Which of the following complex will show maximum ionic conductivity in a aqueous solution?

K_{4} [F e (C N)_{6}]

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

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

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[N i (C O)_{4}]

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Solution

The correct option is A $K_{4} [F e (C N)_{6}]$
(i) $K_{4} [F e (C N)_{6}] a q \to 4 K^{+} + [F e (C N)_{6}]^{4 -} (5 - i o n s)$
(ii) $[C O (N H_{3})_{6}] C l_{3} a q \to [C o (N H_{3})_{6}]^{3 +} (4 - i o n s) + 3 C l^{-}$
(iii) $[C u (N H_{3})_{4}] C l_{2} a q \to [C u (N H_{3})_{4}]^{2 +} (3 - i o n s) + 2 C l^{-}$
(iv) $[N i (C O)_{4}] (a q) \to$ give no ion.

More the ion more is the conductivity.
Hence, option $A$ is correct.

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Similar questions

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. 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}

Q. Among the following complexes, the diamagnetic complexes are:

(K) : K_{3} [F e (C N)_{6}]

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

(M) : N a_{3} [C o (o x a l a t e)_{3}]

(N) : [N i (H_{2} O)_{6}] C l_{2}

(O) : K_{2} [P t (C N)_{4}]

(P) : [Z n (H_{2} O)_{6}] (N O_{3})_{2}

Q. Write IUPAC names of the following coordination compounds.
a)

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

K_{3} [F e (C N)_{6}]

K_{2} [P d (C l)_{4}]

[N i (C o)_{4}]

Q. Among the following, which complex compound is diamagnetic?

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

,
ii)

[N i (N H_{3})_{6}] C l_{2}

,
iii)

[C r (H_{2} O)_{6}] C l_{3}

,
iv)

[F e (H_{2} O)_{6}] C l_{2}

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Which of the following complex will show maximum ionic conductivity in a aqueous solution?

The correct option is A K4[Fe(CN)6](i) K4[Fe(CN)6]aq→4K++[Fe(CN)6]4−(5−ions)(ii) [CO(NH3)6]Cl3aq→[Co(NH3)6]3+(4−ions)+3Cl−(iii) [Cu(NH3)4]Cl2aq→[Cu(NH3)4]2+(3−ions)+2Cl−(iv) [Ni(CO)4](aq)→ give no ion.More the ion more is the conductivity.Hence, option A is correct.