Highest conductivity is seen in:

K2 [PtCl6]

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K4 [Fe(CN)6]

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[Co (NH3)6] Cl3

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[ Co( NH3)3 (NO2)3]

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Solution

The correct option is B K4 [Fe(CN)6]
The complex $K_{4} [F e (C N)_{6}]$ will show the higher conductivity because among the given option it will dissociate into maximum 5 ions as $4 K^{+}$ and a $[F e (C N)_{6}]^{4 -}$ . Hence the correct option will be B.

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Similar questions

Highest molar conductivity will be in the case of

a)[CO(NH3)6Cl3]

b)[CO(NH3)3Cl3]

K_{2} [P t {C l}_{6}]

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

K [A g {(C N)}_{2}]

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

[A g {(N H_{3})}_{2}] C l

[C r {(N H_{3})}_{6}] {C l}_{3}

IUPAC nomenclature of following co-ordination compounds:

1) K₄[Fe(CN)₆]

2)[Co(NH₃)₅ cl] cl₂

3)[Co(NH₃)₅(NO₂)] cl₂

4)[Cr(Co)₅]

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

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

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

K_{2} [N i (C N)_{4}]

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