Which of the following orders are correct as per the properties mentioned against each?

a. As2O3 < SiO2 < P2O3 < SO2                                        Acid strength.

b. AsH3 < PH3 < NH3                                                      Enthalpy of vaporisation.

c. S < O < Cl < F                                                             More negative electron gain enthalpy.

d. H2O > H2S > H2Se > H2Te                                          Thermal stability.

1. (a, b)
2. (b, c)
3. (c, d)
4. (a, d)



Hint: Enthalpy of vaporization increases down the group for the 15th group hydride.


As the electronegativity of the central atom increases, the acidic character also increases.

Increases order is  As2O3 < SiO2 < P2O3 < SO2

Hence, first is the correct statement.

(b) The  enthalpy of vaporization in correct order is as follows:

AsH3 > PH3 > NH3

Hence, statement second is incorrect.

The negative electron gain enthalpy value is as follows:

S = -200 kJ mol-1 
O = – 141 kJ mol-1 
Cl = – 349  kJ mol-1 
F = – 328 kJ mol-1 
The correct order of negative electron gain enthalpy is Cl > F > S > O.

Hence, third statement is incorrect. 

Thermal stability decreases on moving from top to bottom due to an increase in its bond length.


This is related to the bond dissociation enthalpies of the E-H bonds where E stands for the element.
E-H bond: O−H, S−H, Se−H, and Te−H,
Bond dissociation enthalpy : (kJ mol−1) 463, 347, 276, 238

Based on bond dissociation enthalpy, H2O is maximum stable thermally while H2Te is the least stable. Hence, statement fourth is correct.