Question

How would you order the following elements with respect to electron Affinity $C{l^ - },Ar,{K^ + },Ti,T{i^{2 + }}$?

Answer 1

When an electron is added to a neutral atom then its energy is changed and it acquires negative charge. The change in the energy is known as electron affinity. In other words, the amount of energy is released when an electron is added to a neutral atom or any molecule which is in gaseous state. We can arrange the given set of elements in increasing order with the help of their electronic configuration.

Complete step-by-step answer:
To find out the trend of electron affinity, we have to write their electronic configuration.
$C{l^ - } = \left[ {Ne} \right]3{s^2}3{p^6}$
$Ar = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}$
${K^ + } = [Ar]4{s^1}$
$Ti = [Ar]3{d^2}4{s^2}$
$T{i^{2 + }} = [Ar]3{d^2}$
Electron affinity is dependent upon nuclear charge, atomic size, electronic configuration. Electron affinity increases with increase in nuclear charge and decreases with increase in atomic size. Electron affinity increases as we move from left to right and decreases when we move from top to down group.
Group 18 elements have 8 valence electrons and they do not want to share electrons with other atoms and they do not react with other elements. It means it has high Ionization energy, that is more energy is required to remove an electron. Therefore, noble gas has zero electron affinity as compared to other elements in the periodic elements. Argon belongs to group 18 therefore its electron affinity is very less compared to any elements.
Now, compare $C{l^ - },Ti,T{i^{2 + }},{K^ + }$
Chloride ion has an extra electron already that means it is negative already. Therefore the electron affinity of $C{l^ - }$ is less than the electron affinity of ${K^ + },$ $Ti$,$T{i^{2 + }}$.
Now, compare ${K^+}$ ,$Ti$,$T{i^{2 + }}$
If we add an electron to Titanium, it acquires negative charge but due to instability it releases the electron and becomes positive. So, its electron affinity is less than ${K^+}$,$T{i^{2 + }}$.
Now, compare and ${K^+}$,$T{i^{2 + }}$.
Potassium has $+ 1$ charge whereas $T{i^{2 + }}$ has $+ 2$ charge. Therefore $T{i^{2 + }}$ has more electron affinity as compared to potassium because electron affinity depends upon the nuclear charge. As nuclear charges increase the electron affinity increases.
So, the arrangement of electron affinity of the given set is:
$Ar < C{l^ - } < Ti < {K^ + } < T{i^ + 2}$

Note: Always keep in mind that the electron affinity is zero in the case of noble gases. Remember that the electron gains enthalpy and electron affinity both are the same thing except their signs. It means that $\Delta {H_{eg}} = - E.A.$ While doing this type of question, keep in mind that on what factors electron affinity depends.