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The Hybridisiation of [NiCl4]2- , [CoCl4]2- , MnO4 is.?
  • a)
    sp3, sp3, sd3
  • b)
    sd3, sp3, sp3,
  • c)
    sp3, sp3, sp3
  • d)
    sp3,sd3 sp3,
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The Hybridisiation of [NiCl4]2-, [CoCl4]2-, MnO4-is.?a)sp3,sp3,sd3b)sd...
  • [NiCl4]2-: The nickel ion Ni2+ has an electron configuration of [Ar]3d8. In this complex, it can be considered to utilize 3d, 4s, and two 4p orbitals to form four sp3 hybrid orbitals which are half-filled and will form bonds with four chloride ions. So the hybridization state is sp3.
  • Cl- is weak ligand 
  • MnO4-: The manganese ion Mn7+ has an electron configuration of [Ne]3p6, which means it has no unpaired electron in the 3d subshell. However, for the bonding purposes with four oxygen atoms, it needs to engage the vacant 3d orbital, along with 4s and two 4p orbitals for hybridization. This would give the reference of the 3d subshell's involvement in hybridization, hence we use sd3 notation to specify it, instead of the general sp3 for tetrahedral geometry.
So, The Hybridisiation of [NiCl4]2- , [CoCl4]2- , MnO4-  is sp3, sp3, sd3
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