Looking for classes? Ksquare Career Institute, Bengaluru →
Cobalt (III) chloride forms several octahedral complexes with ammonia. Which of the following will not give test for chloride ions with silver nitrate at $25^\circ C$?
A
$CoCl_3\cdot 3NH_3$
B
$CoCl_3\cdot 4NH_3$
C
$CoCl_3\cdot 5NH_3$
D
$CoCl_3\cdot 6NH_3$
Detailed Solution
$CoCl_3\cdot 3NH_3$ is $[Co(NH_3)_3Cl_3]$: all three chlorides are coordinated inside the sphere.
No chloride ion is ionisable, so $[Co(NH_3)_3Cl_3] \xrightarrow{AgNO_3}$ no precipitate.
No chloride ion is ionisable, so $[Co(NH_3)_3Cl_3] \xrightarrow{AgNO_3}$ no precipitate.