Complex X of composition Cr(H2O)6Cln has a spin only magnetic moment of 3.83 B.M. It reacts with AgNO₃ and shows…

🧠 Three Clues, One Formula

We have $\mathrm{Cr(H_2O)_6Cl_n}$ and three pieces of data:

1. μ = 3.83 BM → $n(n+2) = 14.7 \Rightarrow n = 3$ unpaired electrons → Cr in $\mathrm{d^3}$ → Cr(III).
2. Reacts with $\mathrm{AgNO_3}$ → has at least one outer-sphere (ionisable) Cl⁻.
3. Shows geometrical isomerism → has cis/trans forms → must be $[\mathrm{Cr(H_2O)_4Cl_2}]^+$ (the only Cr(III) octahedral aqua-chloro arrangement that supports cis-trans).

So $\mathrm{Cr(H_2O)_6Cl_n}$ resolves as $[\mathrm{Cr(H_2O)_4Cl_2}]\mathrm{Cl} \cdot 2\mathrm{H_2O}$ → simplified to $[\mathrm{Cr(H_2O)_4Cl_2}]\mathrm{Cl}$.

🗺️ IUPAC Name

Inside bracket: 4 aqua + 2 chlorido + Cr(III).

• Alphabetical order in IUPAC 2005: aqua before chlorido.
• Name the inner ligands first, then metal with OS in Roman numerals, then outer counter-ion.

→ Tetraaquadichloridochromium(III) chloride.

(The outer Cl⁻ is named "chloride" at the end, as a separate counter-ion.)

⚡ The μ → n → d-Count Chain

Chain of inferences from magnetic moment alone:

$\mu \to n \to d{-}\text{count} \to \text{oxidation state}.$

For $\mathrm{\mu = 3.83}$: $n = 3$, $\mathrm{d^3}$, Cr(III), strictly. No other Cr oxidation state gives 3 unpaired in an octahedral aqua field.

⚠️ Don't Use "Chloro" in IUPAC 2005

The IUPAC 2005 recommendation replaced "chloro" with "chlorido" for $\mathrm{Cl^-}$ as a ligand. Old name "tetraaquadichlorochromium(III)" — wrong by current rules. Same for "fluorido", "bromido", "iodido", "hydrido", "oxido".

$\boxed{\text{Answer: (2) Tetraaquadichloridochromium(III) chloride}}$