Cobalt (III) chloride forms several octahedral complexes with ammonia. Which of the following will not give test for chloride ions with silver nitrate at $25^{\circ}$ ?

$C o C l_{3} .3 N H_{3}$

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$C o C l_{3} .4 N H_{3}$

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$C o C l_{3} .5 N H_{3}$

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$C o C l_{3} .6 N H_{3}$

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Solution

The correct option is A
$C o C l_{3} .3 N H_{3}$

$[C o (N H_{3})_{6}] C l_{3} \to [C o (N H_{3})_{6}]^{3 +} + 3 C l^{-} [C o (N H_{3})_{3} C l_{3}] \to [C o (N H_{3})_{3} C l_{3}] [C o (N H_{3})_{4} C l_{2}] C l \to [C o (N H_{3})_{4} C l_{2}]^{+} + C l^{-} [C o (N H_{3})_{5} C l] C l_{2} \to [C o (N H_{3})_{5} C l]^{2 +} + 2 C l^{-}$
So, $[C o (N H_{3})_{3} C l_{3}]$ does not ionise so does not give test for chloride ions.

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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∘?

The correct option is A CoCl3.3NH3[Co(NH3)6]Cl3→[Co(NH3)6]3++3Cl−[Co(NH3)3Cl3]→[Co(NH3)3Cl3][Co(NH3)4Cl2]Cl→[Co(NH3)4Cl2]++Cl−[Co(NH3)5Cl]Cl2→[Co(NH3)5Cl]2++2Cl− So, [Co(NH3)3Cl3] does not ionise so does not give test for chloride ions.

Cobalt (III) chloride forms several octahedral complexes with ammonia. Which of the following will not give test for chloride ions with silver nitrate at $25^{\circ}$ ?