Question

The atomic number of Vanadium $\left(V\right)$, chromium $\left(Cr\right)$, manganese $\left(Mn\right)$ and iron $\left(Fe\right)$ 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

Answer: (A)

$Cr$

$Cr$ is expected to have the highest second ionization enthalpy as after removal of $4s$ electrons, the electrons will be removed from d-orbital. $C{r}^{+}$ have stable half-filled d-orbital with maximum spin multiplicity.

$Cr$ configuration $=\left[Ar\right]4s^{1}3d^5$

$C{r}^{+}$ configuration $=\left[Ar\right]3d^5$