Question

The atomic number 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

Hint: Ionization energy is defined as the amount of energy required to remove an electron from an isolated gaseous atom and is expressed in the unit of kJ/ mol. Similarly, second ionization energy is defined as the amount of energy required to remove the second loose bounded electron.

Complete step by step answer: In the case of vanadium (v), the atomic number is 23. Hence its outer electronic configuration is $3d^{3}4s^2$. Similarly, the outer electronic configuration for chromium (Cr) with atomic number 24 is:
$3d^{5}4s^1$. In the case of manganeses (Mn), the atomic number is 25. Hence electronic configuration is
$3d^{5}4s^2$ and the electronic configuration for iron (Fe) with atomic number 26 the electronic configuration is $3d^{6}4s^2$
After removing one electron in each of the above case the electronic configuration of each atom will be:
${\text{V}}^{+}=3d^{3}4s^1$
$C{r}^{+}=3d^5$
$M{n}^{+}=3d^{5}4s^1$
$F{e}^{+}=3d^{6}4s^1$
After removal the one electron, chromium gets a half-filled 3d electronic configuration which is more stable as compared to the partial filled electronic configuration so it is quite difficult to remove the second electron in the case of chromium hence the chromium have maximum second ionization energy as compared to vanadium, manganese, and iron.
Hence the correct answer is option A.

Additional Information: he ionization energy is always positive because energy is absorbed for the removal of an electron.

Note: Generally the second ionization energy will be higher than the first ionization energy because it is more difficult to remove an electron from a positively charged ion than from a neutral atom due increase in effective nuclear charge on valence electrons.