Arrange the following compounds in the order of decreasing molar conductivity in aqueous solution.
a. K[Co $(N H_{3})_{2} (N O_{2})_{4}$ ]
b. [Cr $(N H_{3})_{3} (N O_{2})_{3}$ ]
c. $[C r (N H_{3})_{5} (N O_{2})]_{3} [C o (N O_{2})_{6}]_{2}$
d. Mg[Cr $(N H_{3}) (N O_{2})_{5}$ ]

c>d>a>b

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c>a>d>b

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c>b>a>d

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b>a>d>c

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Solution

The correct option is A c>d>a>b
Conductance or conductivity in an aqueous solution depends on:
* Greater the number of ions in solution, more is the conductance.
* For same number of ions, greater the charge on the ion, more is the conductance.
* If the number of ions and charge on the ions are equal, then conductance is inversely proportional to its size in the aqueous solution.
* For example, the size of $C s^{\oplus}$ > $R b^{\oplus}$ > $K^{\oplus}$ > $N a^{\oplus}$ > $L i^{\oplus}$
Thus, smallest ion should move more rapidly, however, ionic mobility shows the reverse trend because smaller ions are more extensively hydrated (due to high charge density) and thus hydrated ion size for these ions is:
$L i^{\oplus} (a q) > N a^{\oplus} (a q) > K^{\oplus} (a q) > R b^{\oplus} (a q) > C s^{\oplus} (a q)$
i. a. $_{K}^{\oplus} [\begin{matrix} + 3 & 0 & - 1 \times 4 C o & (N H_{3})_{2} & (N O_{2})_{4} \end{matrix}] + a q ⇌ K^{\oplus} + [C o (N H_{3})_{2} (N O_{2})_{4}]^{\oplus}$
(Two ions with charge +1 or – 1)
b. $[\begin{matrix} + 3 & 0 & - 1 \times 3 C r & (N H_{3})_{2} & (N O_{2})_{3} \end{matrix}] + a q ⇌$ No ionization (no ions with zero charge)
c. ${[\begin{matrix} + 3 & 0 & - 1 C r & (N H_{3})_{5} & (N O_{2}) \end{matrix}]}_{3} {[\begin{matrix} + 3 & - 1 \times 6 C o & (N O_{2})_{6} \end{matrix}]}_{2} + a q ⇌ 3 {[\begin{matrix} + 3 & 0 & - 1 C r & (N H_{3})_{5} & (N O_{2}) \end{matrix}]}^{2 +} + 2 {[\begin{matrix} + 3 & - 1 \times 6 C o & (N O_{2})_{6} \end{matrix}]}^{3 -}$
(Five ions with charge = +6 or -6)
d. $M g^{+ 2} {[\begin{matrix} + 3 & 0 & - 1 \times 5 C r & (N H_{3})_{5} & (N O_{2})_{5} \end{matrix}]}^{2 -} + a q ⇌ M g^{2 +} + {[\begin{matrix} + 3 & 0 & - 1 \times 5 C r & (N H_{3})_{5} & (N O_{2})_{5} \end{matrix}]}^{2 -}$
(Two ions with charge = +2 or -2
So the decreasing order of conductance is:
c>d>a>b

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