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.
The atomic numbers of Cr and Fe are 24 and 26, respectively. Which of the following is paramagnetic?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \([Cr(NH_3)_6]^{3+}\)
A complex is paramagnetic if it has unpaired electrons. By valence bond theory, strong ligands such as \( \ce{CO} \) and \( \ce{CN^-} \) pair up the d electrons to vacate d orbitals for hybridisation, but only when the metal has more than three d electrons.
\( \ce{[Cr(CO)6]} \): CO is neutral, so Cr is 0. Cr (Z = 24) is [Ar]3d54s1: six electrons. Strong CO pairs them into three 3d orbitals, leaving two 3d orbitals free for d2sp3 → 0 unpaired, diamagnetic.
\( \ce{Fe(CO)5} \): Fe is 0. Fe (Z = 26) is [Ar]3d64s2: eight electrons. Strong CO pairs them into four 3d orbitals, leaving one 3d orbital for dsp3 → 0 unpaired, diamagnetic.
\( \ce{[Fe(CN)6]^4-} \): x + 6(−1) = −4, so Fe is +2. Removing 4s2 gives \( \ce{Fe^2+} \): 3d6. Strong \( \ce{CN^-} \) pairs the six electrons into three 3d orbitals (d2sp3) → 0 unpaired, diamagnetic.
\( \ce{[Cr(NH3)6]^3+} \): NH3 is neutral, so Cr is +3. Removing 4s1 and two 3d electrons gives \( \ce{Cr^3+} \): 3d3. With only three d electrons, two 3d orbitals are already empty, so no pairing is needed (d2sp3) → 3 unpaired, paramagnetic.
NCERT (2026-27 edition):
5.5.2 Magnetic Properties of Coordination Compounds – “… For metal ions with upto three electrons in the d orbitals, like Ti3+ (d1); V3+ (d2); Cr3+ (d3); two vacant d orbitals are available for octahedral hybridisation with 4s and 4p orbitals. The magnetic behaviour of these free ions and their coordination entities is similar.”