Write orbital diagrams for each ion and determine if the ion is diamagnetic or paramagnetic. a. Cd2+ b. Au+ c. Mo3+ d. Zr2+
The Correct Answer and Explanation is:
Here’s the correct orbital diagrams and magnetic properties for each ion:
a. Cd²⁺ (Cadmium ion)
Atomic number of Cd = 48
Electron configuration of Cd = [Kr] 4d¹⁰ 5s²
Cd²⁺ loses two electrons → [Kr] 4d¹⁰
Orbital diagram (Cd²⁺):
makefileCopyEdit4d: ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
All electrons are paired → Diamagnetic
b. Au⁺ (Gold(I) ion)
Atomic number of Au = 79
Electron configuration of Au = [Xe] 4f¹⁴ 5d¹⁰ 6s¹
Au⁺ loses one electron → [Xe] 4f¹⁴ 5d¹⁰
Orbital diagram (Au⁺):
makefileCopyEdit5d: ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
All electrons are paired → Diamagnetic
c. Mo³⁺ (Molybdenum(III) ion)
Atomic number of Mo = 42
Electron configuration of Mo = [Kr] 4d⁵ 5s¹
Mo³⁺ loses 3 electrons → [Kr] 4d³
Orbital diagram (Mo³⁺):
makefileCopyEdit4d: ↑ ↑ ↑ _ _
3 unpaired electrons → Paramagnetic
d. Zr²⁺ (Zirconium(II) ion)
Atomic number of Zr = 40
Electron configuration of Zr = [Kr] 4d² 5s²
Zr²⁺ loses two electrons → [Kr] 4d²
Orbital diagram (Zr²⁺):
makefileCopyEdit4d: ↑ ↑ _ _ _
2 unpaired electrons → Paramagnetic
Explanation
The magnetic properties of an ion—whether it’s diamagnetic or paramagnetic—depend on the presence of unpaired electrons. Diamagnetic substances have all electrons paired and are not attracted to a magnetic field, while paramagnetic substances have unpaired electrons and are attracted to a magnetic field.
We begin by determining the ground-state electron configuration of the neutral atom and then remove electrons according to ionization (typically starting from the outermost s orbital, then d).
- For Cd²⁺, cadmium loses two electrons from its 5s orbital, resulting in a completely filled 4d subshell (4d¹⁰). Since all electrons are paired, Cd²⁺ is diamagnetic.
- For Au⁺, gold loses its single 6s electron, leaving a filled 5d subshell (5d¹⁰). This full set of paired electrons also makes Au⁺ diamagnetic.
- For Mo³⁺, molybdenum normally has 4d⁵5s¹, and upon losing three electrons, we are left with a 4d³ configuration. These three electrons each occupy separate orbitals due to Hund’s rule, leading to three unpaired electrons, making it paramagnetic.
- For Zr²⁺, zirconium (4d² 5s²) loses the two 5s electrons, resulting in a 4d² configuration. These two d electrons are unpaired, so Zr²⁺ is also paramagnetic.
Understanding these magnetic properties is essential in chemistry and materials science, as they influence how substances interact with magnetic fields and can be used in technologies like MRI or spintronic devices.
