Among the following,the pair of paramagnetic | vedclass

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(A) The correct answer is $A$.$1$. $K_{3}[Fe(CN)_{6}]$: The central metal ion is $Fe^{3+}$,which has a $[Ar] 3d^{5}$ configuration as a free ion.In th

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$K_{3}[Fe(CN)_{6}]$ and $K_{3}[CoF_{6}]$

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(A) The correct answer is $A$.$1$. $K_{3}[Fe(CN)_{6}]$: The central metal ion is $Fe^{3+}$,which has a $[Ar] 3d^{5}$ configuration as a free ion.In the strong ligand field of $CN^{-}$,the $d$-orbitals split to become $t_{2g}^{5} e_{g}^{0}$.There is one unpaired electron,therefore it is paramagnetic.$2$. $K_{3}[CoF_{6}]$: The central metal ion is $Co^{3+}$.The free metal ion configuration is $[Ar] 3d^{6}$.In the weak ligand field of $F^{-}$,the $d$-orbitals split to become $t_{2g}^{4} e_{g}^{2}$.There are four unpaired electrons,therefore it is paramagnetic.$3$. In $K_{4}[Fe(CN)_{6}]$,the central ion is $Fe^{2+}$ $(d^{6})$,and in $[Co(NH_{3})_{6}]Cl_{3}$,the central ion is $Co^{3+}$ $(d^{6})$.Both $CN^{-}$ and $NH_{3}$ are strong field ligands,so $\Delta_{0} > P$.Thus,the $d^{6}$ configuration splits to $t_{2g}^{6} e_{g}^{0}$.There are no unpaired electrons,therefore both are diamagnetic.

List $I$ (Complex)	List $II$ (Type of isomerism)
$A$. $[Pt(NH_3)_2Cl_2]$	$I$. Optical
$B$. $[Co(en)_3]^{3+}$	$II$. Solvate
$C$. $[Co(NH_3)_5NO_2]Cl_2$	$III$. Geometrical
$D$. $[Cr(H_2O)_6]Cl_3$	$IV$. Linkage

Among the following,the pair of paramagnetic complexes is

Similar Questions

Which one of the following species does not have a magnetic moment of $1.73 \ BM$ (spin only value)?

Which of the following complex ions shows geometrical isomerism?

Match List $I$ with List $II$ :
List $I$ (Complex) List $II$ (Type of isomerism)
$A$. $[Pt(NH_3)_2Cl_2]$ $I$. Optical
$B$. $[Co(en)_3]^{3+}$ $II$. Solvate
$C$. $[Co(NH_3)_5NO_2]Cl_2$ $III$. Geometrical
$D$. $[Cr(H_2O)_6]Cl_3$ $IV$. Linkage

Choose the correct answer from the options given below :

Which of the following complexes will not show optical isomerism?

Which of the following does not have optical isomers?

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