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Home/Practice questions/Chemistry
ChemistryInorganic: Coordination Compounds

What is the hybridization of [Co(NH3)6]3+ ?

Coordination Compounds

Alfred Werner (1866-1919), a Swiss chemist was the first to formulate his ideas about the structures of coordination compounds. He prepared and characterised a large number of coordination compounds and studied their physical and chemical behaviour by simple experimental techniques. Werner proposed the concept of a primary valence and a secondary valence for a metal ion. The formulas and names adopted for coordination entities are based on the recommendations of the International Union of Pure and Applied Chemistry (IUPAC). Many approaches have been put forth to explain the nature of bonding in coordination compounds viz. Valence Bond Theory (VBT), Crystal Field Theory (CFT), Ligand Field Theory (LFT) and Molecular Orbital Theory (MOT). The Valence Bond Theory (VBT) explains with reasonable success, the formation, magnetic behaviour and geometrical shapes of coordination compounds. It, however, fails to provide a quantitative interpretation of magnetic behaviour and has nothing to say about the optical properties of these compounds. The Crystal Field Theory (CFT) to coordination compounds is based on the effect of different crystal fields (provided by the ligands taken as point charges), on the degeneracy of d orbital energies of the central metal atom/ion. The splitting of the d orbitals provides different electronic arrangements in strong and weak crystal fields. The treatment provides for quantitative estimations of orbital separation energies, magnetic moments and spectral and stability parameters. However, the assumption that ligands constitute point charges creates many theoretical difficulties.

What is the hybridization of [Co(NH3)6]3+ ?

Answer & explanation+

Correct answer: option 3

The correct answer is Option (3) → d2sp3

By valence bond theory, the metal ion uses a set of hybrid orbitals to accept one lone pair from each ligand. For six ligands (octahedral), two d orbitals are needed:

  • d2sp3 (inner orbital complex): uses the inner (n−1)d orbitals. This is possible when the d electrons pair up and leave two inner d orbitals empty; the complex is usually low spin.

  • sp3d2 (outer orbital complex): uses the outer nd orbitals, so the inner d electrons stay unpaired; the complex is high spin.

In \( \ce{[Co(NH3)6]^{3+}} \), cobalt is in the +3 state:

    \( \displaystyle \ce{Co}\ (Z=27):\ [\ce{Ar}]\,3d^{7}\,4s^{2} \)

    \( \displaystyle \ce{Co^3+}:\ [\ce{Ar}]\,3d^{6} \)

\( \ce{NH3} \) pairs the six 3d electrons, leaving two inner 3d orbitals free, so the complex is an inner orbital, diamagnetic octahedral complex.

So the hybridisation of \( \ce{[Co(NH3)6]^{3+}} \) is d2sp3.

Why the other options are incorrect

  • Option (1) sp3d3 — Incorrect. It needs seven orbitals and gives a pentagonal bipyramid (as in \( \ce{IF7} \)), not six-coordination.

  • Option (2) sp3d2 — Incorrect. This outer orbital hybridisation is found in \( \ce{[CoF6]^{3-}} \), where the 3d electrons stay unpaired (4 unpaired electrons, paramagnetic).

  • Option (4) sp3d — Incorrect. It uses five orbitals and gives a trigonal bipyramid, which fits five ligands, not six.

NCERT (2026-27 edition):
Valence Bond Theory – …

Textbook extract

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