[Cr(H_2O)_6]Cl_3 (At. text{ } no. text{ } of | vedclass

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(D) Magnetic moment $(\mu) = \sqrt{n(n+2)} 	ext{ } B.M.$Given $\mu = 3.83 	ext{ } B.M.$$3.83 = \sqrt{n(n+2)} \implies n \approx 3$This indicates the

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

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(D) Magnetic moment $(\mu) = \sqrt{n(n+2)} 	ext{ } B.M.$Given $\mu = 3.83 	ext{ } B.M.$$3.83 = \sqrt{n(n+2)} \implies n \approx 3$This indicates there are $3$ unpaired electrons.For the complex $[Cr(H_2O)_6]Cl_3$,the oxidation state of $Cr$ is $+3$.The electronic configuration of $Cr$ $(Z=24)$ is $[Ar] 3d^5 4s^1$.Thus,$Cr^{3+}$ is $[Ar] 3d^3$.According to Hund's rule,the $3$ electrons in the $3d$ orbitals occupy the $t_{2g}$ set $(d_{xy}, d_{yz}, d_{xz})$ singly.Therefore,the distribution is $3d_{xy}^1, 3d_{yz}^1, 3d_{xz}^1$.

List-$I$ (Hybridisation)	List-$II$ (Compound/ion)
$A. sp^3d$	$I. [PtCl_4]^{2-}$
$B. sp^3d^2$	$II. SF_6$
$C. dsp^2$	$III. BCl_3$
$D. dsp^3$	$IV. PCl_5$
	$V. ClF_3$

$[Cr(H_2O)_6]Cl_3$ $(At. \text{ } no. \text{ } of \text{ } Cr = 24)$ has a magnetic moment of $3.83 \text{ } B.M.$ The correct distribution of $3d$ electrons in the chromium of the complex is

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The magnetic moment for two complexes of empirical formula $Ni(NH_3)_4(NO_3)_2 \cdot 2H_2O$ is $0$ and $2.84 \ BM$ respectively. The second complex is not a neutral complex. The correct formula and geometry of the first complex is:

Match the following:
List-$I$ (Hybridisation) List-$II$ (Compound/ion)
$A. sp^3d$ $I. [PtCl_4]^{2-}$
$B. sp^3d^2$ $II. SF_6$
$C. dsp^2$ $III. BCl_3$
$D. dsp^3$ $IV. PCl_5$
$V. ClF_3$

The correct match is:

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