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.

(a)

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.
(c) 

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. 
(d)  

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

H2O>H2S>H2Se>H2Te

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.