Advertisements
Advertisements
Question
Give reasons for the following : Oxygen has less electron gain enthalpy with negative sign than sulphur.
Advertisements
Solution
The elements of group 16 have two electrons less than the nearest noble gas configuration. Therefore, they have a high tendency to accept two additional electrons and hence have large negative electron gain enthalpies next only to the halogens. The electron gain enthalpy of oxygen is however least negative in this group. This is due to its small size. As a result of which, the electron – electron repulsions in the relatively small 2p-subshell are comparatively large and hence the incoming electrons are not accepted with the same ease in case of other elements of this group.
APPEARS IN
RELATED QUESTIONS
Which of the following does not react with oxygen directly?
Zn, Ti, Pt, Fe
Why does NH3 form hydrogen bond but PH3 does not?
Knowing the electron gain enthalpy values for \[\ce{O -> O-}\] and \[\ce{O -> O^{2-}}\] as −141 and 702 kJ mol−1 respectively, how can you account for the formation of a large number of oxides having O2− species and not O−?
(Hint: Consider lattice energy factor in the formation of compounds).
The boiling points of hydrides of group 16 are in the order:
Which of the following statements are correct?
(i) \[\ce{CaF2 + H2SO4 -> CaSO4 + 2HF}\]
(ii) \[\ce{2HI + H2SO4 -> I2 + SO2 + 2H2O}\]
(iii) \[\ce{Cu + 2H2SO4 -> CuSO4 + SO2 + 2H2O}\]
(iv) \[\ce{Nacl + H2SO4 -> NaHSO4 + HCl}\]
Out of \[\ce{H2O}\] and \[\ce{H2S}\], which one has higher bond angle and why?
In forming (i) \[\ce{N2 -> N^{+}2}\] and (ii) \[\ce{O2 -> O^{+}2}\]; the electrons respectively are removed from:
The correct order of ΔiHs among the following elements is
Which of the following compound is a peroxide?
Given below are two statements:
Statement I: The boiling point of hydrides of Group 16 elements follows the order:
H2O > H2Te > H2Se > H2S
Statement II: On the basis of molecular mass, H2O is expected to have a lower boiling point than the other members of the group but due to the presence of extensive H-bonding in H2O, it has a higher boiling point.
In the light of the above statements, choose the correct answer from the options given below:
