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The atomic numbers of $V,Cr,Mn$ and $Fe$ are respectively $23,24,25$ and $26$. Which one of these may be expected to have the highest second ionization enthalpy ?

$ (a)\;V \\(b)\; Cr \\ (c)\;Mn \\ (d)\;Fe $
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The electronic configurations are as follows;
$V: (Ar) 3d^34s^2$
$Cr: (Ar) 3d^54s^1$
$Mn: (Ar) 3d^54s^2$
$Fe: (Ar) 3d^64s^2$
The second ionization of $Cr$ means removal of electron from the stable configuration of $3d^5$.
The highest second ionization enthalpy is $Cr$
Hence b is the correct answer.
answered Mar 7, 2014 by meena.p
edited Mar 15, 2014 by balaji.thirumalai

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