Hybridisation of which complex is correctly matched a) AuCl4 - (sp 3) b) [Co(oxalato)3]3- (sp3d2) c) [RhCl(PPh3)3] (dsp2) d) [Fe(NH3)6]2+ (d2sp3) Share with your friends Share 9 Vartika Jain answered this Dear Student, a) Electronic configuration of Au = [Xe]4f14 5d10 6s1Electronic configuration of Au3+= [Xe] 4f14 5d8But as we know that pairing occurs in case of Au, therefore all electrons in 5d will be paired. On orbital in 5d will be vacant, so now 4 electron pair of Cl goes to one 5d,one 6s and two 6p orbitals forming dsp2 hybridisation. b) [Co(oxalato)3]3-Electronic configuration of Co = [Ar] 3d7 6s2Electronic configuration of Co3+= [Ar] 3d6 But as we know that pairing occurs in case of oxalate ion, therefore all electrons in 3d will be paired. Two orbitals in 3d will be vacant, so now 6 electron pair of oxalate goes to two 3d,one 4s and three 4p orbitals forming d2sp3 hybridisation. c) [RhCl(PPh3)3] Electronic configuration of Rh = [Kr] 4d8 5s1 Electronic configuration of Rh+= [Kr] 4d8But as we know that pairing occurs in case of Rh, therefore all electrons in 4d will be paired. One orbital in 4d will be vacant, so now 1 electron pair of Cl goes to one 4d and three electron pairs from PPh3 goes to one 5s and two 5p orbitals forming dsp2 hybridisation. Therefore, this configuration is correct matched. d) [Fe(NH3)6]2+Electronic configuration of Fe = [Ar] 3d6 6s2Electronic configuration of Fe2+= [Ar] 3d6 But as we know that no pairing occurs in case of NH3, therefore one electron will be paired and all other d-orbital will be half filled. No orbitals in 3d will be vacant, so now 6 electron pair of NH3 goes to one 4s,three 4p and two 4d orbitals forming sp3d2 hybridisation. Hence, the correct answer is (c) 22 View Full Answer Gaurang Yadav answered this mh -8
