Atomic Structure and Electronic Configuration Study Guide

Energy Levels in Hydrogen and One-Electron Systems

In a hydrogen atom, the energy of an electron is uniquely determined by its principal quantum number ($n$). Because of this singular dependency, orbitals belonging to the same principal quantum number possess the same energy, regardless of their azimuthal quantum number ($l$). The resulting energy order for orbitals in hydrogen and other one-electron systems (such as $He^+$ or $Li^{2+}$) is as follows:

1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f < \dots

The most stable state for a hydrogen atom occurs when the electron occupies the $1s$ orbital; this is designated as the ground state. In this state, the electron is held most strongly by the nucleus due to its proximity. Conversely, when an electron occupies the $2s$, $2p$, or any orbital with a higher principal quantum number, the atom is said to be in an excited state.

Energy Determination in Multi-Electron Atoms

For atoms containing multiple electrons, the energy of an electron is no longer solely dependent on the principal quantum number ($n$); it also depends on the azimuthal quantum number ($l$). This complexity arises from two primary physical interactions within the atom: the electrostatic attraction between the electrons and the positively charged nucleus, and the mutual repulsion between electrons situated in various subshells.

In these systems, the interaction between the nucleus and the electrons in the outermost orbits is mitigated by the presence of electrons in the innermost orbits. This phenomenon is known as shielding or screening, where inner electrons effectively block some of the nuclear charge. The net positive charge experienced by a valence electron is termed the effective nuclear charge ($Z_{eff}$).

As the atomic number ($Z$) increases, the energy of the valence electrons generally decreases. Furthermore, the energies of orbitals within the same subshell also decrease as the atomic number ($Z$) increases. For example, the energy of the $2s$ orbital across different elements follows the order:

E_{2s}(H) > E_{2s}(Li) > E_{2s}(Na) > E_{2s}(K)

The Aufbau Principle and the $(n+l)$ Rule

The filling of electrons into the various orbitals of multi-electron atoms is governed by three fundamental principles: the Aufbau principle, Pauli's exclusion principle, and Hund's rule of maximum multiplicity. The term "Aufbau" is a German word meaning "building up."

According to the Aufbau principle, orbitals are filled in the ground state of an atom in order of their increasing energies. This means electrons first occupy the lowest energy orbital available to them. Higher energy orbitals are only entered once the lower energy orbitals are completely filled. The relative energies of these orbitals are determined by the $(n+l)$ rule:

1. As the value of $(n+l)$ increases, the energy of the orbital increases.

2. If two orbitals have the same $(n+l)$ value, the orbital with the lower principal quantum number ($n$) has the lower energy and is filled first.

Calculated values for specific orbitals include:

• For $1s$: $n=1$, $l=0$, $(n+0) = 1$

• For $2s$: $n=2$, $l=0$, $(n+0) = 2$

• For $2p$: $n=2$, $l=1$, $(n+1) = 3$

• For $3s$: $n=3$, $l=0$, $(n+0) = 3$

Based on these rules and the assistance of Moeller's diagram, the order of increasing energy for atomic orbitals is defined as:

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s

Pauli's Exclusion Principle and Hund's Rule

Pauli's Exclusion Principle states that "no two electrons in an atom can have the same set of four quantum numbers." This implies that if two electrons share the same $n$, $l$, and $m_l$ values (meaning they occupy the same orbital), they must have different spin quantum number ($m_s$) values. Consequently, a single orbital can accommodate a maximum of two electrons, and these electrons must have opposite spins.

Hund's Rule of Maximum Multiplicity dictates the behavior of electron pairing within degenerate orbitals (orbitals belonging to the same subshell, such as $p$, $d$, or $f$). Pairing of electrons does not occur in these subshells until every orbital in that subshell is occupied by one electron. Since $p$, $d$, and $f$ subshells contain 3, 5, and 7 orbitals respectively, pairing begins with the arrival of the 4th, 6th, and 8th electron in those subshells.

Electron configurations are typically written using the notation $nl^x$, where:

• $n$ represents the shell number.

• $l$ represents the subshell designation ($s, p, d, f$).

• $x$ represents the number of electrons present in that subshell.

