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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

a
A
b
B
c
D
d
E

detailed solution

Correct option is D

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 < SbH3  < BiH3Bond length order: NH3 < PH3 < AsH3 < SbH3  < BiH3Bond energy order :NH3 > PH3 > AsH3 > SbH3 > BiH3Enthalpy of formation order:NH3 > PH3 > AsH3 > SbH3 > BiH3Bond angle order : NH3 > PH3 > AsH3 > SbH3 > BiH3Thermal stability order : NH3 > PH3 > AsH3 > SbH3 > BiH3Basic nature order :NH3 > PH3 > AsH3 > SbH3 > BiH3Reducing power order :NH3 < PH3 < AsH3 < SbH3  < BiH3H- bonding capacity order : NH3 > PH3 > AsH3 > SbH3 > BiH3Central atom electro negativity order :NH3 > PH3 > AsH3 > SbH3 > BiH3S-character percentage at central atom order : NH3 > PH3 > AsH3 > SbH3 > BiH3Melting point order :  PH3 < AsH3  < SbH3  < NH3< BiH3Boiling point order : PH3 < AsH3 AsH3 > SbH3 > NH3 >BiH3Enthalpy of vapourisation :  PH3 < AsH3  < SbH3  < NH3< BiH3

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Similar Questions

Bond dissociation enthalpy of E - H (E = element) bonds is given below, which of the compounds will act as strongest reducing agent?

CompoundNH3PH3AsH3SbH3
dissE-H/kJ mol-1389322297255


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