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A purple complex of $CoCl_3$ with NH₃ gives **2 moles of AgCl** per mole of complex on treatment with excess $AgNO_3$. If Co has CN = 6, the formula of the complex is:
A$[Co(NH_3)_6]Cl_3$
B$[Co(NH_3)_5Cl]Cl_2$
C$[Co(NH_3)_4Cl_2]Cl$
D$[Co(NH_3)_3Cl_3]$
Answer & Solution
Correct answer: B. $[Co(NH_3)_5Cl]Cl_2$
2 ionisable Cl⁻ (give 2 AgCl) + 1 Cl⁻ in coordination sphere = total 3 Cl. CN = 6 means 5 NH₃ + 1 Cl in sphere. Formula: $[Co(NH_3)_5Cl]Cl_2$. (3 AgCl → all Cl outside → $[Co(NH_3)_6]Cl_3$; 1 AgCl → only 1 Cl outside → $[Co(NH_3)_4Cl_2]Cl$.)
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