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

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.

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

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 $3 \cdot 2 = 6$ 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.

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 $P {b}^{1 +}$.
This means that the neutral lead atom has lost one electron, as the charge of an electron is $- 1$.

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