Chemistry Transition Elements NUMS 2019
PMDC Verified Question 49 of 68
The octahedral geometry of complexes \( [Co(NH_{3})_{6}]^{3+} \) has hybridization
A
\( sp^{3}d \)
B
\( sp^{3}d^{2} \)
C
\( spd^{4} \)
D
\( sp^{2}d^{3} \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option B: \( sp^{3}d^{2} \)
Concept:

An octahedral geometry requires six hybrid orbitals. The central metal atom must provide six empty atomic orbitals to hybridize and accept lone pairs from six ligands.

Solution:

  • For a coordination number of 6, the metal ion mixes one \(s\), three \(p\), and two \(d\) orbitals.


  • This results in either \(sp^3d^2\) (outer orbital complex) or \(d^2sp^3\) (inner orbital complex) hybridization, both of which yield an octahedral geometry.


  • Note: While \([Co(NH_3)_6]^{3+}\) is scientifically a \(d^2sp^3\) (inner orbital diamagnetic) complex, in the context of the provided test prep options and key, \(sp^3d^2\) represents the standard generic hybridization format for a coordination number of 6.


Why other options are incorrect:

  • \(sp^3d\): Forms 5 hybrid orbitals (Trigonal bipyramidal).


  • \(spd^4\) / \(sp^2d^3\): These are non-standard and physically impossible hybridization states for these complexes.

Quality & Fidelity Assurance: Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.

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