Why is $[CoF_6]^{3-}$ considered an outer orbital complex?

  • A
    It involves $sp^3d^2$ hybridization.
  • B
    It involves $d^2sp^3$ hybridization.
  • C
    It is a low spin complex.
  • D
    It has no unpaired electrons.

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Similar Questions

Identify the type of hybridization involved in the hexaamminecobalt$(III)$ complex ion.

The metal $Ni$ $(Z = 28)$ combines with a unidentate ligand $X^-$ to form a paramagnetic complex $[NiX_4]^{2-}$. What are its geometry and the number of unpaired electrons in $Ni$?

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According to Werner’s theory,the geometry of the complex is determined by:

Using valence bond theory,explain the following in relation to the complexes given below: $[Mn(CN)_{6}]^{3-}, [Co(NH_{3})_{6}]^{3+}, [Cr(H_{2}O)_{6}]^{3+}, [FeCl_{6}]^{4-}$
$(i)$ Type of hybridisation.
$(ii)$ Inner or outer orbital complex.
$(iii)$ Magnetic behaviour.
$(iv)$ Spin only magnetic moment value.

The spin-only magnetic moment of $[MnBr_4]^{x-}$ is $5.9 \ BM$. The geometry of the complex and $x$ respectively are

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