What is the spin-only magnetic moment of the complex ion \( \ce{[Fe(CN)6]^3-} \)? (Atomic number of Fe = 26)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1.73 BM
The spin-only magnetic moment depends only on the number of unpaired electrons (n): \( \mu = \sqrt{n(n+2)} \) BM. CN− is a strong field ligand, so the 3d electrons of the metal ion pair up as far as they can.
Oxidation state of Fe: x + 6(−1) = −3, so x = +3.
Fe (Z = 26): [Ar] 3d6 4s2; losing the two 4s electrons and one 3d electron gives Fe3+: [Ar] 3d5.
[Fe(CN)6]3− is an inner orbital complex (d2sp3). To free two 3d orbitals, the five 3d electrons pair into the other three orbitals, which leaves one unpaired electron, n = 1.
\( \mu = \sqrt{1(1+2)} = \sqrt{3} = 1.73 \) BM.
NCERT (2026-27 edition):
5.5.2 Magnetic Properties of Coordination Compounds – “…Thus, for d4 (Cr2+, Mn3+), d5 (Mn2+, Fe3+), d6 (Fe2+, Co3+) cases, a vacant pair of d orbitals results only by pairing of 3d electrons which leaves two, one and zero unpaired electrons, respectively.”
“[Fe(CN)6]3– has magnetic moment of a single unpaired electron while [FeF6]3– has a paramagnetic moment of five unpaired electrons.”
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).”