To explain bonding in complexes, the valence bond theory was proposed by Linus Pauling. The main postulates of VB theory are as follows:
(v) In the presence of strong ligands such as CN−, NO, CO, the d-electrons are rearranged vacating some d-orbitals (only when there are more than three d-electrons) which can then take part in hybridisation.
(vi) In the presence of weak ligands such as F−, Cl−, H2O, etc., the d-electrons are not rearranged.
(vii) The d-orbitals involved in hybridisation may be either (n−1)d orbitals or outer d-orbitals.
The complexes formed by the involvement of (n−1)d orbitals in hybridisation are called inner orbital complexes or low spin complexes. The complexes formed by the involvement of outer d-orbitals are called outer orbital complexes or high spin complexes.
(viii) Each ligand contains a lone pair of electrons. A dative bond is formed by the overlap of a vacant hybrid orbital of metal ion and a filled orbital of ligand.
(ix) The complex will be paramagnetic, if unpaired electrons are present; otherwise it is diamagnetic.
Arrange the following in decreasing order of the number of unpaired electrons. (Atomic number of Fe = 26)
(I) \([Fe(H_2O)_6]^{2+}\)
(II) \([Fe(CN)_6]^{3-}\)
(III) \([Fe(CO)_5]\)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → I, II, III
In valence bond theory a strong ligand (\( \ce{CN^-} \), CO) makes the metal d electrons pair up, while a weak ligand (\( \ce{H2O} \)) leaves them unpaired. For each complex, find the d electrons of iron and then fill them.
(I) \( \ce{[Fe(H2O)6]^2+} \) — first place, 4 unpaired electrons.
Oxidation state of Fe: water is neutral, so x + 6(0) = +2 and x = +2.
Configuration: Fe (Z = 26) is [Ar]3d64s2; losing the two 4s electrons gives \( \ce{Fe^2+} \): 3d6.
Ligand: \( \ce{H2O} \) is a weak ligand, so the d electrons are not rearranged: one 3d orbital holds a pair and four hold single electrons.
Unpaired electrons: 4.
(II) \( \ce{[Fe(CN)6]^3-} \) — second place, 1 unpaired electron.
Oxidation state of Fe: x + 6(−1) = −3, so x = +3.
Configuration: losing the two 4s electrons and one 3d electron gives \( \ce{Fe^3+} \): 3d5.
Ligand: \( \ce{CN^-} \) is a strong ligand, so the five electrons are paired into three 3d orbitals (two pairs and one single electron), leaving two 3d orbitals free for d2sp3 hybridisation.
Unpaired electrons: 1.
(III) \( \ce{[Fe(CO)5]} \) — third place, no unpaired electron.
Oxidation state of Fe: CO is neutral and the complex has no charge, so x = 0.
Configuration: Fe(0) is [Ar]3d64s2: eight electrons outside the argon core.
Ligand: CO is a strong ligand, so all eight electrons are paired into four 3d orbitals, leaving one 3d, the 4s and three 4p orbitals for dsp3 hybridisation (trigonal bipyramidal).
Unpaired electrons: 0.
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.”
4.3.4 Oxidation States – “Low oxidation states are found when a complex compound has ligands capable of π-acceptor character in addition to the σ-bonding. For example, in Ni(CO)4 and Fe(CO)5, the oxidation state of nickel and iron is zero.”