Question

Shows that the ionization energy of cadmium$\left(z=48\right)$is larger than that of its neighbors. Why is this?

Short answer

The given statement is prove given below.

Step-by-step solution

Given Information

We need to find the ionization energy of cadmium $\left(z=48\right)$is larger than that of its neighbors.

Explanation

The ground-state configuration of cadmium $\left(Z=48\right)$is:

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}$

Although Cd doesn't end with a closed p-shell like the rare gases, it does have a filled shell since $4d^{10}$completes the $4d$subshell. The ground-state configuration of the next element $\left(indium,\text{Z}=49\right)$is $\mathrm{...5}{s}^{2}4d^{1}05p^1.$electrons are not as tightly bound as $p‐$electron , so completing a $d-shall$shell doesn't cause Cd to be an inert, rare-gas element. Even so, it's harder to remove an electron from the filled $4d^{10}$shell than from the open $5p$shell of $Z=49$. The ground-state configuration of the previous element $(silver,z=47)$has an$4d^{10}$shell than from the unfilled shell. This is why the graph of the ionization energies has a local maximum at incomplete $4d^9$shell. it's harder to remove an electron from the filled$Z=48$.