Explain the formation of on the basis of hybridisation
Asked by kpbhake | 12th Mar, 2018, 11:45: AM
Formation of PCl5 (sp3d Hybridisation)
- The ground state and the excited state outer electronic configurations of phosphorus (Z = 15) are given below:
- The formation of five half-filled sp3d hybrid orbitals involves the intermixing of one 3s, three 3p and one 3d orbitals during excitation.
- They are then organised in a trigonal bipyramidal symmetry. While three orbitals are organised in trigonal planar symmetry, the remaining two are kept perpendicularly above and below this plane.
- These half-filled sp3d orbitals help phosphorus in forming five σsp3d-p bonds with chlorine atoms. Each chlorine atom makes use of half-filled 3pz orbital for the bond formation.
- The PCl5 molecule has a trigonal bipyramidal shape with 120° and 90° of ∠Cl–P–Cl bond angles.
Answered by Varsha | 12th Mar, 2018, 03:54: PM
- what is hybridisation?
- Among H2+ and H2- molecular ions which ion is more stable and why?
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- Name one cation which is isostructural with NH3.
- Name the d- orbitals that are involved in sp3d2 hybridization
- What is the hybrid state of B in BF3, Al in AlCl3, Be in BeCl2, C in CO2 and C2H4; S in SO2 and SO3.
- Differentiate between valence bond theory and Lewis concept with regard to the formation of covalent bond.
- Explain the formation of sigma and pi bond.
- Define hybridization. Show the formation of ammonia (NH3) molecule.
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