Read the passage carefully and answer.
The crystal field theory (CFT) is an electrostatic model which considers the metal-ligand bond to be ionic arising purely from electrostatic interactions between the metal ion and the ligand. Ligands are treated as point charges in case of anions or dipoles in case of neutral molecules. The five d orbitals in an isolated gaseous metal atom/ion have same energy, i.e., they are degenerate. This degeneracy is maintained if a spherically symmetrical field of negative charges surrounds the metal atom/ion. However, when this negative field is due to ligands (either anions or the negative ends of dipolar molecules like \( NH_3 \) and \( H_2O \)) in a complex, it becomes asymmetrical and the degeneracy of the d orbitals is lifted. It results in splitting of the d orbitals. The pattern of splitting depends upon the nature of the crystal field.
The hybridisation and magnetism of the complexes \( [Co(C_2O_4)_3]^{3-} \) and \( [Mn(CN)_6]^{3-} \) respectively is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (\(d^2sp^3\), diamagnetic) and (\(d^2sp^3\), paramagnetic)
In valence bond theory a six-coordinate complex uses either the inner 3d orbitals (\( d^2sp^3 \), an inner orbital complex) or the outer 4d orbitals (\( sp^3d^2 \), an outer orbital complex). Inner 3d orbitals are freed only by pairing the 3d electrons, as happens with the strong field ligand \( \ce{CN-} \); the book also gives \( \ce{[Co(C2O4)3]^3-} \) as an inner orbital, diamagnetic complex. The number of unpaired electrons left decides the magnetism.
\( \ce{[Co(C2O4)3]^3-} \): \( x + 3(-2) = -3 \), so Co is +3. Co (Z = 27) is \( \mathrm{[Ar]\,3d^{7}\,4s^{2}} \), so \( \ce{Co^3+} \) is \( \mathrm{3d^{6}} \). The six electrons pair up in three 3d orbitals, two 3d orbitals are left empty: \( d^2sp^3 \), no unpaired electron, diamagnetic.
\( \ce{[Mn(CN)6]^3-} \): \( x + 6(-1) = -3 \), so Mn is +3. Mn (Z = 25) is \( \mathrm{[Ar]\,3d^{5}\,4s^{2}} \), so \( \ce{Mn^3+} \) is \( \mathrm{3d^{4}} \). The four electrons pair into three 3d orbitals, leaving two empty: \( d^2sp^3 \), but 2 unpaired electrons, paramagnetic.
So the pair is (\( d^2sp^3 \), diamagnetic) and (\( d^2sp^3 \), paramagnetic).
Why the other options are incorrect
Option (1) (\(d^2sp^3\), diamagnetic) and (\(d^2sp^3\), diamagnetic) — Incorrect. A \( \mathrm{3d^{4}} \) ion in three orbitals must keep 2 unpaired electrons, so \( \ce{[Mn(CN)6]^3-} \) cannot be diamagnetic.
Option (2) (\(sp^3d^2\), diamagnetic) and (\(d^2sp^3\), paramagnetic) — Incorrect. The oxalate complex is an inner orbital complex, so its hybridisation is \( d^2sp^3 \), not \( sp^3d^2 \).
Option (4) (\(d^2sp^3\), paramagnetic) and (\(sp^3d^2\), diamagnetic) — Incorrect. Both magnetic labels are reversed, and the cyanide complex is \( d^2sp^3 \).
NCERT (2026-27 edition):
Magnetic Properties of Coordination Compounds – …
