Which one of the following complexes in an outer orbital complex?

• Inner Orbital Complexes: Formed when the central metal ion utilizes its inner $$(n-1)d$$ orbitals for hybridization, resulting in $$d^2sp^3$$ hybridization. This usually occurs with Strong Field Ligands (SFLs) like $$\text{CN}^-$$ or $$\text{NH}_3$$ (with specific metals), which force the unpaired electrons to pair up.
• Outer Orbital Complexes: Formed when the inner $$(n-1)d$$ orbitals are unavailable or completely filled, forcing the metal ion to utilize its outer $$nd$$ orbitals for bonding, resulting in $$sp^3d^2$$ hybridization. This typically occurs with Weak Field Ligands (WFLs) or when the electronic configuration structurally forbids inner d-participation.

• Option A: $$[\text{Fe(CN)}_6]^{4-}$$

  • Iron is in the $$+2$$ oxidation state: $$\text{Fe}^{2+} = 3d^6$$
  • $$\text{CN}^-$$ is a powerful strong field ligand that forces the six electrons in the $$3d$$ subshell to pair up completely, leaving two $$3d$$ orbitals empty.
  • Hybridization: $$d^2sp^3$$ (Inner orbital complex)



• Option B: $$[\text{Ni(NH)}_3)_6]^{2+}$$

  • Nickel is in the $$+2$$ oxidation state: $$\text{Ni}^{2+} = 3d^8$$
  • The electronic arrangement of $$3d^8$$ leaves only one empty orbital in the $$3d$$ subshell, even if pairing occurs. For an octahedral complex, exactly two vacant $$(n-1)d$$ orbitals are strictly required.
  • Because it is mathematically impossible to vacate two inner $$3d$$ orbitals, the system must utilize the outer $$4d$$ orbitals for its six coordination pairs.
  • Hybridization: $$sp^3d^2$$ (Outer orbital complex)



• Option C: $$[\text{Co(NH)}_3)_6]^{3+}$$

  • Cobalt is in the $$+3$$ oxidation state: $$\text{Co}^{3+} = 3d^6$$
  • In the presence of $$\text{Co}^{3+}$$, ammonia ($$\text{NH}_3$$) behaves as a strong field ligand. It causes the six electrons to pair up perfectly in the low-energy $$t_{2g}$$ set, vacating two $$3d$$ orbitals.
  • Hybridization: $$d^2sp^3$$ (Inner orbital complex)



• Option D: $$[\text{Mn(CN)}_6]^{4-}$$

  • Manganese is in the $$+2$$ oxidation state: $$\text{Mn}^{2+} = 3d^5$$
  • $$\text{CN}^-$$ is a strong field ligand that pairs the electrons up to form a low-spin configuration, leaving two internal $$3d$$ orbitals open for bonding.
  • Hybridization: $$d^2sp^3$$ (Inner orbital complex)

Due to the structural constraints of the $$3d^8$$ electronic configuration, $$[\text{Ni(NH)}_3)_6]^{2+}$$ is forced to hybridize via outer shell orbitals.

Answer: Option B — $$[\text{Ni(NH)}_3)_6]^{2+}$$