Write down the electron configuration of neodymium (for Nd Z60). Rewrite it in terms of a noble gas configuration. What family does this element belong to? a) Write down the four quantum numbers of the electrons that are not part of the noble gas configuration. Show your answer in a table. b) Calculate the magnetic moment of the atom. c) Calculate the spectrocopic ground state for the atom. 2. An unknown metal has a density of 21.45 g/ml. It crystallizes with a compact cubic packing and one edge of the unit cell measures 3.9231 A. a) Calculate the atomic weight of the metal. b) Calculate the atomic radius of the metal.
The Correct Answer and Explanation is :
1. Neodymium (Nd, Z = 60)
Electron Configuration:
The full electron configuration for neodymium is:
1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ 6s² 4f⁴.
In terms of a noble gas configuration:
[Xe] 4f⁴ 6s².
Family: Neodymium belongs to the Lanthanide family, which is characterized by partially filled (4f) orbitals.
a) Four Quantum Numbers for Electrons Outside the Noble Gas Core
The electrons outside the [Xe] core are (4f^4) and (6s^2). For these, the quantum numbers ((n, l, m_l, m_s)) are:
| Electron | (n) | (l) | (m_l) | (m_s) |
|---|---|---|---|---|
| (4f^4) | 4 | 3 | -3 to 3 | ±½ |
| (6s^2) | 6 | 0 | 0 | ±½ |
b) Magnetic Moment
The formula for the magnetic moment is:
[
\mu = \sqrt{n(n+2)} \, \mu_B
]
where (n) is the number of unpaired electrons. Neodymium has 4 unpaired electrons in the (4f) subshell:
[
\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \, \mu_B
]
(\mu_B) is the Bohr magneton.
c) Spectroscopic Ground State
The ground state of (4f^4) is determined using Hund’s rules. For (4f^4):
- (L = 6) (maximum orbital angular momentum for 4 electrons in (f)).
- (S = 2) (maximum spin angular momentum for 4 electrons).
The term symbol is:
[
{}^5I_4
]
2. Unknown Metal
a) Calculate Atomic Weight
The metal crystallizes in a face-centered cubic (FCC) structure, where 4 atoms occupy each unit cell.
Density ((\rho)) = 21.45 g/mL, edge length ((a)) = 3.9231 Å = 3.9231 \times 10^{-8} cm.
Volume of the unit cell:
[
V = a^3 = (3.9231 \times 10^{-8})^3 \approx 6.03 \times 10^{-23} \, \text{cm}^3
]
Mass of unit cell:
[
m_{\text{cell}} = \rho \cdot V = 21.45 \cdot 6.03 \times 10^{-23} \approx 1.29 \times 10^{-21} \, \text{g}
]
Mass of one atom:
[
m_{\text{atom}} = \frac{m_{\text{cell}}}{4} = \frac{1.29 \times 10^{-21}}{4} \approx 3.23 \times 10^{-22} \, \text{g}
]
Atomic weight:
[
A = m_{\text{atom}} \cdot N_A = 3.23 \times 10^{-22} \cdot 6.022 \times 10^{23} \approx 195 \, \text{g/mol}
]
The atomic weight is approximately 195 g/mol, corresponding to platinum (Pt).
b) Calculate Atomic Radius
For FCC, the relationship between edge length ((a)) and atomic radius ((r)) is:
[
a = 2\sqrt{2}r \implies r = \frac{a}{2\sqrt{2}} = \frac{3.9231}{2\sqrt{2}} \approx 1.39 \, \text{Å}
]
The atomic radius is approximately 1.39 Å.
Explanation (300 Words)
Neodymium (Nd) is a lanthanide with Z = 60. Its electron configuration highlights the filling of the (4f) orbitals, a characteristic of the lanthanides. The configuration simplifies to ([Xe] 4f^4 6s^2). Lanthanides exhibit unpaired (4f) electrons, contributing to magnetic properties. For Nd, with 4 unpaired electrons, the magnetic moment is (4.90 \, \mu_B). Hund’s rules predict a ground state of ({}^5I_4), reflecting high spin multiplicity and orbital angular momentum.
For the unknown metal, its density and FCC structure allow us to calculate its atomic weight. FCC crystals have 4 atoms per unit cell, and using the unit cell’s dimensions and density, the atomic mass is determined as ~195 g/mol, likely platinum. The atomic radius, derived from FCC geometry, is about 1.39 Å, aligning with typical values for platinum. These calculations showcase the interplay of crystallography, density, and atomic properties in identifying metals.