11th Class Chemistry Chemical Bonding and Molecular Structure

  • 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.

You need to login to perform this action.
You will be redirected in 3 sec spinner