‘Write the full electron configuration and orbital diagram for each ion and indicate whether it is diamagnetic or paramagnetic (10 pts): Cdl+ b. Aut Mo3+ d. Zrt’
The Correct Answer and Explanation is:
Here’s the complete answer and explanation for the ions you’ve listed: Cd⁺, Au⁺, Mo³⁺, and Zr⁺.
1. Cd⁺ (Cadmium ion)
Atomic number of Cd: 48 Neutral Cd configuration: [Kr] 4d¹⁰ 5s² Cd⁺ configuration: [Kr] 4d¹⁰ 5s¹ Orbital diagram (valence shell only): 4d: ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ 5s: ↑ Unpaired electrons: 1 Magnetism: Paramagnetic
2. Au⁺ (Gold ion)
Atomic number of Au: 79 Neutral Au configuration: [Xe] 4f¹⁴ 5d¹⁰ 6s¹ Au⁺ configuration: [Xe] 4f¹⁴ 5d¹⁰ Orbital diagram (valence shell only): 5d: ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ Unpaired electrons: 0 Magnetism: Diamagnetic
3. Mo³⁺ (Molybdenum ion)
Atomic number of Mo: 42 Neutral Mo configuration: [Kr] 4d⁵ 5s¹ Mo³⁺ configuration: [Kr] 4d³ Orbital diagram (valence shell only): 4d: ↑ ↑ ↑ Unpaired electrons: 3 Magnetism: Paramagnetic
4. Zr⁺ (Zirconium ion)
Atomic number of Zr: 40 Neutral Zr configuration: [Kr] 4d² 5s² Zr⁺ configuration: [Kr] 4d² 5s¹ Orbital diagram (valence shell only): 4d: ↑ ↑ 5s: ↑ Unpaired electrons: 3 Magnetism: Paramagnetic
Explanation
Magnetism in transition metal ions hinges on the presence of unpaired electrons in their d or s orbitals. Ions with all electrons paired exhibit diamagnetism, meaning they are weakly repelled by magnetic fields. In contrast, ions with one or more unpaired electrons exhibit paramagnetism, meaning they are attracted by magnetic fields due to their net magnetic moment.
Cd⁺, having lost only one electron, retains the [Kr] 4d¹⁰ 5s¹ configuration. The single 5s electron remains unpaired, making Cd⁺ paramagnetic.
Au⁺ is unusual. Neutral gold has a filled 5d shell and one 6s electron. Upon losing the 6s¹ electron to become Au⁺, the resulting 5d¹⁰ is fully filled. No unpaired electrons remain, hence Au⁺ is diamagnetic.
Mo³⁺ arises when Mo loses its two 5s and one 4d electron. The resulting 4d³ configuration includes three unpaired electrons, firmly qualifying Mo³⁺ as paramagnetic.
Zr⁺ loses one electron, typically from the 5s orbital first, resulting in 4d² 5s¹. Both 4d and 5s orbitals contain unpaired electrons. Zr⁺ thus has three unpaired electrons, making it paramagnetic.
This analysis underscores the importance of electron configuration in predicting chemical and physical behavior, particularly for ions of transition metals. Understanding how orbital occupancy changes with ionization is vital to determining an ion’s magnetic character.
