Question

Account for the following:
a. Bi(V) is a stronger oxidizing agent than Sb (V).
b. N-N single bond is weaker than a P-P single bond.
c. Noble gases have very low boiling points.

Answer 1

Hint:
(i) An oxidizing agent removes the electrons from the other reactants during the redox reaction. To know the reason behind it we have to know about reducing and oxidizing agents.
(ii) We have to understand the basic concept of bond length.
(iii) Noble gases are very stable and do not form chemical bonds.

Complete answer:
(i) Bi has +5 oxidation state and it is a stronger oxidizing agent than Sb because Bi gains electrons from the reducing agent very much faster than the Sb and becomes stable from +5 oxidation state to +3 oxidation state.
(ii) The N-N single bond is weaker than the P-P single bond because of the smaller size of the N atom as compared to the P atom. The smaller size of N leads to the smaller bond length of N-N due to which lone pairs present on N repel each other and weaken the N-N bond. On the other hand, in P-P bond, the repulsion of lone pairs is less. That’s why P-P bonds are stronger than N-N single bonds.
(iii) The noble gases are the group 18 elements and located in the right of the periodic table. They are called inert gases because their octets make them non-reactive. Their outer shells are full, so they are very much stable and not form chemical bonds and also do not lose or gain electrons. They have very low boiling points because of the weak interatomic forces and they exist in monatomic form and it is difficult to liquify the noble gases.

Note:
(i) In Bi, a self-reduction process takes place in which Bi becomes stable from +5 to +3.
(ii) In nitrogen, there is greater repulsion between N-N so that’s why it is weaker.
(iii) Low boiling point is due to weak interatomic forces and small size.