Arrange the following complexes in order of increasing electrical conductivity.
(I) $[C o C l_{3} (N H_{3})_{3}]$ (II) $[C o C l (N H_{3})_{5}] C I_{2}$
(III) $[C o (N H_{3})_{6}] C l_{3}$ (IV) $[C o C l_{2} (N H_{3})_{]} 1 C l$

III > II > IV > I

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II > III > IV > I

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II > III > I > IV

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III > IV > II > I

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Solution

The correct option is C III > II > IV > I
Electrical conductivity $\times$ no. / moles of ions present after dissociation.
$I$ $∴ [C o {C l}_{3} {({N H}_{3})}_{3}]$
$I I$ $[C o C l {({N H}_{3})}_{5}] {C l}_{2} \to {[C o C l {({N H}_{3})}_{5}]}^{2 +} + 2 {C l}^{(-)}$
$I V$ $[C o {C l}_{2} {({N H}_{3})}_{4}] C l \to {[C o {C l}_{2} {({N H}_{3})}_{4}]}^{+} + {C l}^{(-)}$
$I I I$ ${[C o {({N H}_{3})}_{6}] C l}_{3} \to {[C o {({N H}_{3})}_{6}]}^{3 +} + 3 {C l}^{(-)}$
Hence the correct answer is,
$I I I > I I > I V > I$

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