Which lanthanide 3+ ion has four unpaired electrons?
D) Ho³⁺ is correct: its ten 4f electrons leave four unpaired. This d- and f-block elements NEET 2026 worked Chemistry question tests two separate steps: forming the ion and filling its orbitals.
The question asks which tripositive lanthanide ion has exactly four unpaired electrons in its ground-state configuration. The complete options and supplied atomic numbers are:
Medium difficulty and 75 seconds are question-bank tags, not an official difficulty rating or a guaranteed solving time.
How do you form each lanthanide ion before counting electrons?
Remove both 6s electrons first. Then remove one 5d electron if that subshell is occupied in the supplied neutral configuration; otherwise, remove one 4f electron. This follows the official solution’s method: establish the tripositive ion before applying Hund’s rule.
The xenon core, written as [Xe], contains 54 electrons. It is a closed shell and contributes no unpaired electrons, so count only the electrons outside that core.
Cerium: The supplied neutral configuration contains one 5d electron. After removing the two 6s electrons, remove that 5d electron, leaving one 4f electron.
Neodymium: No 5d electron is present in the supplied configuration. The third electron therefore comes from 4f.
Terbium: Removing the two 6s electrons leaves nine 4f electrons. Removing one more gives eight.
Holmium: Remove the two 6s electrons, then one 4f electron. This reduces the 4f population from eleven to ten.
Use electron totals only as a check:
- Ce³⁺: 55 electrons, including one outside the xenon core.
- Nd³⁺: 57 electrons, including three outside the core.
- Tb³⁺: 62 electrons, including eight outside the core.
- Ho³⁺: 64 electrons, including ten outside the core.
How does Hund’s rule give the unpaired count for all four options?
The 4f subshell has seven orbitals and holds at most 14 electrons. Its first seven electrons occupy separate orbitals with parallel spins before pairing begins. Up to seven electrons, the unpaired count equals the electron count; beyond seven, each added electron reduces the unpaired count by one.
For a general 4f configuration, use:
The second expression follows directly from pairing. Starting from seven singly occupied orbitals, each extra electron pairs with one existing electron:
Applying this rule to each ionic configuration gives the complete option check:
- A) Nd³⁺:
- B) Ce³⁺:
- C) Tb³⁺:
- D) Ho³⁺:
For Ho³⁺, seven electrons first occupy seven separate orbitals. The remaining three create three pairs, leaving four orbitals singly occupied:
The final answer is D) Ho³⁺. Count singly occupied orbitals, not the total number of 4f electrons.
Why does counting neutral neodymium lead to the wrong option A?
Neutral Nd has four unpaired 4f electrons, while its two 6s electrons are paired. That count belongs to the neutral configuration:
Selecting A) Nd³⁺ from this count silently replaces the ion in the question with the neutral atom. Hund’s rule has been applied to the wrong species, even if the orbital counting itself was correct.
Repair the calculation by removing two 6s electrons and one 4f electron:
The three remaining 4f electrons occupy separate orbitals, giving three unpaired electrons, not four. Write the charge and derive the ionic configuration before applying Hund’s rule.
How can I practise ion formation and unpaired-electron counting?
Use these three original related practice questions, not additional verified NEET PYQs, to test electron removal and pairing separately. Attempt all three before reading the numbered worked answers. For each, write the configuration you are actually counting.
- How does neodymium’s unpaired-electron count change when it forms Nd³⁺? Use the supplied neutral configuration:
- How many 4f orbitals in Tb³⁺ are doubly occupied, singly occupied and empty? Start from:
- Can the following configurations have the same number of unpaired electrons? Derive both counts:
Worked answers
- The count falls from four to three. Removing the paired 6s electrons does not change the unpaired count. Removing one 4f electron then reduces it by one:
- One orbital is doubly occupied, six are singly occupied and zero are empty. Seven electrons first occupy all seven orbitals separately; the eighth pairs in one of them. Both totals check:
- Yes, both have four unpaired electrons. Use the first branch of the rule for four electrons and the second for ten:
Equal unpaired counts do not mean identical configurations. The first has four singles and three empty orbitals; the second has four singles and three pairs.
If your answer differs, check electron removal first, then check whether pairing has started. Correct the ionic configuration before recounting its unpaired electrons.
Next step: the past-paper archive on NEET JEEnius AI and search past NEET papers by year, subject or chapter, each with a worked solution (100 free searches a month).
Related on NEET JEEnius AI: Adiabatic Process NEET 2026: Why the Answer Is 124.5 J.
Frequently asked questions
Which lanthanide 3+ ion has four unpaired electrons?
Among Nd³⁺, Ce³⁺, Tb³⁺ and Ho³⁺, the correct answer is Ho³⁺. Its configuration is [Xe] 4f¹⁰, with three doubly occupied and four singly occupied 4f orbitals.
How do you count unpaired electrons in a 4f subshell?
The 4f subshell has seven orbitals, which fill singly before pairing begins. For 4fⁿ, the unpaired count is n when n is between 0 and 7, and 14 − n when n is between 8 and 14. Apply this rule to the ionic configuration, not the neutral atom.
Which electrons are removed when a lanthanide forms a 3+ ion?
For the supplied configurations of Ce, Nd, Tb and Ho, remove both 6s electrons first. Then remove one 5d electron if that subshell is occupied; otherwise, remove one 4f electron. This gives Ce³⁺ as [Xe] 4f¹, Nd³⁺ as [Xe] 4f³, Tb³⁺ as [Xe] 4f⁸ and Ho³⁺ as [Xe] 4f¹⁰.
Why does Nd³⁺ have three unpaired electrons instead of four?
Neutral Nd has the supplied configuration [Xe] 4f⁴ 6s² and four unpaired electrons. Forming Nd³⁺ removes the two paired 6s electrons and one 4f electron, leaving [Xe] 4f³. Its three 4f electrons occupy separate orbitals, so the ion has three unpaired electrons.