The magnetic moment of Cr in [Cr(H_2O)_6]Cl_3 | vedclass

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(D) The magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \, BM$,where $n$ is the number of unpaired electrons.Given $\mu = 3.88 \, BM$,we

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$3d_{xy}^1, 3d_{yz}^1, 3d_{xz}^1$

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(D) The magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \, BM$,where $n$ is the number of unpaired electrons.Given $\mu = 3.88 \, BM$,we have $\sqrt{n(n+2)} \approx 3.88$,which implies $n = 3$.In $[Cr(H_2O)_6]Cl_3$,the oxidation state of $Cr$ is $+3$.The electronic configuration of $Cr^{3+}$ is $[Ar] 3d^3$.Since there are $3$ unpaired electrons in the $3d$ subshell,they will occupy three different $d$-orbitals according to Hund's rule.The $d$-orbitals are $d_{xy}, d_{yz}, d_{xz}, d_{x^2-y^2},$ and $d_{z^2}$.Any combination of three distinct $d$-orbitals is possible for the $3d^3$ configuration,but $d_{xy}, d_{yz},$ and $d_{xz}$ are the standard set of $t_{2g}$ orbitals.

List-$I$ $(Compounds)$	List-$II$ $(Hybridization)$
$(a)$ $[Ni(NH_3)_6]^{+2}$	$(i)$ $sp^3 d^2$
$(b)$ $[PtCl_4]^{-2}$	$(ii)$ $dsp^2$
$(c)$ $[Ni(CO)_4]$	$(iii)$ $sp^3$
$(d)$ $[Co(OX)_3]^{-3}$	$(iv)$ $d^2 sp^3$

The magnetic moment of $Cr$ in $[Cr(H_2O)_6]Cl_3$ is $3.88 \, BM$. Which of the following is the possible orbital distribution for $Cr$?

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