Statement I: The inner orbital complex [Mn(CN)6]3− is diamagnetic.
Statement II: An inner orbital complex may contain unpaired electrons.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Statement I is incorrect but Statement II is correct.
An inner orbital complex uses inner (n−1)d orbitals in d2sp3 hybridisation. A complex is diamagnetic only when it has no unpaired electron.
Oxidation state of Mn: in \( \ce{[Mn(CN)6]^3-} \), x + 6(−1) = −3, so x = +3.
Configuration: Mn (Z = 25) is [Ar]3d54s2; losing the two 4s electrons and one 3d electron gives \( \ce{Mn^3+} \): 3d4.
Hybridisation: for d2sp3, two 3d orbitals must be empty.
Filling the other three 3d orbitals: the four 3d electrons form one pair and two single electrons.
Unpaired electrons: 2.
Statement I: The inner orbital complex [Mn(CN)6]3− is diamagnetic. — False. With 2 unpaired electrons the complex is paramagnetic, even though it is an inner orbital complex.
Statement II: An inner orbital complex may contain unpaired electrons. — True. Using inner d orbitals forces pairing only until two d orbitals are free; any electrons left over stay unpaired, as in \( \ce{[Mn(CN)6]^3-} \) (two) and \( \ce{[Fe(CN)6]^3-} \) (one).
NCERT (2026-27 edition):
5.5.2 Magnetic Properties of Coordination Compounds – “… [Mn(CN)6]3– has magnetic moment of two unpaired electrons while [MnCl6]3– has a paramagnetic moment of four unpaired electrons. … [Mn(CN)6]3–, [Fe(CN)6]3– and [Co(C2O4)3]3– are inner orbital complexes involving d2sp3 hybridisation, the former two complexes are paramagnetic and the latter diamagnetic.”