Dear student,

I will tell u the general method of finding unpaired electrons...

An orbital is made up of 2 electrons. When you figure out the electron configuration by looking at the atomic mass of an element, any orbitals left with only 1 electron is considered unpaired.

For example, the element P, has an atomic mass of 15. So the electron configuration is 1s^2, 2s^2, 2p^6, 3s^2, 3p^3 (Notice how the exponents add up tp 15).

Once you figure out the electron configuration, you fill up the corresponding orbitals with electrons, any left with one is conidered unpaired. Since 1s can only hold 2 electrons, and P has 15, thats obviously filled and has no unpaired electrons. The same is for 2s which holds 2, 2p which holds 6, 3s which holds 2.

However3p can hold 6 electrons and in order for that to be filled up you would need to have an element of 18 electrons. So you fill up as much as you can in 3p by first adding 1 electron to each energy level. 3p has 3 energy levels and there are only 3 electrons left to distribute, so each of those energy levels only gets 1, because you have to fill them all with one before you can start adding a second.

So since you are only able to fill one electron in each of the three energy levels of the 3p orbital, that leaves the orbital open for 1 more electron in each of its energy levels. So there are 3 unpaired electrons in P.

Dear Gundoos 1234,

It is hard to tell exactly what you mean based on how you wrote them, however assuming you mean KOO, AlOO-, BaO and NOO+, then only KOO would have an unpaired electron. KOO's valance number would be 13 and there is no way to make bonds even with 13. AlOO- would have 14 valence electrons and you could fill the orbits by forming 2 double bonds, one O=O and the other Al=O. The BaO has 8 valence electrons, so that makes it full and NOO+ would have 18 and you can fill the orbits by making a double bond O=O.

Hope this helped you dear.........

All The Best........

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Dear Gundoos 1234,

Among these KO₂ has unpaired electrons in it !

The ideal way of approaching the problem is to see how many electrons are present in the compounds ...

If the number of electrons present are in even numbers then it well may be completely paired(but not sure though) but if it is odd in number then there surely are unpaired electron(s) .

Here in this case only KO₂ had odd(35) electrons in it & so it clearly has unpaired ellctrons in it !!

Hope this helped you immensely..

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