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प्रश्न
Give reasons for the following : H2Te is the strongest reducing agent amongst all the hydrides of Group 16 elements.
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उत्तर
As the size of the elements increases in the order O < S < Se < Te, the X−H bond strength decreases from H2O to H2Te and therefore, the bond dissociation enthalpy decreases. Hence, due to the increase in the tendency to release proton, the element's reducing tendency also increases. Therefore, H2Te is the strongest reducing agent amongst all the hydrides of Group 16 elements.
संबंधित प्रश्न
Give reasons for the following : Oxygen has less electron gain enthalpy with negative sign than sulphur.
List the important sources of sulphur.
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).
Explain why inspite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not.
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Arrange the following in the order of the property indicated against set :
H2O, H2S, H2Se, H2Te − increasing acidic character.
The formation of \[\ce{O^+_2[PtF6]^-}\] is the basis for the formation of first xenon compound. This is because ____________.
Match the items of Columns I and II and mark the correct option.
| Column I | Column II |
| (A) \[\ce{H2SO4}\] | (1) Highest electron gain enthalpy |
| (B) \[\ce{CCl3NO2}\] | (2) Chalcogen |
| (C) \[\ce{Cl2}\] | (3) Tear gas |
| (D) Sulphur | (4) Storage batteries |
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) \[\ce{Nacl + H2SO4 -> NaHSO4 + HCl}\]
______ is a radioactive element in group 16 elements.
