Atomic Structure, Quantum Numbers, and Periodic Trends
Before you can master chemical bonding—whether it’s the lattice energy of ionic compounds or the hybridization of covalent bonds—you must understand the "DNA" of the atom: its electron configuration. Why does copper conduct electricity? Why is the chloride ion larger than the chlorine atom? The answers do not lie in memorization, but in the quantum mechanical rules that dictate electron behavior.
This guide bridges the gap between basic atomic theory and the rigorous understanding required for advanced exams. We will dismantle the atom into its quantum numbers, build electron configurations from the ground up (including the tricky exceptions), and use the concept of Effective Nuclear Charge ($Z_{eff}$) to predict periodic trends.
The Quantum Mechanical Model and Atomic Orbitals
The Bohr model, with its planetary orbits, is a useful stepping stone, but it fails to explain the complex behaviors of multi-electron atoms. The modern Quantum Mechanical Model replaces precise "orbits" with orbitals—three-dimensional regions of space where there is a high probability (over 90%) of finding an electron.
A mathematical function that describes the wave-like behavior of an electron. Each orbital is defined by a specific set of three quantum numbers ($n, l, m_l$) and can hold a maximum of two electrons.
These orbitals are not just empty boxes; they have specific shapes and energies determined by quantum numbers. Understanding these numbers is the first step to writing a valid electron configuration.
Understanding Quantum Numbers ($n, l, m_l, m_s$)
Think of quantum numbers as the "address" of an electron. Just as a postal address requires a state, city, street, and house number, an electron is described by four distinct values. According to the Pauli Exclusion Principle, no two electrons in the same atom can have the exact same set of all four quantum numbers.
1. Principal Quantum Number ($n$)
- Significance: Indicates the main energy level (shell) and the relative size of the orbital.
- Rule: Must be a positive integer ($n = 1, 2, 3, \dots$).
- Trend: As $n$ increases, the orbital becomes larger and the electron has higher energy.
2. Azimuthal Quantum Number ($l$)
- Significance: Determines the shape of the orbital (subshell).
- Rule: Can be any integer from $0$ to $n-1$.
- Codes:
- $l=0 \rightarrow s$ orbital (spherical)
- $l=1 \rightarrow p$ orbital (dumbbell)
- $l=2 \rightarrow d$ orbital (cloverleaf)
- $l=3 \rightarrow f$ orbital (complex)
3. Magnetic Quantum Number ($m_l$)
- Significance: Determines the orientation of the orbital in space.
- Rule: Can be any integer from $-l$ to $+l$.
4. Spin Quantum Number ($m_s$)
- Significance: Describes the electron's magnetic spin.
- Rule: Only two possible values: $+\frac{1}{2}$ (spin up) or $-\frac{1}{2}$ (spin down).
Illustrative Example: Calculating Subshell Capacity
Let's determine the properties of the subshell defined by $n=4$ and $l=2$.
- Identify the subshell: Since $n=4$ and $l=2$ corresponds to a $d$ shape, this is the 4d subshell.
- Count the orbitals: The number of orbitals is determined by the possible values of $m_l$. For $l=2$, $m_l$ can be $-2, -1, 0, +1, +2$. This gives us 5 distinct orbitals. Alternatively, use the formula $2l + 1$.
- Calculate max electrons: Since each orbital holds 2 electrons, the total capacity is $5 \times 2 = 10$ electrons.
Practice: Identifying Invalid Quantum Numbers
One of the most common exam questions asks to identify which sets of quantum numbers cannot occur together. You must strictly check the rules above.
- Set A ($n=3, l=3, m_l=0$): INVALID. $l$ must be at least one less than $n$. Here $l=n$, which is forbidden.
- Set B ($n=5, l=-4, m_l=2$): INVALID. $l$ cannot be negative.
- Set C ($n=2, l=1, m_l=-1$): VALID. $n$ is an integer, $0 \le l < n$, and $-l \le m_l \le +l$.
Building Electron Configurations
To map out where electrons live in a neutral atom, we follow the Aufbau Principle ("building up"), which states that electrons occupy the lowest energy orbitals first.
The (n + l) Rule and Filling Order
Orbitals are not always filled in numerical order (e.g., 3d fills after 4s). The energy is determined by the sum $(n + l)$.
- Low $(n+l)$ fills first.
- If $(n+l)$ is tied, the lower $n$ fills first.
Case Study: What fills around the 4s subshell?
