The elements 57 through 71 (Lanthanum and the Lanthanoids) all form Ln+3 ions in aqueous solution. These all have an electron configuration is [Xe]4fn, where n increases from zero to fourteen over the series from Lanthanum to Lutetium
Select the two ions that have 4 unpaired spins.
To determine the two ions that have 4 unpaired spins out of the Ln+3 ions, we need to identify the electron configuration for each ion and count the number of unpaired electrons.
The electron configuration for the Ln+3 ions is [Xe]4fn, where n increases from zero to fourteen over the series from Lanthanum (La) to Lutetium (Lu).
To find the number of unpaired electrons, we need to consider the number of electrons in the f orbitals and determine their electron occupancy.
In the f orbital, there are a total of 14 electrons that can be distributed across the 7 available orbitals. Each orbital can accommodate a maximum of 2 electrons.
To determine the number of unpaired spins, we need to identify the value of n in [Xe]4fn for each ion and subtract it from 14.
Let's calculate the number of unpaired spins for each of the Ln+3 ions:
- Lanthanum (La): [Xe]4f0 (n = 0) → 14 - 0 = 14 unpaired spins
- Cerium (Ce): [Xe]4f1 (n = 1) → 14 - 1 = 13 unpaired spins
- Praseodymium (Pr): [Xe]4f2 (n = 2) → 14 - 2 = 12 unpaired spins
- Neodymium (Nd): [Xe]4f3 (n = 3) → 14 - 3 = 11 unpaired spins
- Promethium (Pm): [Xe]4f4 (n = 4) → 14 - 4 = 10 unpaired spins
- Samarium (Sm): [Xe]4f5 (n = 5) → 14 - 5 = 9 unpaired spins
- Europium (Eu): [Xe]4f6 (n = 6) → 14 - 6 = 8 unpaired spins
- Gadolinium (Gd): [Xe]4f7 (n = 7) → 14 - 7 = 7 unpaired spins
- Terbium (Tb): [Xe]4f8 (n = 8) → 14 - 8 = 6 unpaired spins
- Dysprosium (Dy): [Xe]4f9 (n = 9) → 14 - 9 = 5 unpaired spins
- Holmium (Ho): [Xe]4f10 (n = 10) → 14 - 10 = 4 unpaired spins
- Erbium (Er): [Xe]4f11 (n = 11) → 14 - 11 = 3 unpaired spins
- Thulium (Tm): [Xe]4f12 (n = 12) → 14 - 12 = 2 unpaired spins
- Ytterbium (Yb): [Xe]4f13 (n = 13) → 14 - 13 = 1 unpaired spin
- Lutetium (Lu): [Xe]4f14 (n = 14) → 14 - 14 = 0 unpaired spins
Based on the calculations, the two ions that have 4 unpaired spins are:
1. Holmium (Ho): [Xe]4f10 (n = 10) → 14 - 10 = 4 unpaired spins
2. Erbium (Er): [Xe]4f11 (n = 11) → 14 - 11 = 3 unpaired spins
Therefore, the two ions with 4 unpaired spins in the Ln+3 series are Holmium (Ho) and Erbium (Er).