Assertion (A): [Co(en)3]3+ is paramagnetic.
Reason (R): [Co(en)3]3+ is an inner orbital complex.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) is false but (R) is true.
Oxidation state of Co: en (ethane-1,2-diamine) is neutral, so in \( \ce{[Co(en)3]^3+} \) x + 3(0) = +3 and x = +3.
Configuration: Co (Z = 27) is [Ar]3d74s2; losing the two 4s electrons and one 3d electron gives \( \ce{Co^3+} \): 3d6.
Type of ligand: en lies above NH3 in the spectrochemical series, so it is a strong field ligand (Δo > P).
Filling of the d orbitals: the six 3d electrons pair up in three 3d orbitals (t2g6eg0), leaving two 3d orbitals empty.
Unpaired electrons: 0, so the complex is diamagnetic.
Hybridisation: the two empty inner 3d, one 4s and three 4p orbitals give d2sp3: an inner orbital (low spin) complex, as for \( \ce{[Co(NH3)6]^3+} \).
Assertion (A): [Co(en)3]3+ is paramagnetic. — False. All six 3d electrons are paired, so there is no unpaired electron; the complex is diamagnetic.
Reason (R): [Co(en)3]3+ is an inner orbital complex. — True. The hybridisation uses the inner 3d orbitals (d2sp3).
NCERT (2026-27 edition):
5.5.1 Valence Bond Theory – “… Thus, the complex has octahedral geometry and is diamagnetic because of the absence of unpaired electron. In the formation of this complex, since the inner d orbital (3d) is used in hybridisation, the complex, [Co(NH3)6]3+ is called an inner orbital or low spin or spin paired complex.”
5.5.4 Crystal Field Theory – “… In general, ligands can be arranged in a series in the order of increasing field strength as given below:
I– < Br– < SCN– < Cl– < S2– < F– < OH– < C2O42– < H2O < NCS– < edta4– < NH3 < en < CN– < CO
Such a series is termed as spectrochemical series.”