Question

The atomic number of vanadium (V), chromium (Cr), manganese (Mn) and iron (Fe) are respectively $23,24,25,26$. Which out of these may be expected to have the jump in second ionization enthalpy?
A.Mn
B.Fe
C.V
D.Cr

Answer 1

Hint:Check for the electronic configuration of all the elements given. Find out the one with half- filled or fully filled subshell for second ionization enthalpy. In a group while moving from top to bottom, ionization energy decreases. In a period while moving from left to right it increases.

Complete answer:
 Given are four elements- Vanadium (V) - atomic no. $23$
  Chromium (Cr) – atomic no $24$
 Manganese (Mn) – atomic no $25$
  Iron (Fe) – atomic no $26$
Electronic configuration of V $=\left[Ar\right]3d^{3}4s^2$
Here, for first ionization enthalpy one electron from $4s^2$is removed and then for second ionization enthalpy we have $4s^1$
Electronic configuration of Cr$=\left[Ar\right]3d^{5}4s^1$
Here, is an exception in the configuration of chromium it is $\left[Ar\right]3d^{5}4s^1$instead of $\left[Ar\right]3d^{4}4s^2$because half-filled d-orbitals have extra stability. So, one electron from the 4s orbital goes to the 3d orbital to attain extra stability.
So, for the first ionization enthalpy electron from $4s^1$is removed. Now, for the second ionization enthalpy electron has to be removed from $3d^5$ which is half-filled, which means it is quite stable. So, one requires a jump in the second ionization enthalpy for chromium to remove the electron from the stable$3d^5$state.
For manganese we have - $\left[Ar\right]3d^{5}4s^2$
For iron we have - $\left[Ar\right]3d^{6}4s^2$
So, from all the four given elements comparatively there is a jump in second ionization enthalpy for chromium.

Therefore, the correct choice is D.
Additional Information:
 Ionization Enthalpy: Ionization enthalpy or ionization energy of elements is defined as the amount of energy required by an isolated gaseous atom to lose an electron in its ground state. The first ionization energy of element B is defined as the energy required by an atom to form $B^{2+}$ ion. The units of Ionization energy are KJ per mole. Similarly, second ionization energy is described as the energy needed to remove the second electron from its valence shell. So, $B^{+}$becomes $B^{2+}$ by losing one more electron.

Note:
The factors affecting ionization energy are- force of attraction between the electron and nucleus and force of repulsion between electrons. A specific amount of energy is required to remove the electron hence the ionization enthalpies are always positive. The second most outer electron will be more attracted by the nucleus as compared to the first outer electron. So, the second ionization energy will be greater than the first ionization energy.