Q.14 Which one of the following ions has the maximum number of unpaired electrons?
(A) Cu2+ (B) Na+ (C) Cr3+ (D) Fe3+
Fe³⁺ has the maximum number of unpaired electrons among the given options.
This multiple-choice question tests understanding of transition metal ion electron configurations and Hund’s rule.
Introduction
Determining the maximum number of unpaired electrons in ions like Cu²⁺, Na⁺, Cr³⁺, and Fe³⁺ requires analyzing their electron configurations after ion formation. Unpaired electrons in d-orbitals follow Hund’s rule, where electrons occupy separate orbitals with parallel spins before pairing. This concept appears frequently in CSIR NET Life Sciences and chemistry exams.
Option Analysis
Each option’s atomic number, neutral configuration, ion formation, and unpaired electrons are detailed below. Ions lose 4s electrons first, then d-electrons.
| Ion | Atomic No. | Neutral Config | Ion Config | 3d Electrons | Unpaired Electrons | Orbital Diagram |
|---|---|---|---|---|---|---|
| (A) Cu²⁺ | 29 | [Ar] 4s¹ 3d¹⁰ | [Ar] 3d⁹ | 9 | 1 | ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑ (one unpaired) |
| (B) Na⁺ | 11 | [Ne] 3s¹ | [Ne] | 0 | 0 | All orbitals paired (noble gas config) |
| (C) Cr³⁺ | 24 | [Ar] 4s¹ 3d⁵ | [Ar] 3d³ | 3 | 3 | ↑ ↑ ↑ (three unpaired) |
| (D) Fe³⁺ | 26 | [Ar] 4s² 3d⁶ | [Ar] 3d⁵ | 5 | 5 | ↑ ↑ ↑ ↑ ↑ (five unpaired) |
Fe³⁺ shows the maximum number of unpaired electrons at 5, as its high-spin 3d⁵ configuration fills all five d-orbitals singly.
Why Fe³⁺ Wins
Na⁺ has zero unpaired electrons due to its noble gas-like [Ne] configuration with fully paired orbitals. Cu²⁺ (3d⁹) has one unpaired electron after pairing nine electrons across five orbitals. Cr³⁺ (3d³) has three unpaired electrons, one per orbital. Fe³⁺ maximizes unpaired electrons at five, making it highly paramagnetic—key for magnetic moment calculations in exams (μ=n(n+2) BM, where n=5).
