The correct order of decreasing second ionisa | vedclass

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Question

(C) The energy required to remove an electron from a unipositive ion is called the second ionisation enthalpy. Electronic configurations of $M^+$ ions

Vedclass · Accepted Answer

$Cr > Mn > V > Ti$

Vedclass · Answer

(C) The energy required to remove an electron from a unipositive ion is called the second ionisation enthalpy. Electronic configurations of $M^+$ ions are: $Ti^+: [Ar] 3d^2 4s^1$ $V^+: [Ar] 3d^3 4s^1$ $Cr^+: [Ar] 3d^5$ $Mn^+: [Ar] 3d^5 4s^1$ In $Cr^+$,the $3d^5$ configuration is exactly half-filled,which is highly stable. Removing an electron from this stable configuration requires significantly more energy. Therefore,$Cr$ has a higher second ionisation enthalpy than $Mn$. The overall decreasing order is $Cr > Mn > V > Ti$.

The correct order of decreasing second ionisation enthalpy of $Ti(22), V(23), Cr(24)$ and $Mn(25)$ is

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