Question

When $CoCl_3 \cdot 6NH_3$ is mixed with excess $AgNO_3$ solution, 3 moles of $AgCl$ are precipitated per mole. Write its structural formula, IUPAC name, hybridisation and magnetic behaviour (VBT).

Hint:

Strong field ligands ($NH_3$, $CN^-$, CO) force pairing $\rightarrow$ low spin $\rightarrow$ inner orbital complex $\rightarrow$ $d^2sp^3$ for octahedral.

Answer 1

Step 1: AgCl test interpretation.
3 mol AgCl precipitated per mol means 3 free $Cl^-$ ions outside the coordination sphere. Since formula is $CoCl_3 \cdot 6NH_3$, all $Cl^-$ are counter ions and all $NH_3$ are coordinated. Structural formula: $[Co(NH_3)_6]Cl_3$.
Step 2: IUPAC name.
Six ammine ligands around $Co$ in $+3$ state (charge: $6 \times 0 + Co = +3$ to balance $3Cl^-$). IUPAC: Hexaamminecobalt(III) chloride.
Step 3: Hybridisation.
$Co^{3+}$ has $3d^6$. $NH_3$ is a strong field ligand, pairs all 6 electrons into 3 $t_{2g}$ orbitals, leaving two inner $3d$ orbitals empty. Hybridisation: $d^2sp^3$ (two $3d$, one $4s$, three $4p$).
Step 4: Magnetic behaviour.
All 6 electrons are paired ($t_{2g}^6 e_g^0$); zero unpaired electrons. The complex is diamagnetic.