1.

The hydrogen bonds are encountered in HF, H_2 O, NH_3 and HF_2^-. The relative order of energies of hydrogen bonds is

Answer»

`HF gt H_(2)O gt H_(3)N gt HF_(2)^(-)`
`H_(2)O gt HF_(2)^(-) gt HF gt NH_(3)`
`HF gt HF_(2)^(-) gt H_(2)O gt NH_(3)`
`HF_(2)^(-) gt HF gt H_(2)O gt NH_(3)`

Solution :Higher is the electronegativity, stronger is the H-bond. Fluorine has more electronegativity among the GIVEN compound. In `HF_(2)^(-)` there are TWO fluorine atoms that results into EVEN stronger H-bonding.


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