The ionic conductance of the following cations in a given concentration are in the order

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${Li}^{+} > {Na}^{+} > K^{+} > {Rb}^{+}$

${Li}^{+} = {Na}^{+} < K^{+} < {Rb}^{+}$

${Li}^{+} < {Na}^{+} > K^{+} > {Rb}^{+}$

${Li}^{+} < {Na}^{+} < K^{+} < {Rb}^{+}$

answer is A.

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Detailed Solution

$c o n d u c \tan c e α \frac{1}{s i z e o f i o n}$

The ionic size of the given cations at given concentration are $L {i^{+}}_{(a q)}^{} > N a_{(a q)}^{+} > K_{(a q)}^{+} > R b_{(a q)}^{+}$

Note : it is mentioned that “ at given concentration” so we can assume that this is aqeous solution.

Therefore the conductance order is : $L i_{(a q)}^{+} < N a_{(a q)}^{+} < K_{(a q)}^{+} < R b_{(a q)}^{+}$

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