JEE Main 2020 — Chemistry Inorganic Chemistry

$$\mathrm{Cr}{({H}_{2}O)}_{6}{\mathrm{Cl}}_{n}{({\mu }_{\mathrm{complex}})}_{\mathrm{spin}}=3.83B.M.$$Cr(H2​O)6​Cln​(μcomplex​)spin​=3.83B.M.

From data of magnetic moment oxidation number of Cr should be $+3$+3

Hence, complex is $\mathrm{Cr}{({H}_{2}O)}_{6}{\mathrm{Cl}}_{3}.$Cr(H2​O)6​Cl3​.

The above complex formula can have following different types of complexes-

$1.[\mathrm{Cr}{({H}_{2}O)}_{6}]{\mathrm{Cl}}_{3}$1.[Cr(H2​O)6​]Cl3​ 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

$$2.[\mathrm{Cr}{({H}_{2}O)}_{5}\mathrm{Cl}]{\mathrm{Cl}}_{2}.{H}_{2}O$$2.[Cr(H2​O)5​Cl]Cl2​.H2​O

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

$$3.[\mathrm{Cr}{({H}_{2}O)}_{4}{\mathrm{Cl}}_{2}]\mathrm{Cl}.2{H}_{2}O$$3.[Cr(H2​O)4​Cl2​]Cl.2H2​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.

$$4.[\mathrm{Cr}{({H}_{2}O)}_{3}{\mathrm{Cl}}_{3}]3{H}_{2}O$$4.[Cr(H2​O)3​Cl3​]3H2​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 ).