The hexaquo manganese $(II)$ ion contains five unpaired electrons,while the hexacyano manganese $(II)$ ion contains only one unpaired electron. Explain using Crystal Field Theory.

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$[Mn(H_{2}O)_{6}]^{2+}$$[Mn(CN)_{6}]^{4-}$
$Mn$ is in the $+2$ oxidation state.$Mn$ is in the $+2$ oxidation state.
The electronic configuration is $d^{5}$.The electronic configuration is $d^{5}$.

The crystal field is octahedral. Water $(H_{2}O)$ is a weak field ligand,while cyanide $(CN^-)$ is a strong field ligand.
In $[Mn(H_{2}O)_{6}]^{2+}$,the weak field ligand does not cause pairing of electrons,resulting in the configuration $t_{2g}^{3}e_{g}^{2}$,which has $5$ unpaired electrons.
In $[Mn(CN)_{6}]^{4-}$,the strong field ligand causes pairing of electrons,resulting in the configuration $t_{2g}^{5}e_{g}^{0}$,which has $1$ unpaired electron.

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Similar Questions

Match List-$I$ with List-$II$.
List-$I$ (Coordination entity) List-$II$ (Wavelength of light absorbed in $nm$)
$A$. $[CoCl(NH_3)_5]^{2+}$ $I$. $310$
$B$. $[Co(NH_3)_6]^{3+}$ $II$. $475$
$C$. $[Co(CN)_6]^{3-}$ $III$. $535$
$D$. $[Cu(H_2O)_4]^{2+}$ $IV$. $600$

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