# What is the ground state electron configuration of a ""_27 "Co" atom in the gas phase?

Mar 7, 2016

Cobalt is atomic number $27$, which means it has $27$ electrons as a neutral atom (protons and electrons cancel out when charge = $0$).

If you looked at the periodic table, it's a $3 d$ metal, in the fourth period, so the ground-state electron configuration is:

$\textcolor{b l u e}{1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 4 {s}^{2} 3 {d}^{7}}$

That adds up to $2 + 2 + 6 + 2 + 6 + 2 + 7 = 27$. It's not $\text{Cu}$ or $\text{Cr}$, so it's not a strange configuration. It's implied that it's in the gas phase so it's not really necessary to state that.

In this configuration, three $3 d$ orbitals are singly occupied.

CHALLENGE: Is cobalt paramagnetic or diamagnetic as an atom? How do you know? What is its closest-in-energy first-excited configuration (it cannot promote into the $4 s$, and it cannot promote into the $4 d$ in one step)?