Animated Solution for Chemistry - Coordination Compounds: Complex $X$ of composition ${\text{Cr(H}}_2{\text{O)}}_6{\text{Cl}}_n$ has a spin only magnetic moment of $3.83\text{ BM}$. It reacts with ${\text{AgNO}}_3$ and shows geometrical isomerism. The IUPAC nomenclature of $X$ is

The journey to unraveling the identity of our mystery complex $X$ is a beautiful exercise in chemical deduction. We are given a few seemingly disconnected clues: a magnetic moment, a reaction with silver nitrate, and the ability to show geometrical isomerism. Let's piece them together step by step.

Our first and most powerful clue is the spin-only magnetic moment, given as $3.83\text{ BM}$. In coordination chemistry, the magnetic moment $\mu$ is directly tied to the number of unpaired electrons $n$ through the formula:

By substituting our given value, we get:

Solving this equation reveals that $n=3$. This tells us that the central metal ion in our complex possesses exactly three unpaired electrons.

Now, let's look at our central metal, Chromium (Cr). The atomic number of Chromium is 24, and its ground-state electronic configuration is $[\text{Ar}]3d^{5}4s^1$.

To be left with exactly three unpaired electrons in its $d$-orbitals, Chromium must lose three electrons—one from the $4s$ orbital and two from the $3d$ orbitals. This places Chromium in a $+3$ oxidation state, giving it a $3d^3$ configuration.

Since the overall complex must be electrically neutral and water (${\text{H}}_2\text{O}$) is a neutral ligand, the $+3$ charge of the Chromium ion must be balanced by exactly three chloride (${\text{Cl}}^{-}$) ions. Therefore, the empirical formula of our complex is ${\text{CrCl}}_3\cdot 6{\text{H}}_2\text{O}$.

The problem states that the complex exhibits geometrical isomerism. For an octahedral complex to show geometrical isomerism (like cis and trans forms), it cannot have all identical ligands (type ${\text{MA}}_6$) or just one different ligand (type ${\text{MA}}_5\text{B}$). It must have at least two of one type of ligand and at least two of another, such as ${\text{MA}}_4{\text{B}}_2$ or ${\text{MA}}_3{\text{B}}_3$.

This constraint narrows down our possible coordination spheres to two candidates: 1. $[{\text{Cr(H}}_2{\text{O)}}_4{\text{Cl}}_2]\text{Cl}\cdot 2{\text{H}}_2\text{O}$ 2. $[{\text{Cr(H}}_2{\text{O)}}_3{\text{Cl}}_3]\cdot 3{\text{H}}_2\text{O}$

Here is where the final piece of the puzzle falls into place. The complex reacts with ${\text{AgNO}}_3$. This reaction is a classic test for the presence of ionizable halide ions located outside the coordination sphere.

If our complex were $[{\text{Cr(H}}_2{\text{O)}}_3{\text{Cl}}_3]\cdot 3{\text{H}}_2\text{O}$, all the chloride ions would be locked tightly inside the coordination sphere, and no reaction with silver nitrate would occur.

Therefore, our complex must be $[{\text{Cr(H}}_2{\text{O)}}_4{\text{Cl}}_2]\text{Cl}\cdot 2{\text{H}}_2\text{O}$, which has one ionizable chloride ion ready to precipitate as $\text{AgCl}$. This ${\text{MA}}_4{\text{B}}_2$ type complex perfectly satisfies the condition for geometrical isomerism, existing in both cis (chlorides adjacent) and trans (chlorides opposite) forms.

Finally, we apply the IUPAC rules to name our deduced complex, $[{\text{Cr(H}}_2{\text{O)}}_4{\text{Cl}}_2]\text{Cl}\cdot 2{\text{H}}_2\text{O}$.

Inside the coordination sphere, we list the ligands alphabetically: four water molecules become tetraaqua, and two chloride ions become dichlorido. The central metal is chromium(III). Outside the sphere, we have a chloride counter-ion and two water molecules of crystallization, termed dihydrate.

Stitching it all together, the official IUPAC name is tetraaquadichlorido - chromium (III) chloride dihydrate.