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[Co(cl)2(NH3)4]^(+) ..... tetraammoniadichloridocobalt(|||) is a AB6L type of molecule , shape is octahedron. But how?? plz explain the hybridisation of the compound

[Co(cl)2(NH3)4]^(+) ..... tetraammoniadichloridocobalt(|||) is a  AB6L type of molecule , shape is octahedron. But how?? plz explain the hybridisation of the compound 

Grade:12th pass

2 Answers

GAURAV SINGH
askIITians Faculty 36 Points
7 years ago
tetraammoniadichloridocobalt(|||), in this compound the oxidation state of cobalt is +3, which means its configuration will become [Ar] 4s0 3d6
So now the hybridisation of cobalt will d2sp3

Gaurav Singh
askIITians Faculty
BS-MS, IISER bhopal
shubhangi modi
39 Points
7 years ago
so u are doing hybridization first by strong ligand and than by weak ligand. ok fine.thanks

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