What is the s,p,d,f configuration of Co?

Jan 16, 2014

The s,p,d,f configuration for cobalt (Co) is $1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 4 {s}^{2} 3 {d}^{7}$, determined by the position of the element on the periodic table.

Cobalt is an inner transition metal which means the electron configuration will end in a d block. Cobalt is in the 7th column of the d block and therefore has 7 d electrons ${d}^{7}$. The element cobalt can be found in the 4th row or 4th energy level of the periodic table. However, the inner transition metals or d block are always one level lower than the energy level that the element is found on meaning the electron configuration for cobalt must end as $3 {d}^{7}$.

All electron levels must be filled above that point. So the entire configuration is as follows:
$1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 4 {s}^{2} 3 {d}^{7}$

As another example, Argon is a noble gas and has each of the electron levels filled. Using Noble Gas Notation you can write the s,p,d,f configuration in an abbreviated form.

[Ar] can represent $1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6}$

Making the configuration:
[Ar] $4 {s}^{2} 3 {d}^{7}$

Here is a You Tube video on determining electron configurations:

Nov 14, 2015

$1 {s}^{2} , 2 {s}^{2} , 2 {p}^{6} , 3 {s}^{2} , 3 {p}^{6} , 4 {s}^{2} , 3 {d}^{7}$