Answer:
σ: Xe( Sp3d2) - F(p)
Explanation:
Step one: write the ground state configuration and the excited state configuration of the xenon atom.
Ground state= [Kr] 5s2. 4d10. 5p6.
The Sp3d2 hybrid will be the impaired electrons will bond with the fluorine atom.
For pi, π bonding we have;
Xe(Sp3d2) - F(p), Xe(sp3d) - F(s),
Xe(sp3d) - F(p) and Xe(Sp3d2) - F(s).
For sigma, σ bonding we have;
Xe(sp3d) - F(s) and Xe(Sp3d2) - F(p).
Because the participation in hybridization involves only p orbital electron, in the pi bonding scheme;Xe(sp3d) - F(s), and Xe(Sp3d2) - F(s) are ruled out. Also, in the sigma bonding scheme; Xe(sp3d) - F(s) is ruled out.
Hence, we have;
σ: Xe( Sp3d2) - F(p).
