Question

Describe hybridisation in the case of $PC{l}_5\text{and}S{F}_6$. The axial bonds are longer as compared to equatorial bonds in $PC{l}_5$ whereas in $S{F}_6$ axial bonds and equatorial bonds have the same bond length. Explain.

Answer 1

Hybridisation and structure of $PC{l}_5$

• Five $s{p}^3$d hybrid orbitals of Phosphorus atom in $PC{l}_5$ are filled by five $Cl$ atoms.
• Shape of $PC{l}_5$ is trigonal bipyramidal.

Hybridisation and structure of $S{F}_6$:

• Six $s{p}^3{d}^2$ hybridised orbitals of $S\text{in}S{F}_6$ are filled by six fluorine atoms.
• Shape of $S{F}_6$ is octahedral.

Axial and equatorial bonds in $PC{l}_5$ molecule

• Three $P-Cl$ bond lie in one plane and make an angle of ${120}^{\circ }$ with each other; these bonds are termed as equatorial bonds. The remaining two $P-Cl$ bonds, one lying above and the other lying below the equatorial plane, make an angle of ${90}^{\circ }$ with the plane. These bonds are called axial bonds.
• The axial bond pairs suffer more repulsive interaction than the equatorial bond pairs, therefore axial bonds have been found to be slightly longer than the equatorial bonds.

Axial and equatorial bonds in $S{F}_6$ molecule

• The shape of the $S{F}_6$ molecule is octahedral. In this case all bonds experience similar repulsion. So, all bonds are of the same length