What are the spin-only magnetic moments of \( \ce{[CoF6]^3-} \) and \( \ce{[Co(NH3)6]^3+} \), respectively? (Atomic number of Co = 27)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 4.90 BM and 0 BM
The spin-only magnetic moment depends only on the number of unpaired electrons (n): \( \mu = \sqrt{n(n+2)} \) BM. Cobalt has the same oxidation state in both complexes; the ligand decides whether the d electrons pair up.
Oxidation state of Co: in [CoF6]3−, x + 6(−1) = −3; in [Co(NH3)6]3+, x + 6(0) = +3. So x = +3 in both.
Co (Z = 27): [Ar] 3d7 4s2; Co3+: [Ar] 3d6.
[CoF6]3−: outer orbital (sp3d2), high spin; 3d6 keeps four unpaired electrons, n = 4.
\( \mu = \sqrt{4(4+2)} = \sqrt{24} = 4.90 \) BM.
[Co(NH3)6]3+: inner orbital (d2sp3), low spin; the six electrons pair up in three 3d orbitals, n = 0, so μ = 0 (diamagnetic).
NCERT (2026-27 edition):
5.5.1 Valence Bond Theory – “Six pairs of electrons, one from each NH3 molecule, occupy the six hybrid orbitals. Thus, the complex has octahedral geometry and is diamagnetic because of the absence of unpaired electron. … The paramagnetic octahedral complex, [CoF6]3– uses outer orbital (4d) in hybridisation (sp3d2). It is thus called outer orbital or high spin or spin free complex.”
4.3.9 Magnetic Properties – “…For these, the magnetic moment is determined by the number of unpaired electrons and is calculated by using the ‘spin-only’ formula, i.e., μ = \( \sqrt{n(n+2)} \) where n is the number of unpaired electrons and μ is the magnetic moment in units of Bohr magneton (BM).”