• # question_answer 107)   Describe hybridisation in case of $PC{{l}_{5}}$ and $S{{F}_{6}}$. The axial bonds are longer as compared to equatorial bonds in $PC{{l}_{5}}$ whereas in $S{{F}_{6}}$both axial bonds and equatorial bonds have same length. Explain.

1.$\mathbf{s}{{\mathbf{p}}^{\mathbf{3}}}$d-Hybridization : It involves the mixing of one s-, three p- and one d-orbitals resulting in the formation of trigonal bipyramidal arrangement. In this arrangement, the three hybrid orbitals forming a plane, are directed towards the corners of an equilateral triangle while the other two hybrid orbitals are perpendicular to the plane of a triangle lying above and below it. $PC{{l}_{5}}$ is an example of this type of hybridization. $\mathbf{PC}{{\mathbf{l}}_{\mathbf{5}}}$ molecule: P-atom has configuration $1{{s}^{2}},\text{ }2{{s}^{2}},\text{ }2{{p}^{6}},$$3{{s}^{2}}3p_{x}^{1},3p_{y}^{1},3p_{z}^{1}$In ground state, it can form three covalent bonds as three unpaired orbitals are present in the valency shell. To get pentavalency, 3s-orbital is unpaired and the electron is shifted to 3d-orbital. Now in the excited state the five orbitals involving one s-, one d- and three p-orbitals undergo hybridization giving birth to five hybrid orbitals which overlap with five chlorine atoms forming five sigma bonds. Out of five $\sigma$-bonds, three bonds which are located at$120{}^\circ$ angle are equitorial and the remaining two are axial. Axial bond length is greater.     Each P-Cl axial bond length =219 pm and each P-Cl equatorial bond length = 204 pm. Axial bonds are thus slightly weaker than equatorial bonds. $P{{F}_{5}}$ molecule has similar hybridization and geometry. 2.$\mathbf{s}{{\mathbf{p}}^{\mathbf{3}}}{{\mathbf{d}}^{\mathbf{2}}}$-Hybridization: It involves the mixing of one$s$-, three p- and two d-orbitals resulting in the formation of octahedral arrangement, $S{{F}_{6}}$ is an example of this type of hybridization. In$S{{F}_{6}}$, the central sulphur atom has the ground state configuration $3{{s}^{2}}3p_{x}^{2}3p_{y}^{1}3p_{z}^{1}$to   account   for hexavalency, the paired 3s and paired $3{{p}_{x}}$ are unpaired and electrons are shifted to d-orbitals. These six orbitals undergo hybridization giving six hybrid orbitals directed towards the comers of a regular octahedron. Each of these hybrid orbitals overlaps with 2p- orbital of fluorine to form S?F sigma bond. Thus, $S{{F}_{6}}$ molecule has octahedral structure.                 $S{{F}_{6}}$ is a symmetrical molecule and therefore, is stable and far less reactive. Four S-F bonds are in the same plane at right angles to one another and are directed towards four corners of a square. The other two F-atoms lie at right angle above and below the plane.