Why is $[CoF_6]^{3-}$ considered an outer orbital complex?
- AIt involves $sp^3d^2$ hybridization.
- BIt involves $d^2sp^3$ hybridization.
- CIt is a low spin complex.
- DIt has no unpaired electrons.
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Similar Questions
The spin only magnetic moment of $[MnBr_{4}]^{2-}$ is $5.9 \ BM$. The geometry of this complex ion is
MediumAIIMS 2018
View SolutionMatch the following: Column-$I$ Complex Column-$II$ Structure / Geometry / Property $A$. $[Ni(CN)_4]^{2-}$ $I$. Tetrahedral / Paramagnetic $B$. $[Ni(CO)_4]$ $II$. Tetrahedral / Diamagnetic $C$. $[NiCl_4]^{2-}$ $III$. Square planar / Diamagnetic
The correct match is:
| Column-$I$ Complex | Column-$II$ Structure / Geometry / Property |
|---|---|
| $A$. $[Ni(CN)_4]^{2-}$ | $I$. Tetrahedral / Paramagnetic |
| $B$. $[Ni(CO)_4]$ | $II$. Tetrahedral / Diamagnetic |
| $C$. $[NiCl_4]^{2-}$ | $III$. Square planar / Diamagnetic |
DifficultTS EAMCET 2021
View SolutionMatch each set of hybrid orbitals from $LIST-I$ with complex(es) given in $LIST-II$. The correct option is
$LIST-I$ $LIST-II$ $P. dsp^2$ $1. [FeF_6]^{4-}$ $Q. sp^3$ $2. [Ti(H_2O)_3Cl_3]$ $R. sp^3d^2$ $3. [Cr(NH_3)_6]^{3+}$ $S. d^2sp^3$ $4. [FeCl_4]^{2-}$ $5. Ni(CO)_4$ $6. [Ni(CN)_4]^{2-}$
| $LIST-I$ | $LIST-II$ |
| $P. dsp^2$ | $1. [FeF_6]^{4-}$ |
| $Q. sp^3$ | $2. [Ti(H_2O)_3Cl_3]$ |
| $R. sp^3d^2$ | $3. [Cr(NH_3)_6]^{3+}$ |
| $S. d^2sp^3$ | $4. [FeCl_4]^{2-}$ |
| $5. Ni(CO)_4$ | |
| $6. [Ni(CN)_4]^{2-}$ |
MediumIIT 2018
View SolutionWhich of the following complexes is an outer orbital complex?
Difficult
View SolutionHow many unpaired electrons are present in $Ni^{2+}$?
Easy
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