The atomic numbers of Vanadium (V),Chromium ( | vedclass

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(B) The electronic configurations of the given elements are:$V (Z=23): [Ar] 3d^3 4s^2$$Cr (Z=24): [Ar] 3d^5 4s^1$$Mn (Z=25): [Ar] 3d^5 4s^2$$Fe (Z=26)

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$Cr$

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(B) The electronic configurations of the given elements are:$V (Z=23): [Ar] 3d^3 4s^2$$Cr (Z=24): [Ar] 3d^5 4s^1$$Mn (Z=25): [Ar] 3d^5 4s^2$$Fe (Z=26): [Ar] 3d^6 4s^2$To find the second ionization enthalpy,we remove one electron from each to form the $M^+$ ion:$V^+: [Ar] 3d^3 4s^1$$Cr^+: [Ar] 3d^5$$Mn^+: [Ar] 3d^5 4s^1$$Fe^+: [Ar] 3d^6 4s^1$The second ionization enthalpy involves removing an electron from the $M^+$ ion. For $Cr^+$,the configuration is $[Ar] 3d^5$,which is a stable half-filled $d$-orbital configuration. Removing an electron from this stable state requires a significantly higher amount of energy compared to the others. Therefore,$Cr$ has the highest second ionization enthalpy.

The atomic numbers of Vanadium $(V)$,Chromium $(Cr)$,Manganese $(Mn)$,and Iron $(Fe)$ are $23, 24, 25$,and $26$ respectively. Which of the following has the highest second ionization enthalpy?

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