The magnetic moment of the high spin complex | vedclass

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(C) The magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons. Given $\mu = 5.92 \ BM$,we h

Vedclass · Accepted Answer

$t_{2g}^3 e_{g}^2$

Vedclass · Answer

(C) The magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons. Given $\mu = 5.92 \ BM$,we have $\sqrt{n(n+2)} = 5.92$. Solving for $n$,we get $n(n+2) \approx 35$,which implies $n = 5$. For a high spin octahedral complex with $5$ unpaired electrons,the electrons occupy the $t_{2g}$ and $e_g$ orbitals according to Hund's rule. The configuration is $t_{2g}^3 e_g^2$.

$LIST-I$	$LIST-II$
$P. dsp^2$	$1. [FeF_6]^{4-}$
$Q. sp^3$	$2. [Ti(H_2O)_3Cl_3]$
$R. sp^3d^2$	$3. [Cr(NH_3)_6]^{3+}$
$S. d^2sp^3$	$4. [FeCl_4]^{2-}$
	$5. Ni(CO)_4$
	$6. [Ni(CN)_4]^{2-}$

The magnetic moment of the high spin complex is $5.92 \ BM$. What is the electronic configuration?

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Which of the following matches is incorrect?
Complex compounds $-$ Type of hybridization

Match each set of hybrid orbitals from $LIST-I$ with complex(es) given in $LIST-II$. The correct option is

$LIST-I$ $LIST-II$

$P. dsp^2$ $1. [FeF_6]^{4-}$

$Q. sp^3$ $2. [Ti(H_2O)_3Cl_3]$

$R. sp^3d^2$ $3. [Cr(NH_3)_6]^{3+}$

$S. d^2sp^3$ $4. [FeCl_4]^{2-}$

$5. Ni(CO)_4$

$6. [Ni(CN)_4]^{2-}$

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