To determine the IUPAC name of the complex $Cr(H_2O)_6Cl_n$, we need to consider the given information:
Let us analyze each piece of information step by step:
The spin-only magnetic moment is given by the formula:
where n is the number of unpaired electrons. Given that the magnetic moment is 3.83 BM, we can solve for n:
Upon solving, n approximately equals 3. This indicates there are 3 unpaired electrons.
The presence of 3 unpaired electrons suggests that it is a d^3 configuration. In the case of chromium, this configuration stems from the oxidation state +3 (as Cr^{3+}). Given the hint of geometrical isomerism and the magnetic data, the probable form of the complex is [Cr(H_2O)_4Cl_2]Cl\cdot2H_2O.
The complex will react with $AgNO_3$ to precipitate 1 mole of $AgCl$, indicating the presence of one chloride ion outside the coordination sphere. Also, the complex must be cis if it shows geometrical isomerism, which is only possible if it is octahedral.
The given information correctly aligns with the option:
Tetraaquadichlorido chromium(III) chloride dihydrate
This name indicates:
| Column-I (Complex compound) | Column-II ($\Delta_0$ (CFSE) $\text{cm}^{-1}$) |
| (i) $[Cr(CN)_6]^{3-}$ | (P) 17000 |
| (ii) $[Cr(H_2O)_6]^{3+}$ | (Q) 15000 |
| (iii) $[Cr(en)_3]^{3+}$ | (R) 12000 |
| (iv) $[CrF_6]^{3-}$ | (S) 20000 |