Stability of Half-Filled and Completely Filled Subshells

Subshells that are either half-filled or completely filled exhibit enhanced stability. For instance, configurations such as $p^3$ or $p^6$, d^5$ or d^{10}$, and$f^7$or$f^{14} are more stable than other configurations. There are two primary reasons for this increased stability:\n\n1. Symmetrical Distribution of Electrons: Symmetry is inherently linked to stability. Half-filled and completely filled subshells possess a symmetrical distribution of electron density around the nucleus, which leads to lower energy states.\n\n2. Exchange Energy: When multiple electrons with parallel spins are present in degenerate orbitals, they can exchange their positions. Each such exchange releases a specific amount of energy called exchange energy. The total stability of the subshell increases with the number of possible exchanges. The maximum number of exchanges occurs when the subshell is exactly half-filled or completely filled.\n\nThe number of possible exchanges can be calculated using the formula:\n\n\text{Number of exchanges} = \frac{N(N-1)}{2}\n\nIn this formula, N$is the number of electrons having parallel spin. For example, in Nitrogen ($N: 1s^2 2s^2 2p^3$), the three electrons in the$2p$subshell have parallel spins, allowing for maximal exchange stability relative to other non-half-filled configurations like Oxygen ($O: 1s^2 2s^2 2p^4).\n\n# Anomalous Electronic Configurations\n\nIn certain elements, the actual electronic configuration deviates from the predictions of the Aufbau principle. This occurs most frequently when two subshells (typically an ns$and an$(n-1)d subshell) are very close in energy. An electron may shift from the lower energy subshell to the higher energy subshell if that shift results in a half-filled or completely filled state, thereby gaining stability from exchange energy and symmetry.\n\nSignificant examples include:\n- Chromium (Cr, Z=24$): The expected configuration is$[Ar] 3d^4 4s^2$, but the actual configuration is$[Ar] 3d^5 4s^1.\n- Copper (Cu, Z=29$): The expected configuration is$[Ar] 3d^9 4s^2$, but the actual configuration is$[Ar] 3d^{10} 4s^1.\n\nOther elements exhibiting similar anomalous configurations include:\n- Molybdenum (Mo, Z=42$):$5s^1 4d^5\n- Palladium (Pd, Z=46$):$5s^0 4d^{10}\n- Silver (Ag, Z=47$):$5s^1 4d^{10}\n- Gadolinium (Gd, Z=64$):$6s^2 5d^1 4f^7\n- Gold (Au, Z=79$):$6s^1 5d^{10} 4f^{14}\n\n# Questions & Discussion\n\n**Question 1: A neutral atom of an element has 2 K, 8 L, 11 M and 2 N electrons. The number of p-electrons in the atom are?**\nAnswer: To find the total number of p-electrons, we look at the subshells within each shell. K shell (n=1$) has$1s^2$. L shell ($n=2$) has$2s^2 2p^6$(6 p-electrons). M shell ($n=3$) has$3s^2 3p^6 3d^3$(6 p-electrons). The N shell ($n=4$) has$4s^2$. Total p-electrons =$6 (from 2p) + 6 (from 3p) = 12. The correct option is 2.\n\n**Question 2: An atom has 2 electrons in K-shell, 8 electrons in L-shell & 8 electrons in M-shell. The number of p-electrons present in the element is?**\nAnswer: K shell (1s^2$) has 0 p-electrons. L shell ($2s^2 2p^6$) has 6 p-electrons. M shell ($3s^2 3p^6$) has 6 p-electrons. Total p-electrons =$6 + 6 = 12. The correct option is 3.\n\n**Question 3: The maximum number of such electrons in an atom with quantum number n = 3, l = 2 is?**\nAnswer: The quantum numbers n=3$and$l=2$describe the$3d$subshell. A$d$subshell has 5 orbitals, and each orbital can hold 2 electrons. Therefore, the maximum number of electrons is$5 \times 2 = 10. The correct option is 3.\n\n**Question 4: The number of orbitals in n = 3 are?**\nAnswer: For a given principal quantum number n$, the total number of orbitals is given by$n^2$. For$n=3$, the number of orbitals is$3^2 = 9$(one$3s$, three$3p$, and five$3d orbitals). The correct option is 3.\n\n**Question 5: In potassium, the probable order of energy level for the 19th electron is?**\nAnswer: Potassium (Z=19$) has the configuration$1s^2 2s^2 2p^6 3s^2 3p^6 4s^1$. According to the$(n+l)$rule, the$4s$orbital ($4+0=4$) is lower in energy than the$3d$orbital ($3+2=5$). Therefore, the 19th electron enters the$4s$orbital because$4s < 3d$$. The correct option is 2.