From which atom of each of the following pairs is it easier to remove an electron:

(a) krypton or Bromine,

(b) rubidium or Cerium,

(c) helium or Hydrogen?

The given pairs are:

1. Krypton or Bromine
2. Rubidium or Cerium
3. Helium or Hydrogen

Using the concept of ionization energy size and binding energy, we can predict the atom that is easier to give up an electron or is easier to remove an atom from.

Similarly, a bigger size or far distance of the outermost electron from the nucleus results in low binding energy; thus, it is easier to remove an electron. Again, atoms with low ionization energy are easier to remove the electron.

Electronic configuration of Bromine:$\left[\mathrm{Ar}\right]4s^{2}3d^{10}4p^5$

Electronic configuration of Krypton:$\left[\mathrm{Ar}\right]4s^{2}3d^{10}4p^6$

Now, from the electronic configurations, it can be clearly seen that Bromine needs only one electron to achieve a noble gas configuration, while Krypton is a noble gas. Thus, the binding energy of Krypton is more and will need more ionization energy to remove one electron from that structure. But, in the case of Bromine, the atom requires low ionization energy than Krypton to remove an electron.

Hence, it is easier to remove an electron from Bromine.

Electronic configuration of Rubidium:$\left[Kr\right]5s^1$

Electronic configuration of Cerium:$\left[Xe\right]4f^{1}5d^{1}6s^2$

Here, the atom Cerium has a bigger size than that of Rubidium. Thus, the outermost electron is far from the binding of the nucleus for the atom Cerium.

Hence, it is easier to remove an electron from Cerium.

Electronic configuration of Hydrogen:$1s^1$

Electronic configuration of Helium:$1s^2$

Similar to the concept of part (a), Helium is a noble gas and hence requires more energy to disturb and remove the electron from the surface.

Hence, it is easier to remove an electron from Hydrogen.