Can anyone help with this question: “Assign 4 quantum numbers to the 3rd, 10th, and 18th electrons in cobalt”?

Nov 28, 2017

Here's what I get,

Explanation:

The electron configuration of cobalt is

$\text{1s"^2color(white)(l) "2s"^2 "2p"^6color(white)(l) "3s"^2 "3p"^6 color(white)(l)"4s"^2 "3d"^7}$

The rules for quantum numbers are

• n = 1, 2, 3, 4, …
• l = 0, 1, 2, …, n-1
• m_l = l, l-1, …, 0, …, 1-l, -l
• m_s = ±½

We fill the orbitals using the Aufbau Principle, Hund's Rule, and tthe Pauli Exclusion Principle.

This gives the following table.

$\underline{\boldsymbol{\text{No.} \textcolor{w h i t e}{m} n \textcolor{w h i t e}{m} l \textcolor{w h i t e}{m} {m}_{l} \textcolor{w h i t e}{m} {m}_{s}}}$
$\textcolor{w h i t e}{l l} 1 \textcolor{w h i t e}{m m} 1 \textcolor{w h i t e}{m l} 0 \textcolor{w h i t e}{m} 0 \textcolor{w h i t e}{m l} \text{+½}$
$\textcolor{w h i t e}{l l} 2 \textcolor{w h i t e}{m m} 1 \textcolor{w h i t e}{m l} 0 \textcolor{w h i t e}{m} 0 \textcolor{w h i t e}{m l l} \text{-½}$

$\textcolor{red}{\textcolor{w h i t e}{l l} 3 \textcolor{w h i t e}{m m} 2 \textcolor{w h i t e}{m l} 0 \textcolor{w h i t e}{m} 0 \textcolor{w h i t e}{m l} \text{+½}}$
color(white)(ll)4color(white)(mm)2color(white)(ml)0color(white)(m)0color(white)(mll)"-½"

$\textcolor{w h i t e}{l l} 5 \textcolor{w h i t e}{m m} 2 \textcolor{w h i t e}{m l} 1 \textcolor{w h i t e}{m} 1 \textcolor{w h i t e}{m l} \text{+½}$
$\textcolor{w h i t e}{l l} 6 \textcolor{w h i t e}{m m} 2 \textcolor{w h i t e}{m l} 1 \textcolor{w h i t e}{m} 0 \textcolor{w h i t e}{m l} \text{+½}$
$\textcolor{w h i t e}{l l} 7 \textcolor{w h i t e}{m m} 2 \textcolor{w h i t e}{m l} 1 \textcolor{w h i t e}{l l} \text{-1"color(white)(ml)"+½}$

$\textcolor{w h i t e}{l l} 8 \textcolor{w h i t e}{m m} 2 \textcolor{w h i t e}{m l} 1 \textcolor{w h i t e}{m} 1 \textcolor{w h i t e}{m l l} \text{-½}$
$\textcolor{w h i t e}{l l} 9 \textcolor{w h i t e}{m m} 2 \textcolor{w h i t e}{m l} 1 \textcolor{w h i t e}{m} 0 \textcolor{w h i t e}{m l l} \text{-½}$
$\textcolor{red}{\textcolor{w h i t e}{l} 10 \textcolor{w h i t e}{m l l} 2 \textcolor{w h i t e}{m l} 1 \textcolor{w h i t e}{l l} \text{-1"color(white)(mll)"-½}}$

$\textcolor{w h i t e}{l} 11 \textcolor{w h i t e}{m m} 3 \textcolor{w h i t e}{m l} 0 \textcolor{w h i t e}{m} 0 \textcolor{w h i t e}{m l} \text{+½}$
color(white)(l)12color(white)(mm)3color(white)(ml)0color(white)(m)0color(white)(mll)"-½"

$\textcolor{w h i t e}{l} 13 \textcolor{w h i t e}{m m} 3 \textcolor{w h i t e}{m l} 1 \textcolor{w h i t e}{m} 1 \textcolor{w h i t e}{m l} \text{+½}$
$\textcolor{w h i t e}{l} 14 \textcolor{w h i t e}{m m} 3 \textcolor{w h i t e}{m l} 1 \textcolor{w h i t e}{m} 0 \textcolor{w h i t e}{m l} \text{+½}$
$\textcolor{w h i t e}{l} 15 \textcolor{w h i t e}{m m} 3 \textcolor{w h i t e}{m l} 1 \textcolor{w h i t e}{l l} \text{-1"color(white)(ml)"+½}$

$\textcolor{w h i t e}{l} 16 \textcolor{w h i t e}{m m} 3 \textcolor{w h i t e}{m l} 1 \textcolor{w h i t e}{m} 1 \textcolor{w h i t e}{m m} \text{-½}$
$\textcolor{w h i t e}{l} 17 \textcolor{w h i t e}{m m} 3 \textcolor{w h i t e}{m l} 1 \textcolor{w h i t e}{m} 0 \textcolor{w h i t e}{m m} \text{-½}$
$\textcolor{red}{\textcolor{w h i t e}{l} 18 \textcolor{w h i t e}{m m} 3 \textcolor{w h i t e}{m l} 1 \textcolor{w h i t e}{l l} \text{-1"color(white)(mm)"-½}}$

The third, tenth, and eighteenth sets of quantum numbers are marked in red.