volatility∝1b.pBoiling point trend is: H2S H2Se>H2Te>H2O

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$H_{2} S < H_{2} S e < H_{2} T e < H_{2} O ......$ volatility

$H_{2} O < H_{2} S < H_{2} S e < H_{2} T e ......$ acidic strength

$H_{2} O < H_{2} S < H_{2} S e < H_{2} T e ......$ reducing power

$H_{2} O > H_{2} S > H_{2} S e > H_{2} T e ......$ bond angle

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Detailed Solution

$volatility \propto \frac{1}{b . p}$

Boiling point trend is:

$H_{2} S < H_{2} S e < H_{2} T e < H_{2} O$

Water has high boiling point due to hydrogen bonding. After that boiling point decreases from H₂Te to H₂S because of decrease of van der Waal forces.

As boiling point inversely proportional to volatility. Therefore, order of volatility is

$H_{2} S > H_{2} S e > H_{2} T e > H_{2} O$

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