First slide
Chemical properties
Question

In which of the fallowing molecule have fourth highest enthalpy of formation (Hf ) of   in 15th group hydrides( EH3)

A=NH3, B= PH3, C=AsH3, D= SbH E= BiH3

Easy
Solution

All the elements of Group 15 form hydrides of the type EH3 where E = N, P, As, Sb or Bi. The hydrides show regular gradation in their properties. The stability of hydrides decreases from NH3 to BiH3 which can be observed from their bond dissociation enthalpy. Consequently, the reducing character of the hydrides increases. Ammonia is only a mild reducing agent while BiH3 is the strongest reducing agent amongst all the hydrides. Basicity also decreases in the order NH3 > PH3 > AsH3 > SbH3 > BiH3. Due to high electronegativity and small size of nitrogen, NH3 exhibits hydrogen bonding in solid as well as liquid state. Because of this, it has higher melting and boiling points than that of PH3 .

Central atom size order :NH3 < PH3 < AsH3 < SbH < BiH3

Bond length order: NH3 < PH3 < AsH3 < SbH < BiH3

Bond energy order :NH3 > PH3 > AsH3 > SbH3 > BiH3

Enthalpy of formation order:NH3 > PH3 > AsH3 > SbH3 > BiH3

Bond angle order : NH3 > PH3 > AsH3 > SbH3 > BiH3

Thermal stability order : NH3 > PH3 > AsH3 > SbH3 > BiH3

Basic nature order :NH3 > PH3 > AsH3 > SbH3 > BiH3

Reducing power order :NH3 < PH3 < AsH3 < SbH < BiH3

H- bonding capacity order : NH3 > PH3 > AsH3 > SbH3 > BiH3

Central atom electro negativity order :NH3 > PH3 > AsH3 > SbH3 > BiH3

S-character percentage at central atom order : NH3 > PH3 > AsH3 > SbH3 > BiH3

Melting point order :  PH3 < AsH3  < SbH < NH3< BiH3

Boiling point order : PH3 < AsH3 <NH3  < SbH < BiH3

vapourisation order : PH3 > AsH3 > SbH3 > NH3 >BiH3

Enthalpy of vapourisation :  PH3 < AsH3  < SbH < NH3< BiH3

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