The atomic numbers of V, Cr, Mn and Fe are 23 | 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$$Fe (Z=26): [Ar] 3d^6 4s^2$For the second ionization enthalpy,we remove one electron from the $M^+$ ion:$V^+: [Ar] 3d^3 4s^1$$Cr^+: [Ar] 3d^5$$Mn^+: [Ar] 3d^5 4s^1$$Fe^+: [Ar] 3d^6 4s^1$Removing the second electron from $Cr^+$ is difficult because it has a stable half-filled $3d^5$ configuration. Therefore,$Cr$ has the highest second ionization enthalpy.

The atomic numbers of $V, Cr, Mn$ and $Fe$ are $23, 24, 25$ and $26$ respectively. Which of these elements will have the highest second ionization enthalpy?

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