Question

Why is the boiling point of water higher than would be expected from the trend of Group 6 hydrides? Why is the boiling point of `H_2Te` higher than `H_2Se`?

Answer 1

Because of hydrogen bonding for the first series.....and for the telluride and selenide, probably dispersion forces....

Explanation:

Hydrogen bonding occurs where hydrogen is bound to a strongly electronegative element, which polarizes electron density towards itself. The result? Charge separation occurs, and in say water, or `HF`, or `NH_3`, hydrides in which the heteroatom is strongly electronegative, we could represent this polarity by....

`stackrel(delta+)H-stackrel(delta-)O-stackrel(delta+)H`; `"normal boiling point"` `100` `""^@C`

`stackrel(delta+)H-stackrel(delta-)F;` `"normal boiling point"` `19.5` `""^@C`

`stackrel(delta+)H-stackrel(delta-)N(-stackrel(delta+)H)_2;` `"normal boiling point"` `-33.3` `""^@C`

In bulk solution, the dipoles line up appropriately, and this constitutes a potent force of INTERMOLECULAR interaction. And complare the normal boiling points of these hydrides with say `PH_3`, or `H_2S`, or `HCl`...

`H-S-H`; `"normal boiling point"` `-60` `""^@C`

`stackrel(delta+)H-stackrel(delta-)Cl;` `"normal boiling point"` `-85.1` `""^@C`

`PH_3;` `"normal boiling point"` `-87.7` `""^@C`

Note that hydrogen sulfide, phosphine, etc. are bigger molecules than their congeners....and with more electrons, might be facilely expected to more involatile. The opposite is true, and ths may be attributed to the extent of hydrogen bonding that acts as intermolecular force in the first row hydrides....

For `H_2Te`, `"normal boiling point"` `-2.2` `""^@C` and `H_2Se` `"normal boiling point"` `-41.2` `""^@C` (both of which SMELL really FOUL), dispersion forces seem to dominate....many electron tellurium, has the greater possibility of intermolecular interaction..

Anyway, please check on the quoted boiling points to see if I have got the right data (for once!).