Which of the following species is diamagnetic? (Atomic numbers: V = 23, Fe = 26, Co = 27, Cu = 29)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A low spin octahedral Co3+ complex
A diamagnetic species has no unpaired electrons.
Configuration of Co: Z = 27, [Ar]3d74s2.
Ion formed: losing the two 4s electrons and one 3d electron gives \( \ce{Co^3+} \): 3d6.
Low spin octahedral: the ligands are strong field (Δo > P), so all six electrons pair in the lower set: t2g6eg0.
Unpaired electrons: 0, so the complex is diamagnetic, like \( \ce{[Co(NH3)6]^3+} \).
Why the other options are incorrect
Option (1) An isolated gas-phase V3+ ion — Incorrect. V (Z = 23) is [Ar]3d34s2, so \( \ce{V^3+} \) is 3d2. By Hund’s rule the two electrons stay in separate d orbitals: 2 unpaired electrons, paramagnetic.
Option (2) A high spin octahedral Fe2+ complex — Incorrect. Fe (Z = 26) is [Ar]3d64s2, so \( \ce{Fe^2+} \) is 3d6. High spin (Δo < P) gives t2g4eg2: 4 unpaired electrons, paramagnetic.
Option (3) An isolated gas-phase Cu2+ ion — Incorrect. Cu (Z = 29) is [Ar]3d104s1, so \( \ce{Cu^2+} \) is 3d9, with 1 unpaired electron; paramagnetic.
NCERT (2026-27 edition):
5.5.1 Valence Bond Theory – “In the diamagnetic octahedral complex, [Co(NH3)6]3+, the cobalt ion is in +3 oxidation state and has the electronic configuration 3d6. … Thus, the complex has octahedral geometry and is diamagnetic because of the absence of unpaired electron. In the formation of this complex, since the inner d orbital (3d) is used in hybridisation, the complex, [Co(NH3)6]3+ is called an inner orbital or low spin or spin paired complex.”