# Describe all 4 quantum numbers for last electron of 3d^5?

Jun 6, 2018

see below

#### Explanation:

it is Mn (manganese)
N= 3 third energetic level
l= 3 shape of orbitals of type d : (quatrefoil)
5 = the 5 orbitals d are all full means with only one electron in every orbitals (M,magnetic orbitatals = -2,-1, 0 +1 +2 what is the last, you don't know because are isoenergetic)
s =intrinsic magnetic moment $= \pm \frac{h}{2 \times \pi}$(what is the last, you don't know, for convention the negative one)

Jun 6, 2018

The four quantum numbers are 3, 2, +2, +½.

#### Explanation:

The four quantum numbers are

• $n \textcolor{w h i t e}{l l} =$ principal quantum number; determines energy and size of orbital
• $l \textcolor{w h i t e}{m} =$ secondary quantum number; determines shape of orbital
• ${m}_{\textrm{l}} =$magnetic quantum number; determines orientation of orbital in magnetic field
• ${m}_{\textrm{s}} =$ spin quantum number; determines direction of electron spin

The rules for quantum numbers are:

• $n \textcolor{w h i t e}{l l} =$ integers from 1 to ∞.
• $l \textcolor{w h i t e}{m} =$ integers from 0 to $n - 1$
• ${m}_{\textrm{l}} =$ integers from $- l$ to $+ l$
• ${m}_{\textrm{s}} =$ +½ or -½

Here is a table of quantum numbers for the first five $\text{3d}$ electrons. The quantum numbers for the fifth electron are red.

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

Here's how the quantum numbers fit in the Periodic Table