Discuss the complex species $[Co(NH_3)_6]^{3+}$ based on Valence Bond Theory.
(N/A) $1$. The central metal ion $Co^{3+}$ has an electronic configuration of $[Ar]3d^6$.
$2$. In the presence of the strong field ligand $NH_3$,the electrons in the $3d$ orbitals pair up,leaving two $3d$ orbitals vacant.
$3$. These two $3d$ orbitals,along with one $4s$ and three $4p$ orbitals,undergo $d^2sp^3$ hybridization to form six equivalent hybrid orbitals.
$4$. Six pairs of electrons from six $NH_3$ ligands occupy these hybrid orbitals,resulting in an octahedral geometry.
$5$. Since there are no unpaired electrons,the complex is diamagnetic.
$6$. Because the inner $3d$ orbitals are involved in hybridization,it is called an inner orbital complex or a low-spin complex.
$2$. In the presence of the strong field ligand $NH_3$,the electrons in the $3d$ orbitals pair up,leaving two $3d$ orbitals vacant.
$3$. These two $3d$ orbitals,along with one $4s$ and three $4p$ orbitals,undergo $d^2sp^3$ hybridization to form six equivalent hybrid orbitals.
$4$. Six pairs of electrons from six $NH_3$ ligands occupy these hybrid orbitals,resulting in an octahedral geometry.
$5$. Since there are no unpaired electrons,the complex is diamagnetic.
$6$. Because the inner $3d$ orbitals are involved in hybridization,it is called an inner orbital complex or a low-spin complex.
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Similar Questions
$Mn^{2+}$ forms a complex with $Br^{-}$ ion. The magnetic moment of the complex is $5.92 \ B.M.$ What could not be the probable formula and geometry of the complex?
Difficult
View SolutionMatch List-$I$ with List-$II$:
List-$I$ (Tetrahedral Complex) List-$II$ (Electronic configuration) $A$. $TiCl_4$ $I$. $e^2, t_2^0$ $B$. $[FeO_4]^{2-}$ $II$. $e^4, t_2^3$ $C$. $[FeCl_4]^{-}$ $III$. $e^0, t_2^0$ $D$. $[CoCl_4]^{2-}$ $IV$. $e^2, t_2^3$
Choose the correct answer from the options given below:
| List-$I$ (Tetrahedral Complex) | List-$II$ (Electronic configuration) |
|---|---|
| $A$. $TiCl_4$ | $I$. $e^2, t_2^0$ |
| $B$. $[FeO_4]^{2-}$ | $II$. $e^4, t_2^3$ |
| $C$. $[FeCl_4]^{-}$ | $III$. $e^0, t_2^0$ |
| $D$. $[CoCl_4]^{2-}$ | $IV$. $e^2, t_2^3$ |
Choose the correct answer from the options given below:
Match the hybridisation in Column $I$ with the complexes in Column $II$. The options represent the matches for $(A), (B), (C), (D)$ respectively.
Column $I$ Column $II$ $(A)$ $sp^3$ $(i)$ $[Co(NH_3)_6]^{3+}$ $(B)$ $dsp^2$ $(ii)$ $[Ni(CO)_4]$ $(C)$ $sp^3d^2$ $(iii)$ $[Pt(NH_3)_2Cl_2]$ $(D)$ $d^2sp^3$ $(iv)$ $[CoF_6]^{3-}$ $(v)$ $[Fe(CO)_5]$
| Column $I$ | Column $II$ |
|---|---|
| $(A)$ $sp^3$ | $(i)$ $[Co(NH_3)_6]^{3+}$ |
| $(B)$ $dsp^2$ | $(ii)$ $[Ni(CO)_4]$ |
| $(C)$ $sp^3d^2$ | $(iii)$ $[Pt(NH_3)_2Cl_2]$ |
| $(D)$ $d^2sp^3$ | $(iv)$ $[CoF_6]^{3-}$ |
| $(v)$ $[Fe(CO)_5]$ |
The geometry possessed by $[Ni(CO)_4]$ is
Easy
View SolutionThe geometries of $[Ni(CO)_4]$,$[PtCl_4]^{2-}$ and $[Co(NH_3)_6]^{3+}$ respectively are
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