The hybridisation of metals involved in the f | vedclass

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Question

(D) In $[Mn(CN)_6]^{3-}$,$Mn$ is in $+3$ oxidation state ($3d^4$ configuration). Since $CN^-$ is a strong field ligand,it causes the pairing of $3d$ e

Vedclass · Accepted Answer

$d^2sp^3, sp^3d^2$

Vedclass · Answer

(D) In $[Mn(CN)_6]^{3-}$,$Mn$ is in $+3$ oxidation state ($3d^4$ configuration). Since $CN^-$ is a strong field ligand,it causes the pairing of $3d$ electrons,making two $3d$-orbitals vacant. These two $3d$,one $4s$,and three $4p$-orbitals hybridise to give $d^2sp^3$ hybridisation. In $[CoF_6]^{3-}$,$Co$ is in $+3$ oxidation state ($3d^6$ configuration). Since $F^-$ is a weak field ligand,it cannot cause the pairing of unpaired $3d$ electrons. Thus,one $4s$,three $4p$,and two $4d$-orbitals hybridise to give $sp^3d^2$ hybridisation.

List-$I$ (Complexes)	List-$II$ (Hybridisation)
$A$. $[Ni(CO)_4]$	$I$. $sp^3$
$B$. $[Cu(NH_3)_4]^{2+}$	$II$. $dsp^2$
$C$. $[Fe(NH_3)_6]^{2+}$	$III$. $sp^3d^2$
$D$. $[Fe(H_2O)_6]^{2+}$	$IV$. $d^2sp^3$

The hybridisation of metals involved in the following complexes,$[Mn(CN)_6]^{3-}$ and $[CoF_6]^{3-}$,are respectively:

Similar Questions

Which of the following is an outer orbital complex? (Atomic numbers: $Mn = 25, Fe = 26, Co = 27, Ni = 28$)

Match List-$I$ with List-$II$:

List-$I$ (Complexes) List-$II$ (Hybridisation)

$A$. $[Ni(CO)_4]$ $I$. $sp^3$

$B$. $[Cu(NH_3)_4]^{2+}$ $II$. $dsp^2$

$C$. $[Fe(NH_3)_6]^{2+}$ $III$. $sp^3d^2$

$D$. $[Fe(H_2O)_6]^{2+}$ $IV$. $d^2sp^3$

Chromium hexacarbonyl,$Cr(CO)_6$,is an octahedral compound involving which type of hybridization?

Which statement is incorrect regarding the complex $[Cr(NH_3)_6]Cl_3$?

Assertion : $[FeF_6]^{3-}$ is a low spin complex.
Reason : Low spin complexes have a lesser number of unpaired electrons.

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