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
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 $L$.
$J=L+S=1+1=2(*ℏ)$ 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 $L_z$: $L=L_z=(3+2+1+0-1-2-3)+3+2+1=6$