The atomic numbers of vanadium (V),chromium ( | vedclass

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(B) The electronic configurations of the neutral atoms are:$V: [Ar] 3d^3 4s^2$$Cr: [Ar] 3d^5 4s^1$$Mn: [Ar] 3d^5 4s^2$$Fe: [Ar] 3d^6 4s^2$To find the

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$Cr$

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(B) The electronic configurations of the neutral atoms are:$V: [Ar] 3d^3 4s^2$$Cr: [Ar] 3d^5 4s^1$$Mn: [Ar] 3d^5 4s^2$$Fe: [Ar] 3d^6 4s^2$To find the second ionization enthalpy,we remove one electron to form the $+1$ ion and then remove the second electron:$V^+: [Ar] 3d^3 4s^1 \rightarrow V^{2+}: [Ar] 3d^3$$Cr^+: [Ar] 3d^5 \rightarrow Cr^{2+}: [Ar] 3d^4$$Mn^+: [Ar] 3d^5 4s^1 \rightarrow Mn^{2+}: [Ar] 3d^5$$Fe^+: [Ar] 3d^6 4s^1 \rightarrow Fe^{2+}: [Ar] 3d^6$Among these,$Cr^+$ has a stable half-filled $d^5$ configuration. Removing an electron from this stable configuration requires a very high amount of energy. 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 one of these is expected to have the highest second ionization enthalpy?

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