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(B) To determine the magnetic moment,we calculate the number of unpaired electrons $(n)$ in each complex:$1$. $[Cr(H_2O)_6]^{2+}$: $Cr^{2+}$ is $3d^4$

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$[Cr(H_2O)_6]^{2+} , [Fe(H_2O)_6]^{2+}$

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(B) To determine the magnetic moment,we calculate the number of unpaired electrons $(n)$ in each complex:$1$. $[Cr(H_2O)_6]^{2+}$: $Cr^{2+}$ is $3d^4$. With a weak field ligand $(H_2O)$,$n = 4$.$2$. $[Fe(H_2O)_6]^{2+}$: $Fe^{2+}$ is $3d^6$. With a weak field ligand $(H_2O)$,$n = 4$.$3$. $[CoCl_4]^{2-}$: $Co^{2+}$ is $3d^7$. With a weak field ligand $(Cl^-)$,$n = 3$.$4$. $[Mn(H_2O)_6]^{2+}$: $Mn^{2+}$ is $3d^5$. With a weak field ligand $(H_2O)$,$n = 5$.Since $[Cr(H_2O)_6]^{2+}$ and $[Fe(H_2O)_6]^{2+}$ both have $n = 4$,they will have the same magnetic moment.

List-$I$ (Complex)	List-$II$ (Spin only Magnetic moment)
$A$. $[CoF_6]^{3-}$	$I$. $0$
$B$. $[Co(C_2O_4)_3]^{3-}$	$II$. $\sqrt{24}$
$C$. $[FeF_6]^{3-}$	$III$. $\sqrt{8}$
$D$. $[Mn(CN)_6]^{3-}$	$IV$. $\sqrt{35}$
	$V$. $\sqrt{15}$

Which of the following pairs will have the same value of magnetic moment?

Similar Questions

Match the following:

List-$I$ (Complex) List-$II$ (Spin only Magnetic moment)

$A$. $[CoF_6]^{3-}$ $I$. $0$

$B$. $[Co(C_2O_4)_3]^{3-}$ $II$. $\sqrt{24}$

$C$. $[FeF_6]^{3-}$ $III$. $\sqrt{8}$

$D$. $[Mn(CN)_6]^{3-}$ $IV$. $\sqrt{35}$

$V$. $\sqrt{15}$

The magnetic moment of the complexes is zero in:

Which of the following does not have optical isomers?

The complexes $[Co(NH_3)_6][Cr(CN)_6]$ and $[Cr(NH_3)_6][Co(CN)_6]$ are the examples of which type of isomerism?

$[(C_6H_5)_2 Pd (SCN)_2]$ and $[(C_6H_5)_2 Pd (NCS)_2]$ are :

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