What is Unpaired electron?

Example 1.18 (Carbon, nitrogen, oxygen, iron)

The method gives, for four atoms:

atom	configuration	last subshell	unpaired
$\ce{C}$ ($Z = 6$)	$[\ce{He}]\,2\text{s}^2 2\text{p}^2$	2p:	2
$\ce{N}$ ($Z = 7$)	$[\ce{He}]\,2\text{s}^2 2\text{p}^3$	2p:	3
$\ce{O}$ ($Z = 8$)	$[\ce{He}]\,2\text{s}^2 2\text{p}^4$	2p:	2
$\ce{Fe}$ ($Z = 26$)	$[\ce{Ar}]\,3\text{d}^6 4\text{s}^2$	3d:	4

For iron the Klechkowski order fills 4s ($Z = 19, 20$) before 3d ($Z = 21$ to 30); the configuration is written with 3d first.

Example 1.22 (Copper and its ions)

$\ce{Cu}$ is $[\ce{Ar}]\,3\text{d}^{10} 4\text{s}^1$; $\ce{Cu+}$ is $[\ce{Ar}]\,3\text{d}^{10}$, with no unpaired electron; $\ce{Cu^{2+}}$ is $[\ce{Ar}]\,3\text{d}^9$, with one. $\ce{Zn^{2+}}$, $[\ce{Ar}]\,3\text{d}^{10}$, has none.