Answer
See the detailed answer below.
Work Step by Step
The image explains the ground-state electron configuration of cadmium (Cd), which has an atomic number of 48. Its configuration is:
$$ 1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{10} 5s^2 $$
The cadmium does not end with a closed $ p $-shell like the noble gases, it has a filled $ 4d^{10} $ subshell. This filled $ 4d $-shell plays a role in stabilizing the element but doesn't make cadmium as chemically inert as noble gases because it lacks a fully closed $ p $-shell.
For the next element, indium (In), with atomic number 49, the electron configuration is:
$$ 5s^2 4d^{10} 5p^1 $$
The $ d $-electrons are generally not as tightly bound as $ p $-electrons, so the completion of the $ d $-shell in cadmium does not lead to the same inert properties as a rare-gas element.
Furthermore, it is noted that it is more difficult to remove an electron from the completely filled $ 4d^{10} $ subshell in cadmium compared to removing an electron from the $ 5p $-shell of indium. Similarly, for silver (Ag), with atomic number 47, the configuration includes an incomplete $ 4d^9 $ subshell, making it easier to remove an electron from this incomplete subshell compared to cadmium’s filled $ 4d $-shell.
This observation explains why there is a local maximum in the ionization energy graph at cadmium (Z = 48). The filled $ 4d^{10} $ subshell provides a relatively stable configuration, making it harder to ionize compared to elements around it.
