3+}$ has the electronic configuration $[Ar] 3d^5$. Since $Cl^-$ is a weak field ligand,it forms a high-spin tetrahedral complex with $sp^3$ hybridization.$(C)$ $[FeCl_4]^-$ has $5$ unpaired electrons,so its spin-only magnetic moment is $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$. In $[Co(en)(NH_3)_2Cl_2]^+$,$Co^{3+}$ has the configuration $[Ar] 3d^6$. Due to the presence of strong field ligands ($en$ and $NH_3$),the electrons pair up,resulting in $0$ unpaired electrons $(n=0, \mu=0)$. Therefore,$[FeCl_4]^-$ has a higher magnetic moment.$(B)$ is incorrect because $[Co(en)(NH_3)_2Cl_2]^+$ has $3$ geometrical isomers (trans,cis-cis,and cis-trans).$(D)$ is incorrect because the hybridization of $Co^{3+}$ in this low-spin octahedral complex is $d^2sp^3$.

$[Ni(H_2O)_6]^{2+} : en$ molar ratio	Colour of the product
$A. 1:1$	$I. \text{Pale blue}$
$B. 1:2$	$II. \text{Blue/Purple}$
$C. 1:3$	$III. \text{Violet}$

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