The value of the 'spin only' magnetic moment | vedclass

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Question

(A) The 'spin only' magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.For $\mu = 2.84 \

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$d^4$ (in strong ligand field)

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(A) The 'spin only' magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.For $\mu = 2.84 \ BM$,we have $\sqrt{n(n+2)} = 2.84$,which implies $n(n+2) \approx 8$,so $n = 2$.In a $d^4$ configuration with a strong ligand field,the electrons pair up in the $t_{2g}$ orbitals,resulting in $2$ unpaired electrons $(t_{2g}^4, e_g^0)$.Thus,$n = 2$ and $\mu = \sqrt{2(2+2)} = \sqrt{8} = 2.828 \approx 2.84 \ BM$.

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 value of the 'spin only' magnetic moment for one of the following configurations is $2.84 \ BM$. The correct one is:

Similar Questions

Among $(a) - (d)$,the complexes that can display geometrical isomerism are:
$(a) [Pt(NH_3)_3 Cl]^+$
$(b) [Pt(NH_3) Cl_5]^-$
$(c) [Pt(NH_3)_2 Cl(NO_2)]$
$(d) [Pt(NH_3)_4 ClBr]^{2+}$

$A$ complex compound which can show optical isomerism but not geometrical isomerism is:

Match the following:

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

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

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

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

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

$V$. $\sqrt{15}$

If the magnetic moment of $[Ag(CN)_2]^{-1}$ is zero,then the number of unpaired electrons will be

Which coordination number is associated with geometric isomerism in coordination compounds?

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