Advertisements
Advertisements
प्रश्न
Give reasons: SO2 is reducing while TeO2 is an oxidising agent.
Advertisements
उत्तर
In case of sulphur, because of the presence of empty d-orbital, it can expand its oxidation state from the + 4 to the +6 oxidation state. Hence, it acts as a reducing agent.
Te is a heavy element and so because of the inert pair effect, the lower oxidation state is more stable. Hence, it acts as an oxidising agent.
APPEARS IN
संबंधित प्रश्न
Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuration, oxidation state and hydride formation.
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).
Draw the structures of `H_3PO_2`
Give reactions for the following:
O – O single bond is weaker than S – S single bond.
Give a reason for the following:
Fluorine gives only one oxide but chlorine gives a series of oxides.
The boiling points of hydrides of group 16 are in the order:
Which of the following statement is incorrect?
The correct order of ΔiHs among the following elements is
______ is a radioactive element in group 16 elements.
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:
