Question

The atomic numbers of $V,Cr,Mn$ and $Fe$ are respectively $23,24,25\text{and}26$. Which one of these is expected to have the highest second ionization energy ?

• A. $Cr$
• B. $Mn$
• C. $Fe$
• D. $V$

Answer 1

The correct option is A $Cr$

The electronic configuration of the following elements would be:
$\text{V}=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^3$
$\text{Cr}=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^5$
$\text{Mn}=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^5$
$\text{Fe}=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^6$

$Cr$ after the loss of one electron acquires stable half-filled $3d^5$ configuration. Thus, its second ionization enthalpy is highest among the given elements.