Which of the following represents the decreas | vedclass

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(A) The electronic configurations of the neutral atoms are: $Ti ([Ar] 3d^2 4s^2)$,$V ([Ar] 3d^3 4s^2)$,$Cr ([Ar] 3d^5 4s^1)$,and $Mn ([Ar] 3d^5 4s^2)$

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$Cr > Mn > V > Ti$

Vedclass · Answer

(A) The electronic configurations of the neutral atoms are: $Ti ([Ar] 3d^2 4s^2)$,$V ([Ar] 3d^3 4s^2)$,$Cr ([Ar] 3d^5 4s^1)$,and $Mn ([Ar] 3d^5 4s^2)$.After the removal of the first electron $(I.P_1)$,the configurations become: $Ti^+ ([Ar] 3d^2 4s^1)$,$V^+ ([Ar] 3d^3 4s^1)$,$Cr^+ ([Ar] 3d^5)$,and $Mn^+ ([Ar] 3d^5 4s^1)$.The second ionization enthalpy $(I.P_2)$ involves removing an electron from these ions.$Cr^+$ has a stable half-filled $d^5$ configuration,making it very difficult to remove the second electron,resulting in the highest $I.P_2$.$Mn^+$ has a $3d^5 4s^1$ configuration,where the $4s$ electron is relatively easier to remove than the $d^5$ electron of $Cr^+$.Comparing the trends,the order of $I.P_2$ is $Cr > Mn > V > Ti$.

Which of the following represents the decreasing order of the second ionization enthalpy $(I.P_2)$?

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