1.

Which one of the following is the correct decreasing order or boiling point ?

Answer»

`H_(2)Te GT H_(2)O gt H_(2)Se gt H_(2)S`
`H_(2)O gt H_(2)S gt H_(2)Se gt H_(2)Te`
`H_(2)Te gt H_(2)Se gt H_(2)S gt H_(2)O`
`H_(2)O gt H_(2)Te gt H_(2)Se gt H_(2)S`

Solution :Among the hydrides of group 16 elements, boiling point of `H_(2)O` is higher than `H_(2)S` (difference in boiling points of `H_(2)O and H_(2)S` is around `200^(@)C`) because of strong inter molecular hydrogen bonding. After the decrease in boiling point from `H_(2)O` to `H_(2)S`, from `H_(2)S " to " H_(2)Te`, it INCREASES due to increase in size of the atoms from S to Te which increases the magnitude of van der WAAL's forces among the molecules. So, the CORRECT ORDER of boiling points is
`H_(2)O gt H_(2)Te gt H_(2)Se gt H_(2)S`


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