Identify transition metal complexes which are not octahedral in shape
(A) \([Co(NH_3)_6]^{3+}\)
(B) \([Ni(CO)_4]\)
(C) \([CoCl(NH_3)_5]^{2+}\)
(D) \([CoCl_2(NH_3)_4]^{+}\)
(E) \([PtCl_4]^{2-}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (B) and (E) only
(A) \( \ce{[Co(NH3)6]^3+} \) — Octahedral. Six ligands, d2sp3.
(B) \( \ce{[Ni(CO)4]} \) — Not octahedral. Four ligands, sp3: tetrahedral.
(C) \( \ce{[CoCl(NH3)5]^2+} \) — Octahedral. Five \( \ce{NH3} \) + one \( \ce{Cl-} \) = six donor atoms.
(D) \( \ce{[CoCl2(NH3)4]+} \) — Octahedral. Four \( \ce{NH3} \) + two \( \ce{Cl-} \) = six donor atoms.
(E) \( \ce{[PtCl4]^2-} \) — Not octahedral. Four ligands, dsp2: square planar.
NCERT (2026-27 edition):
Werner’s Theory of Coordination Compounds – “… 4. The ions/groups bound by the secondary linkages to the metal have characteristic spatial arrangements corresponding to different coordination numbers.
He further postulated that octahedral, tetrahedral and square planar geometrical shapes are more common in coordination compounds of transition metals. Thus, [Co(NH3)6]3+, [CoCl(NH3)5]2+ and [CoCl2(NH3)4]+ are octahedral entities, while [Ni(CO)4] and [PtCl4]2– are tetrahedral and square planar, respectively.”