The correct order of decreasing third ionisat | vedclass

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(A) The third ionisation enthalpy $(IE_3)$ corresponds to the energy required to remove an electron from the $M^{2+}$ ion.The electronic configuration

Vedclass · Accepted Answer

$Mn > Cr > V > Ti$

Vedclass · Answer

(A) The third ionisation enthalpy $(IE_3)$ corresponds to the energy required to remove an electron from the $M^{2+}$ ion.The electronic configurations of $M^{2+}$ ions are:$Ti^{2+} ([Ar] 3d^2)$$V^{2+} ([Ar] 3d^3)$$Cr^{2+} ([Ar] 3d^4)$$Mn^{2+} ([Ar] 3d^5)$Removing an electron from $Mn^{2+}$ $(3d^5)$ is difficult due to the stable half-filled $d$-orbital configuration.Removing an electron from $Cr^{2+}$ $(3d^4)$ is relatively easier than $Mn^{2+}$,but $Cr^{2+}$ has a higher $IE_3$ than $V^{2+}$ and $Ti^{2+}$ due to the exchange energy and stability factors.The correct order of decreasing $IE_3$ is $Mn > Cr > V > Ti$.

The correct order of decreasing third ionisation enthalpy of $Ti$,$V$,$Cr$,and $Mn$ is:

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