The correct order of decreasing third ionisat | vedclass

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Question

(A) The third ionisation enthalpy involves the removal of an electron from the $M^{2+}$ ion.The electronic configurations of the $M^{2+}$ ions are:$Ti

Vedclass · Accepted Answer

$Mn > Cr > V > Ti$

Vedclass · Answer

(A) The third ionisation enthalpy involves the removal of an electron from the $M^{2+}$ ion.The electronic configurations of the $M^{2+}$ ions are:$Ti^{2+}: [Ar] 3d^2$$V^{2+}: [Ar] 3d^3$$Cr^{2+}: [Ar] 3d^4$$Mn^{2+}: [Ar] 3d^5$For $Mn^{2+}$,the $3d^5$ configuration is half-filled and highly stable,making the removal of the third electron very difficult (highest third ionisation enthalpy).For $Cr^{2+}$,the removal of the third electron leads to the stable $3d^3$ configuration ($t_{2g}^3$ in octahedral field),which is also quite stable.Comparing the values,the order of decreasing third ionisation enthalpy is $Mn > Cr > V > Ti$.

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

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