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

##### 1 Answer

I got

Well, the **ground-state electron configuration** is:

#[Ne] 3s^2 3p^1#

which looks like this:

#ul(uarr color(white)(darr)) " " ul(color(white)(uarr darr)) " " ul(color(white)(uarr darr))#

#ul(uarr darr)#

If you recall, an **atomic term symbol** is written like so:

#\mathbf(""^(2S + 1) L_J)# where:

#S# is thetotal spin angular momentum(#|sum m_s|# ) , and#2S+1# is the spin multiplicity.#L# is thetotal orbital angular momentum, which is the maximum sum of#m_l# values (#|sum m_l|# ) of occupied orbitals, made into a positive quantity.#J = {|L - S|, L - S + 1, . . . , 0, . . . , L + S - 1, |L + S|}# is thetotal angular momentum, of which there can be more than one possible value if#S > 0# . The ground term though, has only one of these#J# values (differing inintegerincrements).

Starting by knowing the *ground-state electron configuration*, finding the term symbols are a bit easier. Then we'll have to narrow the ground term down with Hund's Rule.

**TERM SYMBOL**

#ul(uarr darr)#

- Since the
#3s# electrons are paired,#S = |sum m_s| = |+1/2 + (-1/2)| = color(green)(0)# , and therefore,#2S + 1 = color(green)(1)# . #S = 0# , so there is only one value of#J# .- Since the maximum
#L# possible would be if the electrons are in#m_l = 0# (there is only one#3s# orbital),#L_max = 0# . That corresponds to the letter#color(green)(S)# in the term symbol, just as#l = 0 -> s# . - Therefore,
#J = L + S = color(green)(0)#

That means the **(ground) term symbol** is *any* configuration that has *fully-occupied subshells* (

**TERM SYMBOLS**

For this, we don't have to consider all *all* possible term symbols, and you only want the ground term symbol.

However, since the ground term requires that we understand how *Hund's Rule* works, we'll look at the **two primary possibilities**.

#ul(uarr color(white)(darr)) " " ul(color(white)(uarr darr)) " " ul(color(white)(uarr darr))#

- Since there is only one electron,
#S = |sum m_s| = color(green)(+1/2)# . - That means
#2S + 1 = color(green)(2)# . - The maximum
#L# is when the electron is in the#m_l = pm1# orbital, so#L = |pm1| = 1# , and the letter is#color(green)(P)# , just as#l = 1 -> p# . - With
#S > 0# , we have*more than one*#J# value (differing in integer increments). #J = {L - S, . . . , L + S}# cannot be#0# , and we should get:

#J = L pm S = 1 pm "1/2" = color(green)("1/2, 3/2")# . The value of#J# due to#S# cannot use a greater#S# than#1/2# , and#J# must be positive, so we havetwo valuesof#J# .

So, our

#color(darkblue)(""^2 P_"1/2")# ,#color(darkblue)(""^2 P_"3/2")#

**GROUND TERM SYMBOL**

To find the ground term, we'll need to look at Hund's Rule. **Hund's Rule** (for Physical Chemists) states:

- The ground term with the
**largest**#\mathbf(S)# is most stable,*unless*all have the*same*#S# . - If all terms have the
*same*#S# , the**largest**#\mathbf(L)# is the most stable,*unless*they all have the*same*#L# as well. - For terms with the
*same*#S# AND#L# , the**ground term**for a configuration where the subshell is*less than half-filled*has the*lowest*#J# , or vice versa. - If the subshell is EXACTLY half-filled, then
#L = 0# , and thus,#J = S_"max"# .

Since **less than half-filled** (half-filled is

#=> \mathbf(color(blue)(""^2 P_"1/2"))#