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The first, second and third ionization potentials of an element are 100, 150 and 1500 eV respectively. If the number of moles of $H C l$ and $H_{2} S O_{4}$ required separately to neutralize one mole of the hydroxide of that element is X and Y respectively then $\frac{X}{Y}$ is ……

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answer is 2.

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Detailed Solution

The difference between 2^ndand 3^rd I.E is more so element belongs to 2^nd group.

$M {(O H)}_{2} + 2 H C l \to M C l_{2} + 2 H_{2} O$ $; M {(O H)}_{2} + H_{2} S O_{4} \to M S O_{4} + 2 H_{2} O$

$X = 2; Y = 1$ then $\frac{X}{Y} = \frac{2}{1} = 2$

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