Unit 1: Electronic Configuration
Understanding how electrons are arranged within atoms and how this relates to the Periodic Table.
1.4 Energy Levels, Orbitals & Shells
Electrons within an atom do not orbit the nucleus randomly. Instead, they occupy specific regions of space called orbitals, each associated with a distinct energy. These orbitals are grouped into sub-levels (or subshells), which in turn are organized into larger energy levels (or principal shells). This quantized nature of electron energy is supported by experimental evidence, most notably the observation of line spectra. When atoms absorb energy (e.g., from heat or electricity), electrons jump to higher energy levels. When they fall back to lower energy levels, they emit light of specific wavelengths, creating a unique "fingerprint" line spectrum for each element. This demonstrates that electrons can only exist at discrete energy states.
Principal energy levels are designated by integers ($n = 1, 2, 3, \dots$) or by letters (K, L, M, ...). The lowest energy level is $n=1$ (K shell), closest to the nucleus. As $n$ increases, the energy levels become higher and are further from the nucleus.
Within each principal energy level, there are one or more sub-levels (subshells), denoted by letters: s, p, d, f. Each sub-level contains a specific number of orbitals:
- s-orbitals: Spherical in shape. Each principal energy level ($n \ge 1$) has one s-orbital.
- p-orbitals: Dumbbell-shaped. Each principal energy level ($n \ge 2$) has three p-orbitals, oriented along the x, y, and z axes ($p_x, p_y, p_z$).
- d-orbitals: More complex shapes. Each principal energy level ($n \ge 3$) has five d-orbitals.
- f-orbitals: Even more complex shapes. Each principal energy level ($n \ge 4$) has seven f-orbitals.
The energy of sub-levels within a given principal energy level generally increases in the order: $s < p < d < f$. However, due to interactions between electrons, some sub-levels from higher principal energy levels can have lower energy than d or f sub-levels from lower principal energy levels (e.g., 4s is filled before 3d).
Solved Examples:
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How many sub-levels are there in the third principal energy level ($n=3$), and what are they?
Solution: For $n=3$, there are three sub-levels: 3s, 3p, and 3d. -
Describe the shape of a 2p orbital and state how many 2p orbitals exist.
Solution: A 2p orbital is dumbbell-shaped. There are three 2p orbitals (2p$_x$, 2p$_y$, 2p$_z$), each oriented along a different axis. -
Which orbital is filled first: 3d or 4s? Explain why.
Solution: The 4s orbital is filled before the 3d orbital. This is due to the Aufbau principle, which states that electrons fill the lowest energy orbitals first. Despite being in a higher principal energy level, the 4s orbital has a slightly lower energy than the 3d orbitals. -
What information do atomic line spectra provide about atomic structure?
Solution: Line spectra provide experimental evidence that electrons in atoms can only exist at specific, discrete energy levels. The unique set of lines for each element corresponds to the specific energy differences between these allowed energy levels. -
How many orbitals are present in the second principal energy level ($n=2$)?
Solution: For $n=2$, there are an s-sublevel (one 2s orbital) and a p-sublevel (three 2p orbitals). So, there are a total of $1 + 3 = 4$ orbitals in the second principal energy level.
1.5 Electronic Configuration of Atoms and Ions
The electronic configuration describes the distribution of electrons among the orbitals of an atom or ion. This arrangement is governed by three fundamental principles:
- Pauli Exclusion Principle: No two electrons in an atom can have the same set of four quantum numbers. This means that an atomic orbital can hold a maximum of two electrons, and these two electrons must have opposite spins (one spin-up, one spin-down).
- Aufbau Principle: Electrons fill atomic orbitals of the lowest available energy levels before occupying higher energy levels. This "building up" principle dictates the order of orbital filling (e.g., 1s, 2s, 2p, 3s, 3p, 4s, 3d, etc.).
- Hund's Rule of Maximum Multiplicity: For degenerate orbitals (orbitals of the same energy, like the three p-orbitals), electrons will first occupy each orbital singly with parallel spins before any orbital is doubly occupied. This maximizes the total spin and minimizes electron-electron repulsion.
Electronic configurations can be written in two main ways:
- Orbital Notation (Spectroscopic Notation): This method uses the principal quantum number, the letter for the sub-level, and a superscript for the number of electrons in that sub-level (e.g., $1s^2 2s^2 2p^6$).
- Orbital Diagram (Arrow-in-Box Notation): This method uses boxes or lines to represent orbitals and arrows to represent electrons, with the direction of the arrow indicating spin.
For ions, electrons are added or removed. For cations (positive ions), electrons are removed from the highest principal energy level first. If multiple subshells exist in the highest level, electrons are removed from p-orbitals, then s-orbitals, then d-orbitals. For transition metals, 4s electrons are typically removed before 3d electrons when forming ions. For anions (negative ions), electrons are added to the lowest available energy orbital according to the Aufbau principle.
