Question

What is the ground-state term symbol for the aluminium atom in a magnetic field?

Answer 1

Hint: The term symbol in quantum mechanics is an abbreviated description of the angular momentum quantum numbers in a multi-electron system. Every energy level is not only described by its configuration but also its term symbol. The term symbol usually assumes LS coupling.

Complete Step By Step Answer:
The term symbol has a form of: ${}^{2S+1}{L}_J$
Where $2S+1$ is the spin multiplicity, L is the orbital quantum number having values S, P, D, F, G, etc. and J is the total angular momentum quantum number. The value of J ranges from $J_{max}-J_{min}$ (max to min) . The value of $J_{max}=|L+S|$ and $J_{min}=|L-S|$
The spin multiplicity or the total spin angular momentum can be given as: $S=|M_S|=|\sum _i{m}_{s,i}|$ for I no. of electrons. And total orbital angular momentum quantum number L can be given as: $L=|M_L|=|\sum _i{m}_{l,i}|$ for I no. of electrons. If the value of L =0,1,2,3,4, etc. it corresponds to L = S,P,D,F,G, etc, respectively.
We are given the atom Aluminum. The electronic configuration of Aluminum is given as: $Al:[Ne]3s^{2}3p^1$
Diagrammatically it is given as:

The 3s orbital has two paired electrons and 3p has one unpaired electron. Let us find the term symbols for each orbital one by one.
TERM SYMBOL FOR 3S ORBITAL:
Now, since we know the electronic configuration let us find the term symbols.
The total spin angular momentum can be given as: $S=|M_S|=|\sum _i{m}_{s,i}|$
For the given configuration of electrons the value of $S={\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{2}}=0$
The spin multiplicity will be equal to $S_m=2S+1=2(0)+1=1$ . Spin multiplicity = 1 indicates Singlet state.
The value of total orbital angular momentum quantum number L can be given as: $L=|M_L|=|\sum _i{m}_{l,i}|$
The doubly occupied 3s orbital will have a $m_l=0$ . The total angular momentum quantum number L will be: $L=|0|=0\to S$
The term symbol until now can be written as ${}^{1}S$
The value of J will be from $J_{max}=|L+S|$ to $J_{min}=|L-S|$ i.e. from $J_{min}=|0-0|=0$ to $J_{max}=|0-0|=0$ . Therefore, the value of J will be $J=0$ . The term symbol for 3s orbital will be ${}^1{S}_0$
TERM SYMBOL FOR 3p ORBITAL:
Now, since we know the electronic configuration let us find the term symbols.
The total spin angular momentum can be given as: $S=|M_S|=|\sum _i{m}_{s,i}|$
For the given configuration of electrons the value of $S={\displaystyle \frac{1}{2}}={\displaystyle \frac{1}{2}}$
The spin multiplicity will be equal to $S_m=2S+1=2\left({\displaystyle \frac{1}{2}}\right)+1=2$ . Spin multiplicity = 2 indicates Doublet state.
The value of total orbital angular momentum quantum number L can be given as: $L=|M_L|=|\sum _i{m}_{l,i}|$
The singly occupied orbital will have a $m_l=+1$ . The total angular momentum quantum number L will be: $L=|+1|=1\to P$
The term symbol until now can be written as ${}^{2}P$
The value of J will be from $J_{max}=|L+S|$ to $J_{min}=|L-S|$ i.e. from $J_{min}=|1-{\displaystyle \frac{1}{2}}|={\displaystyle \frac{1}{2}}$ to $J_{max}=|1+{\displaystyle \frac{1}{2}}|={\displaystyle \frac{3}{2}}$ . Therefore, the value of J will be $J={\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{2}}$
The term symbols for 3p orbitals will thus will have two values: ${}^2{P}_{{\displaystyle \frac{1}{2}}}{,}^2{P}_{{\displaystyle \frac{3}{2}}}$
We are asked to find the Ground state term symbol, according to Hund’s rule:
- The term with the largest S is more stable, unless all have the same value of S.
- For terms having the same value of S and L, the subshell that has less than half filled electrons will have the smallest J and vice versa. If it has exactly half-filled electrons J will be 0.
In the given configuration the values of S and L are same, and 3p is less than half filled orbital, therefore 1/2 is more stable than 3/2. The final ground state term symbol is ${}^2{P}_{1/2}$ . This is the required answer.

Note:
If we are asked the ground state term symbol, the value of J will be $J_{min}=|L-S|$ for less than half filled orbitals and $J_{max}=|L+S|$ for more than half filled orbitals. In this case the orbital is less than half filled, hence the value of J will be $J_{min}=|5-1|=4$ and the ground state term symbol will be ${}^3{H}_4$