Let's analyze the energies of 3p, 4s, and 3d:
- 3p: $n=3, l=1 \rightarrow \text{Sum} = 4$
- 4s: $n=4, l=0 \rightarrow \text{Sum} = 4$ (Tied with 3p, but higher $n$, so it fills after 3p).
- 3d: $n=3, l=2 \rightarrow \text{Sum} = 5$
Result: The subshell filled immediately before 4s is the 3p. The subshell filled immediately after 4s is the 3d.
Exceptions to the Aufbau Principle
Certain electron configurations are energetically favorable despite violating the standard filling order. The most notable examples occur in transition metals like Chromium and Copper, where half-filled ($d^5$) or fully-filled ($d^{10}$) subshells offer extra stability.
Example: Copper ($Z=29$)
Predicted configuration: $[Ar] 4s^2 3d^9$
Actual configuration: $[Ar] 4s^1 3d^{10}$
By promoting one electron from the 4s to the 3d, Copper achieves a fully filled d-subshell, which is lower in energy than the predicted state.
Writing Configurations for Ions (The Transition Metal Trap)
When forming cations from transition metals, electrons are removed from the orbital with the highest principal quantum number ($n$) first, not necessarily the last one filled.
Detailed Example: How many d-electrons does $Ni^{2+}$ have?
Many students incorrectly simply remove the last electrons added. Let's do this correctly.
- Write Neutral Nickel ($Z=28$): $[Ar] 4s^2 3d^8$.
- Ionize: We need to remove 2 electrons. Even though 3d was filled after 4s, the 4s electrons are in the outermost shell (n=4). Once the orbitals are occupied, the 4s electrons are higher in energy and easier to remove.
- Remove from 4s: $[Ar] 4s^0 3d^8$.
- Final Count: $Ni^{2+}$ has 8 d-electrons.
Core Periodic Trends Explained by Effective Nuclear Charge
Why do atoms get smaller as you add more protons and electrons across a row? The answer lies in Effective Nuclear Charge ($Z_{eff}$).
What is Effective Nuclear Charge?
$Z_{eff}$ is the net positive pull experienced by an electron. It is calculated roughly as:
$$Z_{eff} = Z - S$$Where $Z$ is the number of protons (nuclear charge) and $S$ is the shielding constant (repulsion from inner electrons).
Trend 1: Atomic Radius
Across a Period (Left to Right): Radius Decreases.
As you move across a period, you add protons ($Z$ increases) and electrons to the same energy shell. These outer electrons do not shield each other effectively. Therefore, $Z_{eff}$ increases significantly, pulling the electron cloud tighter toward the nucleus.
Down a Group (Top to Bottom): Radius Increases.
Although $Z$ increases, you are adding entire new electron shells (increasing $n$). The distance effect dominates the increased charge, making the atom larger.
Practice Problem: List elements in order of increasing atomic radius
Elements: Cl, Ni, Ga, Y, Cs
- Cs (Period 6, Group 1): Largest. Highest shell ($n=6$), lowest $Z_{eff}$ for its period.
- Y (Period 5): Second largest. Shell $n=5$.
- Ni vs Ga (Period 4): Both are $n=4$. Nickel is Group 10, Gallium is Group 13. Since radius decreases to the right, Ga is smaller than Ni? Wait, check the trend carefully.
Actually, Gallium ($Z=31$) is to the right of Nickel ($Z=28$). Generally, size decreases rightward. However, the d-block contraction makes this area tricky. Standard trend logic suggests $Ga < Ni$. (Note: In strictly trend-based questions, Ga is to the right of Ni, so it has higher $Z_{eff}$ and is smaller). - Cl (Period 3): Smallest. Lowest shell ($n=3$).
Correct Order (Smallest to Largest): $Cl < Ga < Ni < Y < Cs$.
Summary: The "Why" Behind the Trends
If asked to explain why atomic size decreases across a period, avoid saying "because of interactions between electrons." The precise chemical reason is: "Electrons are pulled closer to the nucleus by the increased number of protons (higher $Z_{eff}$) without a corresponding increase in shielding."
Key Takeaways
- Orbitals: Defined by $n, l, m_l$. Only 2 electrons per orbital (spin up/down).
- Aufbau & Exceptions: Fill by lowest energy $(n+l)$. Watch out for Cu ($s^1d^{10}$) and Cr ($s^1d^5$).
- Ionization: For transition metals, remove $s$ electrons before $d$ electrons.
- Periodic Trends: Driven by $Z_{eff}$. Higher $Z_{eff}$ means smaller radius and higher ionization energy.