Question

Arrange in decreasing order of melting and boiling pouints of hydrides of groups `15,16` and 17
(b) Give the decreasing order of melting and boiling points of `H_(2)ONH_(3)` and `HF`
(c ) Give the decreasing order of boiling points (I) `C_(2)H_(5)OH` (II) `(CH_(3))_(3)NH` (III) `C_(2)H_(5)NH_(2)`
(d) Give the decreasing of solubility in `H_(2)O` (I) `PhNH_(2)` (II) `(C_(2)H_(5))_(2)NH` , (III) `C_(2)H_(5)NH_(2)` .

Answer 1

(a) (i) Melting point of group 15 hydries `NH_(3) gt BiH_(3) gt SbH_(3) gt AsH_(3) gt PH_(3)`
Boiling point of group 15 hydries `BiH_(3) gt SbH_(3) gt NH_(3) gt AsH_(3) gt PH_(3)`
(ii) Melting point of group 16 hybries `H_(2)O gt H_(2)Te gt H_(2)Se gt H_(2)S`
Boiling point of group 16 hybries `H_(2)O gt H_(2)Te gt H_(2)Se gt H_(2)S`
(iii) Melting point of group 17 hydries `HI gt HF gt HBr gt HCI`
Boiling point of group 17 hydries `HF gt HI gt HBr gt HCI`
(b) `H_(2)O gt NH_(3) gt HF` Boiling point `H_(2)O gt HF gt NH_(3)`
(c ) Boiling point of alcohols gt boiling points of amines (due to EN of `O gt ENofN)` (ii) Boiling points of `1^(@) gt2^(@)` amine gt `3^(@)` amine due to the three H-bonds, two H-bonds in `1^(@)` and `2^(@)` amines, `3^(@)` amine do not form H-bonds, since no H-atoms is present Therefore, decreasing boiling order is
`C_(2)H_(5)OH gt C_(2)H_(5)NH_(2) gt (C_(2)H_(5))_(2)NH(I gt III gt II)`
(d) Solubility in `H_(2)O` of `1^(@)` amine `gt 2^(@)` amine `gt 3^(@)` amine due to three H-bonds and two H-bonds in `1^(@)` and `2^(@)` amines respectively Aromatic amines are less solubel in `H_(2)O` due to larger non polar part than aliphatic amines Therefore, decreasing solubility order is
`C_(2)H_(5)NH_(2) gt (C_(2)H_(5))_(2)NH:PhNH_(2)(III gt II gt I)` .