The complex ion which has no d electrons in t | vedclass

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(D) To determine the number of $d$ electrons,we calculate the oxidation state of the central metal atom:$1$. In $[Co(NH_3)_6]^{3+}$,$Co$ is in $+3$ st

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$[MnO_4]^-$

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(D) To determine the number of $d$ electrons,we calculate the oxidation state of the central metal atom:$1$. In $[Co(NH_3)_6]^{3+}$,$Co$ is in $+3$ state. $Co$ $(Z=27)$ is $[Ar] 3d^7 4s^2$,so $Co^{3+}$ is $3d^6$.$2$. In $[Fe(CN)_6]^{3-}$,$Fe$ is in $+3$ state. $Fe$ $(Z=26)$ is $[Ar] 3d^6 4s^2$,so $Fe^{3+}$ is $3d^5$.$3$. In $[Cr(H_2O)_6]^{3+}$,$Cr$ is in $+3$ state. $Cr$ $(Z=24)$ is $[Ar] 3d^5 4s^1$,so $Cr^{3+}$ is $3d^3$.$4$. In $[MnO_4]^-$,$Mn$ is in $+7$ state. $Mn$ $(Z=25)$ is $[Ar] 3d^5 4s^2$,so $Mn^{7+}$ is $3d^0$.Therefore,$[MnO_4]^-$ has no $d$ electrons.

List-$I$	List-$II$
$A$. Butane $\to$ Isobutane	$a$. Cracking
$B$. Butane $\to$ Lower hydrocarbons	$b$. Isomerisation
$C$. $n$-Heptane $\to$ Toluene	$c$. Reed reaction
$D$. Propane $\to$ $CH_3CH_2CH_2SO_2Cl$	$d$. Aromatisation

The complex ion which has no $d$ electrons in the central metal atom is:

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