Question:
Which one of the following complexes is an outer orbital complex? (Atomic numbers Mn=25, Fe=26, Co=27, Ni=28)
Which one of the following complexes is an outer orbital complex? (Atomic numbers Mn=25, Fe=26, Co=27, Ni=28)
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Which one of the following complexes is an outer orbital complex? (Atomic numbers Mn=25, Fe=26, Co=27, Ni=28)
Updated On: Apr 15, 2026
- $[Fe(CN)_{6}]^{4-}$
- $[Mn(CN)_{6}]^{4-}$
- $[Co(NH_{3})_{6}]^{3+}$
- $[Ni(NH_{3})_{6}]^{2+}$
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The Correct Option is D
Solution and Explanation
Step 1: Concept
An outer orbital complex uses higher energy 'nd' orbitals for hybridization (usually $sp^3d^2$).
Step 2: Analysis of (A), (B), (C)
- $[Fe(CN)_{6}]^{4-}$, $[Mn(CN)_{6}]^{4-}$, and $[Co(NH_{3})_{6}]^{3+}$ use $(n-1)d$ orbitals ($3d$) and undergo $d^2sp^3$ hybridization, making them inner orbital complexes.
Step 3: Analysis of (D)
- In $[Ni(NH_{3})_{6}]^{2+}$, $Ni^{2+}$ has a $3d^8$ configuration. The two vacant orbitals required for octahedral geometry cannot be provided by the $3d$ subshell even with pairing. Thus, it uses $4s$, $4p$, and $4d$ orbitals ($sp^3d^2$ hybridization).
Step 4: Conclusion
$[Ni(NH_{3})_{6}]^{2+}$ is an outer orbital complex.
Final Answer: (D)
An outer orbital complex uses higher energy 'nd' orbitals for hybridization (usually $sp^3d^2$).
Step 2: Analysis of (A), (B), (C)
- $[Fe(CN)_{6}]^{4-}$, $[Mn(CN)_{6}]^{4-}$, and $[Co(NH_{3})_{6}]^{3+}$ use $(n-1)d$ orbitals ($3d$) and undergo $d^2sp^3$ hybridization, making them inner orbital complexes.
Step 3: Analysis of (D)
- In $[Ni(NH_{3})_{6}]^{2+}$, $Ni^{2+}$ has a $3d^8$ configuration. The two vacant orbitals required for octahedral geometry cannot be provided by the $3d$ subshell even with pairing. Thus, it uses $4s$, $4p$, and $4d$ orbitals ($sp^3d^2$ hybridization).
Step 4: Conclusion
$[Ni(NH_{3})_{6}]^{2+}$ is an outer orbital complex.
Final Answer: (D)
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