Question

Complex X of composition $Cr(H_{2}O{)}_{6}C{l}_{n,}$ has a spin only magnetic moment of 3.83 BM. It reacts with $AgN{O}_3$ and shows geometrical isomerism. The IUPAC nomenclature of X is :

• A. Hexaaqua chromium(III) chloride
• B. Tetraaquadichlorido chromium(III) chloride dihydrate
• C. Hexaaqua chromium(IV) chloride
• D. Tetraaquadichlorido chromium(IV) chloride dihydrate

Answer 1

The correct option is B Tetraaquadichlorido chromium(III) chloride dihydrate

Spin only magnetic moment = 3.8 B.M. This implies, $\mu =\sqrt{\left(n\right(n+2\left)\right)}B.M.$
$\sqrt{16}=4$ implies that $\sqrt{15}$ should be less than four.
This means, $n=3$ as $\sqrt{15}=\sqrt{\left(3\right(3+2\left)\right)}Cr\left(24\right)=\left[Ar\right]4s^{1}3d_{(g.s)}^5$
For 3 unpaired electrons, the oxidation state of Cr should be +3
Since the complex also shows geometrical isomerism and reacts with $AgN{O}_3$, the complex is: $\left[Cr\right(H_{2}O{)}_{4}C{l}_2]Cl.2H_{2}O$
It has the IUPAC name: Tetraaquadichloridochromium (III) chloride dihydrate