Discuss the $[CoF_6]^{3-}$ complex species based on Valence Bond Theory $(VBT)$.
(N/A) $1$. The central metal ion is $Co^{3+}$,which has an electronic configuration of $[Ar] 3d^6$.
$2$. In the presence of the weak field ligand $F^-$,the $d$-electrons do not pair up.
$3$. The $Co^{3+}$ ion uses one $4s$,three $4p$,and two $4d$ orbitals to undergo $sp^3d^2$ hybridization.
$4$. Six pairs of electrons from six $F^-$ ions occupy these six $sp^3d^2$ hybrid orbitals.
$5$. Since the outer $4d$ orbitals are used,it is an outer orbital complex (high spin complex).
$6$. Due to the presence of four unpaired electrons in the $3d$ orbitals,the complex is paramagnetic.
$2$. In the presence of the weak field ligand $F^-$,the $d$-electrons do not pair up.
$3$. The $Co^{3+}$ ion uses one $4s$,three $4p$,and two $4d$ orbitals to undergo $sp^3d^2$ hybridization.
$4$. Six pairs of electrons from six $F^-$ ions occupy these six $sp^3d^2$ hybrid orbitals.
$5$. Since the outer $4d$ orbitals are used,it is an outer orbital complex (high spin complex).
$6$. Due to the presence of four unpaired electrons in the $3d$ orbitals,the complex is paramagnetic.
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Similar Questions
The pair$(s)$ of complexes wherein both exhibit tetrahedral geometry is(are):
Note: $py = \text{pyridine}$
Given: Atomic numbers of $Fe, Co, Ni$ and $Cu$ are $26, 27, 28$ and $29$,respectively.
$(A)$ $[FeCl_4]^-$ and $[Fe(CO)_4]^{2-}$
$(B)$ $[Co(CO)_4]^-$ and $[CoCl_4]^{2-}$
$(C)$ $[Ni(CO)_4]$ and $[Ni(CN)_4]^{2-}$
$(D)$ $[Cu(py)_4]^+$ and $[Cu(CN)_4]^{3-}$
Note: $py = \text{pyridine}$
Given: Atomic numbers of $Fe, Co, Ni$ and $Cu$ are $26, 27, 28$ and $29$,respectively.
$(A)$ $[FeCl_4]^-$ and $[Fe(CO)_4]^{2-}$
$(B)$ $[Co(CO)_4]^-$ and $[CoCl_4]^{2-}$
$(C)$ $[Ni(CO)_4]$ and $[Ni(CN)_4]^{2-}$
$(D)$ $[Cu(py)_4]^+$ and $[Cu(CN)_4]^{3-}$
MediumIIT 2021
View SolutionMatch the following:
List-$I$ (Hybridisation) List-$II$ (Compound/ion) $A. sp^3d$ $I. [PtCl_4]^{2-}$ $B. sp^3d^2$ $II. SF_6$ $C. dsp^2$ $III. BCl_3$ $D. dsp^3$ $IV. PCl_5$ $V. ClF_3$
The correct match is:
| List-$I$ (Hybridisation) | List-$II$ (Compound/ion) |
|---|---|
| $A. sp^3d$ | $I. [PtCl_4]^{2-}$ |
| $B. sp^3d^2$ | $II. SF_6$ |
| $C. dsp^2$ | $III. BCl_3$ |
| $D. dsp^3$ | $IV. PCl_5$ |
| $V. ClF_3$ |
The correct match is:
Which of the following ions is not tetrahedral?
Medium
View SolutionAssertion $:$ Complex ion $[Co(NH_3)_6]^{2+}$ is readily oxidized to $[Co(NH_3)_6]^{3+}$.
Reason $:$ Unpaired electron in complex ion $[Co(NH_3)_6]^{2+}$ is present in $4p$ orbital.
Reason $:$ Unpaired electron in complex ion $[Co(NH_3)_6]^{2+}$ is present in $4p$ orbital.
Medium
View SolutionFor octahedral $Mn(II)$ and tetrahedral $Ni(II)$ complexes,consider the following statements:
$(I)$ Both the complexes can be high spin.
$(II)$ $Ni(II)$ complex can very rarely be low spin.
$(III)$ With strong field ligands,$Mn(II)$ complexes can be low spin.
$(IV)$ Aqueous solution of $Mn(II)$ ions is yellow in color.
$(I)$ Both the complexes can be high spin.
$(II)$ $Ni(II)$ complex can very rarely be low spin.
$(III)$ With strong field ligands,$Mn(II)$ complexes can be low spin.
$(IV)$ Aqueous solution of $Mn(II)$ ions is yellow in color.
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