Question

Choose the correct answer from the alternative given.
$E_{M{n}^{3+}}^o/M{n}^{2+}$ is highly positive than that of $E_{C{r}^{3+}}^o/C{r}^{2+}$or $E_{F{e}^{3+}}^o/F{e}^{2+}$because:

• A. $M{n}^{2+}$ $\left(d^5\right)$ can be easily oxidised to $M{n}^{3+}\left(d^4\right)$ due to low ionisation enthalpy
• B. third ionization enthalpy of Mn is much larger due to stable half filled $d^5$ electronic configuration of $M{n}^{2+}$
• C. $M{n}^{3+}$ is more stable than $M{n}^{2+}$ due to higher oxidation state
• D. second ionisation enthalpy of Mn is higher than third ionisation enthalpy

Answer 1

Answer: (B)

third ionization enthalpy of Mn is much larger due to stable half filled $d^5$ electronic configuration of $M{n}^{2+}$

Electronic configuration of Mn is: [Ar] 3d${}^{5}4s^2$

For Mn${}^{3+}$ : [Ar] 3d${}^{4}4s^0$

For Mn${}^{2+}$ : [Ar] 3d${}^{5}4s^0$

Since in Mn${}^{2+}$, half filled d${}^5$ electronic configuration is attained so, it will be now much difficult to remove e${}^{-}$ from this stable configuration, so, E${}^0$ Mn${}^{3+}$/Mn${}^{2+}$ is highly positive as the 3rd ionisation enthalpy of Mn is very high.