Question

(a) Draw the molecular structoms of following compounds :
(i) $Xe{F}_6$
(ii) $H_2{S}_2{O}_8$
(b) Explain the following observations :
(i) The molecules $N{H}_3$ and $N{F}_3$, have dipole moments which are of opposite direction.
(ii) All the bonds of $PC{l}_5$ molecule are not equivalent.
(iii) Sulphur in vapour state exhibits paramagnetism.

Answer 1

(a) (i) $Xe{F}_6$

Distorted octahedral

(ii) $H_2{S}_2{O}_8$

(b) (i) This is due to the fact that in $N{H}_3$, the orbital dipole due to lone pair is in the same direction as the resultant dipole moment of the three N-H bonds. On the other hand, in $N{F}_3$ the orbital dipole due to lone pairs is in opposite direction as the resultant dipole moment. The diagram given below illustrates this.

(ii) In a gaseous and liquid state, $PC{l}_5$ has a trigonal bipyramidal structure. In this structure, the two axial P-Cl bonds are longer and less stable than the three equatorial $P-Cl$ bonds. This is because of the greater bond pair repulsion in the axial bonds. Hence, all the bonds in $PC{l}_5$ are not equivalent.

(iii) In vapour state sulphur partly exists as $S_2$ molecule which has two unpaired electrons in the antibonding ${\pi }^{\ast }$ orbitals like $O_2$. Hence it exhibits paramagnetism.