Chemistry - Coordination Chemistry Question with Solution | TestHub

$\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6{\mathrm{Cl}}_{\mathrm{n}}{\left({\mathrm{\mu }}_{\mathrm{complex}}\right)}_{\mathrm{spin}}=3.83 \mathrm{B}.\mathrm{M}.$
From data of magnetic moment oxidation number of Cr should be $+3$
Hence, complex is $\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6{\mathrm{Cl}}_3.$

The above complex formula can have following different types of complexes-
$1. \left[\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]{\mathrm{Cl}}_3$ As chlorine atoms are outside of co-ordination sphere hence it can react with Silver nitrate to form silver chloride, but can not show Geometrical Isomerism because all 6 ligands are same i.e. aqua

$2. \left[\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_5\mathrm{Cl}\right]{\mathrm{Cl}}_2.{\mathrm{H}}_2\mathrm{O}$ 

This complex can react with silver nitrate as chlorine atoms are outside of co-ordination sphere, but can not show Geometrical Isomerism.

$3. \left[\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4{\mathrm{Cl}}_2\right]\mathrm{Cl}.2{\mathrm{H}}_2\mathrm{O}$

This complex can react with silver nitrate as one chlorine atom is outside of co-ordination sphere and also show Geometrical isomerism. The IUPAC name of this complex is Tetraaquadichloridochromium(III) chloride dihydrate.

$4. \left[\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_3{\mathrm{Cl}}_3\right]3{\mathrm{H}}_2\mathrm{O}$

This complex can not react with silver nitrate because not chlorine atom is outside of co-ordination sphere, but this complex can show Geometrical isomerism(Facial and meridional ).