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Assuming $100 %$ ionization of the following in aqueous solutions,
(i) $[Pt {({NH}_{3})}_{6}] {Cl}_{4}$ (ii) $[Cr {({NH}_{3})}_{6}] {Cl}_{3}$ (iii) $[Co {(H_{2} O)}_{5} Cl] {Cl}_{2}$ (iv) NaCl
The increasing order of conductivity is

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All have equal conductivity

$I < II < III < IV$

$I < IV < II < I$

$IV < III < II < I$

answer is C.

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Detailed Solution

(i) $[Pt {({NH}_{3})}_{6}] {Cl}_{4}$ Gives five ions
(ii) $[Cr {({NH}_{3})}_{6}] {Cl}_{3}$ Gives four ions
(iii) $[Co {(H_{2} O)}_{5} Cl] {Cl}_{2}$ Given three ions
(iv) NaCl Given two ions

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