
What is the electronic configuration of $ {P^{3 - }} $ and $ M{o^{3 + }} $ ions?
Answer
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Hint : $ {P^{3 - }} $ ion is formed when it gains three electrons and $ 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 $ 15 $ . So, the electronic configuration of the ground species $ {}_{15}P $ can be found by filling the $ 15 $ electrons using the Aufbau’s principle, Pauli’s exclusion principle and Hund’s rule. We will get the electronic configuration as $ 1{s^2}2{s^2}2{p^6}3{s^2}3{p^3} $ . We can observe that the ionic species $ {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 $ 2p $ orbitals are half-filled, the new three electrons will be paired up to completely fill them. Hence, the electronic configuration of $ {P^{3 - }} $ is $ 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6} $ .
We know that the atomic number of molybdenum is $ 42 $ . We will find the electronic configuration of its ground state $ {}_{42}Mo $ first using the above-mentioned principles. 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^4} $ . We know that the ionic species $ 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 $ 5s $ orbital. So, the two electrons of $ 5s $ orbital and an electron from the pen-ultimate shell $ 4d $ is lost. Hence, the electronic configuration of $ M{o^{3 + }} $ 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}4{d^3} $ .
Note :
It is common to get confused and remove the three electrons from the $ 4d $ orbital instead of removing from $ 5s $ 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.
Complete Step By Step Answer:
We know that the atomic number of phosphorus is $ 15 $ . So, the electronic configuration of the ground species $ {}_{15}P $ can be found by filling the $ 15 $ electrons using the Aufbau’s principle, Pauli’s exclusion principle and Hund’s rule. We will get the electronic configuration as $ 1{s^2}2{s^2}2{p^6}3{s^2}3{p^3} $ . We can observe that the ionic species $ {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 $ 2p $ orbitals are half-filled, the new three electrons will be paired up to completely fill them. Hence, the electronic configuration of $ {P^{3 - }} $ is $ 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6} $ .
We know that the atomic number of molybdenum is $ 42 $ . We will find the electronic configuration of its ground state $ {}_{42}Mo $ first using the above-mentioned principles. 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^4} $ . We know that the ionic species $ 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 $ 5s $ orbital. So, the two electrons of $ 5s $ orbital and an electron from the pen-ultimate shell $ 4d $ is lost. Hence, the electronic configuration of $ M{o^{3 + }} $ 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}4{d^3} $ .
Note :
It is common to get confused and remove the three electrons from the $ 4d $ orbital instead of removing from $ 5s $ 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.
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