# What is the term symbol for #"Cr"# in #["Cr"("CN")_6]^(4-)#?

##### 1 Answer

By recognizing that

What that means is that

Therefore, since it's a strong-field ligand and there are six *octahedral field* in which the **far apart in energy**, relative to

This means the electron configuration of **low-spin**.

For low-spin, we thus have this

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

As for the term symbol, I assume since this is a molecule, a molecular term symbol would be appropriate. But apparently, since you want the *atomic* term symbol, I'll give that instead...

#""^(2S+1) L_J#

#S# is the total spin angular momentum, and is calculated as#S = |M_S| = |sum_i m_(s,i)|# for electron#i# .#L# is the total orbital angular momentum, and is calculated as#L = |M_L| = |sum_i m_(l,i)|# for electron#i# .#L = 0,1,2,3,4,5, . . . # corresponds to#S,P,D,F,G,H, . . . # .#J = {|L+S|, |L+S-1|, . . . , |L-S+1|, |L-S|}# is the total angular momentum, and ranges from the magnitude of#L-S# to the magnitude of#L+S# , including integer values in between.

From the above configuration, we take the doubly-occupied orbital to have

#S = 1/2-1/2+1/2+1/2 = 1#

#2S+1 = 2(1) + 1 = 3# #=># triplet

#L = |sum_i m_(l,i)| = |-2-2-1+0| = 5# #=> H#

Thus, we do have

We don't need

#J = {5-1, 5+0, 5+1} = {4,5,6}#

That would give us, in a magnetic field, three levels, each *individually* triply-degenerate:

#""^(3) H_4, ""^(3) H_5, ""^(3) H_6#

From Hund's rules, we maximize

As an approximation, these orbitals have significant metal contributions (significant ionic character), so we **assume** they are *essentially equivalent* to metal *degeneracies*.

Therefore, for this determination, we ** approximate** them as:

#underbrace(ul(uarr darr)" "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr)))_(3d)#

As a result, it's **less than half**-filled, so we take the smallest

Had this been high-spin