Question

The ground state of dysprosium (element 66, in the 6th row of the Periodic Table)

is listed as ${}^{\mathbf{5}}{\mathbf{I}}_{\mathbf{s}}$. What are the total spin, total orbital, and grand total angular

momentum quantum numbers? Suggest a likely electron configuration for

dysprosium.

Short answer

The total spin, orbital spin and total angular momentum are:

$S=2;L=6;J=8.$

Electronic configuration:

$\underset{\mathbf{d}\mathbf{e}\mathbf{f}\mathbf{i}\mathbf{n}\mathbf{i}\mathbf{t}\mathbf{e}\left(36electrons\right)}{\underbrace{{\left(1s\right)}^{\mathbf{2}}{\left(2s\right)}^{\mathbf{2}}{\left(2p\right)}^{\mathbf{6}}{\left(3s\right)}^{\mathbf{2}}{\left(3p\right)}^{\mathbf{6}}{\left(3d\right)}^{\mathbf{10}}{\left(4s\right)}^{\mathbf{2}}{\left(4p\right)}^{\mathbf{6}}}}\underset{\mathbf{l}\mathbf{i}\mathbf{k}\mathbf{e}\mathbf{l}\mathbf{y}\left(30electrons\right)}{\underbrace{{\left(4d\right)}^{\mathbf{10}}{\left(5s\right)}^{\mathbf{2}}{\left(5p\right)}^{\mathbf{6}}{\left(4f\right)}^{\mathbf{10}}{\left(6s\right)}^{\mathbf{2}}}}$

Step-by-step solution

Definition of total spin, total orbital and total angular momentum quantum number

Spin quantum number, the fourth quantum number introduced to explain the rotation and direction of the electron spin in space. By combining a particle's orbital and intrinsic angular momentum, the total angular momentum quantum number in quantum mechanics parameterizes the total angular momentum of a particular particle.

Determining the total spin, total orbital and grand total angular momentum quantum numbers

Ground state of dysprosium is ⁵l₈. That means 2s + 1 = 5 and J = 8. So total spin , orbital spin and total angular momentum are:

S = 2, L = 6, J = 8.

Electronic configuration :

$\underset{\text{definite (36 electrons)}}{\underbrace{{\left(1s\right)}^2{\left(2s\right)}^2{\left(2p\right)}^6{\left(3s\right)}^2{\left(3p\right)}^6{\left(3d\right)}^{10}{\left(4s\right)}^2{\left(4p\right)}^6}}\underset{\text{likely (30 electrons)}}{\underbrace{{\left(4d\right)}^{10}{\left(5s\right)}^2{\left(5p\right)}^6{\left(4f\right)}^{10}{\left(6s\right)}^2}}$

30 electrons which fill all orbits from$4d^{10}$ to$6s^2$ are likely there.