$[FeF_6]^{3-}$ has $Fe$ atom . . . . . . hybridised with unpaired . . . . . . electrons.
- A$d^2sp^3, 4$
- B$d^2sp^3, 5$
- C$sp^3d^2, 5$
- D$sp^3d^2, 3$
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Similar Questions
Match List-$I$ with List-$II$ and select the correct answer using the codes given below the lists $:$
List-$I$ $(Compounds)$ List-$II$ $(Hybridization)$ $(a)$ $[Ni(NH_3)_6]^{+2}$ $(i)$ $sp^3 d^2$ $(b)$ $[PtCl_4]^{-2}$ $(ii)$ $dsp^2$ $(c)$ $[Ni(CO)_4]$ $(iii)$ $sp^3$ $(d)$ $[Co(OX)_3]^{-3}$ $(iv)$ $d^2 sp^3$
| List-$I$ $(Compounds)$ | List-$II$ $(Hybridization)$ |
| $(a)$ $[Ni(NH_3)_6]^{+2}$ | $(i)$ $sp^3 d^2$ |
| $(b)$ $[PtCl_4]^{-2}$ | $(ii)$ $dsp^2$ |
| $(c)$ $[Ni(CO)_4]$ | $(iii)$ $sp^3$ |
| $(d)$ $[Co(OX)_3]^{-3}$ | $(iv)$ $d^2 sp^3$ |
Medium
View SolutionThe magnetic moment of the high spin complex is $5.92 \ BM$. What is the electronic configuration?
For $K_3[Co(CN)_6]$,which of the following statements is not correct?
Medium
View Solution$d^{2}sp^{3}$ hybridisation of the atomic orbitals gives
Considering $H_2O$ as a weak field ligand,the number of unpaired electrons in $[Mn(H_2O)_6]^{2+}$ will be (Atomic number of $Mn = 25$).
MediumAIPMT 2004
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