Question

What is the electronic configuration of $ S{r^{2 + }} $ ?

Answer 1

Hint : $ S{r^{2 + }} $ ion is formed when an $ Sr $ 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 strontium is $ 38 $ . We will find the electronic configuration of its ground state $ {}_{38}Sr $ 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} $ . In the noble gas configuration notation, it is $ \left[ {Kr} \right]5{s^2} $ . We know that the ionic species $ S{r^{2 + }} $ is formed when $ Sr $ 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 $ 5s $ orbital. So, the two electrons of $ 5s $ orbital are lost. Hence, the electronic configuration of $ S{r^{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^0} $ . 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} $ . It is easy to observe the noble gas configuration of $ S{r^{2 + }} $ as $ \left[ {Kr} \right] $ . This implies that $ S{r^{2 + }} $ is very much unreactive and $ + 2 $ is the highest oxidation state of strontium.

Note :
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.