Question

3. Knowing that the magnesium and aluminum atoms possesses 12 and 13 electrons respectively:

(a) Write down the electron configuration (nl) of the ground state. Explain your answer.

(b) We focus on the electron configuration determined in (a). Considering only the electrons from the outer subshell, add their electronic spins and their orbital angular momenta separately. Label the total spin and the total orbital momentum quantum numbers by s and l, respectively. Write down the possible values spanned by s and l.

(c) Determine the possible values spanned by the total angular momentum, labeled by j.

(d) Determine the ground state: you must minimize the total angular momentum J=L+S, if needed (with corresponding the quantum number j). Denote that state using the corresponding spectroscopic notation ^2s+1A_J , where A is a capital letter for the orbital corresponding to the value of l (i.e., A can take the letters S, P, D, F , etc.,). Remember to also consider the Pauli exclusion principle. Note: The ascending atomic orbital energy ordering relevant to this problem is 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d

Answer 1

Mg: 18€ 25+ 2p% 35? AL: 18² 23² 206 38² 3p' Electronac spin = + 1/2 - 1/2 = 0 199 : th orbital and ar moment bon, 33 Leo +1 0 1 - AL A TT Electronic spin = % Orbital angular momentum 33 3P post ( Grocend «tate) 0 © mgjs Lts=0+ 0 0 Alij -STELEE to Other possible ralues 14-31 to 14+5) are. % to 3% nie. Y214,379 S I @ The ground state is 25+1 ang: So AP: 2P CS Scanned with CamScanner