A coordination compound $\mathrm{Cr}\mathrm{~Cl}_3 \cdot 4 \mathrm{H}_2 0$ precipitates silver chloride when treated with silver nitrate. The molar conductance of its solution corresponds to a total of two ions. Write structural formula of the compound and name it.
Formation of white precipitate with $\mathrm{AgNO}_3$ shows that atleast one Cl ion is present outside the coordination sphere. Moreover only two ions are obtained in solution, so only one $\mathrm{Cl}^{-}$is present outside the sphere. Thus, the formula of the complex is $\left[\mathrm{Co}\left(\mathrm{H}_2\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl}$ and its IUPAC name is Tetraaquadichloridocobalt (III) chloride.
A complex of the type $\left[M(A A)_2 X_2\right]^{n+}$ is known to be optically active. What does this indicate about the structure of the complex? Give one example of such complex.
An optically active complex of the type $\left[M(A A)_2 X_2\right]^{++}$indicates cis-octahedral structure, e.g., cis- $\left[\mathrm{Pt}(\mathrm{en})_2 \mathrm{Cl}_2\right]^{2+}$ or cis- $\left[\mathrm{Cr}(\mathrm{en})_2 \mathrm{Cl}_2\right]^{+}$because its mirror image isomers are non-superimposable.
Non-superimposable isomers of $\left[\mathrm{Pt}(\mathrm{en})_2 \mathrm{Cl}_2\right]^{2+}$.
Magnetic moment of $\left[\mathrm{MnCl}_4\right]^{2-}$ is 5.92 BM . Explain giving reason present.
The magnetic moment 5.92 BM shows that there are five unpaired electrons present in the $d$-orbitals of $\mathrm{Mn}^{2+}$ ion. As a result, the hybridisation involved is $s p^3$ rather than $d s p^2$. Thus tetrahedral structure of $\left[\mathrm{MnCl}_4\right]^{2-}$ complex will show 5.92 BM magnetic moment value.
On the basis of crystal field theory explain why Co(III) forms paramagnetic octahedral complex with weak field ligands whereas it forms diamagnetic octahedral complex with strong field ligands.
With weak field ligands; $\Delta_0< P$, (pairing energy) so, the electronic configuration of Co (III) will be $t_{2 g}^4 e_g^2$ i.e., it has 4 unpaired electrons and is paramagnetic.
With strong field ligands, $\Delta_0>P$ (pairing energy), so pairing occurs thus, the electronic configuration will be $t_{2 g}^6 e_g^0$. It has no unpaired electrons and is diamagnetic.
Why are low spin tetrahedral complexes not formed?
In tetrahedral complex, the $d$-orbital is splitting to small as compared to octahedral. For same metal and same ligand $\Delta_t=\frac{4}{9} \Delta_0$.
Hence, the orbital splitting energies are not enough to force pairing. As a result, low spin configurations are rarely observed in tetrahedral complexes.