What is the correct order of paramagnetism of the following complex ions?
P: [FeF6]3−
Q: [CoF6]3−
R: [V(H2O)6]3+
S: [Ti(H2O)6]3+
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → P > Q > R > S
Paramagnetism grows with the number of unpaired electrons (\( \mu=\sqrt{n(n+2)} \) BM). F− is a weak field ligand (Δo < P), so its octahedral complexes are high spin.
P: \( \ce{[FeF6]^3-} \) — first (5 unpaired).
Oxidation state: x + 6(−1) = −3, so x = +3.
Configuration: Fe (Z = 26) is [Ar]3d64s2; \( \ce{Fe^3+} \) is 3d5.
Filling (weak field, high spin): t2g3eg2, so n = 5 and μ = 5.92 BM.
Q: \( \ce{[CoF6]^3-} \) — second (4 unpaired).
Oxidation state: x + 6(−1) = −3, so x = +3.
Configuration: Co (Z = 27) is [Ar]3d74s2; \( \ce{Co^3+} \) is 3d6.
Filling (weak field, high spin): t2g4eg2, so n = 4 and μ = 4.90 BM.
R: \( \ce{[V(H2O)6]^3+} \) — third (2 unpaired).
Oxidation state: water is neutral, so x = +3.
Configuration: V (Z = 23) is [Ar]3d34s2; \( \ce{V^3+} \) is 3d2.
Filling: t2g2, so n = 2 and μ = 2.83 BM.
S: \( \ce{[Ti(H2O)6]^3+} \) — last (1 unpaired).
Oxidation state: x = +3.
Configuration: Ti (Z = 22) is [Ar]3d24s2; \( \ce{Ti^3+} \) is 3d1.
Filling: t2g1, so n = 1 and μ = 1.73 BM.
NCERT (2026-27 edition):
5.5.2 Magnetic Properties of Coordination Compounds – “… [Fe(CN)6]3– has magnetic moment of a single unpaired electron while [FeF6]3– has a paramagnetic moment of five unpaired electrons. [CoF6]3– is paramagnetic with four unpaired electrons while [Co(C2O4)3]3– is diamagnetic.”
5.5.5 Colour in Coordination Compounds – “… Consider, for example, the complex [Ti(H2O)6]3+, which is violet in colour. This is an octahedral complex where the single electron (Ti3+ is a 3d1 system) in the metal d orbital is in the t2g level in the ground state of the complex.”