Question

A Complex P of composition $Cr \left( H _{2} O \right)_{6} Br _{ n }$ has a spin only magnetic moment of 3.83 BM. It reacts with $AgNO_{3}$ and shows geometrical isomerism. The IUPAC nomenclature of P is

• A. Hexaaqua chromium (III) bromide
• B. Dibromidotetraaqua chromium (IV) bromide dihydrate
• C. Tetraaquadibromido chromium (IV) bromide dihydrate
• D. Tetraaquadibromido chromium (III) bromide dihydrate

Answer 1

Answer: (D)

This compound can be identified and named as follows

This compound can be identified and named as follows $Cr \left( H _{2} O \right)_{6} Br _{ n }$

As $\left(\mu_{\text {complex }}\right)_{\text {spin }}=3.8$ B.M. $=\sqrt{n(n+2)}$ On solving $n =3$ Hence the complex is $\left[ Cr \left( H _{2} O \right)_{6}\right] Br _{3}$

oxidation number of $Cr =+3$.

As compound also shows geometrical isomerism so it will be $\left[ Cr \left( H _{2} O \right)_{4} Br _{2}\right] Br .2 H _{2} O$

It is named as $\rightarrow$ Tetraaquadibromido chromium (III) bromide

dihydrate