The f-block elements are those in which the differentiating electron enters the (n-2)f orbitals. There are two series of f-Block elements corresponding to filling of 4f and 5f-orbitals. The series of 4f-orbitals is called lanthanides. Lanthanides show different oxidation states depending upon stability of f0, f7 and f14 configurations, though the most common oxidation state is +3. There is a regular decrease in the size of lanthanide ions with increase in atomic number which is known as lanthanide contraction.
Neodymium (Z = 60) is a lanthanoid that shows the +4 oxidation state, but only in its oxide, NdO2. How many 4f electrons are present in the Nd4+ ion?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2
In a lanthanoid, the electrons outside the xenon core (Z = 54) are in the 4f, 5d and 6s orbitals. When the atom forms a cation, the two 6s electrons are lost first, and then electrons are removed from the 4f level.
Electrons outside the [Xe] core in Nd = 60 − 54 = 6.
Nd (Z = 60): [Xe] 4f4 6s2.
Nd4+: four electrons removed – the two 6s electrons and then two 4f electrons.
Nd4+: [Xe] 4f2, so it has 2 4f electrons.
NCERT (2026-27 edition):
4.5.3 Oxidation States – “…Pr, Nd, Tb and Dy also exhibit +4 state but only in oxides, MO2.”
Table 4.9: Electronic Configurations and Radii of Lanthanum and Lanthanoids (row for neodymium)
| Name | Symbol | Electronic configurations* | Radii/pm | ||||
|---|---|---|---|---|---|---|---|
| Ln | Ln2+ | Ln3+ | Ln4+ | Ln | Ln3+ | ||
| Neodymium | Nd | 4f46s2 | 4f4 | 4f3 | 4f2 | 181 | 99 |
* Only electrons outside [Xe] core are indicated