Solved Examples:
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Write the full electronic configuration for a neutral Nitrogen (N) atom using orbital notation.
Solution: Nitrogen has 7 electrons. Following the Aufbau principle and Hund's rule: $1s^2 2s^2 2p^3$. -
Draw the orbital diagram for Oxygen (O).
Solution: Oxygen has 8 electrons.1s 2s 2p [↑↓] [↑↓] [↑↓][↑ ][↑ ] -
Write the shorthand electronic configuration for Iron (Fe).
Solution: Iron has 26 electrons. The noble gas before Fe is Argon (Ar), which has 18 electrons ($[Ar] = 1s^2 2s^2 2p^6 3s^2 3p^6$). So, Fe: $[Ar] 3d^6 4s^2$. (Note: 4s is filled before 3d, but often written after 3d for convention). -
Determine the electronic configuration for the Fe²⁺ ion.
Solution: A neutral Fe atom is $[Ar] 3d^6 4s^2$. When forming a cation, electrons are removed from the highest principal energy level first. In this case, the 4s electrons are removed before the 3d electrons. Fe²⁺: $[Ar] 3d^6$. -
Write the full electronic configuration for a Bromide (Br⁻) ion.
Solution: A neutral Bromine (Br) atom has 35 electrons. Its configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^5$. When Br gains one electron to become Br⁻, the electron is added to the 4p subshell. Br⁻: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6$. This is isoelectronic with Krypton (Kr).
1.6 The Periodic Table & Blocks
The Periodic Table is a systematic arrangement of elements based on their increasing atomic number, which directly correlates with their electronic configuration. This arrangement highlights recurring patterns in chemical properties, known as periodicity.
Elements are organized into:
- Periods (Rows): Elements in the same period have the same number of principal energy shells occupied by electrons. For example, elements in Period 3 (Na, Mg, Al, Si, P, S, Cl, Ar) all have electrons in their first, second, and third principal shells.
- Groups (Columns): Elements in the same group generally have the same number of valence electrons (electrons in the outermost shell) and similar outer electronic configurations. This similarity in valence electrons is why elements within a group exhibit similar chemical properties.
The Periodic Table can also be divided into four main blocks based on the type of sub-level being filled with electrons:
- s-block: Groups 1 and 2 (and Helium). Their outermost electrons are in an s-orbital ($ns^1$ or $ns^2$). These are highly reactive metals.
- p-block: Groups 13 to 18. Their outermost electrons are in p-orbitals ($ns^2 np^{1-6}$). This block contains metals, metalloids, and nonmetals.
- d-block (Transition Metals): Groups 3 to 12. These elements are characterized by having their d-orbitals being filled. Their general electronic configuration involves $(n-1)d$ and $ns$ orbitals.
- f-block (Inner Transition Metals): Lanthanides and Actinides. Their f-orbitals are being filled. These are typically placed below the main body of the periodic table for convenience.
Solved Examples:
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Identify the period and group for Silicon (Si), and state its block.
Solution: Silicon has atomic number 14. Its electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^2$. The highest principal energy level is 3, so it is in Period 3. It has 4 valence electrons ($3s^2 3p^2$), placing it in Group 14. Since its outermost electrons are in a p-orbital, it belongs to the p-block. -
An element has the electronic configuration $[Ne] 3s^2 3p^5$. To which period, group, and block does it belong?
Solution: The highest principal energy level is 3, so it is in Period 3. It has $2+5=7$ valence electrons (in $3s^2 3p^5$), placing it in Group 17 (Halogens). Since the last electrons are in p-orbitals, it is in the p-block. This element is Chlorine (Cl). -
Why do elements in Group 1 (Alkali Metals) exhibit similar chemical properties?
Solution: Elements in Group 1 all have one valence electron in an s-orbital ($ns^1$). This similar outer electronic configuration means they tend to lose this single electron readily to form a +1 cation, which is the basis for their similar highly reactive metallic properties. -
To which block does Copper (Cu) belong, and why is its electronic configuration unusual?
Solution: Copper belongs to the d-block (transition metals). Its electronic configuration is unusual because it is $[Ar] 3d^{10} 4s^1$ instead of the expected $[Ar] 3d^9 4s^2$. This anomaly occurs because a completely filled d-subshell ($3d^{10}$) is more stable than a partially filled one ($3d^9$), so one electron from the 4s orbital is promoted to fill the 3d subshell. -
An element is in Period 4 and Group 2. What is its electronic configuration?
Solution: Being in Period 4 means its outermost electrons are in the 4th principal energy level. Being in Group 2 means it has 2 valence electrons in an s-orbital. Therefore, its electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2$ or shorthand $[Ar] 4s^2$. This element is Calcium (Ca).
Knowledge Check (20 Questions)
1s 2s 2p
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