Question

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

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

Answer 1

The correct option is A $Cr$

$V\left(23\right)=\left[Ar\right]{3d}^3{4S}^2$

$Cr\left(24\right)=\left[Ar\right]{3d}^5{4S}^1$

$Mn\left(25\right)=\left[Ar\right]{3d}^5{4S}^2$

$Fe\left(26\right)=\left[Ar\right]3d^6{4S}^2$

$2^{nd}IE$ highest of Cr since ${Cr}^{+}$ has $d^5$ $\leftarrow$ half filled stable.