Question

What is the electronic configuration of $H{g^{2 + }}$ ?

Answer 1

Hint : $H{g^{2 + }}$ ion is formed when an $Hg$ atom loses two electrons. When losing electrons, the electrons in the outermost (valence) shells are lost first. But while filling up the orbitals we have to follow the Aufbau principle and other rules.

Complete Step By Step Answer:
Well, we know that the atomic number of mercury is $80$ . We will find the electronic configuration of its ground state ${}_{80}Hg$ first using the Aufbau’s principle, Pauli’s exclusion principle and Hund’s rule. We can easily find its electronic configuration to be $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^{10}}4{p^6}5{s^2}4{d^{10}}5{p^6}6{s^2}4{f^{14}}5{d^{10}}$ . We know that the ionic species $H{g^{2 + }}$ is formed when it loses two electrons. We have to remember that an element loses electrons from its outermost or valence shell. Here, the valence electrons are in the $6s$ orbital. So, the two electrons of $6s$ orbital are lost. Hence, the electronic configuration of $H{g^{2 + }}$ is $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^{10}}4{p^6}5{s^2}4{d^{10}}5{p^6}6{s^0}4{f^{14}}5{d^{10}}$ . Now re-arranging it we get the electronic configuration to be $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^{10}}4{s^2}4{p^6}4{d^{10}}4{f^{14}}5{s^2}5{p^6}5{d^{10}}6{s^0}$ . It is easy to observe the noble gas configuration of ${}_{80}Hg$ to be $\left[ {Xe} \right]4{f^{14}}5{d^{10}}6{s^2}$ and so that of $H{g^{2 + }}$ as $\left[ {Xe} \right]4{f^{14}}5{d^{10}}$ .

Note :
It is common to get confused and remove the electrons from the $5d$ orbital instead of removing from $6s$ orbital as $5d$ orbitals are filled up last. Always keep in mind that while filling up orbitals we have to use the Aufbau’s principle and fill up the less energy orbitals first. But in case of removing electrons, the electrons in the valence shells are easy to lose, so we have to remove them from the valence shells first, and then the pen-ultimate shells and so on.