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Question: What is the electronic configuration of \( {P^{3 - }} \) and \( M{o^{3 + }} \) ions?...

What is the electronic configuration of P3{P^{3 - }} and Mo3+M{o^{3 + }} ions?

Explanation

Solution

Hint : P3{P^{3 - }} ion is formed when it gains three electrons and Mo3+M{o^{3 + }} in is formed when it loses three 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.

Complete Step By Step Answer:
We know that the atomic number of phosphorus is 1515 . So, the electronic configuration of the ground species 15P{}_{15}P can be found by filling the 1515 electrons using the Aufbau’s principle, Pauli’s exclusion principle and Hund’s rule. We will get the electronic configuration as 1s22s22p63s23p31{s^2}2{s^2}2{p^6}3{s^2}3{p^3} . We can observe that the ionic species P3{P^{3 - }} is formed when it gains three electrons. So, we have to include three electrons in its electronic configuration by following the usual principles used to fill up electrons. Since, the 2p2p orbitals are half-filled, the new three electrons will be paired up to completely fill them. Hence, the electronic configuration of P3{P^{3 - }} is 1s22s22p63s23p61{s^2}2{s^2}2{p^6}3{s^2}3{p^6} .
We know that the atomic number of molybdenum is 4242 . We will find the electronic configuration of its ground state 42Mo{}_{42}Mo first using the above-mentioned principles. We can easily find its electronic configuration to be 1s22s22p63s23p64s23d104p65s24d41{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^4} . We know that the ionic species Mo3+M{o^{3 + }} is formed when it loses three electrons. We have to remember that an element loses electrons from its outermost or valence shell. Here, the valence electrons are in the 5s5s orbital. So, the two electrons of 5s5s orbital and an electron from the pen-ultimate shell 4d4d is lost. Hence, the electronic configuration of Mo3+M{o^{3 + }} is 1s22s22p63s23p64s23d104p65s04d31{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}4{d^3} .

Note :
It is common to get confused and remove the three electrons from the 4d4d orbital instead of removing from 5s5s orbital first. 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.