Question

When a coordination compound \( CoCl_{3} \cdot 6NH_{3} \) is mixed with excess of \( AgNO_{3} \) solution, 3 moles of \( AgCl \) are precipitated per mole of the compound. Write the structural formula of the complex, IUPAC name, its hybridisation and magnetic behaviour on the basis of valence bond theory.

Hint:

Strong field ligands (\( CN^{-}, CO, NH_{3}, en \)) often cause pairing in d6 systems, leading to diamagnetic complexes.

Answer 1

Concept:

• Werner's Theory states that ions outside the coordination sphere are ionizable and react with reagents like \( AgNO_{3} \).

• Since 3 moles of \( AgCl \) are formed, there must be 3 chloride ions outside the brackets.

• Valence Bond Theory (VBT) uses orbital hybridization to explain geometry and magnetism.
Step 1: Determine the structural formula
Total components: 1 Co, 3 Cl, 6 \( NH_{3} \).
If 3 Cl are outside, the 6 \( NH_{3} \) must be inside as ligands.
Formula: \( [Co(NH_{3})_{6}]Cl_{3} \)

Step 2: Determine the IUPAC name
Metal: Cobalt. Ligand: Ammine (6). Counter ion: Chloride.
Oxidation state of Co: \( x + 6(0) + 3(-1) = 0 \Rightarrow x = +3 \).
Name: Hexaamminecobalt(III) chloride

Step 3: Determine hybridisation and geometry
\( Co^{3+} \) configuration: \( [Ar] 3d^{6} 4s^{0} 4p^{0} \).
\( NH_{3} \) is a strong field ligand, causing the 6 electrons in \( 3d \) to pair up. This leaves two \( 3d \) orbitals empty.
Hybridization: \( d^{2}sp^{3} \) (Inner orbital complex).
Geometry: Octahedral.

Step 4: Determine magnetic behaviour
Since all 6 electrons in the \( 3d \) subshell are paired due to the strong field ligand, there are zero unpaired electrons.
Behaviour: Diamagnetic. The final answer is [Co(NH3)6]Cl3, Hexaamminecobalt(III) chloride, d2sp3, diamagnetic.