Question

The second IP of $Cr$ is high due to?
A.$3{d^0}$
B.$3{d^5}$
C.$3{d^{10}}$
D.$3{d^4}$

Answer 1

The ionisation potential energy or IP can be defined as the minimum energy required to remove the most loosely bound electron from the orbital of the isolated neutral gaseous atom or molecule. So the Ionisation potential energy of an isolated gaseous atom or molecule depends on the electronic configuration of that particular atom or molecule.

Complete answer:
So , as above we understand the ionisation energy, let's also have a look at what is the difference between the first ionisation potential energy and the second ionisation potential energy.
Now as the IP depends on the electronic configuration of the particular element, so let’s have a look at the electronic configuration of the chromium.
Chromium has the atomic number $- 24$
$Cr{\text{ (Z = 24)}} \to {\text{1}}{{\text{s}}^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^1}3{d^5}$
So, as we can see, in the first IP the electron will be removed from the less stable $4{s^1}$ orbital.
So in the next step i.e. in the second IP the electron has to be removed from the half-filled $3{d^5}$ orbital.
And the half- filled orbitals are more stable and symmetrical and that’s why they need more energy to remove their valence electrons.
So, from the above discussion we can easily say that the correct answer is option B. i.e. $3{d^5}$ Orbital.
Thus, because of the better symmetry and more stability of $3{d^5}$ orbital, the second IP of $Cr$ is higher.

Note:
As we know from above, the ‘first’ ionisation energy is the minimum energy required to remove the electrons from the gaseous neutral atom or the molecule but the second ionisation energy is the energy which is required to remove the loosely bound electron from the ions.