Question

Complex $\mathrm{X}$ of composition$\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6{\mathrm{Cl}}_{\mathrm{n}}$, has a spin-only magnetic moment of $3.83$ $\mathrm{BM}$. It reacts with ${\mathrm{AgNO}}_3$and shows geometrical isomerism. The IUPAC nomenclature of $\mathrm{X}$ is

• A. Hexaaquachromium(III) chloride
• B. Tetraaquadichloridochromium(III) chloride dihydrate
• C. Hexaaquachromium(IV) chloride
• D. Tetraaquadichloridochromium(IV) chloride dihydrate

Answer 1

The correct option is B

Tetraaquadichloridochromium(III) chloride dihydrate

The explanation for the correct option:

(B) Here spin only magnetic moment $=3.8\mathrm{BM}$.

So, $\sqrt{n\left(n+2\right)}=3.8$ ($\sqrt{16}=4$ that means$\sqrt{15}$ should be less than $4$)

So, $\sqrt{n\left(n+2\right)}=\sqrt{15}$

Or, $\sqrt{n\left(n+2\right)}=\sqrt{3\left(3+2\right)}$

Hence, $n=3$

For $3$ unpaired electrons, the oxidation state of $\mathrm{Cr}$ should be $+3$

${\mathrm{Cr}}^{3+}$can be attained if the complex has a structure that looks like: $\left[\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4{\mathrm{Cl}}_2\right]\mathrm{Cl}.2{\mathrm{H}}_2\mathrm{O}$ has the IUPAC name: Tetraaquadichloridochromium(III) chloride dehydrate.

In the ionosphere zone, there is a chloride anion that's why it reacts with silver nitrate to form $\mathrm{AgCl}$ precipitation.

The explanation for the incorrect options:

In Hexaaquachromium(IV) chloride and Tetraaquadichlorido chromium(IV) chloride dihydrate, in both cases, chromium is in an $+4$oxidation state.

In Hexaaqua chromium(III) chloride, there is an absence of $3$ unpaired electrons.

Hence, option (B) is correct.