Advertisements
Advertisements
Question
Give reactions for the following:
O – O single bond is weaker than S – S single bond.
Advertisements
Solution
Due to smaller size of ‘O’ as compared to ‘S’. Smaller size of O leads to smaller O–O bond length. As a result, the lone pair of electrons on the both O atoms repel each other leading to instability or weakening of O–O bond. S is relatively larger so an repulsion between lone pair electrons.
APPEARS IN
RELATED QUESTIONS
Account for the following : There is large difference between the melting and boiling points of oxygen and sulphur.
Give reasons for the following : H2Te is the strongest reducing agent amongst all the hydrides of Group 16 elements.
Which of the following does not react with oxygen directly?
Zn, Ti, Pt, Fe
The HNH angle value is higher than HPH, HAsH and HSbH angles. Why? [Hint: Can be explained on the basis of sp3 hybridisation in NH3 and only s−p bonding between hydrogen and other elements of the group].
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.
Draw the structures of `H_3PO_2`
Arrange the following in the order of the property indicated against set :
H2O, H2S, H2Se, H2Te − increasing acidic character.
Arrange the following in order of the property indicated set.
HF, HCl, HBr, HI - decreasing bond enthalpy.
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 |
In forming (i) \[\ce{N2 -> N^{+}2}\] and (ii) \[\ce{O2 -> O^{+}2}\]; the electrons respectively are removed from:
