When one mole of each of the following complex is treated with an excess of $A g N O_{3}$ .which will give maximum amount of $A g C l$ ?

[C o (N H_{3})_{6}] C l_{3}

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[C o (N H_{3})_{5} C l] C l_{2}

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[C o (N H_{3}) C l_{2}] C l

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[C o (N H_{3})_{3} C l_{3}]

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Solution

The correct option is A $[C o (N H_{3})_{6}] C l_{3}$
In coordination compounds, the counter ions can react to give another product but the complex ion will not split in a solution.

Here, when one mole of a complex given when reacted with $A g N O_{3}$ then the maximum counter ion compound will give the maximum amount of $A g C l$ .

Hence, $[C o (N H_{3})_{6}] C l_{3}$ having a maximum number of 3 counter ions in the given options will produce more amount of $A g C l$ .

$[C o (N H_{3})_{6}] C l_{3} \to [C o (N H_{3})_{6}]^{3 +} (a q) + 3 C l^{-} (a q)$
$A g N O_{3} + C l^{-} (a q) \to A g C l + N O_{3}^{-} (a q)$

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Similar questions

Q. When one mole of each of the following complexes is treated with excess of

A g N O_{3}

, which will give maximum amount of

A g C l

Q. One mole of the complex with molecular formula

[C o (N H_{3})_{5} C l_{3}]

gives two moles of

A g C l

with excess of

A g N O_{3}

. The complex is

Arrange the following complexes in the increasing order of conductivity of their solution

$[C o (N H_{3})_{3} C l_{3}], [C o (N H_{3})_{4} C l_{2}] C l, [C o (N H_{3})_{6}] C l_{3}, [C o (N H_{3})_{5} C l] C l_{2}$

Q. Which of the following does not give a precipitate of

A g C l

with

A g N O_{3}

solution?

Q. The correct order of the following complexes according to increasing conductivity of their solutions will be:
(i)

[C o (N H_{3})_{3} C l_{3}]

(ii)

[C o (N H_{3})_{4} C l_{2}] C l

(iii)

[C o (N H_{3})_{6}] C l_{3}

(iv)

[C o (N H_{3})_{5} C l] C l_{2}

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When one mole of each of the following complex is treated with an excess of AgNO3.which will give maximum amount of AgCl?

When one mole of each of the following complex is treated with an excess of $A g N O_{3}$ .which will give maximum amount of $A g C l$ ?