The theoretical value of the magnetic moment | vedclass

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Question

(B) The complex formed by $Fe^{3+}$ and $CN^{-}$ is $[Fe(CN)_6]^{3-}$.In this complex,the oxidation state of $Fe$ is $+3$,which has an electronic conf

Vedclass · Accepted Answer

$1.73$

Vedclass · Answer

(B) The complex formed by $Fe^{3+}$ and $CN^{-}$ is $[Fe(CN)_6]^{3-}$.In this complex,the oxidation state of $Fe$ is $+3$,which has an electronic configuration of $[Ar] 3d^5$.$CN^{-}$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals.For $d^5$ configuration with strong field ligand,the electrons are arranged as $t_{2g}^5 e_g^0$,resulting in $1$ unpaired electron $(n = 1)$.The magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$Substituting $n = 1$,we get $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ B.M.$

List-$I$	List-$II$
$A$. $[Fe(H_2O)_6]^{2+}$	$P$. $0$
$B$. $[Fe(H_2O)_6]^{3+}$	$Q$. $1$
$C$. $[Fe(CN)_6]^{4-}$	$R$. $2$
$D$. $[Fe(CN)_6]^{3-}$	$S$. $4$
$E$. $[Ni(H_2O)_4]^{2+}$	$T$. $5$

The theoretical value of the magnetic moment of the complex formed by the reaction of $Fe^{3+}$ ion with $CN^{-}$ ligand is .......... $B.M.$

Similar Questions

The magnetic moment of a transition element is $\sqrt{15} \, B.M.$ Determine the number of unpaired electrons present in it.

$\mu = \sqrt{15}$ is true for the pair :-

Match the complex ion (in List-$I$) with the number of unpaired electrons (in List-$II$):

List-$I$ List-$II$

$A$. $[Fe(H_2O)_6]^{2+}$ $P$. $0$

$B$. $[Fe(H_2O)_6]^{3+}$ $Q$. $1$

$C$. $[Fe(CN)_6]^{4-}$ $R$. $2$

$D$. $[Fe(CN)_6]^{3-}$ $S$. $4$

$E$. $[Ni(H_2O)_4]^{2+}$ $T$. $5$

Which of the following isomerism is not possible for complexes having molecular formulas:
$(i) \ Pt(SCN)_2 \cdot 3PEt_3$
$(ii) \ CoBr \cdot SO_4 \cdot 5NH_3$
$(iii) \ FeCl_3 \cdot 6H_2O$

The spin-only magnetic moment of the complex present in Fehling's reagent is $B.M.$ (Nearest integer).

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