Magnetic Properties (Continued)

Please find L, S, and J for the following different atoms using Hund’s and Aufbau principles

1. Sulfur (S) Z= 16

2. Vanadium (V), Z = 23

3. Zirconium (Zr), Z = 40

4. Xenon (Xe), Z = 54

5. Dysprosium (Dy), Z = 66


Hund’s rules

1.   Electrons try to align their spins.

2.   Electron align their spins such that the angular momentum L is maximum.

3.    The total angular momentum J is such that J=L+S for more than half filled orbitals and $J=L-S$ for less than half filled orbitals.

For Sulphur $Z=16$ the electron configuration is $1s^2,2s^2,2p^6,3s^2,3p^4$  .

Last shell is $3p^4$ which means 4 electrons on p orbitals  [possible $L_z=(-1,0,1)ℏ$].

4th electron is in $L_z=+1$ to maximize $L$.

Total $L_z=L=(1+1+0-1)=1(*ℏ)$

Total $S=3*1/2-1/2=1(*ℏ)$

$J=L+S=1+1=2(*ℏ)$   more than half filled $p$ orbital

For Vanadium $Z=23$  we have  $1s^2$,   $2s^2 2p^6$    $3s^2 3p^6$    $4s^2 3d^3$

Last shell is $3d^3$ which means there are 3 electrons for possible $l_z=-2.-1,0,1,2$. To maximize $L$ electrons are on $l_z=2.1,0$  (with spins aligned).



$J=L-S=3-3/2=3/2$     (less than half filled d orbital)

For Zirconium $Z=40$  we have  $1s^2$   $2s^2 2p^6$    $3s^2 3p^6$    $4s^2 3d^10 4p^6$   $5s^2 4d^2$

Last shell is $4d^2$ ; 2 electrons on possible $l_z=-2,-1,0,1,2$. To maximize $L=L_z$ the electrons are on $l_z=2,1$ so


$S=1/2+1/2=1$  (parallel spin electrons)


For Xenon =54 we have  $1s^2$   $2s^2 2p^6$   $3s^2 3p^6$   $4s^2 3d^10 4p^6$   $5s^2 4d^10 5p^6$

Last shell is $5p^6$ completely filled $p$ orbital.  Possible $l_z=-1,0,1$ . There are two electrons (of opposite spins) ineach orbital



$J=L+S=0+0=0(ℏ)$       (more than half filled)

For Dyprosium =66 we have  $1s^2$   $2s^2$  $2p^6$   $3s^2 3p^6$  $4s^2 3d^10 4p^6$   $5s^2 4d^10 5p^6$    $6s^2 4f^10$
Last shell is $4f^10$. Possible $l_z=-3,-2,-1,0,1,2,3$ (there are 7 suborbitals in f). To maximize $L_z$:  $L=L_z=(3+2+1+0-1-2-3)+3+2+1=6$