Bohr-Bury Rules & Electron Distribution - Ninth Grade Science
Having established that electrons inhabit discrete, quantized energy levels (shells) surrounding the atomic nucleus, a fundamental structural question arises: how are electrons organized within these shells? Can a hundred electrons crowd into the innermost K shell? Does an atom fill its shells randomly, or does nature enforce a strict architectural blueprint? In 1921, Niels Bohr and English chemist Charles Rugeley Bury formulated the Bohr-Bury Rules of electron distribution. These rules govern the electronic configuration of elements—the specific spatial arrangement of electrons across successive quantum shells. Electronic configuration is the Rosetta Stone of chemistry; it explains the structure of the periodic table, dictates chemical reactivity, and explains why sodium is an explosive metal while neon is an inert gas. In this chapter, you will master the Bohr-Bury rules, write electronic configurations for the first twenty elements, and construct two-dimensional atomic shell diagrams.
The Bohr-Bury Rules for Electron Distribution
The arrangement of electrons across the successive energy shells of an atom is governed by three fundamental rules formulated by Niels Bohr and Charles Bury:
+--------------------------------------------------------------------------+
| The Three Bohr-Bury Rules |
+--------------------------------------------------------------------------+
| Rule 1: The maximum number of electrons that can be accommodated in any |
| shell is given by the formula: 2 * n^2 (where n is shell number).|
| Rule 2: The outermost shell cannot hold more than 8 electrons, even if |
| its theoretical 2n^2 capacity is larger (Octet Rule). |
| Rule 3: Electrons do not enter a new outer shell until the inner shells |
| are completely filled (Step-wise shell filling). |
+--------------------------------------------------------------------------+
Maximum Capacity of Shells via 2n^2
K Shell (n = 1): 2 * (1)^2 = 2 * 1 = 2 electrons maximum
L Shell (n = 2): 2 * (2)^2 = 2 * 4 = 8 electrons maximum
M Shell (n = 3): 2 * (3)^2 = 2 * 9 = 18 electrons maximum
N Shell (n = 4): 2 * (4)^2 = 2 * 16 = 32 electrons maximum
Rule 1: Maximum Shell Capacity ($2n^2$)
The maximum number of electrons that any given energy level can physically accommodate is given by the mathematical formula:
\text{Maximum Capacity} = 2n^2
Where $n$ represents the Principal Quantum Number (the shell number counted outward from the nucleus):
- For the K Shell ($n = 1$): Maximum capacity is $2(1)^2 = 2 imes 1 = 2 ext{ electrons}$.
- For the L Shell ($n = 2$): Maximum capacity is $2(2)^2 = 2 imes 4 = 8 ext{ electrons}$.
- For the M Shell ($n = 3$): Maximum capacity is $2(3)^2 = 2 imes 9 = 18 ext{ electrons}$.
- For the N Shell ($n = 4$): Maximum capacity is $2(4)^2 = 2 imes 16 = 32 ext{ electrons}$.
Rule 2: The Outermost Shell Limit (The Octet Rule)
The maximum number of electrons that can be accommodated in the outermost shell (the valence shell) of any atom is strictly eight (8), with the sole exception of the K shell, which can hold a maximum of two (2) electrons (a duplet). Even though the third (M) shell has a theoretical capacity of 18 electrons, an atom cannot place nine electrons into its M shell if that M shell is the outermost layer of the atom. Once the outer shell acquires eight electrons, it forms a stable octet, and subsequent electrons must begin filling the next outer shell.
Rule 3: Step-Wise Progressive Shell Filling
Electrons occupy shells in a step-wise, progressive manner: electrons do not enter a higher-energy outer shell until the lower-energy inner shells beneath it have been filled to capacity. The K shell must take its full complement of two electrons before the L shell can receive electrons; the L shell must take its full complement of eight electrons before the M shell can receive electrons.
The Potassium and Calcium Case Study
A critical test of the Bohr-Bury rules arises when examining elements with atomic numbers 18, 19, and 20:
Argon (Z = 18):
Electrons = 18.
Configuration: K=2, L=8, M=8.
The M shell has 8 electrons; it is the outermost shell!
Potassium (Z = 19):
Does the 19th electron enter the M shell to make it K=2, L=8, M=9?
NO! Rule 2 forbids more than 8 electrons in the outermost shell!
===> The 19th electron ENTERS THE N SHELL!
Configuration of Potassium: 2, 8, 8, 1
Calcium (Z = 20):
Configuration of Calcium: 2, 8, 8, 2
For argon ($Z = 18$), the eighteen electrons distribute as $2, 8, 8$. The outer M shell holds eight electrons, forming a stable octet. For potassium ($Z = 19$), the nineteenth electron does not enter the M shell (which would create a configuration of $2, 8, 9$, violating Rule 2). Instead, the nineteenth electron enters the fourth (N) shell, giving potassium the configuration $2, 8, 8, 1$. Similarly, calcium ($Z = 20$) has the configuration $2, 8, 8, 2$. Only after the N shell has received two electrons does subsequent electron filling resume in the interior M shell for transition metals ($Z = 21$ to $30$).
