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Molar conductivities of Li^(-),Na^(+),K^...

Molar conductivities of `Li^(-),Na^(+),K^(+)` and `Rb^(+)` ions in aqueous solutions are in the following order.

A

`Li^(+) gt Na^(+) = K^(+) lt Rb^(+)`

B

`Li^(+) gt Na^(+) gt K^(+) = Rb^(+)`

C

`Rb^(+) gt K^(+) gt Na^(+) gt Li^(+)`

D

`Li^(+) gt Rb^(+) gt K^(+) gt Na^(+)`

Text Solution

Verified by Experts

The correct Answer is:
C
The smaller the size of the ion, the greater is the degree of hydration. Therefore, the extent of hydration decreases from `Li^(+) to Rb^(+)` So, hydrated ionic radii decreases in the order: `Li^(+) gt Na^(+) gt K^(+) gt Rb^(+)`
As a result, hydrated Li ion being largest in ionic size, has lowest mobility in aqueous solution. On the other hand, hydrated `Rb^(+)` ion being smallest in size has the highest mobility in aqueous solution. Higher the ionic mobility, greater is the molar conductivity. Thus, the decreasing order of molar conductivity is,`Rb^(+) gt K^(+) gt Na^(+) gt Li^(+)`
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