# Question #6706a

Aug 31, 2017

$n = 4 , l = 0 , m = 0 , s = \frac{1}{2}$

#### Explanation:

For Chromium,

$Z = 24$

Electronic configuration

$1 {s}^{2} , 2 {s}^{2} , 2 {p}^{6} , 3 {s}^{2} , 3 {p}^{6} , 3 {d}^{5} , 4 {s}^{1}$

So, for the last electron,

$n = 4$

$l = 0$

$m = 0$

$s = \frac{1}{2}$

Aug 31, 2017

The quantum numbers are $n = 3 , l = 2 , {m}_{l} = \text{+2", m_s = "+½}$.

#### Explanation:

Chromium has atomic number 24.

You can write its electron configuration as

${\text{[Cr] 1s"^2 color(white)(l)"2s"^2 "2p"^6 color(white)(l)"3s"^2 "3p"^6color(white)(l) "4s" color(white)(l)"3d}}^{5}$ or

$\text{[Cr] 1s"^2color(white)(l) "2s"^2 "2p"^6color(white)(l) "3s"^2 "3p"^6 "3d"^5color(white)(l)"4s}$

However you write it, you must remember that the $\text{3d}$ orbitals have the highest energy and are filled last in the Aufbau process.

The quantum numbers of the first five $\text{3d}$ electrons are

$\boldsymbol{\underline{n \textcolor{w h i t e}{m} l \textcolor{w h i t e}{m} {m}_{l} \textcolor{w h i t e}{m} {m}_{s}}}$
$3 \textcolor{w h i t e}{m} 2 \textcolor{w h i t e}{m l} \text{-2"color(white)(m)"+½}$
$3 \textcolor{w h i t e}{m} 2 \textcolor{w h i t e}{m l} \text{-1"color(white)(m)"+½}$
$3 \textcolor{w h i t e}{m} 2 \textcolor{w h i t e}{m l l} 0 \textcolor{w h i t e}{m} \text{+½}$
$3 \textcolor{w h i t e}{m} 2 \textcolor{w h i t e}{m} \text{+1"color(white)(m)"+½}$
$\textcolor{red}{3 \textcolor{w h i t e}{m} 2 \textcolor{w h i t e}{m} \text{+2"color(white)(m)"+½}}$

Thus, the quantum numbers of the last electron added to chromium are

$3 , 2 , \text{+2, +½}$.