Chemistry · D And F Block Elements · NEET
First write the ion in the 4f-to-the-power-n form. For most Ln3+ ions you take the neutral atom and remove 3 electrons (usually the two 6s and one from 4f or 5d), leaving [Xe] 4f^n. Then draw 7 boxes for the seven f-orbitals and fill them singly first (Hund's rule), only pairing after all 7 have one electron. The number of boxes with a single electron = number of unpaired electrons. Example: Nd3+ is 4f3, so 3 boxes are filled singly = 3 unpaired electrons.
Only ions with a totally empty (4f0) or totally full (4f14) 4f subshell are diamagnetic, because then every electron is paired. The common diamagnetic ones are La3+ (4f0), Ce4+ (4f0), Yb2+ (4f14) and Lu3+ (4f14). All other Ln3+ ions have some unpaired 4f electrons, so they are paramagnetic. Memory hook: 'Empty or Full = no pull'.
Cerium neutral is [Xe] 4f1 5d1 6s2. Ce3+ loses 3 electrons and keeps one 4f electron, so it is [Xe] 4f1 - that single unpaired electron makes it paramagnetic. Ce4+ loses one more electron and becomes [Xe] 4f0 (empty 4f, same as noble gas xenon). With zero unpaired electrons, Ce4+ is diamagnetic. This empty-4f stability is also why Ce shows the +4 state.
The spin-only formula mu = root of n(n+2) works for d-block ions because their orbital motion is 'quenched' by ligands. But in lanthanoids the 4f orbitals are deep inside the atom and shielded, so orbital angular momentum is NOT quenched. Their real magnetic moment comes from both spin AND orbital motion, so you must use mu = g times root of J(J+1). For NEET, you still use spin-only counting to decide paramagnetic vs diamagnetic and to count unpaired electrons - just know the exact moment value needs the J formula.
Both are diamagnetic. La3+ has an empty 4f (4f0) and Lu3+ has a completely filled 4f (4f14). In both, all electrons are paired, so neither is attracted by a magnetic field. They sit at the two ends of the lanthanoid series and are the 'colourless, non-magnetic' bookends.
For counting unpaired electrons in the common Ln3+ ions, always place the electrons in 4f as 4f^n. The 5d and 6s electrons are removed first when the ion forms. Exceptions like Gd ([Xe] 4f7 5d1 6s2) matter for the neutral atom, but Gd3+ is still simply 4f7 with 7 unpaired electrons - the maximum in the series.
The pair of lanthanoid ions which are diamagnetic is:
The lanthanide ion having four unpaired electrons is: (Atomic numbers: Ce = 58, Nd = 60, Tb = 65, Ho = 67)
Statement II: The number of unpaired electrons in a Cr2+ ion (Z = 24) is the same as that of a Nd3+ ion (Z = 60). Is Statement II true or false?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Most trivalent lanthanoid ions (except La3+ and Lu3+) have partly filled 4f orbitals. These unpaired 4f electrons make the ions paramagnetic, and f-f electronic transitions give them colour. So paramagnetism and colour go together in the middle of the series.
Gd3+ and Eu2+, both 4f7, have 7 unpaired electrons - the highest possible in the 4f subshell. This half-filled arrangement is very stable and gives the largest spin contribution.
For deciding paramagnetic versus diamagnetic and for counting unpaired electrons, use simple Hund's-rule filling. NEET rarely asks for the exact numerical moment of a lanthanoid; if it does, the correct expression is mu = g times root of J(J+1) because 4f orbital motion is not quenched. Spin-only root of n(n+2) is the d-block tool.
No. La3+ is 4f0 (empty 4f), so it has zero unpaired electrons and is diamagnetic. It is one of the two colourless, non-magnetic ends of the series (the other being Lu3+, 4f14).
In d-block ions the 3d orbitals are outer and their orbital motion is quenched by ligands, so spin-only works. In lanthanoids the 4f orbitals are buried deep and shielded, so both spin and orbital motion contribute. This is the key difference tested in comparison questions like Cr2+ versus Nd3+.