Electronic Configurations of the First Twenty Elements
+-------------------+--------+--------+-----------------------+--------------+
| Element Name | Symbol | Atomic | Electronic Config. | Valence |
| | | Number | K L M N | Electrons |
+-------------------+--------+--------+-----------------------+--------------+
| Hydrogen | H | 1 | 1 | 1 |
| Helium | He | 2 | 2 | 2 (Duplet) |
| Lithium | Li | 3 | 2 1 | 1 |
| Beryllium | Be | 4 | 2 2 | 2 |
| Boron | B | 5 | 2 3 | 3 |
| Carbon | C | 6 | 2 4 | 4 |
| Nitrogen | N | 7 | 2 5 | 5 |
| Oxygen | O | 8 | 2 6 | 6 |
| Fluorine | F | 9 | 2 7 | 7 |
| Neon | Ne | 10 | 2 8 | 8 (Octet) |
| Sodium | Na | 11 | 2 8 1 | 1 |
| Magnesium | Mg | 12 | 2 8 2 | 2 |
| Aluminum | Al | 13 | 2 8 3 | 3 |
| Silicon | Si | 14 | 2 8 4 | 4 |
| Phosphorus | P | 15 | 2 8 5 | 5 |
| Sulfur | S | 16 | 2 8 6 | 6 |
| Chlorine | Cl | 17 | 2 8 7 | 7 |
| Argon | Ar | 18 | 2 8 8 | 8 (Octet) |
| Potassium | K | 19 | 2 8 8 1 | 1 |
| Calcium | Ca | 20 | 2 8 8 2 | 2 |
+-------------------+--------+--------+-----------------------+--------------+
Constructing Two-Dimensional Bohr Shell Diagrams
To visualize an atom's electronic architecture, chemists draw two-dimensional Bohr shell diagrams:
Bohr Diagram for Sodium (Na, Z = 11, Config: 2, 8, 1)
( . ) <-- 1 Valence Electron in M Shell
/ \
/ ( . ) \
| / \ |
| | (11p+)| | | <-- Nucleus: 11 Protons, 12 Neutrons
| | (12n0)| | | <-- K Shell: 2 Electrons (innermost)
| \ / | <-- L Shell: 8 Electrons (middle)
\ ( . ) / <-- M Shell: 1 Electron (outermost)
\ /
( . )
Step-by-Step Drawing Procedure
- Nucleus: Draw a small central circle and write the number of protons and neutrons inside (e.g., for sodium: $11 ext{p}^+, 12 ext{n}^0$).
- K Shell: Draw the first concentric circle around the nucleus and plot up to two electrons as dots or small crosses.
- L Shell: Draw the second concentric circle and plot up to eight electrons, typically paired at the cardinal compass points (north, south, east, west).
- M Shell: Draw the third concentric circle and plot the remaining electrons.
Chapter Exercises
Exercise 1: State the three Bohr-Bury rules governing the distribution of electrons in an atom. Using the $2n^2$ formula, calculate the theoretical maximum electron capacity of the fourth ($n = 4$, N) shell and the fifth ($n = 5$, O) shell.
Exercise 2: Write the complete electronic configuration across the K, L, M, and N shells for each of the following elements: carbon ($Z = 6$), phosphorus ($Z = 15$), chlorine ($Z = 17$), and potassium ($Z = 19$). Identify the number of valence electrons in each element.
Exercise 3: Explain why the electronic configuration of potassium ($Z = 19$) is written as $2, 8, 8, 1$ rather than $2, 8, 9$, even though the third (M) shell has a theoretical maximum capacity of 18 electrons according to the $2n^2$ rule.
Solutions and Explanations
Solution 1: The three Bohr-Bury rules are: first, the maximum number of electrons in any shell is given by $2n^2$, where $n$ is the shell number; second, the outermost valence shell of an atom cannot accommodate more than eight electrons (the octet rule); and third, inner shells must be filled to capacity before electrons can enter a new outer shell. For the fourth shell ($n = 4$, N shell), maximum capacity is $2 imes (4)^2 = 2 imes 16 = 32$ electrons. For the fifth shell ($n = 5$, O shell), maximum theoretical capacity is $2 imes (5)^2 = 2 imes 25 = 50$ electrons.
Solution 2: The electronic configurations and valence electron counts are: Carbon ($Z = 6$): configuration is $2, 4$ (K=2, L=4); valence electrons $= 4$. Phosphorus ($Z = 15$): configuration is $2, 8, 5$ (K=2, L=8, M=5); valence electrons $= 5$. Chlorine ($Z = 17$): configuration is $2, 8, 7$ (K=2, L=8, M=7); valence electrons $= 7$. Potassium ($Z = 19$): configuration is $2, 8, 8, 1$ (K=2, L=8, M=8, N=1); valence electrons $= 1$.
Solution 3: Although the M shell ($n = 3$) has a theoretical maximum capacity of 18 electrons under the $2n^2$ formula ($2 imes 3^2 = 18$), the Bohr-Bury rules contain a second constraint: the outermost shell of an atom cannot accommodate more than eight electrons. If the nineteenth electron of potassium were placed into the M shell, the resulting configuration would be $2, 8, 9$. In this arrangement, the M shell would be the outermost shell of the atom, holding nine electrons, violating the octet rule. To maintain electronic stability, once the M shell acquires eight electrons (as in argon, $2, 8, 8$), it forms a temporarily stable configuration; the next electron must enter the fourth (N) shell, giving potassium the configuration $2, 8, 8, 1$.