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 the total spin angular momentum (|sum m_s| ) , and2S+1 is the spin multiplicity.L is the total orbital angular momentum, which is the maximum sum ofm_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 the total angular momentum, of which there can be more than one possible value ifS > 0 . The ground term though, has only one of theseJ values (differing in integer increments).
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.
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 ofJ .- Since the maximum
L possible would be if the electrons are inm_l = 0 (there is only one3s orbital),L_max = 0 . That corresponds to the lettercolor(green)(S) in the term symbol, just asl = 0 -> s . - Therefore,
J = L + S = color(green)(0)
That means the
For this, we don't have to consider all
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 them_l = pm1 orbital, soL = |pm1| = 1 , and the letter iscolor(green)(P) , just asl = 1 -> p . - With
S > 0 , we have more than oneJ value (differing in integer increments). J = {L - S, . . . , L + S} cannot be0 , and we should get:
J = L pm S = 1 pm "1/2" = color(green)("1/2, 3/2") . The value ofJ due toS cannot use a greaterS than1/2 , andJ must be positive, so we have two values ofJ .
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 sameS . - If all terms have the same
S , the largest\mathbf(L) is the most stable, unless they all have the sameL as well. - For terms with the same
S ANDL , the ground term for a configuration where the subshell is less than half-filled has the lowestJ , or vice versa. - If the subshell is EXACTLY half-filled, then
L = 0 , and thus,J = S_"max" .
Since
=> \mathbf(color(blue)(""^2 P_"1/2"))