Write the electron configuration for the following particles S V Cl Al3+ Fe3+
The Correct Answer and Explanation is:
Electron Configurations:
- Sulfur (S):
Atomic number = 16
Electron configuration:
→ 1s² 2s² 2p⁶ 3s² 3p⁴ - Vanadium (V):
Atomic number = 23
Electron configuration:
→ 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d³ - Chlorine (Cl):
Atomic number = 17
Electron configuration:
→ 1s² 2s² 2p⁶ 3s² 3p⁵ - Aluminum ion (Al³⁺):
Atomic number = 13
Neutral Al: 1s² 2s² 2p⁶ 3s² 3p¹
Remove 3 electrons for Al³⁺ → Remove from 3p and 3s
Electron configuration:
→ 1s² 2s² 2p⁶ - Iron ion (Fe³⁺):
Atomic number = 26
Neutral Fe: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶
Remove 3 electrons → 2 from 4s, 1 from 3d
Electron configuration:
→ 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵
Explanation
Electron configuration represents the distribution of electrons among atomic orbitals. It follows specific rules: Aufbau principle (fill lower energy orbitals first), Pauli exclusion principle (no two electrons in an atom can have the same set of four quantum numbers), and Hund’s rule (maximize unpaired electrons in orbitals of the same energy).
For sulfur (S) with 16 electrons, electrons fill the 1s to 3p orbitals in order. The 3p sublevel holds 4 electrons, which aligns with its position in group 16 of the periodic table.
Vanadium (V) has 23 electrons. After the 4s orbital is filled, electrons begin filling the 3d orbitals. This explains why vanadium’s configuration includes 3d³, consistent with its position among transition metals.
Chlorine (Cl), with 17 electrons, fills up to the 3p sublevel. The 3p orbital holds 5 electrons, just one short of a full p-orbital (which holds 6), explaining chlorine’s high reactivity—it tends to gain one electron to achieve a full outer shell.
For Al³⁺, we remove three electrons from aluminum’s neutral state. Since electrons are lost from higher energy levels first, the two 3s electrons and the single 3p electron are removed, leaving a stable noble gas configuration (same as neon).
Fe³⁺ is more complex because it is a transition metal. Although 4s fills before 3d, electrons are lost from 4s first upon ionization. So, Fe³⁺ has lost 2 electrons from 4s and one from 3d, resulting in a 3d⁵ configuration. This half-filled 3d sublevel is relatively stable and common among transition metal ions.
Understanding electron configurations is essential for predicting chemical reactivity, bonding behavior, and properties of elements and ions.
