The calculated spin-only magnetic moment valu | vedclass

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Question

(A) For both complexes, the central metal ion is $Fe^{3+}$.Electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$.In $[FeCl_4]^-$, $Cl^-$ is a weak fiel

Vedclass · Accepted Answer

$5.9 \ BM, 1.732 \ BM$

Vedclass · Answer

(A) For both complexes, the central metal ion is $Fe^{3+}$.Electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$.In $[FeCl_4]^-$, $Cl^-$ is a weak field ligand, so no pairing of electrons occurs. The number of unpaired electrons $(n)$ is $5$.The spin-only magnetic moment is $\mu = \sqrt{n(n+2)} = \sqrt{5(5+2)} = \sqrt{35} \approx 5.9 \ BM$.In $[Fe(CN)_6]^{3-}$, $CN^-$ is a strong field ligand, causing pairing of electrons. The configuration becomes $t_{2g}^5 e_g^0$, leaving $n = 1$ unpaired electron.The spin-only magnetic moment is $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.732 \ BM$.Thus, the values are $5.9 \ BM$ and $1.732 \ BM$.

Compound	Total stereoisomers
$A$. $[Co(en)_2Cl_2]^+$	$P$. $2$
$B$. $[Co(en)_3]Cl_3$	$Q$. $3$
$C$. $[Pt(NH_3)_2Cl_2]$	$R$. $4$
$D$. $[Pt(NH_3)_3Cl_3]^+$	$S$. $5$

The calculated spin-only magnetic moment values in $BM$ for $[FeCl_4]^-$ and $[Fe(CN)_6]^{3-}$ are

Similar Questions

Compound Total stereoisomers
$A$. $[Co(en)_2Cl_2]^+$ $P$. $2$
$B$. $[Co(en)_3]Cl_3$ $Q$. $3$
$C$. $[Pt(NH_3)_2Cl_2]$ $R$. $4$
$D$. $[Pt(NH_3)_3Cl_3]^+$ $S$. $5$

Match the compounds with their total number of stereoisomers and select the correct code:

Which of the following complex ions is not expected to absorb visible light?

Which of the following has a magnetic moment of $1.73 \, B.M.$?

What is the reason for the deviation between the experimental and theoretical values of magnetic moment?

Which of the following is expected to be a paramagnetic complex?

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