How many unpaired electrons are there in lead(I)?

1 Answer
Jan 17, 2018

#1# unpaired electron.

Explanation:

Let's figure out the number of unpaired electrons in a lead atom first, before moving onto the lead ion.

Lead is in the #p# block, which means that its highest orbital series is #p#. That's the orbital series that have unpaired electrons.

KnowledgeDoor (pink elements are the #p# block)

We know that in a #p# orbital series, there are a total of #3# sublevels or orbitals.

me

Therefore, because of the Pauli Exclusion Principle which basically states that the maximum number of electrons that may occupy any orbital is #2#, there are #3*2=6# total vacant spaces for electrons in the valence orbital series of lead.

Counting from left to right within the #p# block, lead is second. This means that it has #2# electrons in the #p# orbital series.
This is because there's a trend in the periodic table where the number of valence electrons is equal to the number we get when we count from left to right within its block.

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This means there are #2# unpaired electrons in a neutral lead atom. Now, from this, we can find the number of unpaired electrons in a lead(I) ion.

First, we should understand that lead(I) is #Pb^(1+)#.
This means that the neutral lead atom has lost one electron, as the charge of an electron is #-1#.

me

Removing one electron from #2# unpaired electrons results in #1# unpaired electron.