# Question #9326b

Dec 30, 2017

It's not possible until you identify the element and assume that it is a NEUTRAL atom, by which point it doesn't matter what notation you use at all.

KLMN notation, or "X-ray notation", is an old way of denoting electron configurations, and is not very informative.

Here is how it corresponds to electron script configurations:

${\overbrace{1 s}}^{\text{K shell")color(white)(.)overbrace(2s2p)^("L shell")color(white)(.)overbrace(3s3p3d)^("M shell")color(white)(.)overbrace(4s4p4d4f)^("N shell}} \cdots$

and so on. Take the following unstated atom as an example:

${\overbrace{1 {s}^{2}}}^{K} {\overbrace{2 {s}^{2} 2 {p}^{6}}}^{L} {\overbrace{3 {s}^{2} 3 {p}^{6} 3 {d}^{0}}}^{M} {\overbrace{4 {s}^{2}}}^{N}$

One would write this in X-ray notation as:

$2 , 8 , 8 , 2$

See the problem?

...Which electrons are which? All that can be determined is how many electrons there are. One is then stuck with saying "there are 8 electrons in the M shell"... whatever that means.

To determine where the element is, we know it has $2 + 8 + 8 + 2 = 20$ electrons, which indicates that, assuming a NEUTRAL atom,

• it must be atomic number $20$ (but we can only know that by assuming a neutral atom!).
• it must be calcium as a consequence.
• it must be on column $2$, row $4$, by inspection.

Now consider the following:

${\overbrace{1 {s}^{2}}}^{K} {\overbrace{2 {s}^{2} 2 {p}^{6}}}^{L} {\overbrace{3 {s}^{2} 3 {p}^{6} 3 {d}^{10}}}^{M} {\overbrace{4 {s}^{2} 4 {p}^{6} 4 {d}^{5}}}^{N} {\overbrace{5 {s}^{2}}}^{O}$

One would write this, rather confusingly, in X-ray notation as:

$2 , 8 , 18 , 13 , 2$

and clearly, there are $2 + 8 + 18 + 13 + 2 = 43$ electrons. That's the best information we can get from this... we cannot know which electrons are in which orbital subshells and which orbitals.

If we only know the number of electrons, we must assume a NEUTRAL atom, and only after that we can say

• it must be atomic number $43$.
• it must be technetium as a consequence.
• it must be on column $7$, row $5$ by inspection.

In this case, the average student might wonder why technetium has more than $8$ electrons in one shell, because they know of a so-called "octet rule", but not when it does not apply.

In general, the "octet rule" only applies to $s$ and $p$ orbitals on the same energy level. But how would we know that? We don't know of angular momentum quantum numbers in X-ray notation, which allow the construction of the octet rule itself...