The hybridization and magnetic nature of [Mn( | vedclass

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(C) For $[Mn(CN)_6]^{4-}$:$Mn$ is in $+2$ oxidation state ($3d^5$ configuration).$CN^-$ is a strong field ligand $(SFL)$,so it causes pairing of elect

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$d^2sp^3$ and paramagnetic

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(C) For $[Mn(CN)_6]^{4-}$:$Mn$ is in $+2$ oxidation state ($3d^5$ configuration).$CN^-$ is a strong field ligand $(SFL)$,so it causes pairing of electrons.Thus,the configuration becomes $t_{2g}^5 e_g^0$,leaving one unpaired electron.The hybridization is $d^2sp^3$ and it is paramagnetic.For $[Fe(CN)_6]^{3-}$:$Fe$ is in $+3$ oxidation state ($3d^5$ configuration).$CN^-$ is a strong field ligand $(SFL)$,causing pairing of electrons.Thus,the configuration becomes $t_{2g}^5 e_g^0$,leaving one unpaired electron.The hybridization is $d^2sp^3$ and it is paramagnetic.Therefore,both complexes have $d^2sp^3$ hybridization and are paramagnetic.

$A$. $[CoF_6]^{3-}$	$M$. $sp^3d^2$,paramagnetic
$B$. $[Mn(CN)_6]^{3-}$	$N$. $sp^3d^2$,diamagnetic
$C$. $[Co(C_2O_4)_3]^{3-}$	$O$. $d^2sp^3$,paramagnetic
$D$. $[Zn(H_2O)_6]^{2+}$	$P$. $d^2sp^3$,diamagnetic

The hybridization and magnetic nature of $[Mn(CN)_6]^{4-}$ and $[Fe(CN)_6]^{3-}$,respectively are

Similar Questions

Which of the following complexes is paramagnetic,an inner orbital complex,and coloured?

Match the correct option.
$A$. $[CoF_6]^{3-}$ $M$. $sp^3d^2$,paramagnetic
$B$. $[Mn(CN)_6]^{3-}$ $N$. $sp^3d^2$,diamagnetic
$C$. $[Co(C_2O_4)_3]^{3-}$ $O$. $d^2sp^3$,paramagnetic
$D$. $[Zn(H_2O)_6]^{2+}$ $P$. $d^2sp^3$,diamagnetic

The number of species from the following that are involved in $sp^3 d^2$ hybridization is $[Co(NH_3)_6]^{3+}, SF_6, [CrF_6]^{3-}, [CoF_6]^{3-}, [Mn(CN)_6]^{3-}$ and $[MnCl_6]^{3-}$.

The complex $[Pt(NH_3)_4]^{2+}$ has $.....$ structure.

$K_3[CoF_6]$ is a high-spin complex. The hybridization state of $Co$ in this complex is .....................

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